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MODULE 4
FEEDBACK AND
OSCILLATOR CIRCUITS
ANALOG ELECTRONIC CIRCUITS
(ECE34)
16/10/2017
after
2Aravinda K., Dept. of E&C, NHCE
16/10/2017 3Aravinda K., Dept. of E&C, NHCE
Theory of sinusoidal oscillation
If the amplifier is provided with positive feedback,
and if the feedback signal is in correct phase with the
previous input, then there will be an output signal
without any external input signal.
16/10/2017
With respect to the input, the gain
of the amplifier is –A, due to the
phase shift of 180°. If the feedback
circuit also produces an additional
phase shift of 180°, then B is +ve,
and the overall gain equation is:
Af=A/(1-AB)
4Aravinda K., Dept. of E&C, NHCE
16/10/2017
AB < 1 AB > 1 AB = 1
Barkhausen criterion for oscillation: Magnitude of
Loop gain should be unity, at a phase angle of 180°.
The starting signal is the infinitesimal noise voltage,
which is always present, and contains all the frequencies,
due to the random movement of the free electrons. The
amplified noise drives the resonant feedback circuit, and
hence oscillations build up at the desired frequency.
5Aravinda K., Dept. of E&C, NHCE
• Bypass circuit, Lag circuit, Low pass circuit
• Coupling circuit, Lead circuit, High pass circuit
• These are examples of phase-shifting circuits,
from 0° till 90°, depending on the frequency.
16/10/2017 6Aravinda K., Dept. of E&C, NHCE
PHASE SHIFT OSCILLATOR
Here, B = 1/29 and hence A ≥ 29. Each RC stage
provides a phase-shift of 60° at a specific frequency,
provided the loading effect is taken care of.
16/10/2017
IC phase shift
oscillator
7Aravinda K., Dept. of E&C, NHCE
FET phase-shift
oscillator
In general, Ri is much
larger. Further, impedance
of the feedback network
is also much larger when
compared to RL.
16/10/2017 8Aravinda K., Dept. of E&C, NHCE
Exercise - 1
• An FET has gm = 5mS and rd = 50 KΩ. Obtain
the value of C and RD for the phase-shift
oscillator’s frequency of 10 KHz, choosing
R = 4.7 KΩ.
Solution:
C = 1/(2x3.14x2.45x4.7Kx10K) = 1.38 nF.
As A ≥ 29, let A = 40.  RL = 40/5m = 8 KΩ.
Hence, RD = rd RL / (rd - RL) = 9.52 KΩ ≈ 10 KΩ.
16/10/2017 9Aravinda K., Dept. of E&C, NHCE
BJT phase-shift
oscillator
Here, Ri is much lesser.
Hence, one solution is to
use an Emitter-follower
at the input stage, and
the other solution is to
use an additional resistor
R′ in series with Ri, such
that, R′ + Ri = R.
16/10/2017 10Aravinda K., Dept. of E&C, NHCE
Wien-Bridge
Oscillator
• This is the standard oscillator circuit used for general
applications, in the range of 5 Hz till 1 MHz.
• It uses the lead-lag circuit as the resonant feedback.
The series capacitor blocks LF and the parallel capacitor
bypasses HF, and hence, the phase angle varies
between +90° till -90°.
• The output remains maximum in the mid-range, at the
resonant frequency, where the phase angle is 0°.
16/10/2017 11Aravinda K., Dept. of E&C, NHCE
When R = XC, resonance occurs, and B = 1/3.
16/10/2017 12Aravinda K., Dept. of E&C, NHCE
16/10/2017
WIEN-
BRIDGE
OSCILLATOR
CIRCUIT
USING
OP-AMP
13Aravinda K., Dept. of E&C, NHCE
• It uses both feedback types: +ve through the
lead-lag circuit, and –ve through the voltage-
divider circuit.
• At power up, there is more +ve feedback, and
AB > 1, and hence oscillations build up.
• Slowly when the Tungsten lamp heats up, its
resistance increases, and when it equals R′,
• At this point, the oscillations become stable,
and the output voltage will have a constant
peak-to-peak value.
16/10/2017 14Aravinda K., Dept. of E&C, NHCE
• The bridge analysis yields,
• When R1=R2 and C1=C2, then R3/R4 = 2.
16/10/2017 15Aravinda K., Dept. of E&C, NHCE
Exercise - 2
• Obtain the RC values for a Wien-Bridge
Oscillator with an output frequency of 20 KHz.
Solution:
Let R = 100 KΩ.
Then, C = 1/(2x3.14x100Kx20K) = 79.62 pF.
If R4 = 100 KΩ, then R3 ≥ 200 KΩ.
16/10/2017 16Aravinda K., Dept. of E&C, NHCE
• Different types of oscillators can be obtained by choosing the
reactance elements in feedback network.
Oscillator Type
Reactance Elements
X1 X2 X3
Colpitts C C L
Hartley L L C
Tuned input
Tuned output
LC LC ------
Fig: Basic Configuration of Tuned Oscillators
TUNED OSCILLATOR CIRCUITS
16/10/2017 17Aravinda K., Dept. of E&C, NHCE
Colpitts Oscillator (BJT –CE Configuration)
Fig: Colpitts oscillator using BJT CE config.
• The voltage divider
bias set up the Q-point
for biasing.
• RF CHOKE has high
inductive reactance.
• C1, C2, L provide the
feedback for
oscillations.
• The feedback voltage
is across C2 drives the
base and sustain the
oscillations.
Fig: AC equivalent Circuit
16/10/2017 18Aravinda K., Dept. of E&C, NHCE
Colpitts Oscillator (BJT Configuration)
Resonant Frequency:
 Exact frequency
𝑓𝑟 =
1
2∏ 𝐿𝐶𝑒𝑞
𝑄2
𝑄2 + 1
 Approximate Frequency
𝑓𝑟 =
1
2∏ 𝐿𝐶 𝑒𝑞
for Q >10
𝑄 =
𝑓𝑟
𝐵𝑊
, 𝐶𝑒𝑞 =
𝐶1 𝐶2
𝐶1 + 𝐶2
16/10/2017
Starting Condition for
Oscillation
• AB >1
• A>1/B
• Feedback fraction to the
oscillator ‘B’
𝐵 =
𝐶1
𝐶2
• Minimum voltage gain Av
𝐴𝑣 =
𝐶2
𝐶1
19Aravinda K., Dept. of E&C, NHCE
Coupling Load to Oscillator
Capacitive coupling Link coupling
16/10/2017
• Directly connecting load lowers Q , hence may not initiate
oscillation process.
• Coupling should not lower the Q of tank circuit.
20Aravinda K., Dept. of E&C, NHCE
16/10/2017
Colpitts Oscillator (BJT –CB Configuration)
• CB Configuration oscillate at high
frequency.
• Frequency of Oscillation
• 𝑓𝑟 =
1
2∏ 𝐿𝐶 𝑒𝑞
𝑄2
𝑄2+1
𝐶𝑒𝑞=
𝐶1 𝐶2
𝐶1 + 𝐶2
• Feedback factor
• Minimum Gain
Fig:Colpitts oscillator BJT-CE confg.
21Aravinda K., Dept. of E&C, NHCE
Colpitts Oscillator-JFET,OPAMP
16/10/2017
Frequency of Oscillation 𝑓𝑟 =
1
2∏ 𝐿𝐶𝑒𝑞
Fig: Colpitts oscillator using OPAMP.Fig: Colpitts oscillator using JFET.
22Aravinda K., Dept. of E&C, NHCE
Problem
• What is the frequency of oscillation for Colpitts
oscillator ? What is the feedback fraction ? How
much voltage gain does circuit needs to start
oscillating ?
Given C1=0.001 uF, C2= 0.01uF, L=15 uF
Solution
fr= 1.36 MHz, B= 0.1, Av= 10
16/10/2017 23Aravinda K., Dept. of E&C, NHCE
16/10/2017
Hartley Oscillator
Fig: Hartley Oscillator using BJT. Fig: Hartley Oscillator using JFET.
24Aravinda K., Dept. of E&C, NHCE
Crystal Oscillator
16/10/2017
Characteristics of Quartz crystal
• Piezoelectric effect
 R represents the internal friction.
 L represents vibrating mass.
 Cs represents crystal compliance.
 CM represents capacitance due to
mechanical mounting of crystal.
Used for accurate and stable frequency of oscillations.
25Aravinda K., Dept. of E&C, NHCE
Characteristics of Quartz crystal
16/10/2017
Exhibits2 resonant frequencies
1. Series resonance frequency
-> When impedance is low (R)
2. Parallel resonance frequency
->When reactance of series
resonance equals the
reactance of CM.
26Aravinda K., Dept. of E&C, NHCE
16/10/2017
Series Resonant Circuit Parallel Resonant Circuit
Crystal Oscillator
27Aravinda K., Dept. of E&C, NHCE
16/10/2017
Problem
• What is the series and parallel resonant frequency of
crystal? Given L=3H, Cs=0.05 pF, R= 2KΩ, Cm=10
pF.
Solution
fs=411 KHz, Cp=0.0498 pF, fp=412 KHz.
28Aravinda K., Dept. of E&C, NHCE
END
OF
MODULE 4
16/10/2017 29Aravinda K., Dept. of E&C, NHCE

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ECE34 Module 4 - Feedback and Oscillator Circuits Notes

  • 1. MODULE 4 FEEDBACK AND OSCILLATOR CIRCUITS ANALOG ELECTRONIC CIRCUITS (ECE34)
  • 3. 16/10/2017 3Aravinda K., Dept. of E&C, NHCE
  • 4. Theory of sinusoidal oscillation If the amplifier is provided with positive feedback, and if the feedback signal is in correct phase with the previous input, then there will be an output signal without any external input signal. 16/10/2017 With respect to the input, the gain of the amplifier is –A, due to the phase shift of 180°. If the feedback circuit also produces an additional phase shift of 180°, then B is +ve, and the overall gain equation is: Af=A/(1-AB) 4Aravinda K., Dept. of E&C, NHCE
  • 5. 16/10/2017 AB < 1 AB > 1 AB = 1 Barkhausen criterion for oscillation: Magnitude of Loop gain should be unity, at a phase angle of 180°. The starting signal is the infinitesimal noise voltage, which is always present, and contains all the frequencies, due to the random movement of the free electrons. The amplified noise drives the resonant feedback circuit, and hence oscillations build up at the desired frequency. 5Aravinda K., Dept. of E&C, NHCE
  • 6. • Bypass circuit, Lag circuit, Low pass circuit • Coupling circuit, Lead circuit, High pass circuit • These are examples of phase-shifting circuits, from 0° till 90°, depending on the frequency. 16/10/2017 6Aravinda K., Dept. of E&C, NHCE
  • 7. PHASE SHIFT OSCILLATOR Here, B = 1/29 and hence A ≥ 29. Each RC stage provides a phase-shift of 60° at a specific frequency, provided the loading effect is taken care of. 16/10/2017 IC phase shift oscillator 7Aravinda K., Dept. of E&C, NHCE
  • 8. FET phase-shift oscillator In general, Ri is much larger. Further, impedance of the feedback network is also much larger when compared to RL. 16/10/2017 8Aravinda K., Dept. of E&C, NHCE
  • 9. Exercise - 1 • An FET has gm = 5mS and rd = 50 KΩ. Obtain the value of C and RD for the phase-shift oscillator’s frequency of 10 KHz, choosing R = 4.7 KΩ. Solution: C = 1/(2x3.14x2.45x4.7Kx10K) = 1.38 nF. As A ≥ 29, let A = 40.  RL = 40/5m = 8 KΩ. Hence, RD = rd RL / (rd - RL) = 9.52 KΩ ≈ 10 KΩ. 16/10/2017 9Aravinda K., Dept. of E&C, NHCE
  • 10. BJT phase-shift oscillator Here, Ri is much lesser. Hence, one solution is to use an Emitter-follower at the input stage, and the other solution is to use an additional resistor R′ in series with Ri, such that, R′ + Ri = R. 16/10/2017 10Aravinda K., Dept. of E&C, NHCE
  • 11. Wien-Bridge Oscillator • This is the standard oscillator circuit used for general applications, in the range of 5 Hz till 1 MHz. • It uses the lead-lag circuit as the resonant feedback. The series capacitor blocks LF and the parallel capacitor bypasses HF, and hence, the phase angle varies between +90° till -90°. • The output remains maximum in the mid-range, at the resonant frequency, where the phase angle is 0°. 16/10/2017 11Aravinda K., Dept. of E&C, NHCE
  • 12. When R = XC, resonance occurs, and B = 1/3. 16/10/2017 12Aravinda K., Dept. of E&C, NHCE
  • 14. • It uses both feedback types: +ve through the lead-lag circuit, and –ve through the voltage- divider circuit. • At power up, there is more +ve feedback, and AB > 1, and hence oscillations build up. • Slowly when the Tungsten lamp heats up, its resistance increases, and when it equals R′, • At this point, the oscillations become stable, and the output voltage will have a constant peak-to-peak value. 16/10/2017 14Aravinda K., Dept. of E&C, NHCE
  • 15. • The bridge analysis yields, • When R1=R2 and C1=C2, then R3/R4 = 2. 16/10/2017 15Aravinda K., Dept. of E&C, NHCE
  • 16. Exercise - 2 • Obtain the RC values for a Wien-Bridge Oscillator with an output frequency of 20 KHz. Solution: Let R = 100 KΩ. Then, C = 1/(2x3.14x100Kx20K) = 79.62 pF. If R4 = 100 KΩ, then R3 ≥ 200 KΩ. 16/10/2017 16Aravinda K., Dept. of E&C, NHCE
  • 17. • Different types of oscillators can be obtained by choosing the reactance elements in feedback network. Oscillator Type Reactance Elements X1 X2 X3 Colpitts C C L Hartley L L C Tuned input Tuned output LC LC ------ Fig: Basic Configuration of Tuned Oscillators TUNED OSCILLATOR CIRCUITS 16/10/2017 17Aravinda K., Dept. of E&C, NHCE
  • 18. Colpitts Oscillator (BJT –CE Configuration) Fig: Colpitts oscillator using BJT CE config. • The voltage divider bias set up the Q-point for biasing. • RF CHOKE has high inductive reactance. • C1, C2, L provide the feedback for oscillations. • The feedback voltage is across C2 drives the base and sustain the oscillations. Fig: AC equivalent Circuit 16/10/2017 18Aravinda K., Dept. of E&C, NHCE
  • 19. Colpitts Oscillator (BJT Configuration) Resonant Frequency:  Exact frequency 𝑓𝑟 = 1 2∏ 𝐿𝐶𝑒𝑞 𝑄2 𝑄2 + 1  Approximate Frequency 𝑓𝑟 = 1 2∏ 𝐿𝐶 𝑒𝑞 for Q >10 𝑄 = 𝑓𝑟 𝐵𝑊 , 𝐶𝑒𝑞 = 𝐶1 𝐶2 𝐶1 + 𝐶2 16/10/2017 Starting Condition for Oscillation • AB >1 • A>1/B • Feedback fraction to the oscillator ‘B’ 𝐵 = 𝐶1 𝐶2 • Minimum voltage gain Av 𝐴𝑣 = 𝐶2 𝐶1 19Aravinda K., Dept. of E&C, NHCE
  • 20. Coupling Load to Oscillator Capacitive coupling Link coupling 16/10/2017 • Directly connecting load lowers Q , hence may not initiate oscillation process. • Coupling should not lower the Q of tank circuit. 20Aravinda K., Dept. of E&C, NHCE
  • 21. 16/10/2017 Colpitts Oscillator (BJT –CB Configuration) • CB Configuration oscillate at high frequency. • Frequency of Oscillation • 𝑓𝑟 = 1 2∏ 𝐿𝐶 𝑒𝑞 𝑄2 𝑄2+1 𝐶𝑒𝑞= 𝐶1 𝐶2 𝐶1 + 𝐶2 • Feedback factor • Minimum Gain Fig:Colpitts oscillator BJT-CE confg. 21Aravinda K., Dept. of E&C, NHCE
  • 22. Colpitts Oscillator-JFET,OPAMP 16/10/2017 Frequency of Oscillation 𝑓𝑟 = 1 2∏ 𝐿𝐶𝑒𝑞 Fig: Colpitts oscillator using OPAMP.Fig: Colpitts oscillator using JFET. 22Aravinda K., Dept. of E&C, NHCE
  • 23. Problem • What is the frequency of oscillation for Colpitts oscillator ? What is the feedback fraction ? How much voltage gain does circuit needs to start oscillating ? Given C1=0.001 uF, C2= 0.01uF, L=15 uF Solution fr= 1.36 MHz, B= 0.1, Av= 10 16/10/2017 23Aravinda K., Dept. of E&C, NHCE
  • 24. 16/10/2017 Hartley Oscillator Fig: Hartley Oscillator using BJT. Fig: Hartley Oscillator using JFET. 24Aravinda K., Dept. of E&C, NHCE
  • 25. Crystal Oscillator 16/10/2017 Characteristics of Quartz crystal • Piezoelectric effect  R represents the internal friction.  L represents vibrating mass.  Cs represents crystal compliance.  CM represents capacitance due to mechanical mounting of crystal. Used for accurate and stable frequency of oscillations. 25Aravinda K., Dept. of E&C, NHCE
  • 26. Characteristics of Quartz crystal 16/10/2017 Exhibits2 resonant frequencies 1. Series resonance frequency -> When impedance is low (R) 2. Parallel resonance frequency ->When reactance of series resonance equals the reactance of CM. 26Aravinda K., Dept. of E&C, NHCE
  • 27. 16/10/2017 Series Resonant Circuit Parallel Resonant Circuit Crystal Oscillator 27Aravinda K., Dept. of E&C, NHCE
  • 28. 16/10/2017 Problem • What is the series and parallel resonant frequency of crystal? Given L=3H, Cs=0.05 pF, R= 2KΩ, Cm=10 pF. Solution fs=411 KHz, Cp=0.0498 pF, fp=412 KHz. 28Aravinda K., Dept. of E&C, NHCE
  • 29. END OF MODULE 4 16/10/2017 29Aravinda K., Dept. of E&C, NHCE