op-amp circuits: part 6 - iit bombaysequel/ee101/mc_opamp_6.pdf · as the barkhausen criterion. *...

Op-Amp Circuits: Part 6

M. B. Patilmbpatil@ee.iitb.ac.in

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

X′iXi

Consider an amplifier with feedback.

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

→ Af ≡Xo

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

X′iXi

→ Af ≡Xo

1− Aβ.

Af (jω) =A (jω)

1− A (jω)β (jω).

oscillators.

Amplifier

network

XoAX′i

X′iXi

→ Af ≡Xo

1− Aβ.

Af (jω) =A (jω)

1− A (jω)β (jω).

oscillators.

Amplifier

network

XoAX′i

X′iXi

→ Af ≡Xo

1− Aβ.

Af (jω) =A (jω)

1− A (jω)β (jω).

oscillators.

Amplifier

network

XoAX′i

X′iXi

→ Af ≡Xo

1− Aβ.

Af (jω) =A (jω)

1− A (jω)β (jω).

oscillators.

Amplifier

network

XoAX′i

X′iXi

→ Af ≡Xo

1− Aβ.

Af (jω) =A (jω)

1− A (jω)β (jω).

oscillators.

Amplifier

network

XoAX′i

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

X′iXi

Amplifier

network

XoAX′i

X′iXi

Amplifier

network

XoAX′i

X′iXi

Amplifier

network

Amplifier

network

gain limiter

AX′i

β Xo β Xo

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

AX′i

β Xo β Xo

X′iXi

Amplifier

network

Amplifier

network

gain limiter

AX′i

β Xo β Xo

X′iXi

Amplifier

network

Amplifier

network

gain limiter

AX′i

β Xo β Xo

X′iXi

Gain limiting network: example

SEQUEL file: ee101 diode circuit 14.sqproj

−2 −1 0 21

V (Volts)

dV dI(kΩ)

−2 −1 0 21

V (Volts)

dV dI(kΩ)

15 k60 k

(VCC = 15V)

−VEE

dV dI(kΩ)

V (Volts)5 10−5 0−10

15 k60 k

(VCC = 15V)

−VEE

8I(mA)

dV dI(kΩ)

V (Volts)5 10−5 0−10

Amplifier

network

gain limiter

* 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

Amplifier

network

gain limiter

Amplifier

network

gain limiter

Wien bridge oscillator

Amplifier

network

gain limiteramplifier

β network

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)2 + 3sRC + 1.

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

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

→ ω =1

RC, A = 3

Amplifier

network

β network

A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

(sRC)2 + 3sRC + 1.

→ ω =1

RC, A = 3

Amplifier

network

β network

A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

(sRC)2 + 3sRC + 1.

→ ω =1

RC, A = 3

Amplifier

network

β network

A(s) β(s) = AZ2

Z1 + Z2= A

R ‖ (1/sC)

R + (1/sC) + R ‖ (1/sC)= A

(sRC)2 + 3sRC + 1.

→ ω =1

RC, A = 3

amplifier

β network

f (Hz)

105104103102101

R=158 kΩ

H(jω) =V2(jω)

V1(jω)=

−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

β network

f (Hz)

105104103102101

R=158 kΩ

H(jω) =V2(jω)

V1(jω)=

−ω2(RC)2 + 3jωRC + 1.

amplifier

β network

f (Hz)

105104103102101

R=158 kΩ

H(jω) =V2(jω)

V1(jω)=

−ω2(RC)2 + 3jωRC + 1.

amplifier

β network

f (Hz)

105104103102101

R=158 kΩ

H(jω) =V2(jω)

V1(jω)=

−ω2(RC)2 + 3jωRC + 1.

Amplifier

network

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

−1.5

2 1t (msec)

β XoR

10 k 22.1 k

158 k 1 nFVo

* ω0 =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

β network

amplifier

−1.5

2 1t (msec)

β XoR

10 k 22.1 k

158 k 1 nFVo

* ω0 =1

(158 k)× (1 nF)→ f0 = 1 kHz.

R1= 3→ R2 = 2 R1.

Amplifier

network

gain limiter

β network

amplifier

−1.5

2 1t (msec)

β XoR

10 k 22.1 k

158 k 1 nFVo

* ω0 =1

(158 k)× (1 nF)→ f0 = 1 kHz.

R1= 3→ R2 = 2 R1.

Amplifier

network

gain limiter

β network

amplifier

−1.5

2 1t (msec)

β XoR

10 k 22.1 k

158 k 1 nFVo

* ω0 =1

(158 k)× (1 nF)→ f0 = 1 kHz.

R1= 3→ R2 = 2 R1.

Amplifier

network

gain limiter

β network

amplifier

−1.5

2 1t (msec)

β XoR

10 k 22.1 k

158 k 1 nFVo

* ω0 =1

(158 k)× (1 nF)→ f0 = 1 kHz.

R1= 3→ R2 = 2 R1.

Phase-shift oscillator

SEQUEL file: ee101_osc_4.sqproj

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

|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

(sRC)3

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

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

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

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

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

(sRC)3

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

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

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

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

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

(sRC)3

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

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

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

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

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

(sRC)3

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

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

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

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

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

(sRC)3

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

C1 C3C2

f (Hz)

105104103102101

10−10

10−2

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

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

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

(sRC)3

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

f (Hz)

105104103

C1 C3C2

102101

10−10

10−2

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)=

(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√

RC→ f0 = 574 Hz .

Note that, at ω = ω0, β(jω0) =1

(j/√

4 j/√

3= −

12 R= −8.33× 10−6.

f (Hz)

105104103

C1 C3C2

102101

10−10

10−2

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)=

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

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

RC→ f0 = 574 Hz .

(j/√

4 j/√

3= −

12 R= −8.33× 10−6.

f (Hz)

105104103

C1 C3C2

102101

10−10

10−2

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)=

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

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

RC→ f0 = 574 Hz .

(j/√

4 j/√

3= −

12 R= −8.33× 10−6.

β network

current−to−voltage

converter

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

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

As seen before, at → ω = ω0 =1√

RC, we have

I (jω)

V (jω)= −

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

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

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

β network

converter

VC3C2C1 C1 C3C2

I (jω)=

0− Rf I (jω)

I (jω)= −Rf .

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

RC, we have

I (jω)

V (jω)= −

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

β network

converter

VC3C2C1 C1 C3C2

I (jω)=

0− Rf I (jω)

I (jω)= −Rf .

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

RC, we have

I (jω)

V (jω)= −

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

β network

converter

VC3C2C1 C1 C3C2

I (jω)=

0− Rf I (jω)

I (jω)= −Rf .

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

RC, we have

I (jω)

V (jω)= −

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

β network

converter

VC3C2C1 C1 C3C2

I (jω)=

0− Rf I (jω)

I (jω)= −Rf .

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

RC, we have

I (jω)

V (jω)= −

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

β network

converter

VC3C2C1 C1 C3C2

I (jω)=

0− Rf I (jω)

I (jω)= −Rf .

(jωRC)3

3(jωRC)2 + 4 jωRC + 1.

RC, we have

I (jω)

V (jω)= −

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

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)

−VEE

β Xo C1 C3C2

16 nF 16 nF 16 nF

10 k 10 k

ω0 =1√

RC→ f0 = 574 Hz, T = 1.74 ms .

Amplitude control using gain limiting network

15 k60 k

R1=R′1= 15 k

8I(mA)

dV dI(kΩ)

V (Volts)

(VCC = 15V)

R1=R′1= 15 k

−VEE

5 10−5 0−10

Inverting amplifier, revisited

f (Hz)

106105104103102101

AV(dB)

SEQUEL file: ee101 inv amp 3.sqproj

* 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.

f (Hz)

106105104103102101

AV(dB)

f (Hz)

106105104103102101

AV(dB)

f (Hz)

106105104103102101

AV(dB)

f (Hz)

106105104103102101

AV(dB)

f (Hz)

106105104103102101

AV(dB)

Frequency response of Op-Amp 741

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

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.

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

A(s) =A0

1 + s/ωc,

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

A(s) =A0

1 + s/ωc,

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

A(s) =A0

1 + s/ωc,

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

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.

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

A(jω) =A0

1 + jω/ωc, ωc ≈ 2π × 10 rad/s.

jω/ωc.

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

A(jω) =A0

1 + jω/ωc, ωc ≈ 2π × 10 rad/s.

jω/ωc.

Op Amp 741

−20 dB/decade

f (Hz)

10610−1

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

Vo VoVi

AV(s) ViAV(s) Vi

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

−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1 + R2.

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

1 + s/ωc, we get

Vs (s)= −

(R1 + R2

≈ −R2

1 + s/ω′c, with ω′c =

1 + R2/R1=

1 + R2/R1.

Vo VoVi

AV(s) ViAV(s) Vi

−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1 + R2.

1 + s/ωc, we get

Vs (s)= −

(R1 + R2

≈ −R2

1 + R2/R1=

1 + R2/R1.

Vo VoVi

AV(s) ViAV(s) Vi

−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1 + R2.

1 + s/ωc, we get

Vs (s)= −

(R1 + R2

≈ −R2

1 + R2/R1=

1 + R2/R1.

Vo VoVi

AV(s) ViAV(s) Vi

−Vi (s) = Vs (s)R2

R1 + R2+ Vo(s)

R1 + R2.

1 + s/ωc, we get

Vs (s)= −

(R1 + R2

≈ −R2

1 + R2/R1=

1 + R2/R1.

145 k 167

gain (dB)R2 fc′ (kHz)

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

1 k145 k 167

f (Hz)

106105104103102101

AV(dB)

2010 k 91 10 k

2825 k 38

3450 k 19.6

Vs (s)= −

1 + s/ω′cω′c =

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

op-amp circuits: part 6 - iit bombaysequel/ee101/mc_opamp_6.pdf · as the barkhausen criterion. *...

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