1. Chapter 3 – Transistor Amplifiers – Part 2
Special Amplifiers
Difference Amplifier
Complementary Symmetry
Cascading

4. Now assume we increase, say vi, by a small amount.
This will mean that the transistor will try to increase its current level, Ic.
Hence lift the voltage present at its emitter.
The emitter voltage Ve increases.
For example : Ib = 10 A,  = 100
Ic =  Ib = 10 x 10-6 A x 100 = 1 mA.
The emitter voltage Ve = 1 mA x 1k = 1V
Now input signal is increased by small amount.
So that Ib = 30 A.
Therefore, the collector current Ic = 30 x 10-6 x 100 = 3 mA.
The emitter voltage now becomes Ve = 3 mA x 1k = 3 V.

5. Difference Amplifier
The two identical pnp transistor amplifiers are coupled together.
They have common–emitter resistor.
The output signal voltage is proportional to the difference between two input voltages.
The two input signals are introduced to the base terminals of a pair.
The output signal is taken between the collector terminals.
The difference amplifier is very popular.
This is often used as the input stage in laboratory instruments.

7. Vc2
Vc1

8. Vc2
Vc1

9. Operation Case - 1 The fixed emitter current Ie = I1 + I2.
If input voltages v1 = v2 , then by symmetry it follows that I1 = I2, hence I1 = I2 = Ie/2.
The output voltage is vo = 0.
This rejects common mode signals.

10. Case - 2
If v1 (say 1.5V) > v2 (say 1.2V).
The current I1 in T1 increases; the voltage at its emitter increases.
The emitter of T2 is also at the same voltage as of T1.
The emitter voltage at the emitter of T2 improves; but the base voltage is fixed. The base-emitter voltage of the transistor T2 (i.e. Vbe of T2) decreases.
Results: The current I1 in T1 increases and the current I2 in T2 falls. But the emitter current Ie = I1 + I2 does not alter much.

11. 6) The balance between two transistors currents/voltages changes.
7) The rise in input voltage v1, keeping v2 fixed, more current flows through RL of T1 than the RL of T2.
8) The voltage drop across RL of Transistor T2 reduces, and the collector to emitter voltage (Vce) increases.
9) The point Vc2 is at higher voltage than Vc1.

12. Check Point – 1 v2 = 1.2V v1 = 1.2V  = 100 Ve = ? I1 = ? I2 = ?
Ib = 10 A
v2 = 1.2V
v1 = 1.2V
 = 100
Ve = ?
I1 = ?
I2 = ?
Vc1 = ?
Vc2 = ?
VRL = ?
Vce = ?

13. Check Point – 2
v1 = v2 , Ie = I1 + I2 , I1 = I2 = Ie/2

14. Check Point – 3
v1 > v2 , still Ie = I1 + I2

15. Mathematical Expression for Output voltage
Vc2
Vc1

17. Exercise – 1
Vc2
Vc1

19. Exercise – 2 Whether output of this amplifier is
inverting or noninverting.
DC voltage

20. Exercise – 3 Whether output of this amplifier is
inverting or noninverting.
DC voltage

21. Exercise – 4 What will happen to output voltages Vout1 and Vout2 as the input voltage Vin decreases.
DC voltage
DC voltage

22. Exercise – 5 What is the difference voltage vo between two transistors’ collector terminals?
AC voltage

23. Application: Wind Mills
To find the wind speed

25. Complementary Symmetry
negative going signal is applied to
the base of pnp transistor
-veV 0V
Emitter of both is at 0V
+veV
positive going signal is applied to the base of npn transistor

26. Combination npn and pnp common emitter transistors.
Emitter of both transistors are grounded.
They have common input and output connections.
The base bias ( i.e. base voltage) on both transistors is zero.
In the absence of input signal, both transistors are in cut-off region.
Current flows only when the input signal voltage biases its emitter junction in the forward direction.
The npn transistor delivers current to the load resistor RL when pnp unit is cut-off, and vice versa.
Each transistor operates only half the time.
The output signal is a replica of the input waveform.
This simple circuit is an efficient power amplifier.

27. Darlington Connection
Current gain of combination = 2

28. Combination two similar transistors (npn or pnp transistors)
The transistor Q2 is directly connected to Q1.
The emitter of Q1 is connected to the base of Q2.
The base-collector potential (Vbe) of Q2 is equal to the emitter-collector potential (Vec) of Q1 (i.e. the emitter of Q1 is at a base-collector potential of Q2).
This circuit is viewed as an emitter follower, Q1, followed by a grounded–emitter amplifier, Q2.
The combination produces a very large current gain, 2.

29. Cascading +
Two stage Cascaded Transistor Amplifier

30. Actual circuit of two stage Cascaded Transistor Amplifier
Note: The output impedance of the stage-1 and input impedance of stage-2 need to match for the optimum output.

31. Gain factor greater than a single-stage amplifier is obtained by cascading several amplifier stages.
The output of one amplifier stage is amplified by another stage until the desired signal voltage level is achieved.
The output of stage-1 is applied to the base of stage-2.
The output voltage, vo = a2vo’ = a1a2vi
where a1 and a2 are the gain factors of first stage and the second stage, respectively.

32. 5) The overall gain of a cascaded amplifier is equal to the product of the gains of the individual stages.
6) If a1 = a2 = a, then vo = a2 vi.
7) For a n stage cascaded amplifier, the output voltage vo = an vi.