Analog Electronic Volt-Ohm-Milliammeters

4 Objectives

You wiU be able to

l Sketch various transistor analog voltmeter circuits and explain the operation of each circuit Calculate circuit currents voltages and input resistance

2 Sketch an input aUenuator circuit as used with an electronic voltmeter Explain its opshyeration and define the circuit input resistance

3 Using illustrations explain the problems that can occur with electronic voltmeter ground terminals when measuring voltages in a circuit

4 Draw the circuit diagrams of various op-amp analog voltmeters Explain the operation of each circuit

5 Draw series shunt and linear ohmmeter circuits as used in electronic instruments and explain the operation of each circuit

6 Sketch the circuit diagrams of various ac electronic voltmeters and explain their opershyation

7 Sketch a circuit to show how current is measured by an electronic voltmeter Explain the circuit operation

8 Draw the front panel of a typical analog electronic voltmeter showing the various conshytrols and meter scales State typical performance specifications for the instrument and discuss its applications

Introduction

Voltmeters constructed of moving-coil instruments and multiplier resistors (see Chapshyter 3) have some important limitations They cannot measure very low voltages and their resistance is too low for measurements in high-impedance circuitry These restrictions

86

are overcome by the use of electronic circuits that offer high input resistance and which amplify low voltages to measurable levels When such circuits are used the instrument becomes an electronic voltmeter

Electronic voltmeters can be analng instruments in which the measurement is indishycated by a pointer moving over a calibrated scale or digital instruments which display the measurement in numerical form (see Chapter 5) As well as amplification transistor and operational amplifier circuits offer advantages in the measurement of resistance dishyrect current and alternating current

4-1 TRANSISTOR VOLTMETER CIRCUITS

Emitter-Follower Voltmeters

Voltmeter loading (see Section 3-4) can be greatly reduced by using an emitter follower An emitter follower offers a high 4input resistance to voltages being measured and proshyvides a low output resistance to drive current through the coil of a deflection meter The basic emitter-follower voltmeter circuit illustrated in Figure 4-1 shows a PMMC instrushyment and a multiplier resistance (Rs) connected in series with the transistor emitter The dc supply is connected-positive to the transistor collector and the negative to the deflecshytion meter The positive terminal of voltage E (to be measured) is supplied to the transisshytor base and its negative is connected to the same terminal as the power supply negative

The transistor base current in Figure 4-1 is substantially lower than the meter curshyrent

where hFpound is the transistor current gain Thus the circuit input resistance is

FlgUre 4-1 An emitter follower offers a high input resistance to a measured voltage and a low output resistance to a dellection voltmeter circuil VSE introduces an error in the measurement

Sec 4- Transistor Voltmeter Circuits

------0+

+o----------~---~

87

ERmiddot=shy

bull 18

which is much larger than the meter circuit resistance (Rs +Rm)

Example 4-1

The simple emitter-follower voltmeter circuit in Figure 4-1 has Vee =20 V Rs + Rm = 93 ill 1m =1 rnA at full scale and transistor hFpound =100 (a) Calculate the meter current when E =10 V (b) Detennine the voltmeter input resistance with and without the transistor

Solution

(a) VE = E - VBi= to V - 07 V

=93 V

VE 93 V Im= Rs+Rm = 93 kfl

=1 rnA

(b) With the transistor IB= ~ = 1 rnA hF 100

= 10 fJ-A

R=i=~ bull IB to fJ-A

= I Mfl

Without the transistor

The transistor base--emitter voltage drop (VBd introduces an error in the simple emitter-follower voltmeter For example when E is 5 V in the circuit in Example 4-1 the meter should read half of full-scale that is 05 rnA However as a simple calculation shows the meter current is actually 046 rnA The error can be eliminated by using a poshytential divider and an additional emitter follower as illustrated in Figure 4-2

The practical emitter-follower circuit in Figure 4-2 uses a plus-and-minus or dualshypolarity supply (typically plusmn12 V) Transistor Q has its base biased to ground via resistor Rio and a potential divider (R4 Rs and R6 ) provides an adjustable bias voltage (Vp) to the base of transistor Q2 Resistors R2 and R3 connect the transistor emitter terminals to the negative supply voltage (-Vpoundd and the meter circuit is connected between the transistor emitters The circuit input resistance is R in parallel with the input resistance at the transhysistor base

88 Analog Electronic Volt-Ohm-Millimeters Chap 4

Ql

r-------------------------------~-----------~+voc

t Rs RIO E

1 I-+-------V---------~

~----------------------------~~---------~ -VEE

Flgure 4-2 Practical emitter-follower voltmeter cffiuit using a second transistor (Q2)

and a potential divider (R4bull Rs and R 6) to eliminate the Vspounderror produced by Qt

When no input voltage is applied (E =0 V) the base voltage of Q2 is adjusted to give zero meter current This makes Vp =0 V VEl =VEl =-07 V and (meter circuit voltage) V= o V Now suppose that a 5 V input is applied to the Q base The meter voltage is

V= VEl - VE2

= (E - VBE1 ) - VE2

=(5 V -07 V)-(-07 V)

=5V

Thus unlike the case of the simple emitter-follower voltmeter all of the voltage to be measured appears across the meter circuit no part of it is lost as transistor VBE

Example 4-2

An emitter-follower voltmeter circuit such as that in Figure 4-2 has R2 =R3 =39 kfi and Vcc=plusmnl2 V (a) Determine 12 and 13 when E =0 V (b) Calculate the meter circuit voltage when E = I V and when E =05 V

Solution

(a) VR2 = VR3 = 0 V - VBE - VEE

=OV-O7V-(-12V)

=113 V

[__ VR2 _ Il3V 2 - 3 - - 39 kflRz

= 29mA

Sec4-1 Transistor Voltmeter Circuits 89

(b) When E= 1 V

VEl =E- V8g =1 V -07 V

=03 V

VEl =Vp- V8g ==0 V -07 V

=-07V

V= Vgl - VE2 =03 V -(-07 V)

==IV

WhenE=O5 V

VIOl ==E- VBg =O5 V -07 V

=-02V

VEl == Vp - VBg 0 V - 07 V

=-07V

V= VIOl - VE2 == -02 V - (-07 V)

=05 V

Ground Terminals and Floating Power Supplies

The circuit in Figure 4-2 shows the input voltage E as being measured with respect to ground However this may not always be convenient For example suppose that the voltshyage across resistor Rb in Figure 4-3(a) were to be measured by a voltmeter with its negashytive terminals grounded The voltmeter ground would short-circuit resistor Rc and serishyously affect the voltage and current conditions in the resistor circuit Clearly the voltmeter should not have one of its terminals grounded

For the circuit in Figure 4-2 to function correctly the lower end of RI must be at zero volts with respect to +Vcc and -Vee The + and - supply voltage may be derived from two batteries [Figure 4-3(b)] or from two de power supply circuits [Figure 4-3c)J [n both cases the negative terminal of the positive supply is connected to the positive tershyminal of the negative supply For plusmn9 V supplies Vee is +9 V with respect to the common terminal and Vee is -9 V with respect to the common terminal [n many electronic cirshycuits the power supply common terminal is grounded In electronic voltmeter circuits this terminal is not grounded simply to avoid the kind of problem already discussed When left without any grounded terminal the voltmeter supply voltages are said to be floating This means that the common terminal assumes the absolute voltage (with respect to ground) of any terminal to which it may be connected An inverted triangular symbol is employed to identify the common terminal or zero voLtage tenninaL in a circuit [see Figure 4-3(b)(c)]

Although the electronic voltmeter supply voltages are allowed to float some inshystlUments have their common terminal connected to ground via a capacitor usually 01 fLF If batteries are used as supply the capacitor is connected to the chassis Where a

90 Analog Electronic Volt-Ohrn-Millimeters Chap 4

Voltmeter

+pound0-----

R

Short circuit

(a) A voltmeter with one of its tenninals grounded can shortdrcuit a component in a circuit in which voltage is being measured

+Vcc~+ Common terminal

Common terminal CirCUilSymbol~forcommom

tenninal +

--VEE~-

+Vee 0---------

VEE o-----~

(b) supply using batteries (c) ~ supply using power supplies

Figure 4-3 Serious measurement errors can result when a grounded voltmeter terminal is incorshyrectly connected to a circuit When a circuit has a plus-and minus supply voltage the voltmeter comshymon terminal should always be connected to the common terminal of the supply

liS V power supply is included in the voltmeter the chassis and the capacitor are grounded Thus when measuring voltage levels in a transistor circuit for example the common terminal introduces a capacitance to ground wherever it is connected in the cirshycuit To avoid any effect on conditions within the circuit (oscillations or phase shifts) the voltmeter common terminal should always be connected to the transistor circuit ground or zero voltage terminal All voltages are then measured with respect to this point

Sec 41 Transistor Voltmeter Circuits

b

91

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

are overcome by the use of electronic circuits that offer high input resistance and which amplify low voltages to measurable levels When such circuits are used the instrument becomes an electronic voltmeter

Electronic voltmeters can be analng instruments in which the measurement is indishycated by a pointer moving over a calibrated scale or digital instruments which display the measurement in numerical form (see Chapter 5) As well as amplification transistor and operational amplifier circuits offer advantages in the measurement of resistance dishyrect current and alternating current

4-1 TRANSISTOR VOLTMETER CIRCUITS

Emitter-Follower Voltmeters

Voltmeter loading (see Section 3-4) can be greatly reduced by using an emitter follower An emitter follower offers a high 4input resistance to voltages being measured and proshyvides a low output resistance to drive current through the coil of a deflection meter The basic emitter-follower voltmeter circuit illustrated in Figure 4-1 shows a PMMC instrushyment and a multiplier resistance (Rs) connected in series with the transistor emitter The dc supply is connected-positive to the transistor collector and the negative to the deflecshytion meter The positive terminal of voltage E (to be measured) is supplied to the transisshytor base and its negative is connected to the same terminal as the power supply negative

The transistor base current in Figure 4-1 is substantially lower than the meter curshyrent

where hFpound is the transistor current gain Thus the circuit input resistance is

FlgUre 4-1 An emitter follower offers a high input resistance to a measured voltage and a low output resistance to a dellection voltmeter circuil VSE introduces an error in the measurement

Sec 4- Transistor Voltmeter Circuits

------0+

+o----------~---~

87

ERmiddot=shy

bull 18

which is much larger than the meter circuit resistance (Rs +Rm)

Example 4-1

The simple emitter-follower voltmeter circuit in Figure 4-1 has Vee =20 V Rs + Rm = 93 ill 1m =1 rnA at full scale and transistor hFpound =100 (a) Calculate the meter current when E =10 V (b) Detennine the voltmeter input resistance with and without the transistor

Solution

(a) VE = E - VBi= to V - 07 V

=93 V

VE 93 V Im= Rs+Rm = 93 kfl

=1 rnA

(b) With the transistor IB= ~ = 1 rnA hF 100

= 10 fJ-A

R=i=~ bull IB to fJ-A

= I Mfl

Without the transistor

The transistor base--emitter voltage drop (VBd introduces an error in the simple emitter-follower voltmeter For example when E is 5 V in the circuit in Example 4-1 the meter should read half of full-scale that is 05 rnA However as a simple calculation shows the meter current is actually 046 rnA The error can be eliminated by using a poshytential divider and an additional emitter follower as illustrated in Figure 4-2

The practical emitter-follower circuit in Figure 4-2 uses a plus-and-minus or dualshypolarity supply (typically plusmn12 V) Transistor Q has its base biased to ground via resistor Rio and a potential divider (R4 Rs and R6 ) provides an adjustable bias voltage (Vp) to the base of transistor Q2 Resistors R2 and R3 connect the transistor emitter terminals to the negative supply voltage (-Vpoundd and the meter circuit is connected between the transistor emitters The circuit input resistance is R in parallel with the input resistance at the transhysistor base

88 Analog Electronic Volt-Ohm-Millimeters Chap 4

Ql

r-------------------------------~-----------~+voc

t Rs RIO E

1 I-+-------V---------~

~----------------------------~~---------~ -VEE

Flgure 4-2 Practical emitter-follower voltmeter cffiuit using a second transistor (Q2)

and a potential divider (R4bull Rs and R 6) to eliminate the Vspounderror produced by Qt

When no input voltage is applied (E =0 V) the base voltage of Q2 is adjusted to give zero meter current This makes Vp =0 V VEl =VEl =-07 V and (meter circuit voltage) V= o V Now suppose that a 5 V input is applied to the Q base The meter voltage is

V= VEl - VE2

= (E - VBE1 ) - VE2

=(5 V -07 V)-(-07 V)

=5V

Thus unlike the case of the simple emitter-follower voltmeter all of the voltage to be measured appears across the meter circuit no part of it is lost as transistor VBE

Example 4-2

An emitter-follower voltmeter circuit such as that in Figure 4-2 has R2 =R3 =39 kfi and Vcc=plusmnl2 V (a) Determine 12 and 13 when E =0 V (b) Calculate the meter circuit voltage when E = I V and when E =05 V

Solution

(a) VR2 = VR3 = 0 V - VBE - VEE

=OV-O7V-(-12V)

=113 V

[__ VR2 _ Il3V 2 - 3 - - 39 kflRz

= 29mA

Sec4-1 Transistor Voltmeter Circuits 89

(b) When E= 1 V

VEl =E- V8g =1 V -07 V

=03 V

VEl =Vp- V8g ==0 V -07 V

=-07V

V= Vgl - VE2 =03 V -(-07 V)

==IV

WhenE=O5 V

VIOl ==E- VBg =O5 V -07 V

=-02V

VEl == Vp - VBg 0 V - 07 V

=-07V

V= VIOl - VE2 == -02 V - (-07 V)

=05 V

Ground Terminals and Floating Power Supplies

The circuit in Figure 4-2 shows the input voltage E as being measured with respect to ground However this may not always be convenient For example suppose that the voltshyage across resistor Rb in Figure 4-3(a) were to be measured by a voltmeter with its negashytive terminals grounded The voltmeter ground would short-circuit resistor Rc and serishyously affect the voltage and current conditions in the resistor circuit Clearly the voltmeter should not have one of its terminals grounded

For the circuit in Figure 4-2 to function correctly the lower end of RI must be at zero volts with respect to +Vcc and -Vee The + and - supply voltage may be derived from two batteries [Figure 4-3(b)] or from two de power supply circuits [Figure 4-3c)J [n both cases the negative terminal of the positive supply is connected to the positive tershyminal of the negative supply For plusmn9 V supplies Vee is +9 V with respect to the common terminal and Vee is -9 V with respect to the common terminal [n many electronic cirshycuits the power supply common terminal is grounded In electronic voltmeter circuits this terminal is not grounded simply to avoid the kind of problem already discussed When left without any grounded terminal the voltmeter supply voltages are said to be floating This means that the common terminal assumes the absolute voltage (with respect to ground) of any terminal to which it may be connected An inverted triangular symbol is employed to identify the common terminal or zero voLtage tenninaL in a circuit [see Figure 4-3(b)(c)]

Although the electronic voltmeter supply voltages are allowed to float some inshystlUments have their common terminal connected to ground via a capacitor usually 01 fLF If batteries are used as supply the capacitor is connected to the chassis Where a

90 Analog Electronic Volt-Ohrn-Millimeters Chap 4

Voltmeter

+pound0-----

R

Short circuit

(a) A voltmeter with one of its tenninals grounded can shortdrcuit a component in a circuit in which voltage is being measured

+Vcc~+ Common terminal

Common terminal CirCUilSymbol~forcommom

tenninal +

--VEE~-

+Vee 0---------

VEE o-----~

(b) supply using batteries (c) ~ supply using power supplies

Figure 4-3 Serious measurement errors can result when a grounded voltmeter terminal is incorshyrectly connected to a circuit When a circuit has a plus-and minus supply voltage the voltmeter comshymon terminal should always be connected to the common terminal of the supply

liS V power supply is included in the voltmeter the chassis and the capacitor are grounded Thus when measuring voltage levels in a transistor circuit for example the common terminal introduces a capacitance to ground wherever it is connected in the cirshycuit To avoid any effect on conditions within the circuit (oscillations or phase shifts) the voltmeter common terminal should always be connected to the transistor circuit ground or zero voltage terminal All voltages are then measured with respect to this point

Sec 41 Transistor Voltmeter Circuits

b

91

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

ERmiddot=shy

bull 18

which is much larger than the meter circuit resistance (Rs +Rm)

Example 4-1

The simple emitter-follower voltmeter circuit in Figure 4-1 has Vee =20 V Rs + Rm = 93 ill 1m =1 rnA at full scale and transistor hFpound =100 (a) Calculate the meter current when E =10 V (b) Detennine the voltmeter input resistance with and without the transistor

Solution

(a) VE = E - VBi= to V - 07 V

=93 V

VE 93 V Im= Rs+Rm = 93 kfl

=1 rnA

(b) With the transistor IB= ~ = 1 rnA hF 100

= 10 fJ-A

R=i=~ bull IB to fJ-A

= I Mfl

Without the transistor

The transistor base--emitter voltage drop (VBd introduces an error in the simple emitter-follower voltmeter For example when E is 5 V in the circuit in Example 4-1 the meter should read half of full-scale that is 05 rnA However as a simple calculation shows the meter current is actually 046 rnA The error can be eliminated by using a poshytential divider and an additional emitter follower as illustrated in Figure 4-2

The practical emitter-follower circuit in Figure 4-2 uses a plus-and-minus or dualshypolarity supply (typically plusmn12 V) Transistor Q has its base biased to ground via resistor Rio and a potential divider (R4 Rs and R6 ) provides an adjustable bias voltage (Vp) to the base of transistor Q2 Resistors R2 and R3 connect the transistor emitter terminals to the negative supply voltage (-Vpoundd and the meter circuit is connected between the transistor emitters The circuit input resistance is R in parallel with the input resistance at the transhysistor base

88 Analog Electronic Volt-Ohm-Millimeters Chap 4

Ql

r-------------------------------~-----------~+voc

t Rs RIO E

1 I-+-------V---------~

~----------------------------~~---------~ -VEE

Flgure 4-2 Practical emitter-follower voltmeter cffiuit using a second transistor (Q2)

and a potential divider (R4bull Rs and R 6) to eliminate the Vspounderror produced by Qt

When no input voltage is applied (E =0 V) the base voltage of Q2 is adjusted to give zero meter current This makes Vp =0 V VEl =VEl =-07 V and (meter circuit voltage) V= o V Now suppose that a 5 V input is applied to the Q base The meter voltage is

V= VEl - VE2

= (E - VBE1 ) - VE2

=(5 V -07 V)-(-07 V)

=5V

Thus unlike the case of the simple emitter-follower voltmeter all of the voltage to be measured appears across the meter circuit no part of it is lost as transistor VBE

Example 4-2

An emitter-follower voltmeter circuit such as that in Figure 4-2 has R2 =R3 =39 kfi and Vcc=plusmnl2 V (a) Determine 12 and 13 when E =0 V (b) Calculate the meter circuit voltage when E = I V and when E =05 V

Solution

(a) VR2 = VR3 = 0 V - VBE - VEE

=OV-O7V-(-12V)

=113 V

[__ VR2 _ Il3V 2 - 3 - - 39 kflRz

= 29mA

Sec4-1 Transistor Voltmeter Circuits 89

(b) When E= 1 V

VEl =E- V8g =1 V -07 V

=03 V

VEl =Vp- V8g ==0 V -07 V

=-07V

V= Vgl - VE2 =03 V -(-07 V)

==IV

WhenE=O5 V

VIOl ==E- VBg =O5 V -07 V

=-02V

VEl == Vp - VBg 0 V - 07 V

=-07V

V= VIOl - VE2 == -02 V - (-07 V)

=05 V

Ground Terminals and Floating Power Supplies

The circuit in Figure 4-2 shows the input voltage E as being measured with respect to ground However this may not always be convenient For example suppose that the voltshyage across resistor Rb in Figure 4-3(a) were to be measured by a voltmeter with its negashytive terminals grounded The voltmeter ground would short-circuit resistor Rc and serishyously affect the voltage and current conditions in the resistor circuit Clearly the voltmeter should not have one of its terminals grounded

For the circuit in Figure 4-2 to function correctly the lower end of RI must be at zero volts with respect to +Vcc and -Vee The + and - supply voltage may be derived from two batteries [Figure 4-3(b)] or from two de power supply circuits [Figure 4-3c)J [n both cases the negative terminal of the positive supply is connected to the positive tershyminal of the negative supply For plusmn9 V supplies Vee is +9 V with respect to the common terminal and Vee is -9 V with respect to the common terminal [n many electronic cirshycuits the power supply common terminal is grounded In electronic voltmeter circuits this terminal is not grounded simply to avoid the kind of problem already discussed When left without any grounded terminal the voltmeter supply voltages are said to be floating This means that the common terminal assumes the absolute voltage (with respect to ground) of any terminal to which it may be connected An inverted triangular symbol is employed to identify the common terminal or zero voLtage tenninaL in a circuit [see Figure 4-3(b)(c)]

Although the electronic voltmeter supply voltages are allowed to float some inshystlUments have their common terminal connected to ground via a capacitor usually 01 fLF If batteries are used as supply the capacitor is connected to the chassis Where a

90 Analog Electronic Volt-Ohrn-Millimeters Chap 4

Voltmeter

+pound0-----

R

Short circuit

(a) A voltmeter with one of its tenninals grounded can shortdrcuit a component in a circuit in which voltage is being measured

+Vcc~+ Common terminal

Common terminal CirCUilSymbol~forcommom

tenninal +

--VEE~-

+Vee 0---------

VEE o-----~

(b) supply using batteries (c) ~ supply using power supplies

Figure 4-3 Serious measurement errors can result when a grounded voltmeter terminal is incorshyrectly connected to a circuit When a circuit has a plus-and minus supply voltage the voltmeter comshymon terminal should always be connected to the common terminal of the supply

liS V power supply is included in the voltmeter the chassis and the capacitor are grounded Thus when measuring voltage levels in a transistor circuit for example the common terminal introduces a capacitance to ground wherever it is connected in the cirshycuit To avoid any effect on conditions within the circuit (oscillations or phase shifts) the voltmeter common terminal should always be connected to the transistor circuit ground or zero voltage terminal All voltages are then measured with respect to this point

Sec 41 Transistor Voltmeter Circuits

b

91

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

r-------------------------------~-----------~+voc

t Rs RIO E

1 I-+-------V---------~

~----------------------------~~---------~ -VEE

Flgure 4-2 Practical emitter-follower voltmeter cffiuit using a second transistor (Q2)

and a potential divider (R4bull Rs and R 6) to eliminate the Vspounderror produced by Qt

When no input voltage is applied (E =0 V) the base voltage of Q2 is adjusted to give zero meter current This makes Vp =0 V VEl =VEl =-07 V and (meter circuit voltage) V= o V Now suppose that a 5 V input is applied to the Q base The meter voltage is

V= VEl - VE2

= (E - VBE1 ) - VE2

=(5 V -07 V)-(-07 V)

=5V

Thus unlike the case of the simple emitter-follower voltmeter all of the voltage to be measured appears across the meter circuit no part of it is lost as transistor VBE

Example 4-2

An emitter-follower voltmeter circuit such as that in Figure 4-2 has R2 =R3 =39 kfi and Vcc=plusmnl2 V (a) Determine 12 and 13 when E =0 V (b) Calculate the meter circuit voltage when E = I V and when E =05 V

Solution

(a) VR2 = VR3 = 0 V - VBE - VEE

=OV-O7V-(-12V)

=113 V

[__ VR2 _ Il3V 2 - 3 - - 39 kflRz

= 29mA

Sec4-1 Transistor Voltmeter Circuits 89

(b) When E= 1 V

VEl =E- V8g =1 V -07 V

=03 V

VEl =Vp- V8g ==0 V -07 V

=-07V

V= Vgl - VE2 =03 V -(-07 V)

==IV

WhenE=O5 V

VIOl ==E- VBg =O5 V -07 V

=-02V

VEl == Vp - VBg 0 V - 07 V

=-07V

V= VIOl - VE2 == -02 V - (-07 V)

=05 V

Ground Terminals and Floating Power Supplies

The circuit in Figure 4-2 shows the input voltage E as being measured with respect to ground However this may not always be convenient For example suppose that the voltshyage across resistor Rb in Figure 4-3(a) were to be measured by a voltmeter with its negashytive terminals grounded The voltmeter ground would short-circuit resistor Rc and serishyously affect the voltage and current conditions in the resistor circuit Clearly the voltmeter should not have one of its terminals grounded

For the circuit in Figure 4-2 to function correctly the lower end of RI must be at zero volts with respect to +Vcc and -Vee The + and - supply voltage may be derived from two batteries [Figure 4-3(b)] or from two de power supply circuits [Figure 4-3c)J [n both cases the negative terminal of the positive supply is connected to the positive tershyminal of the negative supply For plusmn9 V supplies Vee is +9 V with respect to the common terminal and Vee is -9 V with respect to the common terminal [n many electronic cirshycuits the power supply common terminal is grounded In electronic voltmeter circuits this terminal is not grounded simply to avoid the kind of problem already discussed When left without any grounded terminal the voltmeter supply voltages are said to be floating This means that the common terminal assumes the absolute voltage (with respect to ground) of any terminal to which it may be connected An inverted triangular symbol is employed to identify the common terminal or zero voLtage tenninaL in a circuit [see Figure 4-3(b)(c)]

Although the electronic voltmeter supply voltages are allowed to float some inshystlUments have their common terminal connected to ground via a capacitor usually 01 fLF If batteries are used as supply the capacitor is connected to the chassis Where a

90 Analog Electronic Volt-Ohrn-Millimeters Chap 4

Voltmeter

+pound0-----

R

Short circuit

(a) A voltmeter with one of its tenninals grounded can shortdrcuit a component in a circuit in which voltage is being measured

+Vcc~+ Common terminal

Common terminal CirCUilSymbol~forcommom

tenninal +

--VEE~-

+Vee 0---------

VEE o-----~

(b) supply using batteries (c) ~ supply using power supplies

Figure 4-3 Serious measurement errors can result when a grounded voltmeter terminal is incorshyrectly connected to a circuit When a circuit has a plus-and minus supply voltage the voltmeter comshymon terminal should always be connected to the common terminal of the supply

liS V power supply is included in the voltmeter the chassis and the capacitor are grounded Thus when measuring voltage levels in a transistor circuit for example the common terminal introduces a capacitance to ground wherever it is connected in the cirshycuit To avoid any effect on conditions within the circuit (oscillations or phase shifts) the voltmeter common terminal should always be connected to the transistor circuit ground or zero voltage terminal All voltages are then measured with respect to this point

Sec 41 Transistor Voltmeter Circuits

b

91

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

(b) When E= 1 V

VEl =E- V8g =1 V -07 V

=03 V

VEl =Vp- V8g ==0 V -07 V

=-07V

V= Vgl - VE2 =03 V -(-07 V)

==IV

WhenE=O5 V

VIOl ==E- VBg =O5 V -07 V

=-02V

VEl == Vp - VBg 0 V - 07 V

=-07V

V= VIOl - VE2 == -02 V - (-07 V)

=05 V

Ground Terminals and Floating Power Supplies

The circuit in Figure 4-2 shows the input voltage E as being measured with respect to ground However this may not always be convenient For example suppose that the voltshyage across resistor Rb in Figure 4-3(a) were to be measured by a voltmeter with its negashytive terminals grounded The voltmeter ground would short-circuit resistor Rc and serishyously affect the voltage and current conditions in the resistor circuit Clearly the voltmeter should not have one of its terminals grounded

For the circuit in Figure 4-2 to function correctly the lower end of RI must be at zero volts with respect to +Vcc and -Vee The + and - supply voltage may be derived from two batteries [Figure 4-3(b)] or from two de power supply circuits [Figure 4-3c)J [n both cases the negative terminal of the positive supply is connected to the positive tershyminal of the negative supply For plusmn9 V supplies Vee is +9 V with respect to the common terminal and Vee is -9 V with respect to the common terminal [n many electronic cirshycuits the power supply common terminal is grounded In electronic voltmeter circuits this terminal is not grounded simply to avoid the kind of problem already discussed When left without any grounded terminal the voltmeter supply voltages are said to be floating This means that the common terminal assumes the absolute voltage (with respect to ground) of any terminal to which it may be connected An inverted triangular symbol is employed to identify the common terminal or zero voLtage tenninaL in a circuit [see Figure 4-3(b)(c)]

Although the electronic voltmeter supply voltages are allowed to float some inshystlUments have their common terminal connected to ground via a capacitor usually 01 fLF If batteries are used as supply the capacitor is connected to the chassis Where a

90 Analog Electronic Volt-Ohrn-Millimeters Chap 4

Voltmeter

+pound0-----

R

Short circuit

(a) A voltmeter with one of its tenninals grounded can shortdrcuit a component in a circuit in which voltage is being measured

+Vcc~+ Common terminal

Common terminal CirCUilSymbol~forcommom

tenninal +

--VEE~-

+Vee 0---------

VEE o-----~

(b) supply using batteries (c) ~ supply using power supplies

Figure 4-3 Serious measurement errors can result when a grounded voltmeter terminal is incorshyrectly connected to a circuit When a circuit has a plus-and minus supply voltage the voltmeter comshymon terminal should always be connected to the common terminal of the supply

liS V power supply is included in the voltmeter the chassis and the capacitor are grounded Thus when measuring voltage levels in a transistor circuit for example the common terminal introduces a capacitance to ground wherever it is connected in the cirshycuit To avoid any effect on conditions within the circuit (oscillations or phase shifts) the voltmeter common terminal should always be connected to the transistor circuit ground or zero voltage terminal All voltages are then measured with respect to this point

Sec 41 Transistor Voltmeter Circuits

b

91

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

Voltmeter

+pound0-----

R

Short circuit

(a) A voltmeter with one of its tenninals grounded can shortdrcuit a component in a circuit in which voltage is being measured

+Vcc~+ Common terminal

Common terminal CirCUilSymbol~forcommom

tenninal +

--VEE~-

+Vee 0---------

VEE o-----~

(b) supply using batteries (c) ~ supply using power supplies

Figure 4-3 Serious measurement errors can result when a grounded voltmeter terminal is incorshyrectly connected to a circuit When a circuit has a plus-and minus supply voltage the voltmeter comshymon terminal should always be connected to the common terminal of the supply

liS V power supply is included in the voltmeter the chassis and the capacitor are grounded Thus when measuring voltage levels in a transistor circuit for example the common terminal introduces a capacitance to ground wherever it is connected in the cirshycuit To avoid any effect on conditions within the circuit (oscillations or phase shifts) the voltmeter common terminal should always be connected to the transistor circuit ground or zero voltage terminal All voltages are then measured with respect to this point

Sec 41 Transistor Voltmeter Circuits

b

91

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

Voltmeter Range Changing

The potential divider constituted by resistors Rm Rb Re and Rd in Figure 4-4 allows large input voltages to be measured on an emitter-follower voltmeter This network called an input attenuator accurately divides the voltage to be measured before it is applied to the input transistor Calculation shows that the Q3 input voltage (Ed is always I V when the maximum input is applied on any range For example on the 5 V range

EG =5 V X Rb + Rc +Rd Ra + Rb + Rc + Rd

=5 V x looko+60 ko +40 ko 800 kfl + 100 kfi +60kfl+40 kfi

=lV

The input resistance offered by this circuit to a voltage being measured is the total resistance of the attenuator which is 1 Mo A 9 Mfl resistor could be induded in series with the input terminal to raise the input resistance to 10 MO This would further divide the input voltage by a factor of 10 before it is applied to the gate terminal of Q3

FETInput Voltmeter

The input resistance of the transistor voltmeter circuit can be increased further by using an additional emitter follower connected at the base of Q in Figure 4-2 However the use

FET 1 iii Input ~loE input ~111i 1 I

1 1 1

attenuator1 stage 1 I 1bull 1

1 1 1

800k

100 k

E

60k

40k

IV

5V

Emitter-follower voltmeter

~ +

R4 Vee

Rs

VEER6

Figure 4-4 A voltmeter input attenuator is simply a potential divider that accurately divides the voltage to be measured The FET input stage (Q) gives he emitter follower a very high input resistshyance

92 Analog Electronic Volt-Ohm-Millimeters Chap 4

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

of a FET source follower (Q3) as illustrated in Figure 4-4 gives a higher input resistance than can be achieved with a bipolar transistor The PET source terminal is able to supply all of the base current required by Q while the input resistance at the FEr gate is typishycally in excess of 1 MO

Consider the voltage levels in the circuit of Figure 4-4 When E = 0 V the PET gate is at the zero voltage level But the gate of an n-channel PET must always be negative with respect to its source terminal This is the same as stating that the source must be posshyitive with respect to the gate [f Vos is to be -5 V and Eo =0 V the source te~inal voltshyage must be +5 V This means that the base t~al of Q is at +5 V and since Q2 base voltage must be equal to Q base voltage Q2 base must also be at +5 V As in the circuit of Figure 4-2 Rs in Figure 4-4 is used to zero the meter when the input voltage is 0 V

Now consider what occurs when a voltage to be measured is applied to the circuit input With the attenuator shown Ea will be a maximum of 1 yThis causes the PET source terminal to increase until Vas is again -5 Y That is ~ goes from +5 to -ti V to maintain Vas equal to -5 V The ~ increase of 1 V is also a 1 V increase in the base voltshyage of Q bull As already explained all of this (1 V) increase appears across the meter cirshycuit

Example 4-3

Determine the meter reading for the circuit in Figure 4-4 when E =75 V and the meter is set to its 10 V range The PET gate-source voltage is -5 V Vp = +5 V Rs + Rm = 1 ka and 1m = I rnA at full scale

Solution On the 10 V range

=75 V x 60 kfl +40 kfl 800 kfl + tOO kfl + 60 kfl + 40 kfl

=075 V

Vs=EG- VGs=075 V -(-5 V)

=575 V

VEl = Vs - V8E =575 V -07 V = 505 V

Va= Vp - VBE =5 V -07 V

=43 V

V= VEl - VEl 505 V -43 V

=075 V =EG

[= V 075 V m Rs+Rm =lkfi

== 075 rnA (75 of full scale)

Sec 4-1 Transistor Voltmeter Circuits 93

i

On the 10 V range full scale represents 10 V and 75 offull scale would be read as 75 V

Difference Amplifier Voltmeter

The instruments discussed so far can measure a maximum of around 25 V This could be extended further of course simply by modifying the input attenuator The minimum (full-sCale) voltage measurable by the electronic voltmeter circuits already considered is I V This too can be altered to perhaps a minimum of 100 mV by selection of a meter that will give FSD when 100 mV appears across Rs + Rm However for accurate measshyurement of low voltage levels the voltage must be amplified before it is applied to the meter

Transistors QI and Q2 together with RLI Ru and RE in Figure 4-5(a) constitute a differential amplifier or emitter-coupled amplifier The circuit as a whole is known as a difference amplifier voltmeter This is because when the voltage at the base of Q2 is zero and an input voltage (E) is applied to the Q base the difference between the two base voltages is amplified and applied to the meter circuit

When a small positive voltage is applied to the base of QI in Figure 4-5 the current through QI is increased and that through Q2 is decreased An increase in lei causes fetRLI to increase and thus produces a fall in voltage Vet Similarly a decrease in fez proshyduces a rise in Vez The consequence of this is that the voltage across the meter circuit inshycreases positively at the right-hand side and negatively at the left This meter voltage (I) is directly proportional to the input voltage (E)

+Vcc

Ru Ru ----o+vcc

Rs Rm

ICl t fa V

RuVCl VC2

~ ~ (b) Zero control

R2

1pound1----- -shy1pound2

RE

-VEE

(a) Voltmeter circuit

Figure 4-5 A difference amplifier voltmeter amplifies low-level input voltages for measurement on the deflection voltmeter circuit

94 Analog Electronic Volt-Ohm-Millimeters Chap 4

Potentiometer R3 in Figure 4-5(b) is an alternative method of providing meter-zero adjustment Q2 base control as in Figure 4-4 could also be used in the circuit of Figure 4-5 When the movable contact of R3 is adjusted to the right the portion of R3 added to RLI is increased and the portion of R3 added to Ru is reduced When the contact is moved left the reverse is true Thus VC and VC2 can be adjusted differentially by means of R3 bull

and the meter voltage can be set to zero

4-2 OPERATIONAL AMPLIFIER VOLTMETER CIRCUITS

Op-Amp Voltage-Follower Voltmeter

The operational amplifier voltage-follower voltmeter in Figure 4-6 is comparable to the simple emitter-follower circuit However unlike the emitter-follower there is no baseshyemitter voltage drop from input to output The voltage-foLLower also has a much higher input resistance and lower output resistance than the emitter-follower The voltageshyfoLLower input (pound8) is applied to the op-amp noninverting input terminaL and the feedback from the output goes to the inverting input The very high internal voltage gain of the operational amplifier combined with the negative feedback tends to keep the inverting terminal voltage exactly equal to that at the noninverting terminal Thus the output voltage (Vo) exactly follows the input As discussed earlier the attenuator selects the voltmeter range

I I I I I 10( I

lnput attenuator

I ~ I 0( Voltage follower

I Meter --_~+Io(-- _

I CirCUit

I

I I I I I I I I I I I I I I I I I I I

o---~~---- I I

I I

I I I I I I

I

Figure 4middot6 An Ie operational amplifier voltage-follower voltmeter is similar to the emitter middotfollower voltmeter except that the voltage-fOllower input resistance is much higher than that of the emitter follower and there is no base~miner voltage drop

Sec 4-2 Operational Amplifier Voltmeter Circuits 95

Op-Amp Amplifier Voltmeter

Like a transistor amplifier an IC operational amplifier circuit can be used to amplify low voltages to levels measurable by a deflection instrument Figure 4-7 shows a suitable opshyamp circuit for this purpose Input voltage E is applied to the op-amp noninverting input the output voltage is divided across resistors R3 and R4 and VR3 is fed back to the op-amp inverting input terminal The internal voltage gain of the op-amp causes VR3 to always equal E Consequently the output voltage is

v = E_R3_+_R-4 (4-1) o R3

The circuit is known as a noninverting ampLifie 1 because its output is positive when a positi ve input voltage is applied and negative when the input is a negative quantity The noninverting amplifier has a very high input resistance very low output resistance and a voltage gain of

(4-2)

1 1 1 1 10(

Noninverting amplifier

------~fo(f__- Meter circuit

------ 1 1 1 1 1

t E

Figure 4-7 An operational amplifier noninverting amplifier can be used to amplify low input voltages to a level suitable for the deflection meter circuit The voltmeter gain is (R3 + R4 )IR3middot

96 Analog Electronic Volt-Ohm-Millimeters Chap 4

An op-amp noninverting amplifier voltmeter is very easily designed Current 14 through R3 and ~ is first selected very much larger than the op-amp input bias current (I8) Then the resistors are calculated as

and

Example 4-4

An op-amp voltmeter circuit as in Figure 4-7 is required to measure a maximum input of 20 mV The op-amp input current is 02 JLA and the meter circuit has I = 100 fLA FSD and Rm =10 kil Determine suitable resistance values for R3 and R4 bull

Solution

Select [4 = 1000 X In =1000 x 02 -LA

=02mA

At full scale 1m = 100 V-A

and

=IV

= 100 n

R4 = Vo - E = I V - 20 mV 14 02mA

=49 kil

Voltage-to-Current Converter

The circuit shown in Figure 4-8 is essentially a noninverting amplifier as in Figure 4-7 However instead of connecting the meter between the op-amp output and ground it is substituted in place of resistor R4 (in Figure 4-7) Once again VR1 remains equal to the input voltage and as long as IRl is very much greater than lB the meter current is

(4-3)

Sec 4-2 Operational Amplifier Voltmeter Circuits 97

E

1 Figure 4-8 Voltmeter circuit using an op-amp voltage-to-current converter The meter current is FR3

Example 4-5

Calculate the value of R3 for the circuit in Figure 4-8 if E = 1 V is to give FSD on the meter The moving-coil meter has 1= 1 rnA at full scale and Rnr = 100 n Also detennine the maximum voltage at the operational amplifier output terminaL

Solution From Equation 4-3

E 1 V R) = -- = -- = 1 kfl

I(FSD) 1 rnA

Vo = I(R3 + Rm)

= 1 mA(l kfl + 100 fl)

=11 V

=

Many electronic rnultirange instruments do not have any current-measuring facilishyties Those that do measure current generally have very low-level current ranges and some have relatively high resistances when operating as ammeters For example the meter resistance on one instrument is specified as 9 kfl when operating on a 1S fJA range This must be taken into account when the instrument is connected in series with a circuit in which the current is to be measured The instrument terminal voltage drop when used as an ammeter is termed the burden voltage For a 9 kfl resistance on a 15 ILA range the burden voltage is

VB =9 kn x 1S ILA= 135 mV

Other typical burden voltage specifications are 250 mV max 2 Von a 10 A range and 6 mVIrnA These voltages drops mayor may not be important depending on the circuit under test

PROBLEMS

4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V Rm = 1 kfl a 2 rnA meter and a transistor with hFpound = 80 Calculate a suitable resistance for R to give full scale deflection when E = 5 V Also determine the voltmeter input resistance

4-2 An emitter-follower voltmeter circuit as in Figure 4-2 has the following composhynents R =12 kfl R2 =R3 =21 kG R4 =14 =33 kfl Rs =SOO n and R + Rm = 10 kG A 100 f1A meter is used the supply voltage is plusmn9 V and the transistors have hFpound =7S Determine Vp [81 [82 [2 [3 and 14 when E = O Also calculate the range of adjustment for Vp

4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input voltshyage levels are 06 V 075 V and 1 V

4-4 A 3S V input (E) is applied to the input attenuator shown in Figure 4-4 Calculate the voltage EG on each range selection

4-5 The FET input voltmeter circuit in Figure 4-4 has the following components R = 68 kG R2 = R3 =47 kil R4 = 15 kG Rs = 500 il R6 =33 kG R + Rm = 20 kG The meter full-scale current is 50 ILA the supply voltage is plusmn10 V the transistors

Problems 115

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4

Potentiometer R3 in Figure 4-5(b) is an alternative method of providing meter-zero adjustment Q2 base control as in Figure 4-4 could also be used in the circuit of Figure 4-5 When the movable contact of R3 is adjusted to the right the portion of R3 added to RLI is increased and the portion of R3 added to Ru is reduced When the contact is moved left the reverse is true Thus VC and VC2 can be adjusted differentially by means of R3 bull

and the meter voltage can be set to zero

4-2 OPERATIONAL AMPLIFIER VOLTMETER CIRCUITS

Op-Amp Voltage-Follower Voltmeter

The operational amplifier voltage-follower voltmeter in Figure 4-6 is comparable to the simple emitter-follower circuit However unlike the emitter-follower there is no baseshyemitter voltage drop from input to output The voltage-foLLower also has a much higher input resistance and lower output resistance than the emitter-follower The voltageshyfoLLower input (pound8) is applied to the op-amp noninverting input terminaL and the feedback from the output goes to the inverting input The very high internal voltage gain of the operational amplifier combined with the negative feedback tends to keep the inverting terminal voltage exactly equal to that at the noninverting terminal Thus the output voltage (Vo) exactly follows the input As discussed earlier the attenuator selects the voltmeter range

I I I I I 10( I

lnput attenuator

I ~ I 0( Voltage follower

I Meter --_~+Io(-- _

I CirCUit

I

I I I I I I I I I I I I I I I I I I I

o---~~---- I I

I I

I I I I I I

I

Figure 4middot6 An Ie operational amplifier voltage-follower voltmeter is similar to the emitter middotfollower voltmeter except that the voltage-fOllower input resistance is much higher than that of the emitter follower and there is no base~miner voltage drop

Sec 4-2 Operational Amplifier Voltmeter Circuits 95

Op-Amp Amplifier Voltmeter

Like a transistor amplifier an IC operational amplifier circuit can be used to amplify low voltages to levels measurable by a deflection instrument Figure 4-7 shows a suitable opshyamp circuit for this purpose Input voltage E is applied to the op-amp noninverting input the output voltage is divided across resistors R3 and R4 and VR3 is fed back to the op-amp inverting input terminal The internal voltage gain of the op-amp causes VR3 to always equal E Consequently the output voltage is

v = E_R3_+_R-4 (4-1) o R3

The circuit is known as a noninverting ampLifie 1 because its output is positive when a positi ve input voltage is applied and negative when the input is a negative quantity The noninverting amplifier has a very high input resistance very low output resistance and a voltage gain of

(4-2)

1 1 1 1 10(

Noninverting amplifier

------~fo(f__- Meter circuit

------ 1 1 1 1 1

t E

Figure 4-7 An operational amplifier noninverting amplifier can be used to amplify low input voltages to a level suitable for the deflection meter circuit The voltmeter gain is (R3 + R4 )IR3middot

96 Analog Electronic Volt-Ohm-Millimeters Chap 4

An op-amp noninverting amplifier voltmeter is very easily designed Current 14 through R3 and ~ is first selected very much larger than the op-amp input bias current (I8) Then the resistors are calculated as

and

Example 4-4

An op-amp voltmeter circuit as in Figure 4-7 is required to measure a maximum input of 20 mV The op-amp input current is 02 JLA and the meter circuit has I = 100 fLA FSD and Rm =10 kil Determine suitable resistance values for R3 and R4 bull

Solution

Select [4 = 1000 X In =1000 x 02 -LA

=02mA

At full scale 1m = 100 V-A

and

=IV

= 100 n

R4 = Vo - E = I V - 20 mV 14 02mA

=49 kil

Voltage-to-Current Converter

The circuit shown in Figure 4-8 is essentially a noninverting amplifier as in Figure 4-7 However instead of connecting the meter between the op-amp output and ground it is substituted in place of resistor R4 (in Figure 4-7) Once again VR1 remains equal to the input voltage and as long as IRl is very much greater than lB the meter current is

(4-3)

Sec 4-2 Operational Amplifier Voltmeter Circuits 97

E

1 Figure 4-8 Voltmeter circuit using an op-amp voltage-to-current converter The meter current is FR3

Example 4-5

Calculate the value of R3 for the circuit in Figure 4-8 if E = 1 V is to give FSD on the meter The moving-coil meter has 1= 1 rnA at full scale and Rnr = 100 n Also detennine the maximum voltage at the operational amplifier output terminaL

Solution From Equation 4-3

E 1 V R) = -- = -- = 1 kfl

I(FSD) 1 rnA

Vo = I(R3 + Rm)

= 1 mA(l kfl + 100 fl)

=11 V

=

Many electronic rnultirange instruments do not have any current-measuring facilishyties Those that do measure current generally have very low-level current ranges and some have relatively high resistances when operating as ammeters For example the meter resistance on one instrument is specified as 9 kfl when operating on a 1S fJA range This must be taken into account when the instrument is connected in series with a circuit in which the current is to be measured The instrument terminal voltage drop when used as an ammeter is termed the burden voltage For a 9 kfl resistance on a 15 ILA range the burden voltage is

VB =9 kn x 1S ILA= 135 mV

Other typical burden voltage specifications are 250 mV max 2 Von a 10 A range and 6 mVIrnA These voltages drops mayor may not be important depending on the circuit under test

PROBLEMS

4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V Rm = 1 kfl a 2 rnA meter and a transistor with hFpound = 80 Calculate a suitable resistance for R to give full scale deflection when E = 5 V Also determine the voltmeter input resistance

4-2 An emitter-follower voltmeter circuit as in Figure 4-2 has the following composhynents R =12 kfl R2 =R3 =21 kG R4 =14 =33 kfl Rs =SOO n and R + Rm = 10 kG A 100 f1A meter is used the supply voltage is plusmn9 V and the transistors have hFpound =7S Determine Vp [81 [82 [2 [3 and 14 when E = O Also calculate the range of adjustment for Vp

4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input voltshyage levels are 06 V 075 V and 1 V

4-4 A 3S V input (E) is applied to the input attenuator shown in Figure 4-4 Calculate the voltage EG on each range selection

4-5 The FET input voltmeter circuit in Figure 4-4 has the following components R = 68 kG R2 = R3 =47 kil R4 = 15 kG Rs = 500 il R6 =33 kG R + Rm = 20 kG The meter full-scale current is 50 ILA the supply voltage is plusmn10 V the transistors

Problems 115

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4

Op-Amp Amplifier Voltmeter

Like a transistor amplifier an IC operational amplifier circuit can be used to amplify low voltages to levels measurable by a deflection instrument Figure 4-7 shows a suitable opshyamp circuit for this purpose Input voltage E is applied to the op-amp noninverting input the output voltage is divided across resistors R3 and R4 and VR3 is fed back to the op-amp inverting input terminal The internal voltage gain of the op-amp causes VR3 to always equal E Consequently the output voltage is

v = E_R3_+_R-4 (4-1) o R3

The circuit is known as a noninverting ampLifie 1 because its output is positive when a positi ve input voltage is applied and negative when the input is a negative quantity The noninverting amplifier has a very high input resistance very low output resistance and a voltage gain of

(4-2)

1 1 1 1 10(

Noninverting amplifier

------~fo(f__- Meter circuit

------ 1 1 1 1 1

t E

Figure 4-7 An operational amplifier noninverting amplifier can be used to amplify low input voltages to a level suitable for the deflection meter circuit The voltmeter gain is (R3 + R4 )IR3middot

96 Analog Electronic Volt-Ohm-Millimeters Chap 4

An op-amp noninverting amplifier voltmeter is very easily designed Current 14 through R3 and ~ is first selected very much larger than the op-amp input bias current (I8) Then the resistors are calculated as

and

Example 4-4

An op-amp voltmeter circuit as in Figure 4-7 is required to measure a maximum input of 20 mV The op-amp input current is 02 JLA and the meter circuit has I = 100 fLA FSD and Rm =10 kil Determine suitable resistance values for R3 and R4 bull

Solution

Select [4 = 1000 X In =1000 x 02 -LA

=02mA

At full scale 1m = 100 V-A

and

=IV

= 100 n

R4 = Vo - E = I V - 20 mV 14 02mA

=49 kil

Voltage-to-Current Converter

The circuit shown in Figure 4-8 is essentially a noninverting amplifier as in Figure 4-7 However instead of connecting the meter between the op-amp output and ground it is substituted in place of resistor R4 (in Figure 4-7) Once again VR1 remains equal to the input voltage and as long as IRl is very much greater than lB the meter current is

(4-3)

Sec 4-2 Operational Amplifier Voltmeter Circuits 97

E

1 Figure 4-8 Voltmeter circuit using an op-amp voltage-to-current converter The meter current is FR3

Example 4-5

Calculate the value of R3 for the circuit in Figure 4-8 if E = 1 V is to give FSD on the meter The moving-coil meter has 1= 1 rnA at full scale and Rnr = 100 n Also detennine the maximum voltage at the operational amplifier output terminaL

Solution From Equation 4-3

E 1 V R) = -- = -- = 1 kfl

I(FSD) 1 rnA

Vo = I(R3 + Rm)

= 1 mA(l kfl + 100 fl)

=11 V

=

Many electronic rnultirange instruments do not have any current-measuring facilishyties Those that do measure current generally have very low-level current ranges and some have relatively high resistances when operating as ammeters For example the meter resistance on one instrument is specified as 9 kfl when operating on a 1S fJA range This must be taken into account when the instrument is connected in series with a circuit in which the current is to be measured The instrument terminal voltage drop when used as an ammeter is termed the burden voltage For a 9 kfl resistance on a 15 ILA range the burden voltage is

VB =9 kn x 1S ILA= 135 mV

Other typical burden voltage specifications are 250 mV max 2 Von a 10 A range and 6 mVIrnA These voltages drops mayor may not be important depending on the circuit under test

PROBLEMS

4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V Rm = 1 kfl a 2 rnA meter and a transistor with hFpound = 80 Calculate a suitable resistance for R to give full scale deflection when E = 5 V Also determine the voltmeter input resistance

4-2 An emitter-follower voltmeter circuit as in Figure 4-2 has the following composhynents R =12 kfl R2 =R3 =21 kG R4 =14 =33 kfl Rs =SOO n and R + Rm = 10 kG A 100 f1A meter is used the supply voltage is plusmn9 V and the transistors have hFpound =7S Determine Vp [81 [82 [2 [3 and 14 when E = O Also calculate the range of adjustment for Vp

4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input voltshyage levels are 06 V 075 V and 1 V

4-4 A 3S V input (E) is applied to the input attenuator shown in Figure 4-4 Calculate the voltage EG on each range selection

4-5 The FET input voltmeter circuit in Figure 4-4 has the following components R = 68 kG R2 = R3 =47 kil R4 = 15 kG Rs = 500 il R6 =33 kG R + Rm = 20 kG The meter full-scale current is 50 ILA the supply voltage is plusmn10 V the transistors

Problems 115

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4

An op-amp noninverting amplifier voltmeter is very easily designed Current 14 through R3 and ~ is first selected very much larger than the op-amp input bias current (I8) Then the resistors are calculated as

and

Example 4-4

An op-amp voltmeter circuit as in Figure 4-7 is required to measure a maximum input of 20 mV The op-amp input current is 02 JLA and the meter circuit has I = 100 fLA FSD and Rm =10 kil Determine suitable resistance values for R3 and R4 bull

Solution

Select [4 = 1000 X In =1000 x 02 -LA

=02mA

At full scale 1m = 100 V-A

and

=IV

= 100 n

R4 = Vo - E = I V - 20 mV 14 02mA

=49 kil

Voltage-to-Current Converter

The circuit shown in Figure 4-8 is essentially a noninverting amplifier as in Figure 4-7 However instead of connecting the meter between the op-amp output and ground it is substituted in place of resistor R4 (in Figure 4-7) Once again VR1 remains equal to the input voltage and as long as IRl is very much greater than lB the meter current is

(4-3)

Sec 4-2 Operational Amplifier Voltmeter Circuits 97

E

1 Figure 4-8 Voltmeter circuit using an op-amp voltage-to-current converter The meter current is FR3

Example 4-5

Calculate the value of R3 for the circuit in Figure 4-8 if E = 1 V is to give FSD on the meter The moving-coil meter has 1= 1 rnA at full scale and Rnr = 100 n Also detennine the maximum voltage at the operational amplifier output terminaL

Solution From Equation 4-3

E 1 V R) = -- = -- = 1 kfl

I(FSD) 1 rnA

Vo = I(R3 + Rm)

= 1 mA(l kfl + 100 fl)

=11 V

=

Many electronic rnultirange instruments do not have any current-measuring facilishyties Those that do measure current generally have very low-level current ranges and some have relatively high resistances when operating as ammeters For example the meter resistance on one instrument is specified as 9 kfl when operating on a 1S fJA range This must be taken into account when the instrument is connected in series with a circuit in which the current is to be measured The instrument terminal voltage drop when used as an ammeter is termed the burden voltage For a 9 kfl resistance on a 15 ILA range the burden voltage is

VB =9 kn x 1S ILA= 135 mV

Other typical burden voltage specifications are 250 mV max 2 Von a 10 A range and 6 mVIrnA These voltages drops mayor may not be important depending on the circuit under test

PROBLEMS

4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V Rm = 1 kfl a 2 rnA meter and a transistor with hFpound = 80 Calculate a suitable resistance for R to give full scale deflection when E = 5 V Also determine the voltmeter input resistance

4-2 An emitter-follower voltmeter circuit as in Figure 4-2 has the following composhynents R =12 kfl R2 =R3 =21 kG R4 =14 =33 kfl Rs =SOO n and R + Rm = 10 kG A 100 f1A meter is used the supply voltage is plusmn9 V and the transistors have hFpound =7S Determine Vp [81 [82 [2 [3 and 14 when E = O Also calculate the range of adjustment for Vp

4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input voltshyage levels are 06 V 075 V and 1 V

4-4 A 3S V input (E) is applied to the input attenuator shown in Figure 4-4 Calculate the voltage EG on each range selection

4-5 The FET input voltmeter circuit in Figure 4-4 has the following components R = 68 kG R2 = R3 =47 kil R4 = 15 kG Rs = 500 il R6 =33 kG R + Rm = 20 kG The meter full-scale current is 50 ILA the supply voltage is plusmn10 V the transistors

Problems 115

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4

E

1 Figure 4-8 Voltmeter circuit using an op-amp voltage-to-current converter The meter current is FR3

Example 4-5

Calculate the value of R3 for the circuit in Figure 4-8 if E = 1 V is to give FSD on the meter The moving-coil meter has 1= 1 rnA at full scale and Rnr = 100 n Also detennine the maximum voltage at the operational amplifier output terminaL

Solution From Equation 4-3

E 1 V R) = -- = -- = 1 kfl

I(FSD) 1 rnA

Vo = I(R3 + Rm)

= 1 mA(l kfl + 100 fl)

=11 V

=

Many electronic rnultirange instruments do not have any current-measuring facilishyties Those that do measure current generally have very low-level current ranges and some have relatively high resistances when operating as ammeters For example the meter resistance on one instrument is specified as 9 kfl when operating on a 1S fJA range This must be taken into account when the instrument is connected in series with a circuit in which the current is to be measured The instrument terminal voltage drop when used as an ammeter is termed the burden voltage For a 9 kfl resistance on a 15 ILA range the burden voltage is

VB =9 kn x 1S ILA= 135 mV

Other typical burden voltage specifications are 250 mV max 2 Von a 10 A range and 6 mVIrnA These voltages drops mayor may not be important depending on the circuit under test

PROBLEMS

4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V Rm = 1 kfl a 2 rnA meter and a transistor with hFpound = 80 Calculate a suitable resistance for R to give full scale deflection when E = 5 V Also determine the voltmeter input resistance

4-2 An emitter-follower voltmeter circuit as in Figure 4-2 has the following composhynents R =12 kfl R2 =R3 =21 kG R4 =14 =33 kfl Rs =SOO n and R + Rm = 10 kG A 100 f1A meter is used the supply voltage is plusmn9 V and the transistors have hFpound =7S Determine Vp [81 [82 [2 [3 and 14 when E = O Also calculate the range of adjustment for Vp

4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input voltshyage levels are 06 V 075 V and 1 V

4-4 A 3S V input (E) is applied to the input attenuator shown in Figure 4-4 Calculate the voltage EG on each range selection

4-5 The FET input voltmeter circuit in Figure 4-4 has the following components R = 68 kG R2 = R3 =47 kil R4 = 15 kG Rs = 500 il R6 =33 kG R + Rm = 20 kG The meter full-scale current is 50 ILA the supply voltage is plusmn10 V the transistors

Problems 115

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4

Many electronic rnultirange instruments do not have any current-measuring facilishyties Those that do measure current generally have very low-level current ranges and some have relatively high resistances when operating as ammeters For example the meter resistance on one instrument is specified as 9 kfl when operating on a 1S fJA range This must be taken into account when the instrument is connected in series with a circuit in which the current is to be measured The instrument terminal voltage drop when used as an ammeter is termed the burden voltage For a 9 kfl resistance on a 15 ILA range the burden voltage is

VB =9 kn x 1S ILA= 135 mV

Other typical burden voltage specifications are 250 mV max 2 Von a 10 A range and 6 mVIrnA These voltages drops mayor may not be important depending on the circuit under test

PROBLEMS

4-1 A simple emitter-follower voltmeter circuit as in Figure 4-1 has Vee = 12 V Rm = 1 kfl a 2 rnA meter and a transistor with hFpound = 80 Calculate a suitable resistance for R to give full scale deflection when E = 5 V Also determine the voltmeter input resistance

4-2 An emitter-follower voltmeter circuit as in Figure 4-2 has the following composhynents R =12 kfl R2 =R3 =21 kG R4 =14 =33 kfl Rs =SOO n and R + Rm = 10 kG A 100 f1A meter is used the supply voltage is plusmn9 V and the transistors have hFpound =7S Determine Vp [81 [82 [2 [3 and 14 when E = O Also calculate the range of adjustment for Vp

4-3 Calculate the meter deflections for the circuit in Probl~m 4-2 when the input voltshyage levels are 06 V 075 V and 1 V

4-4 A 3S V input (E) is applied to the input attenuator shown in Figure 4-4 Calculate the voltage EG on each range selection

4-5 The FET input voltmeter circuit in Figure 4-4 has the following components R = 68 kG R2 = R3 =47 kil R4 = 15 kG Rs = 500 il R6 =33 kG R + Rm = 20 kG The meter full-scale current is 50 ILA the supply voltage is plusmn10 V the transistors

Problems 115

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4

bull

have hFE = 80 and the FET gate-source voltage is VGS = -3 V Determine Vp In 12

13 and 14 when E = O Also calculate the range of adjustment for Vp

4-6 Calculate the meter deflectio(ls for the circuit in Problem 4-5 when the attenuator is set to its 5 V range and the input voltage levels are I V 3 V and 4 Y

4-7 The difference amplifier voltmeter in Figure 4-5(a) has the following composhynents Rl = R2 = 15 kfl RLi = RL2 =39 kfl Rpound = 33 kfl Rs = 33 kfl and Rm =

750 fl The meter full-scale current is 50 ~A and the supply voltage is plusmn12 Y Calculate the transistor voltage levels when E = o

4-8 The circuit in Problem 4-7 has transistors with hFE = 100 and hi~ = 12 kfl Detershymine the input voltage (E) that will give full-scale deflection on the meter

4-9 An op-amp voltage-follower voltmeter as in Figure 4-6 has Ra = 800 kfl Rb = 100 kfl Rc = 60 kfl and Rd = 40 kfl A 50 ~A meter is used with a resistance of Rm = 750 fl Determine the required resistance for Rs to give full-scale deflection when E = 10 V and the range switch is as illustrated

4-10 The noninverting amplifier voltmeter circuit in Figure 4-7 uses an op-amp with I B = 300 nA and a 50 LA meter with Rm = 100 kfl Determine suitable resistances for R3 and R4 to give full-scale deflection when the input is 300 mY

4-11 The voltage-to-current converter circuit in Figure 4-8 uses a 375 LA (FSD) deflecshytion meter with Rm =900 fl If R) =80 kfl determine the required input voltage levels to give FSD and 05 FSD

4-12 Determine the new resistance for R3 for the circuit in Problem 4-11 to give FSD when E = 1 V Also calculate the op-amp output voltage

4-13 Calculate the resistance scale markings at 25 and 75 of full scale for the series ohmmeter circuit in Figure 4-9

4-14 Determine the percentage meter deflection in the circuit of Figure 4-9 when the 100 kfl standard resistor is switched into the circuit and Rx = 166 kfl

4-15 Calculate the meter deflection for the shunt ohmmeter circuit in Figure 4-10 when Rx =2 kO and when Rx =300 O

4-16 A 1667 kO resistor is substituted for RE in the linear ohmmeter circuit in Figure 4-11 Calculate the measured resistance when the meter indicates 39 V

4-17 The half-wave rectifier electronic voltmeter in Figure 4-12(b) uses a 500 ~A deflecshytion meter with a 460 fl coil resistance If Rs = 450 fl calculate the rms input voltshyage required to give full-scale deflection

4-18 The components used in Problem 4-17 are reconnected as in Figure 4-13(a) with R3 = Rs Determine the new rms input voltage required to give full-scale deflection Also determine the meter deftections when the input is 100 mV and 200 mY

4-19 The ac electronic voltmeter circuit in Figure 4-12(c) uses the following composhynents R =22 kfl R2 = 225 kfl R3 = 68 kfl Rs + Rm = 1 kfl and a 300 JLA meter Calculate the rms input voltages for meter fun-scale deflection and for 05 FSD

4-20 The full-wave rectifier voltmeter circuit in Figure 4-13(b) uses a 500 ~A meter with Rm =460 n together with R) = 450 n (as for Problems 4-17 and 4-18) Detershymine the rms input voltage for FSD on the meter

116 Analog Electronic Volt-Ohm-Millimeters Chap 4