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

1)Now assume we increase, say v i, by a small amount. 2) This will mean that the transistor will try to increase its current level, I c. 3) Hence lift the voltage present at its emitter. 4)The emitter voltage V e increases. 5) For example : I b = 10  A,  = 100 I c =  I b = 10 x A x 100 = 1 mA. The emitter voltage V e = 1 mA x 1k  = 1V Now input signal is increased by small amount. So that I b = 30  A. Therefore, the collector current I c = 30 x x 100 = 3 mA. The emitter voltage now becomes V e = 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

V c2 V c1

V c2 V c1

Case - 1 1)The fixed emitter current I e = I 1 + I 2. 2)If input voltages v 1 = v 2, then by symmetry it follows that I 1 = I 2, hence I 1 = I 2 = I e / 2. 3)The output voltage is v o = 0. 4)This rejects common mode signals. Operation

Case - 2 1)If v 1 (say 1.5V) > v 2 (say 1.2V). 2)The current I 1 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. V be of T2) decreases. 5)Results: 5)Results: The current I 1 in T1 increases and the current I 2 in T2 falls. But the emitter current I e = I 1 + I 2 does not alter much.

6) The balance between two transistors currents/voltages changes. 7) The rise in input voltage v 1, keeping v 2 fixed, more current flows through R L of T 1 than the R L of T 2. 8) The voltage drop across R L of Transistor T2 reduces, and the collector to emitter voltage (V ce ) increases. 9) The point V c2 is at higher voltage than V c1.

Check Point – 1 I b = 10  A v 2 = 1.2V v 1 = 1.2V  = 100 V e = ? I 1 = ? I 2 = ? V c1 = ?V c2 = ? V RL = ? V ce = ?

Check Point – 2 v 1 = v 2, I e = I 1 + I 2, I 1 = I 2 = I e / 2

Check Point – 3 v 1 > v 2, still I e = I 1 + I 2

V c2 V c1 Mathematical Expression for Output voltage

Exercise – 1 V c2 V c1

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 V out1 and V out2 as the input voltage V in decreases. DC voltage

Exercise – 5 What is the difference voltage v o between two transistors’ collector terminals? AC voltage

Application: Wind Mills To find the wind speed

Complementary 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 0V 0V -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 R L 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 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 (V be ) of Q2 is equal to the emitter-collector potential (V ec ) 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 + Two stage Cascaded Transistor Amplifier

Actual circuit of two stage Cascaded Transistor Amplifier Note Note: 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, v o = a 2 v o ’ = a 1 a 2 v i where a 1 and a 2 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 a 1 = a 2 = a, then v o = a 2 v i. 7) For a n stage cascaded amplifier, the output voltage v o = a n v i.