Op Amp Circuit

2102-487 Industrial Electronics

Amplifier Fundamentals

Vi Ri VL-

++- AVi

Ro +

-+- VS

RS +

-RL

Source Amplifier Load

Vi +- AViRi

RoVo

-

+ +

-

Amplifier is a two-port device that accepts an external applied signal, referred to as input, an in turn produces a signal, referred to as output, that is proportional to the input: output = A x input

Voltage amplifier model Voltage amplifier with source and load

Ssi

ii V

RRRV+

= iLo

LL AV

RRRV+

=

si

i

Lo

L

S

L

RRRA

RRR

VV

++=

|||/| AVV SL <This shows that due to the loading effect

Loading effect complicates life because each time we change the source or the load , we need to recompute the overall gain, and we also have signal loss.

Amplifier Fundamentals

Vi +- AViVo

-

+ +

-Vi VL-

++- AVi

+

-+- VS

Rs +

-RL

Source Amplifier Load

Ideal voltage amplifier model Ideal Voltage amplifier with source and load

To eliminate loading effect, the voltage across Rs andRomust be zero regardless of Rs and RL. The only way to achieve this goal is b imposing Ri = ∞ and Ro =0

AVV

S

L = regardless of Rs and RL.

In practice, the loading effect can be eliminated by the conditions of Ri >> Rs and and Ro << RL

Operational Amplifier (Op Amp)The operational amplifier (op amp) is a voltage amplifier having extremely high gain. By combining with external components, op amp could be configured to perform a variety of operations such as addition, subtraction, multiplication, integration, and differentiation etc.

Op amp symbol Simple practical Op amp model

For the popular 741 op amp, Rd = 2 MΩ, a ~ 200,000, and Ro = 75 Ω

Vp = Non-inverting input voltage, Vn = Inverting input voltage and Vo = output voltagea = unloaded voltage gain.

dnp VVV =− )( npdo VVaaVV −==

Vn+-

+-Vp

+-Vo

+

-

+VCC

-VCC

Vn aVd+-Rd

+

-

Vd+

-+-

+-Vp

+-Vo

+VCC

-VCC

Ro

Ideal Op Amp

We define ideal op amp as being an ideal voltage amplifier with infinite gain. For the ideal op amp, a→ ∞, imply Vp = Vn,

Rd = ∞, imply In = Ip = 0 Ro = 0

Ip = Non-inverting input current, In = Inverting input current

Ideal practical Op amp model

Ideal Op Amp Rules:1. No current flows in to either input terminal2. There is no voltage difference between the two input terminals

aVd+-

+

-

Vd+

-Vn+-

+-Vp

+-Vo

+VCC

-VCC

CCoCC VVV ≤≤−

The output voltages are constrained by the following relationship

Operational Amplifier (Op Amp)

Op Amp Characteristic Property Values

∞∞00 (virtual short)0 (virtual short)|Vo| ≤ VCC

>200,000>2 MΩ<75 Ω<0.1 mV<50 pA|Vo| < VCC

Gain, aInput Resistance, RiOutput Resistance, RoInput voltage difference, Vp-VnInput current, i1 or i2Output voltage limits

Ideal Op Amp Typical Op AmpProperty

Analysis of Op Amp Circuit:

-

+Vp

Vn

Vo

+-Vi

i

RinRout

Vn aVd+-Ri

+

-

Vd+

-+-

+-Vp

+-Vo

Ro

Apply KVL: 0)()( =−+++− npoii VVaRRiV

i

np

RVV

i−

=But:

And we have:no VV =

+-

+-

Ri2 MΩ

75 Ω

Ro

a(Vp-Vn)

Vn

VpVi

+

-

Vo

io

i

i

o

RaRR

VV

)1(1

++−=

Voltage gain for the voltage follower

pi VV =and

Find Vo/Vi

Analysis of Op Amp Circuit:i

Apply KVL: 0)()( =−+++− npoii VVaRRiV

But:

And we have: oii

in RRai

VR ++== )1(

Rin

+-

+-

Ri2 MΩ

75 Ω

Ro

a(Vp-Vn)

Vn

VpVi

+

-

Vo

Find Input resistance: iVR i

in =

npi VViR −=

Find output resistance:0=

=iVt

tout i

VR

+-

Ri2 MΩ

75 ΩRo

a(Vp-Vn)

Vn

Vp

+- Vt

Simple equivalent circuit for finding Rout

Equivalent circuit for finding Rin

itA

Apply KCL at A:

it

0)(=

−−+

−+−

o

npt

i

pnt R

VVAVR

VVi

But: 0=nV tp VV =and

tio

iot V

RRRaRi )1( ++

=

And we have:io

io

t

tout RaR

RRiVR

)1( ++==

Analysis of Op Amp Circuit: Ideal Op Amp

aVd+-

+

-

Vd+

-Vn+-

+-Vp

+-Vo

Using Rule 2: (no voltage difference between inverting and non-inverting inputs)

Using Rule 1: (no current flows into the op Amp inputs)

ip

0== np ii ip VV =

onp VVV == io VV =

Rin

Input Impedance and Output Impedance

0 and =∞= outin RR

-

+Vp

Vn

Vo

+-Vi

Rout

Analysis of Op Amp Circuit:

Using the parameters of 741 op amp, Ri = 2 MΩ, a ~ 200,000, and Ro = 75 Ω

11051 6 ≈×−= −

i

o

VVVoltage gain:

Ω×= 10400 9inRInput resistance:

Ω= µ 375outROutput resistance:

Comparison between Ideal and Practical Voltage Follower

Voltage Gain

Input Resistance

Output Resistance

Ideal Op Amp Typical Op AmpProperty

1)1(

1 ≈++

−io

i

RaRR

1

GΩ 400)1( =++ oi RRa ∞

Ω=++

µ 375)1( io

io

RaRRR

0

Inverting Amplifier

KCL

A

Use KCL at point A and apply Rule 1:

1

0A in A out

f

v v v vR R− −

+ =

(no current flows into the inverting input)

Rearrange

1 1

1 1 0in outA

f f

v vvR R R R

+ − + =

Apply Rule 2:(no voltage difference between inverting and non-inverting inputs)

Since V+ at zero volts, therefore V- is also at zero volts too. 0Av =

1

0in out

f

v vR R

+ =1

fout

in

Rvv R

= −

Input Impedance and Output Impedance 0 and 1 == oi RRR

-

+Vout

+- +

-

R1

Rf

Vin

Non-inverting Amplifier

KCL

A

Use KCL at point A and apply Rule 1:

1

0A outA

f

v vvR R

−+ =

Apply Rule 2: in Av v=

1

1 fout

in

Rvv R

= +

Input Impedance and Output Impedance 0 and == ∞ oi RRR

-

+Vout

+-

+

-

R1

Rf

Vin

Basic Application of the Op Amp

-

+Vo

+- +

-

R1

R2

Vi

-

+Vo

+-

+

-

R1

R2

Vi

Inverting amplifier Non-Inverting amplifier

0=oR

1RRi =1

2

RRAv −=

0=oR

∞=iR1

21RRAv +=

Summing Amplifier: Mathematic Operation

Use KCL and apply Rule 1:

31 2 0A A outA A

f

v v v vv v v vR R R R

− −− −+ + + =

_

+

Rf

Ri1

vout

+

-v1

v2

v3

i2

i3

R

R

i

vA

vB

1 2 3i i i i= + +

Since vA = 0 (Rule 2)

( )1 2 3f

out

Rv v v v

R= − + +

Sum of v1, v2 and v3

_

+

Difference Amplifier: Mathematic Operation

Use KCL and apply Rule 1:

R1

vout

+

-v1v2

R3

vA

vB Since vA = vB (Rule 2) and

Substitute eq. (2) into eq. (1), we get

R4

R2

(1)

(2)

If R1 = R2 = R and R3 = R4 = Rf ( )2 1f

out

Rv v v

R= −

Difference of v1and v2

021

1 =−

+−

Rvv

Rvv outAA

243

4 vRR

Rvv BA

+

==

11

22

1

2

43

4 1 vRRv

RR

RRRvout −

+

+=

Differentiator and Integrator: Mathematic Operation

_

+

_

+

Differentiator

Integrator

R

vin

C

C

R

vin

i

vout

+

-

vout

+

-

i

i

i

vc+ -

outv iR= −

But Cdvi Cdt

= in Cv v=and

inout

dvv RCdt

= −

out Cv v= −

But0

1( ) (0)t

C Cv t idt vC

= +∫ inv iR=and

0

1 (0)t

out in Cv v dt vRC

= − +∫

Difference Amplifier: Superposition

-

+Vo

+-

R3+-

+

-

R1 R2

R4

V1

V2

-

+Vo

+-

R3

+

-

R1 R2

R4

V1 -

+VoR3+-

+

-

R1 R2

R4V2

11

2 VRRVo −=

21

2

43

4 1 VRR

RRRVo

+

+=

21

2

43

41

1

2 1 VRR

RRRV

RRVo

+

++−=

0 and 11 == oi RRR 0 and 432 =+= oi RRRR

Commode and Differential Mode

Differential mode input: 12 VVVdm −=

Common mode input: 212 VVVcm

+=

-

+Vo

+-

R3(=R1)+-

+

-

R1 R2

R4(=R2)

V1

V2

-

+Vo

+-

R3(=R1)

+- +

-

R1 R2

R4(=R2)

Vcm +- Vdm/2

Vdm/2

Difference amplifier, in terms of the common and differential-mode inputs

Rearrange: 2/1 dmcm VVV −= 2/2 dmcm VVV +=

Commode and Differential Mode

Output for difference amplifier:

-

+Vo

+-

R3(=R1)+-

+

-

R1 R2

R4(=R2)

V1

V2

Using superposition, the output from difference amp can be expressed as

cmcmdmdmo VAVAV +=

Adm = amplification of differential input Acm = amplification of common mode input

2211 VAVAVo +=

Difference amplifier

( ) ( )1221

2121

1 VVVVV −−+=

( ) ( )1221

2121

2 VVVVV −++=

( )( ) ( ) ( )1221

12121221 VVAAVVAAVo +++−−=

We have

Therefore in this case,

( )1221 AAAdm −= ( )21 AAAcm +=and

Commode mode rejection ratio: cm

dm

AA

10log20CMRR = Quality index

Commode and Differential Mode

-

+Vo

+-

R3(=R1)+-

+

-

R1 R2

R4(=R2)

V1

V2

The design gain of this amplifier is

Ex The difference amplifier is constructed with an ideal Op Amp and 1% tolerance resistors of nominal values 2.2 kΩ and 5.1 kΩ. The resistors were measured and fond to have the following resistance values:

318.2kΩ 2.2kΩ 1.5

1

2 ===RRA

Difference amplifier

4

3

2

1

RR

RR

=

However, this value is based on the assumption of equal resistor ratios:

R1 = 2.195 kΩ, R3 = 2.215 kΩ,R2 = 5.145 kΩ, R4 = 5.085 kΩ

Determine the gain of the differential amplifier and its CMRR

The more exaction expression for the output voltage

11

22

1

2

43

4 1 VRRV

RR

RRRVo −

+

+=

Commode and Differential Mode

+

+= 1

1

2

43

42 R

RRR

RA2211 VAVAVo += Here and1

21 R

RA −=

329.21195.2145.5

085.5215.2085.5

2 =

+

+=A

344.2195.2145.5

1 −=−=A

Therefore01464.0

337.2−=

=

cm

dm

AA

The CMRR is dB 06.446.159log20log20CMRR 10 =−==cm

dm

AA

This is only a moderately good differential amplifier. If physical resistors used for R1 and R3 were exchanged, the resistor ratios in each gain path would be more nearly exact:

321.21215.2145.5

085.5195.2085.5

2 =

+

+=A

323.2215.2145.5

1 −=−=A

Therefore00186.0

322.2−=

=

cm

dm

AA

dB 92.610.1248log20log20CMRR 10 =−==cm

dm

AA

Non-Ideal Op Amp

aVd+-

-

Ro

+

rd

VOS

+-

1/2IOS

Ibias

Ibias

Non-ideal characteristics

•Finite input resistance•Finite voltage gain•Nonzero output resistance•Output Saturation•Maximum output current

•Input offset voltage, VOS•Input bias current, Ibias•Input offset current, IOS

Vos - the difference in voltage between the Op Amp input terminals when the output voltage is zero

Ibias - the average of the two input currents when the output voltage is zero

IOS - the difference between the input currents( )npbias III += 2

1

pnOS III −=

Non-Ideal Op Amp

Apply KCL at the inverting input: ppp RIV −=

-

+Eo+

-

R1 R2

Rp

In

Ip

Vp

Vn

00

21

=+−

+−

nonn I

REV

RV

From Ohm’s Law:

By Op Amp action Vn = Vp , Eliminating Vn and Vp

( )[ ]ppno IRIRRRRE −

+= 21

1

2 //1

Rp can be specified to cancel the two terms in the brackets

Estimating the output error caused by the input bias currents

21 // RRR p =

In this case, this reduces the output error to

( )[ ]OSo IRRRRE 21

1

2 //1

+=

Non-Ideal Op Amp

VSAT++

-

Vd+

-

+-Vo

Vo

VSAT

+

+

-

Vd+

-

+-

aVd+-

+

-

Vd+

-

+-Vo

Vo (V)

Vd (µV)VSATH/a

VSATL/a

VSATL

VSAT

Linear region:

Upper saturation region:

Lower saturation region:

a

Non-Ideal Op AmpEx The 741 inverting amplifier is driven by a ±10 V peak to peak triangular wave. Sketch and label Vi, Vo and Vn. If 741 is supplied with ±15 V and this maximum output ±13 V.

-

+Vo

+-+

-

R1 R2

Vi10 kΩ 20 kΩ

-6.5V < Vi < 6.5V: the op amp is in the linear region Vo =- 2Vi

Vi > 6.5V: the op amp is Saturation Vo = -13 V

Vi < -6.5V: the op amp is Saturation Vo = 13 V

Vn

SATLin VRR

RVRR

RV21

1

21

2

++

+=

SATHin VRR

RVRR

RV21

1

21

2

++

+=

10 V

6.5V

Vi

-10 V

-6.5V

t

Vo

t

13 V

-13 V

Vn2.33 V

-2.33 V