Chapter 3 – Transistor Amplifiers – Part 2Chapter 3 – Transistor Amplifiers – Part 2

Special Amplifiers 1. Difference Amplifier 2. Complementary Symmetry3. Cascading

1) Now assume we increase, say vi, by a small amount.

2) This will mean that the transistor will try to increase its current level, Ic.

3) Hence lift the voltage present at its emitter.

4) The emitter voltage Ve increases.

5) 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.

1) The two identical pnp transistor amplifiers are coupled together.

2) They have common–emitter resistor.

3) The output signal voltage is proportional to the difference between two input voltages.

4) The two input signals are introduced to the base terminals of a pair.

5) The output signal is taken between the collector terminals.

6) The difference amplifier is very popular.

7) This is often used as the input stage in laboratory instruments.

Difference Amplifier Difference Amplifier

Vc2Vc1

Vc2Vc1

Case - 11) The fixed emitter current Ie = I1 + I2.

2) If input voltages v1 = v2 , then by symmetry it follows that I1 = I2, hence I1 = I2 = Ie/2.

3) The output voltage is vo = 0.

4) This rejects common mode signals.

OperationOperation

CaseCase - 2- 21) If v1 (say 1.5V) > v2 (say 1.2V).

2) The current I1 in T1 increases; the voltage at its emitter increases.

3) The emitter of T2 is also at the same voltage as of T1.

4) 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.

5)5) Results: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.

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.

Check Point – 1 Check Point – 1

Ib = 10 A Ib = 10 A

v2 = 1.2Vv1 = 1.2V

= 100 = 100

Ve = ? Ve = ?

I1 = ? I2 = ?

Vc1 = ? Vc2 = ?

VRL = ? VRL = ?

Vce = ? Vce = ?

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

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

Vc2Vc1

Mathematical Expression for Output voltageMathematical Expression for Output voltage

Exercise – 1 Exercise – 1

Vc2Vc1

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

DC voltage

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

DC voltage

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

DC voltageDC voltage

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

AC voltage

Application: Wind Mills Application: Wind Mills To find the wind speed To find the wind speed

Complementary SymmetryComplementary Symmetry

positive going signal is applied to the base of npn transistor

negative going signal is applied to the base of pnp transistor

Emitter of both is at 0V0V

-veV

+veV

1) Combination npn and pnp common emitter transistors.2) Emitter of both transistors are grounded. 3) They have common input and output connections.4) The base bias ( i.e. base voltage) on both transistors is

zero. 5) In the absence of input signal, both transistors are in

cut-off region.6) Current flows only when the input signal voltage biases

its emitter junction in the forward direction.7) The npn transistor delivers current to the load resistor

RL when pnp unit is cut-off, and vice versa.8) Each transistor operates only half the time. 9) The output signal is a replica of the input waveform. 10)This simple circuit is an efficient power amplifier.

Darlington ConnectionDarlington Connection

Current gain of combination = 2

1) Combination two similar transistors (npn or pnp

transistors)

2) The transistor Q2 is directly connected to Q1.

3) The emitter of Q1 is connected to the base of Q2.

4) 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).

5) This circuit is viewed as an emitter follower, Q1, followed by a grounded–emitter amplifier, Q2.

6) The combination produces a very large current gain, 2.

Cascading Cascading

+

Two stage Cascaded Transistor Amplifier

Actual circuit of two stage Cascaded Transistor Amplifier

NoteNote: The output impedance of the stage-1 and input impedance of stage-2 need to match for the optimum output.

1) Gain factor greater than a single-stage amplifier is obtained by cascading several amplifier stages.

2) The output of one amplifier stage is amplified by another stage until the desired signal voltage level is achieved.

3) The output of stage-1 is applied to the base of stage-2.

4) The output voltage, vo = a2vo’ = a1a2vi

where a1 and a2 are the gain factors of first stage and

the second stage, respectively.

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.