input circuits

8/10/2019 Input Circuits

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INPUT CIRCUITS

N. Sreekanth Nayak 08M128

Ashwin Nandagiri 08M107

Suvarna Sohan Mohan 08m152

Harsh L Kansagara 08M120

Shodhan S. 08M144

Undar Shritheja 08M156


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Applications

Used for signal conditioning of transducers

Used in any applications where we need high

resolution equipments


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CURRENT SENSING CIRCUITS

Principle of working

Resistance of a transducer changes when there

is change in physical quantity which is being

measured, there by, causing a change incurrent.


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Working

Rm = internal resistance of the circuit

Rt =maximum resistance of the transducerK =percentage factor ( 0 to 100 %)

Ei =voltage source


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By ohms law, we have the current flowing

in the circuit as

I = Ei/(Rm + kRt)

Maximum current flows when k=0; i,e

Imax= Ei/Rm

I/Imax = 1/{ 1 + k(Rt/Rm)}.

This relation is non-linear, which isundesirable.


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The graph shows variation of I/Imaxfor different

values of k and Rt/Rm.

Higher the value of Rt/Rm, higher is the sensitivity

of the circuit.

Output current depends on voltage. Hence a

constant supply of voltage is needed if calibration is to

be maintained.


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Ballast circuits(Voltage

sensitive circuits)


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It is a simple variation of the current sensitive circuit.

Voltage sensitive device(Voltmeter) is used in place of current

indicator. The ballast circuit Rb is inserted in the manner as Rmwas used in the

previous circuit.

Voltage sensitive circuit


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Ei = Input voltage.

Rb = Ballast resistance.

Rt = Maximum value of transducer resistance.

K = % factor vary between 0 and 1.0 (0% and 100%)depending on the magnitude of the input signal.

Eo = Voltage across KRt.


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It is assumed that the voltage measuring or recording device has

infinite resistance so that it does not draw any current.

Output current, io= ei (1)Rb+KRt

If eo = Voltage across KRt, then

Then output voltage , eo = i1(KRt) = ei(KRt ) (2)

Rb+KRt

Therefore, eo = (KRt )/Rb (3)

ei 1+(KRt/ Rb)


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Fora given circuit, (eo /ei)is a measure of the output and

(KRt/ Rb) is a measure of the input.

It is clear from the previous equation (3) that input outputrelationship for a Voltage sensitive (ballast) circuit is nonlinear.

Curve shows the relationship between the inputoutput for a ballast circuit.

eo

ei

K


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Sensitivity is the ratio of change of change in output to change

in input.

Sensitivity , (S) = de0 = eiRbRt (4)dei (Rb+KRt)

2

From eqn(4) it is clear that the sensitivity is different for

different values of ballast resistance Rb .

In order to design the circuit so that the sensitivity is

maximum ,an optimum value of Rb has to be obtained. This is

obtained by differentiating eqn (4) sensitivity w.r.to Rb ,

dS = eiRt (KRt -Rb) (5)

dRb (Rb+KRt)3


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This derivative may be zero under the following 2 conditions,

For Rb = infinity, which results in minimum sensitivity.

For Rb= KRt , where maximum sensitivity is obtained.

The 2ndrelation indicates that for full range usefulness the

value of the Rbmust be based on compromise, because Rba

constant cannot always have the value of KRt, a variable.

However Rbmay be selected to give maximum sensitivity bysetting its value corresponding to the value of KRt.

Disadvantages:

Input voltage has to be kept absolutely constant for high

sensitivity. Requires a voltage regulating source.


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Bridge Circuits


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The Wheat-stone Bridge

The basic Wheat-stone bridge was developed byS. H. Christie in 1833

It consists of four resistors connected to form a

quadrilateral, a source of excitation (voltage orcurrent) connected across one of the diagonals,and a voltage detector connected across theother diagonal.

The detector measures the difference betweenthe outputs of two voltage dividers connectedacross the excitation


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Voltage Dividers

V2

The Voltage across the resistor R2 is given by


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Working of a bridge

Where Vex is the excitation voltage supplied to bridge


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Working


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Voltage Sensitive Wheat-stone Bridge

When a resistive element changes its resistance in responseto the physical parameter being measured (e.g. a straingauge) it is called the active element, while the remainingresistors are called completion resistors.

If R1 is an active element, then an increase in the resistance ofthe active element R1 increases the output voltage. Adecrease in this resistance will decrease the voltage appearingat the output.

It is conversely true that if R2 is an active element, then an

increase in its resistance would result in a reduction of thevoltage appearing at the output, while a decrease in thisresistance would result in the output voltage increasing.


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Assumptions


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Advantage


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Disadvantage


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The Current Sensitive

Wheatstone Bridgeig = ` ii * (R2R3-R1R4)

-------------------------------------------------

Rg*(R1+R2+R3+R4)+(R2+R4)(R1+R3)

Rg is the galvanometer resistance


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The Constant Current Bridge

eo = ii * (R2R3-R1R4)

--------------------------

(R1+R2+R3+R4)

eo is the output voltage.


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Wheatstone Deflecton type Bridge

Current Sensitivity can be changed Zero offset can be done

Can be calibrated to directly read out the

converted values.


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Wheatstone Deflecton type Bridge

Current Sensitivity can be changed

Zero offset can be done Can be calibrated to directly read out

the converted values.


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Problems

Ballast circuit1. A Ballast circuit has to be used for a thermistor which has a

nominal resistance of 100.

a. What should be the value of ballast resistor Rb, formaximum sensitivity at the nominal transducer resistance?

b. If the ballast resistance as calculated in (a) is used along withan input of 10 volts, what output voltage will be obtainedwhen the transducer has resistance equal to nominalresistance?

c. If the conditions of (b) remain same, except the transducerresistance reduced by 5%, what output voltage will beindicated?

d. What are the sensitivities for (b) and (c)?


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Solutiona) Nominal transducer resistance - Rt= 100

At nominal resistance of transducer, K = 1

For max sensitivity,

Ballast resistance Rb = K*RtRb= 1*100 = 100

b) Output voltage with Rb= 100

eo= K*Rt /Rb * ei ;1+ K*Rt /Rb

= 1*100/100 * 10 = 1 * 10 = 5 volts

1+1*100/100 1+1


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2) A simple ballast circuit is used to measure the output of

pressure pickup. The circuit is designed so that the internal

resistance is equal to 6 times the transducer resistance. A

source of 100V is used to energize the circuit. Calculate the

voltage output for 25, 50, 60 and 80% full load on the

transducer.

Solution

Given Rb= 6R

t

eo= K*Rt /Rb * ei ;

1+ K*Rt /Rb

= K*1/6 * 100 = 16.7K

1+K*1/6 1+.167K

for K= 0.25 eo= 4V; K= 0.5 eo= 7.7V

K=0.6 eo= 9.1V; K=0.8 e

o= 11.78V


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Bridge circuit

A resistance bridge has configuration as shown in figure in

which R1= 120.4 , R2= 119.0 and R3= 119.7

a) What resistance must R4have for resistive balance?

b) If R4 has a value of 121.2

and the input voltage is12V dc, what is the output

voltage of the bridge?

Assume it to be voltage

sensitive bridge.


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Solution

a) For balancing the bridge,

the condition is R1 = R3

R2 R4

where R1= 120.4 , R2= 119.0 and R3= 119.7

R4= R3 * R2 =118.3

R1

b) Given, R4 = 121.2

therefore output voltage is given by

eo= R1 R4 - R2 R3 *ei =(120.4*121.2-119*119.7)* 12(R1 +R3)(R2 +R4) (120.4+119.7)(119+121.2)

= 72.4 mV


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Wheatstone Null type Bridge

=

3

4

Let R2 be the resistive

transducer

R3 and R4 be fixed resistance

R1 be adjustable resistance (Potentiometer)

Adjust R1 till voltmeter reads

zero

2 = 1 4

3


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input circuits

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