1. Chapter 14 Feedback, Stability and Oscillators
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock

2. Chapter Goals Review concepts of negative and positive feedback.
Develop 2-port approach to analysis of negative feedback amplifiers.
Understand topologies and characteristics of series-shunt, shunt-shunt, shunt-series and series-series feedback configurations.
Discuss common errors that occur in applying 2-port feedback theory.
Discuss effects of feedback on frequency response and feedback amplifier stability and interpret stability in in terms of Nyquist and Bode plots.
Use SPICE ac and transfer function analyses on feedback amplifiers.
Determine loop-gain of closed-loop amplifiers using SPICE simulation or measurement.
Discuss Barkhausen criteria for oscillation and amplitude stabilization
Understand basic RC, LC and crystal oscillator circuits and present LCR model of quartz crystal.

3. Feedback Effects
Gain Stability: Feedback reduces sensitivity of gain to variations in values of transistor parameters and circuit elements.
Input and Output Impedances: Feedback can increase or decrease input and output resistances of an amplifier.
Bandwidth: Bandwidth of amplifier can be extended using feedback.
Nonlinear Distortion: Feedback reduces effects of nonlinear distortion.eg: removal of dead zone in class-B amplifiers

4. Classic Feedback Systems
A(s) = transfer function of open-loop amplifier or open-loop gain.
b(s) = transfer function of feedback network.
T(s) = loop gain
For negative feedback: T(s) > 0
For positive feedback: T(s) < 0

5. Voltage Amplifiers: Series-Shunt Feedback (Voltage Gain Calculation)
and ,

6. Voltage Amplifiers: Series-Shunt Feedback (Two-Port Representation)
Gain of amplifier should include effects of , , RI and RL.
Required h-parameters are found from their individual definitions.
Two-port representation of the amplifier is as shown

7. Voltage Amplifiers: Series-Shunt Feedback (Input and Output Resistances)
Series feedback at a port increases input resistance at that port.
For output resistance:
Shunt feedback at a port reduces resistance at that port.

8. Voltage Amplifiers: Series-Shunt Feedback (Example)
Problem: Find A, b, closed-loop gain, input and output resistances.
Given data: R1=10 kW, R2=91 kW, Rid=25 kW, Ro=1 kW, A=104.
Analysis:

9. Voltage Amplifiers: Series-Shunt Feedback (Example contd.)

10. Transresistance Amplifiers: Shunt-Shunt Feedback (Voltage Gain Calculation)
and ,

11. Transresitance Amplifiers: Shunt-Shunt Feedback (Two-Port Representation)
Gain of amplifier should include effects of , , RI and RL.
Required y-parameters are found from their individual definitions.
Two-port representation of the amplifier is as shown.

12. Transresistance Amplifiers: Shunt-Shunt Feedback (Input and Output Resistances)
Shunt feedback at a port reduces resistance at that port.
For output resistance:
Resistance at output port is reduced due to shunt feedback.

13. Transresistance Amplifiers: Shunt-Shunt Feedback (Example)
Problem: Find A, b, closed-loop gain, input and output resistances.
Given data: VA= 50 V, bF= 150
Analysis: From dc equivalent circuit,

14. Transresistance Amplifiers: Shunt-Shunt Feedback (Example contd.)

15. Current Amplifiers: Shunt-Series Feedback (Voltage Gain Calculation)
and ,

16. CurrentAmplifiers: Shunt-Series Feedback (Two-Port Representation)
Gain of amplifier should include effects of , , RI and RL.
Required g-parameters are found from their individual definitions.
Two-port representation of the amplifier is as shown

17. Current Amplifiers: Shunt-Series Feedback (Input and Output Resistances)
Shunt feedback at a port decreases resistance at that port.
For output resistance:
Series feedback at output port increases resistance at that port.

18. Transconductance Amplifiers: Series-Series Feedback (Voltage Gain Calculation)
and ,

19. Transconductance Amplifiers: Series-Series Feedback (Input and Output Resistances)
Gain of amplifier should include effects of , , RI and RL.
Required g-parameters are found from their individual definitions.
Two-port representation of the amplifier is as shown
Series feedback at input and output port increases resistance at both ports.

20. Erroneous Application of 2-port Feedback Theory
Problem: Find A, b, closed-loop gain, input and output resistances.
Given data: VREF = 5 V, bo= 100, VA = 50 V, Ao = 10,000, Rid =25 kW, Ro =0
Analysis: The circuit is redrawn to identify amplifier and feedback networks and appropriate 2-port parameters of feedback network are found.
This case seems to use series-series feedback.
ie is sampled by feedback network instead of io. This assumption is made since ao is approximately 1.

21. Erroneous Application of 2-port Feedback Theory (contd.)
Z-parameters are found as shown. From A-circuit, IE=1 mA

22. Erroneous Application of 2-port Feedback Theory (contd.)
Results for Rout are in error because output of op amp is referenced to ground, base current of BJT is lost from output port and feedback loop and Rout is limited to
3 and 4 are not valid terminals as current entering 3 is not same as that exiting 4. Amplifier can’t be reduced to a 2-port.
SPICE analyses confirm results for Atc and Rin, but results for Rout are in error. For Atc and Rin, amplifier can be properly modeled as a series-shunt feedback amplifier, as collector of Q1 can be directly connected to ground for calculations and a valid 2-port representation exists as shown.

23. Analysis of Shunt-Series Feedback Pair
Problem: Find A, b, closed-loop gain, input and output resistances.
Given data: bo= 100, VA = 100 V, Q-point for Q1:(0.66 mA, 2.3 V), Q-point for Q2:(1.6 mA, 7.5 V)
Analysis: The circuit is redrawn to identify amplifier and feedback networks and appropriate 2-port parameters of feedback network are found.
Shunt-shunt transresistance configuration is used.

24. Analysis of Shunt-Series Feedback Pair (contd.)
Small signal parameters are found from given Q-points.
For Q1, rp=3.79 kW, ro = 155 kW.
For Q2, rp =1.56 kW, ro = 64.8 kW.

25. Analysis of Shunt-Series Feedback Pair (contd.)
Closed-loop current gain is given by:

26. Direct Calculation of Loop Gain
Example:
is added for proper termination of feedback loop.
Original input source is set to zero.
Test source is inserted at the point where feedback loop is broken.

27. Calculation of Loop Gain using Successive Voltage and Current Injection
Current injection: Current source iX is inserted again at P.
Voltage injection: Voltage source vX is inserted at arbitrary point P in circuit.
where
As Ab = T

28. Simplifications to Successive Voltage and Current Injection Method
Technique is valid even if source resistances with vX and iX are included in analysis.
If at P, RB is zero or RA is infinite, T can be found by only one measurement and T = Tv . In ideal op amp, such point exists at op amp input.
If at P, RB is zero, T = Tv . In ideal op amp, such point exists at op amp output.
If RA = 0 or RB is infinite, T = TI .
In practice, if RB >> RA or RA >> RB, the simplified expressions can be used.

29. Blackman’s Theorem
First we select ports where resistance is to be calculated.
Next we select one controlled source in the amplifier’s equivalent circuit and use it to disable the feedback loop and also as reference to find TSC and TOC.
RCL = resistance of closed-loop amplifier looking into one of its ports (any terminal pair)
RD = resistance looking into same pair of terminals with feedback loop disabled.
TSC = Loop gain with a short-circuit applied to selected port
TOC = Loop gain with same port open-circuited.

30. Blackman’s Theorem (Example 1)
Problem: Find input and output resistances.
Given data:VREF =5 V, R =5 kW bo=100, VA=50 V, Ao=10,000, Rid=25 kW, Ro=0
Assumptions: Q-point is known, gm = 0.04 S, rp =25 kW, ro =25 kW.
For output resistance:
For input resistance:

31. Blackman’s Theorem (Example 2)
Problem: Find input and output resistances.
Given data:bo= 100, VA = 100 V, Q-point for Q1:(0.66 mA, 2.3 V), Q-point for Q2:(1.6 mA, 7.5 V). For Q1, rp=3.79 kW, ro = 155 kW, For Q2, rp =1.56 kW, ro = 64.8 kW.

32. Blackman’s Theorem (Example 2 contd.)
For output resistance:
For input resistance:

33. Blackman’s Theorem (Example 3)
Problem: Find expression for output resistance.
Analysis: Feedback loop is disabled by
setting reference source i to zero.
Next, i is set to 1
Assuming gm1= gm2 = gm3 and mf >> bo >>1.

34. Use of Feedback to Control Frequency Response
Assuming ,
Upper and lower cutoff frequencies as well as bandwidth of amplifier are improved, gain is stabilized at
where

35. Use of Nyquist Plot to Determine Stability
If gain of amplifier is greater than or equal to 1 at the frequency where feedback is positive, instability can arise.
Poles are at frequencies where T(s)=-1.
In Nyquist plots, each value of s in s-plane has corresponding value of T(s).
Values of s on jw axis are plotted.
If -1 point is enclosed by boundary, there is some value of s for which T(s)=-1, pole exists in RHP and amplifier is unstable.
If -1 point lies in outside interior of Nyquist plot, all poles of closed-loop amplifier are in LHP and amplifier is stable.

36. First-Order Systems At dc, T(0) = To, but for w >>1,
As w increases, magnitude monotonically approaches zero and phase asymptotically approaches -900.
As b changes, value of T(0) = To is scaled but as T(0) changes, radius of circle changes, but it can never enclose the -1 point, so amplifier is stable regardless of value of To.
For a simple low-pass amplifier,
It can also represent a single-pole op amp with resistive feedback

37. Second-Order Systems In given example,
T(0) =14, but, for high frequencies
As w increases, magnitude monotonically decreases from 14 towards zero and phase asymptotically approaches The transfer function can never enclose the -1 point but can come arbitrarily close to it.

38. Phase Margin
Where
First we determine frequency for which magnitude of loop gain is unity, corresponding to intersection of Nyquist plot with unit circle shown and then determine phase shift at this frequency. Difference between this angle and is phase margin.
Small phase margin causes excessive peaking in closed-loop frequency response and ringing in step response.
Phase Margin is the maximum increase in phase shift that can be tolerated before system becomes unstable.

39. Third-Order Systems In given example,
T(0) = 7, but, for high frequencies
As w increases, polar plot asymptotically approaches zero along positive imaginary axis and plot can enclose the -1 point under any circumstances and system is unstable.

40. Gain Margin
Where
If magnitude of T(jw) is increased by a factor equal to or exceeding gain margin, then closed-loop system becomes unstable, because Nyquist plot then encloses -1 point.
Gain Margin is the reciprocal of magnitude of T(jw) evaluated at frequency for which phase shift is 1800.

41. Bode Plots
At 1.2e+6 rad/s, magnitude of loop gain is unity and corresponding phase shift is 1450, and phase margin is given by = 350. Amplifier can tolerate additional phase shift of 350 before it becomes unstable.
At 3.2e+6 rad/s, phase shift is exactly 1800 and corresponding magnitude of loop gain is -17 dB, and phase margin is given by 17 dB. Gain of amplifier must increase by 17 dB before amplifier becomes unstable.

42. Use of Bode plot to Determine Stability
Frequency at which curves corresponding to magnitudes of open-loop gain and reciprocal of feedback factor intersect is the point at which loop gain is unity, phase margin is found from phase plot.
Assuming feedback is independent of frequency,
For 1/b =80 dB, fm=850, amplifier is stable.
For 1/b =50 dB, fm=150, amplifier is stable, but with significant overshoot and ringing in its step response.
For 1/b =0 dB, fm= -450, amplifier is unstable

43. Operational Amplifier Compensation Example
Problem: Find value of compensation capacitor for fm=700.
Given data: RC1=3.3 kW , RC2 =12 kW ,SPICE parameters-BF=100, VAF=75 V, IS=0.1 fA, RB=250 W, TF=0.75 ns, CJC= 2 pF.
Assumptions: Dominant pole is set by CC and pnp C-E stage. RZ is included to remove zero associated with CC, pnp and npn transistors are identical, quiescent value of Vo =0, VJC=0.75 V, MJC=0.33. Q4 and Q5 are in parallel, small signal resistances of diode-connected Q7 and Q8 can be neglected.

44. Operational Amplifier Compensation Example (contd.)
Analysis: IC1 = IC2 =250 mA. For Vo =0,voltage across RC2 = =11.3 V and IC3 =11.3V/12 kW=938 mA. Q4 and Q5 mirror currents in Q7 and Q8 , so, IC4 = IC5 =938 mA. For Vo =0, VCE4 =12 V, VCE5=12 V, VCE3 =11.3 V. For VI =0, VCE2 =12.8 V, VCE1 = (0.25 mA)+0.75= 11.9 V
Small signal parameters are found using their respective formulae.

45. Operational Amplifier Compensation Example (contd.)
Input stage pole:
Emitter Follower pole: Q4 and Q5 are in parallel, composite parameters are-
gm =0.02 S, rx =125 W, Cp =56.2 pF, Cm =1.60 pF, Rth =1/ gm3 =267 W.
At fT ,dominant pole due to CC contributes phase shift of 900. For fm=700, other 2 poles can contribute more phase margin of 200.
RZ =1/ gm3 =27.5 W is included to remove zero associated with CC.

46. Barkhausen’s Criteria for Oscillation.
For sinusoidal oscillations,
Barkhausen’s criteria state-
Phase shift around feedback loop should be zero degrees and magnitude of loop gain must be unity.
Loop gain greater than unity causes distorted oscillations.
For sinusoidal oscillator, poles of closed-loop amplifier should be at frequency w0 on jw axis.
Use positive feedback through frequency-selective feedback network to ensure sustained oscillation at w0 .
Or even multiples of 3600

47. Oscillators with Frequency-Selective RC Networks: Wien-Bridge Oscillator
Phase shift will be zero if = 0,
At w0 =1/RC
This oscillator is used for frequencies upto few MHz, limited primarily by characteristics of amplifier.

48. Oscillators with Frequency-Selective RC Networks: Phase-Shift Oscillator
Phase shift will be zero if = 0,
At w0

49. Amplitude Stabilization
Loop gain of oscillator changes due to power supply voltage, component value or temperature changes.
If loop gain is too small, desired oscillation decays and if it is too large, waveform is distorted.
Amplitude stabilization or gain control is used to automatically control loop gain and place poles exactly on jw axis.
At power on, loop gain is larger than that required for oscillation.As oscillation builds up, gain is reduced o minimum required to sustain oscillations.

50. Amplitude Stabilization in RC Oscillators: Method 1
R1 is replaced by a lamp. Small-signal resistance of lamp depends on temperature of bulb filament.
If amplitude is large, current is large, resistance of lamp increases, gain is reduced. If amplitude is small, lamp cools, resistance decreases, loop gain increases. Thermal time constant of bulb averages signal current and amplitude is stabilized.

51. Amplitude Stabilization in RC Oscillators: Method 2
Thus, when diodes are off, op amp gain is slightly >3 ensuring oscillation, but, when one diode is on, gain is reduced to slightly<3.
Same method can also be used in phase shift oscillators.
For positive signal at vo, D1 turns on as voltage across R3 exceeds diode turn-on voltage. R4 is in parallel with R3, loop gain is reduced. D2 functions similarly at negative signal peak.

52. LC Oscillators: Colpitts Oscillator
D=0, collect real and imaginary parts and set them to zero.
At w0
Generally more gain is used to ensure oscillation with amplitude stabilization.

53. LC Oscillators: Hartley Oscillator
D=0, collect real and imaginary parts and set them to zero.
G-S and G-D capacitances are neglected, assume no mutual coupling between inductors.
At w0
Generally more gain is used to ensure oscillation with amplitude stabilization.

54. Amplitude Stabilization in LC Oscillators
Inherent nonlinear characteristics of transistors are used to limit oscillation amplitude. Eg: rectification by JFET gate diode or BJT base-emitter diode.
In MOS version, diode and RG form rectifier to establish negative bias on gate, capacitors act as rectifier filter.
Practically, onset of oscillation is accompanied by slight shift in Q-point values as oscillator adjusts to limit amplitude.

55. Crystal Oscillators
Crystal: A piezoelectric device that vibrates is response to electrical stimulus, can be modeled electrically by a very high Q (>10,000) resonant circuit.
L, CS, R represent intrinsic series resonance path through crystal. CP is package capacitance. Equivalent impedance has series resonance where CS resonates with L and parallel resonance where L resonates with series combination of CS and CP.
Below wS and above wP, crystal appears capacitive, between wS and wP it exhibits inductive reactance.

56. Crystal Oscillators: Example
Problem: Find equivalent circuit elements for crystal with given parameters.
Given data: fS=5 MHz, Q=20,000 R =50 W, CP =5 pF
Analysis:

57. Crystal Oscillators: Topologies
Colpitts Crystal Oscillator
Crystal Oscillator using BJT
Crystal Oscillator using CMOS inverter as gain element.
Crystal Oscillator using JFET