Op-Amp Circuits: Part 6 - IIT Bombaysequel/ee101/mc_opamp_6.pdf · as the Barkhausen criterion. * For the circuit to oscillate at != !0, the network is designed such that the Barkhausen

Op-Amp Circuits: Part 6

M. B. [email protected]

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

Sinusoidal oscillators

Amplifier

network

Frequency−sensitive

XoAX′i

β Xo

Xf

X′iXi

Consider an amplifier with feedback.

Xo = AX ′i = A (Xi + Xf ) = A (Xi + βXo) = AXi + Aβ Xo

→ Af ≡Xo

Xi=

A

1− Aβ.

Since A and β will generally vary with ω, we re-write Af as

Af (jω) =A (jω)

1− A (jω)β (jω).

As A (jω)β (jω)→ 1, Af (jω)→∞, and we get a finite Xo ( = Af Xi ) even if Xi = 0.

In other words, we can remove Xi and still get a non-zero Xo . This is the basic principle behind sinusoidal

oscillators.



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi



→ Af ≡Xo

Xi=

A

1− Aβ.


Af (jω) =A (jω)

1− A (jω)β (jω).



oscillators.



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi



→ Af ≡Xo

Xi=

A

1− Aβ.


Af (jω) =A (jω)

1− A (jω)β (jω).



oscillators.



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi



→ Af ≡Xo

Xi=

A

1− Aβ.


Af (jω) =A (jω)

1− A (jω)β (jω).



oscillators.



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi



→ Af ≡Xo

Xi=

A

1− Aβ.


Af (jω) =A (jω)

1− A (jω)β (jω).



oscillators.



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi



→ Af ≡Xo

Xi=

A

1− Aβ.


Af (jω) =A (jω)

1− A (jω)β (jω).



oscillators.



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi

* The condition, A (jω)β (jω) = 1, for a circuit to oscillate spontaneously (i.e., without any input), is knownas the Barkhausen criterion.

* For the circuit to oscillate at ω = ω0, the β network is designed such that the Barkhausen criterion issatisfied only for ω0, i.e., all components except ω0 get attenuated to zero.

* The output Xo will therefore have a frequency ω0 (ω0/2π in Hz), but what about the amplitude?



Amplifier

network


XoAX′i

β Xo

Xf

X′iXi






Amplifier

network


XoAX′i

β Xo

Xf

X′iXi






Amplifier

network


XoAX′i

β Xo

Xf

X′iXi






Amplifier

network


Amplifier

network


gain limiter

Xo Xo

AX′i

β Xo β Xo

Xf

X′iXi

* A gain limiting mechanism is required to limit the amplitude of the oscillations.

* Amplifier clipping can provide a gain limiter mechanism. For example, in an op-amp, the output voltage islimited to ±Vsat, and this serves to limit the gain as the magnitude of the output voltage increases.

* For a more controlled output with low distortion, diode-resistor networks are used for gain limiting.



Amplifier

network


Amplifier

network


gain limiter

Xo Xo

AX′i

β Xo β Xo

Xf

X′iXi






Amplifier

network


Amplifier

network


gain limiter

Xo Xo

AX′i

β Xo β Xo

Xf

X′iXi






Amplifier

network


Amplifier

network


gain limiter

Xo Xo

AX′i

β Xo β Xo

Xf

X′iXi





Gain limiting network: example

R1

R2

D1

D2

V

2 k

2 k

I

SEQUEL file: ee101 diode circuit 14.sqproj

−2 −1 0 21

2

1

2

0

−2

V (Volts)

dV dI(kΩ)

I(mA)



R1

R2

D1

D2

V

2 k

2 k

I


−2 −1 0 21

2

1

2

0

−2

V (Volts)

dV dI(kΩ)

I(mA)



I

V

VCC

D′

D

R

R′1

R′2

R1

R2

20 k

15 k60 k

15 k60 k

(VCC = 15V)

−VEE

1

0

−1

20

8

I(mA)

dV dI(kΩ)

V (Volts)5 10−5 0−10



I

V

VCC

D′

D

R

R′1

R′2

R1

R2

20 k

15 k60 k

15 k60 k

(VCC = 15V)

−VEE

1

0

−1

20

8I(mA)

dV dI(kΩ)

V (Volts)5 10−5 0−10



Amplifier

network


gain limiter

Xo

β Xo

* Up to about 100 kHz, an op-amp based amplifier and a β network of resistors and capacitors can be used.

* At higher frequencies, an op-amp based amplifier is not suitable because of frequency response and slewrate limitations of op-amps.

* For high frequencies, transistor amplifiers are used, and LC tuned circuits or piezoelectric crystals are usedin the β network.



Amplifier

network


gain limiter

Xo

β Xo






Amplifier

network


gain limiter

Xo

β Xo






Amplifier

network


gain limiter

Xo

β Xo





Wien bridge oscillator

Amplifier

network


gain limiteramplifier

A

β network

Z1

Z2

R

R

C

C

Xo

β Xo

Assuming Rin →∞ for the amplifier, we get

A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

sRC

(sRC)2 + 3sRC + 1.

For A β = 1 (and with A equal to a real positive number),

jωRC

−ω2(RC)2 + 3jωRC + 1must be real and equal to 1/A.

→ ω =1

RC, A = 3



Amplifier

network



A

β network

Z1

Z2

R

R

C

C

Xo

β Xo


A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

sRC

(sRC)2 + 3sRC + 1.


jωRC


→ ω =1

RC, A = 3



Amplifier

network



A

β network

Z1

Z2

R

R

C

C

Xo

β Xo


A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

sRC

(sRC)2 + 3sRC + 1.


jωRC


→ ω =1

RC, A = 3



Amplifier

network



A

β network

Z1

Z2

R

R

C

C

Xo

β Xo


A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

sRC

(sRC)2 + 3sRC + 1.


jωRC


→ ω =1

RC, A = 3



amplifier

A

β network

0.1

0.01

90

0

−90

1

f (Hz)

Z1

Z2

R

R

C

C

V2V1

105104103102101

6H

|H|

R=158 kΩ

C=1nF

H(jω) =V2(jω)

V1(jω)=

jωRC

−ω2(RC)2 + 3jωRC + 1.

Note that the condition ∠H = 0 is satisfied only at one frequency, ω0 = 1/RC , i.e., f0 = 1 kHz.

At this frequency, |H| = 0.33, i.e., β(jω) = 1/3.

For A β = 1→ A = 3, as derived analytically.

SEQUEL file: ee101 osc 1.sqproj



amplifier

A

β network

0.1

0.01

90

0

−90

1

f (Hz)

Z1

Z2

R

R

C

C

V2V1

105104103102101

6H

|H|

R=158 kΩ

C=1nF

H(jω) =V2(jω)

V1(jω)=

jωRC

−ω2(RC)2 + 3jωRC + 1.







amplifier

A

β network

0.1

0.01

90

0

−90

1

f (Hz)

Z1

Z2

R

R

C

C

V2V1

105104103102101

6H

|H|

R=158 kΩ

C=1nF

H(jω) =V2(jω)

V1(jω)=

jωRC

−ω2(RC)2 + 3jωRC + 1.







amplifier

A

β network

0.1

0.01

90

0

−90

1

f (Hz)

Z1

Z2

R

R

C

C

V2V1

105104103102101

6H

|H|

R=158 kΩ

C=1nF

H(jω) =V2(jω)

V1(jω)=

jωRC

−ω2(RC)2 + 3jωRC + 1.







Amplifier

network


gain limiter

gain limiter

β network

amplifier

Ref.: S. Franco, "Design with Op Amps and analog ICs"

SEQUEL file: wien_osc_1.sqproj

Block diagram Implementation Output voltage

0

−1.5

1.5

0

2 1t (msec)

R2

R3

R1

Vo

Xo

β XoR

R C

C

10 k 22.1 k

100 k

158 k 1 nFVo

* ω0 =1

RC=

1

(158 k)× (1 nF)→ f0 = 1 kHz.

* Since the amplifier gain is required to be A = 3, we must have 1 +R2

R1= 3→ R2 = 2 R1.

* For gain limiting, diodes have been used. With one of the two diodes conducting, R2 → R2 ‖ R3, and the gain reduces.

* Note that there was no need to consider loading of the β network by the amplifier because of the large input resistance ofthe op-amp. That is why β could be computed independently.



Amplifier

network


gain limiter

gain limiter

β network

amplifier




0

−1.5

1.5

0

2 1t (msec)

R2

R3

R1

Vo

Xo

β XoR

R C

C

10 k 22.1 k

100 k

158 k 1 nFVo

* ω0 =1

RC=

1

(158 k)× (1 nF)→ f0 = 1 kHz.


R1= 3→ R2 = 2 R1.





Amplifier

network


gain limiter

gain limiter

β network

amplifier




0

−1.5

1.5

0

2 1t (msec)

R2

R3

R1

Vo

Xo

β XoR

R C

C

10 k 22.1 k

100 k

158 k 1 nFVo

* ω0 =1

RC=

1

(158 k)× (1 nF)→ f0 = 1 kHz.


R1= 3→ R2 = 2 R1.





Amplifier

network


gain limiter

gain limiter

β network

amplifier




0

−1.5

1.5

0

2 1t (msec)

R2

R3

R1

Vo

Xo

β XoR

R C

C

10 k 22.1 k

100 k

158 k 1 nFVo

* ω0 =1

RC=

1

(158 k)× (1 nF)→ f0 = 1 kHz.


R1= 3→ R2 = 2 R1.





Amplifier

network


gain limiter

gain limiter

β network

amplifier




0

−1.5

1.5

0

2 1t (msec)

R2

R3

R1

Vo

Xo

β XoR

R C

C

10 k 22.1 k

100 k

158 k 1 nFVo

* ω0 =1

RC=

1

(158 k)× (1 nF)→ f0 = 1 kHz.


R1= 3→ R2 = 2 R1.




Phase-shift oscillator

B

SEQUEL file: ee101_osc_4.sqproj

A

R1 R2

IV

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

180

90

270

|I(s)/V(s)| (A/V)

6 (I(s)/V(s)) (deg)

Let R1 = R2 = R = 10 k, G = 1/R, and C1 = C2 = C3 = C = 16 nF .

Using nodal analysis,

sC(VA − V ) + GVA + sC(VA − VB ) = 0 (1)

sC(VB − VA) + GVB + sCVB = 0 (2)

Solving (1) and (2), we get I =1

R

(sRC)3

3 (sRC)2 + 4 sRC + 1V .

STOP



B


A

R1 R2

IV

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

180

90

270

|I(s)/V(s)| (A/V)

6 (I(s)/V(s)) (deg)



sC(VA − V ) + GVA + sC(VA − VB ) = 0 (1)



R

(sRC)3

3 (sRC)2 + 4 sRC + 1V .

STOP



B


A

R1 R2

IV

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

180

90

270

|I(s)/V(s)| (A/V)

6 (I(s)/V(s)) (deg)



sC(VA − V ) + GVA + sC(VA − VB ) = 0 (1)



R

(sRC)3

3 (sRC)2 + 4 sRC + 1V .

STOP



B


A

R1 R2

IV

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

180

90

270

|I(s)/V(s)| (A/V)

6 (I(s)/V(s)) (deg)



sC(VA − V ) + GVA + sC(VA − VB ) = 0 (1)



R

(sRC)3

3 (sRC)2 + 4 sRC + 1V .

STOP



B


A

R1 R2

IV

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

180

90

270

|I(s)/V(s)| (A/V)

6 (I(s)/V(s)) (deg)



sC(VA − V ) + GVA + sC(VA − VB ) = 0 (1)



R

(sRC)3

3 (sRC)2 + 4 sRC + 1V .

STOP



B


A

R1 R2

IV

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

180

90

270

|I(s)/V(s)| (A/V)

6 (I(s)/V(s)) (deg)



sC(VA − V ) + GVA + sC(VA − VB ) = 0 (1)



R

(sRC)3

3 (sRC)2 + 4 sRC + 1V .

STOP



B

f (Hz)


A

R1 R2

IV

105104103

C1 C3C2

102101

10−10

10−2

180

90

270

6 (I(s)/V(s)) (deg)

|I(s)/V(s)| (A/V)

(R1 = R2 = R = 10 k, and C1 = C2 = C3 = C = 16 nF .)

β(jω) =I (jω)

V (jω)=

1

R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

For β(jω) to be a real number, the denominator must be purely imaginary.

→ −3(ωRC)2 + 1 = 0, i.e., 3(ωRC)2 = 1 → ω ≡ ω0 =1√

3

1

RC→ f0 = 574 Hz .

Note that, at ω = ω0, β(jω0) =1

R

(j/√

3)3

4 j/√

3= −

1

12 R= −8.33× 10−6.



B

f (Hz)


A

R1 R2

IV

105104103

C1 C3C2

102101

10−10

10−2

180

90

270

6 (I(s)/V(s)) (deg)

|I(s)/V(s)| (A/V)

(R1 = R2 = R = 10 k, and C1 = C2 = C3 = C = 16 nF .)

β(jω) =I (jω)

V (jω)=

1

R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


→ −3(ωRC)2 + 1 = 0, i.e., 3(ωRC)2 = 1 → ω ≡ ω0 =1√

3

1

RC→ f0 = 574 Hz .


R

(j/√

3)3

4 j/√

3= −

1

12 R= −8.33× 10−6.



B

f (Hz)


A

R1 R2

IV

105104103

C1 C3C2

102101

10−10

10−2

180

90

270

6 (I(s)/V(s)) (deg)

|I(s)/V(s)| (A/V)

(R1 = R2 = R = 10 k, and C1 = C2 = C3 = C = 16 nF .)

β(jω) =I (jω)

V (jω)=

1

R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


→ −3(ωRC)2 + 1 = 0, i.e., 3(ωRC)2 = 1 → ω ≡ ω0 =1√

3

1

RC→ f0 = 574 Hz .


R

(j/√

3)3

4 j/√

3= −

1

12 R= −8.33× 10−6.



A B

β network

A B

current−to−voltage

converter

Rf

R2

I

R1

V V

R1 R2

I

VC3C2C1 C1 C3C2

Note that the functioning of the β network as a stand-alone circuit (left figure) and as a feedback block (right figure) is thesame, thanks to the virtual ground provided by the op-amp.

The amplifier gain is A(jω) ≡V (jω)

I (jω)=

0− Rf I (jω)

I (jω)= −Rf .

→ A(jω)β(jω) = −Rf1

R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

As seen before, at → ω = ω0 =1√

3

1

RC, we have

I (jω)

V (jω)= −

1

12 R.

For the circuit to oscillate, we need Aβ = 1→ −Rf

(−

1

12 R

)= 1, i.e., Rf = 12 R

In addition, we employ a gain limiter circuit to complete the oscillator design.



A B

β network

A B


converter

Rf

R2

I

R1

V V

R1 R2

I

VC3C2C1 C1 C3C2



I (jω)=

0− Rf I (jω)

I (jω)= −Rf .


R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


3

1

RC, we have

I (jω)

V (jω)= −

1

12 R.


(−

1

12 R

)= 1, i.e., Rf = 12 R




A B

β network

A B


converter

Rf

R2

I

R1

V V

R1 R2

I

VC3C2C1 C1 C3C2



I (jω)=

0− Rf I (jω)

I (jω)= −Rf .


R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


3

1

RC, we have

I (jω)

V (jω)= −

1

12 R.


(−

1

12 R

)= 1, i.e., Rf = 12 R




A B

β network

A B


converter

Rf

R2

I

R1

V V

R1 R2

I

VC3C2C1 C1 C3C2



I (jω)=

0− Rf I (jω)

I (jω)= −Rf .


R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


3

1

RC, we have

I (jω)

V (jω)= −

1

12 R.


(−

1

12 R

)= 1, i.e., Rf = 12 R




A B

β network

A B


converter

Rf

R2

I

R1

V V

R1 R2

I

VC3C2C1 C1 C3C2



I (jω)=

0− Rf I (jω)

I (jω)= −Rf .


R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


3

1

RC, we have

I (jω)

V (jω)= −

1

12 R.


(−

1

12 R

)= 1, i.e., Rf = 12 R




A B

β network

A B


converter

Rf

R2

I

R1

V V

R1 R2

I

VC3C2C1 C1 C3C2



I (jω)=

0− Rf I (jω)

I (jω)= −Rf .


R

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.


3

1

RC, we have

I (jω)

V (jω)= −

1

12 R.


(−

1

12 R

)= 1, i.e., Rf = 12 R




0

−6

6

Amplifier

network


gain limiter β network

ImplementationBlock diagram Output voltage

amplifier

(i−to−v converter)

gain limiter

Ref.: Sedra and Smith, "Microelectronic circuits"

0 1 2 3 4

t (msec)

Rf

Vo

VCC

−VEE

Vo

Xo

R1 R2

β Xo C1 C3C2


16 nF 16 nF 16 nF

10 k 10 k

3 k

3 k

1 k

1 k

125 k

ω0 =1√

3

1

RC→ f0 = 574 Hz, T = 1.74 ms .


Amplitude control using gain limiting network

1

0

−1

I

V

VCC

D′

D

R

R′1

R′2

R1

R2

20 k

15 k60 k

15 k60 k

R1=R′1= 15 k

25 k

40 k

20

8I(mA)

dV dI(kΩ)

V (Volts)

(VCC = 15V)

40 k

25 k

R1=R′1= 15 k

−VEE


5 10−5 0−10


Inverting amplifier, revisited

RL

R2

R1Vs

Vo

1 k

20

0

40

f (Hz)

106105104103102101

AV(dB)

SEQUEL file: ee101 inv amp 3.sqproj

R2=5k

10 k

25 k

50 k

* As seen earlier, AV = −R2/R1 → |AV | should be independent of the signal frequency.

* However, a measurement with a real op-amp will show that |AV | starts reducing at higher frequencies.

* If |AV | is increased, the gain “roll-off” starts at lower frequencies.

* This behaviour has to do with the frequency response of the op-amp which we have not considered so far.



RL

R2

R1Vs

Vo

1 k

20

0

40

f (Hz)

106105104103102101

AV(dB)


R2=5k

10 k

25 k

50 k







RL

R2

R1Vs

Vo

1 k

20

0

40

f (Hz)

106105104103102101

AV(dB)


R2=5k

10 k

25 k

50 k







RL

R2

R1Vs

Vo

1 k

20

0

40

f (Hz)

106105104103102101

AV(dB)


R2=5k

10 k

25 k

50 k







RL

R2

R1Vs

Vo

1 k

20

0

40

f (Hz)

106105104103102101

AV(dB)


R2=5k

10 k

25 k

50 k







RL

R2

R1Vs

Vo

1 k

20

0

40

f (Hz)

106105104103102101

AV(dB)


R2=5k

10 k

25 k

50 k






Frequency response of Op-Amp 741

0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

VoVi

The gain of the 741 op-amp starts falling at rather low frequencies, with fc ' 10 Hz!

The 741 op-amp (and many others) are designed with this feature to ensure that, in typical amplifier applications, the overallcircuit is stable (and not oscillatory).

In other words, the op-amp has been internally compensated for stability.

The gain of the 741 op-amp can be represented by,

A(s) =A0

1 + s/ωc,

with A0 ≈ 105 (i.e., 100 dB), ωc ≈ 2π × 10 rad/s.



0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

VoVi





A(s) =A0

1 + s/ωc,




0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

VoVi





A(s) =A0

1 + s/ωc,




0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

VoVi





A(s) =A0

1 + s/ωc,




0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

fc

VoVi

ft

A(jω) =A0

1 + jω/ωc, ωc ≈ 2π × 10 rad/s.

For ω ωc , we have A(jω) ≈A0

jω/ωc.

|A(jω)| becomes 1 when A0 = ω/ωc , i.e., ω = A0ωc .

This frequency, ωt = A0ωc , is called the unity-gain frequency.

For the 741 op-amp, ft = A0fc ≈ 105 × 10 = 106 Hz.

Let us see how the frequency response of the 741 op-amp affects the gain of an inverting amplifier.



0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

fc

VoVi

ft

A(jω) =A0

1 + jω/ωc, ωc ≈ 2π × 10 rad/s.


jω/ωc.







0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

fc

VoVi

ft

A(jω) =A0

1 + jω/ωc, ωc ≈ 2π × 10 rad/s.


jω/ωc.







0

100

Gain

(dB

)

ideal

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

fc

VoVi

ft

A(jω) =A0

1 + jω/ωc, ωc ≈ 2π × 10 rad/s.


jω/ωc.




Let us see how the frequency response of the 741 op-amp affects the gain of an inverting amplifier.M. B. Patil, IIT Bombay


Vi

R2

R1Vs

Vs R1

R2

Vs

R2

Vo VoVi

AV(s) ViAV(s) Vi

Ro

Ri

R1

Vo

Assuming Ri to be large and Ro to be small, we get

−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1

R1 + R2.

Using Vo(s) = AV (s) Vi (s) and AV (s) =A0

1 + s/ωc, we get

Vo(s)

Vs (s)= −

R2

R1

1[1 +

(R1 + R2

R1

)1

A0

]+

(R1 + R2

R1A0

)s

ωc

≈ −R2

R1

1

1 + s/ω′c, with ω′c =

ωcA0

1 + R2/R1=

ωt

1 + R2/R1.



Vi

R2

R1Vs

Vs R1

R2

Vs

R2

Vo VoVi

AV(s) ViAV(s) Vi

Ro

Ri

R1

Vo


−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1

R1 + R2.


1 + s/ωc, we get

Vo(s)

Vs (s)= −

R2

R1

1[1 +

(R1 + R2

R1

)1

A0

]+

(R1 + R2

R1A0

)s

ωc

≈ −R2

R1

1


ωcA0

1 + R2/R1=

ωt

1 + R2/R1.



Vi

R2

R1Vs

Vs R1

R2

Vs

R2

Vo VoVi

AV(s) ViAV(s) Vi

Ro

Ri

R1

Vo


−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1

R1 + R2.


1 + s/ωc, we get

Vo(s)

Vs (s)= −

R2

R1

1[1 +

(R1 + R2

R1

)1

A0

]+

(R1 + R2

R1A0

)s

ωc

≈ −R2

R1

1


ωcA0

1 + R2/R1=

ωt

1 + R2/R1.



Vi

R2

R1Vs

Vs R1

R2

Vs

R2

Vo VoVi

AV(s) ViAV(s) Vi

Ro

Ri

R1

Vo


−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1

R1 + R2.


1 + s/ωc, we get

Vo(s)

Vs (s)= −

R2

R1

1[1 +

(R1 + R2

R1

)1

A0

]+

(R1 + R2

R1A0

)s

ωc

≈ −R2

R1

1


ωcA0

1 + R2/R1=

ωt

1 + R2/R1.



RL

R2

R1Vs

Vo


1 k

145 k 167

gain (dB)R2 fc′ (kHz)

20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k

145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91

10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).



RL

R2

R1Vs

Vo


1 k145 k 167


20

0

40

f (Hz)

106105104103102101

AV(dB)

R2=5k

2010 k 91 10 k

2825 k 38

25 k

3450 k 19.6

50 k

Vo(s)

Vs (s)= −

R2

R1

1

1 + s/ω′cω′c =

ωt

1 + R2/R1, (ft = 1 MHz).


Documents

Op-Amp Circuits: Part 6 - IIT Bombaysequel/ee101/mc_opamp_6.pdf · as the Barkhausen criterion. * For the circuit to oscillate at != !0, the network is designed such that the Barkhausen