1. BJT Common-Emitter Amplifier By: Syahrul Ashikin School of Electrical System Engineering

2. Objectives To understand and familiar with dc analysis of bipolar transistor circuits. To study common-emitter amplifier in term of ac analysis and familiar with general characteristic of this circuit.

3. Basic common-emitter circuit Voltage divider biasing Coupling capacitor -> dc isolation between amplifier and signal source Emitter at ground -> common emitter

4. Rules in dc analysis Replacing all capacitors by open circuit. Replacing all inductors by short circuit. Replacing ac voltage source by short circuit or ground connection. Replacing ac current source by open circuit.

5. Dc analysis The circuit can be analyzed by forming a Thevenin equivalent circuit. C C acts as an open circuit to dc.

6. Thevenin circuit analysis We know that, Thevenin resistance, R TH is: Thevenin voltage, V TH is: Apply KVL around B-E loop; The collector current, I CQ is then:

7. Cont. Thevenin circuit analysis Apply KVL to collector-emitter loop; Thus, Q-point of the amplifier circuit is the coordinate between I CQ and V CEQ.

8. Rules in ac analysis Replacing all capacitors by short circuits Replacing all inductors by open circuits Replacing dc voltage sources by ground connections Replacing dc current sources by open circuits

9. AC analysis -small-signal equivalent circuit- Inside the transistor

10. Small-signal hybrid- π parameters Small-signal input resistance, r π Transconductance, g m Small-signal output resistance, r o Control voltage, V π Output voltage, V o Input resistance, R i

11. Small-signal hybrid- π parameters Output resistance, R o Voltage gain, A v

12. Example 1 Given V CC =12V,R S =0.5kΩ, R 1 =93.7kΩ, R 2 =6.3kΩ, R C =6kΩ, β=100, V BE(on) =0.7V and V A =100V. Determine small-signal voltage gain, input resistance and output resistance of the circuit.

13. Solution Example 1 1 st step: DC solution Find Q-point values. ICQ = 0.95mA VCEQ=6.31V.

14. Cont Solution Example 1 2 nd step: AC solution Small-signal hybrid-π parameters are:

15. Cont Solution Example 1 Small-signal voltage gain is: Input resistance, Ri is:

16. Cont Solution Example 1 O/p resistance, R o -> by setting independent source V s = 0 -->no excitation to input portion, V π =0, so g m V π =0 (open cct).

17. Common-emitter circuit with emitter resistor Why we need to add emitter resistor, R E in the circuit design? Without R E, when β increases or decreases -> I CQ and V CEQ also vary, thus Q-point will be shifted and makes the circuit unstable. By adding R E, there will be not much shift in Q- point is stabilized even with variation of β. Moreover, the voltage gain is less dependent on transistor current gain in ac analysis.

18. Common-emitter circuit with emitter resistor Emitter resistor

19. Thevenin circuit analysis Apply KVL around B-E loop,

20. Thevenin circuit analysis We will get collector current as: Apply KVL around C-E loop to find V CEQ,

21. Ac analysis -small-signal equivalent circuit-

22. Small-signal hybrid-π parameters The ac output voltage is: (if we consider equivalent circuit with current gain β) Input voltage equation: Input resistance looking into the base of BJT, Rib: Input resistance to the amplifier is:

23. Small-signal hybrid-π parameters By voltage divider, we get relate Vin and Vs: Small-signal voltage gain is then: If R i >>R S and if (1+β)R E >> r π, voltage gain is: Exact value Approximate value

24. Example 2 Given V CC =10V, R 1 =56kΩ, R 2 =12.2kΩ, R C =2kΩ, R E =0.4kΩ, RS=0.5kΩ, V BE(on) =0.7V, β=100 and V A = ∞. a) Sketch Thevenin equivalent circuit. b) Determine Q-points. c) Sketch and label small-signal equivalent hybrid-π circuit. d) Find small-signal voltage gain, A V.

25. Common-emitter circuit with positive and negative voltage biasing Biasing with dual supplies in desirable in some applications because: Eliminate coupling capacitor Allow dc input voltages as input signals.

26. Example 3 A simple transistor circuit biased with both +ve and –ve dc voltages is shown in figure below. Given β=100 and V BE(on) =0.7V. Calculate I EQ, I CQ and V CEQ.

27. Solution Example 3 For dc analysis, set v s =0 so that base terminal is at ground potential. KVL around B-E loop, So, emitter current: Collector current:

28. Cont solution example 3 Apply KVL around C-E loop yields Rearrange the equation to find VCEQ;

29. Example 4 Let β=120, R1=175kΩ, R2=250kΩ, RC=10kΩ, RE=20kΩ and VBE(on)=0.7V. For the given circuit, i) Find RTH, VTH and Q-points. ii) Sketch dc load line

30. Solution Example 4 1 st : Sketch Thevenin equivalent circuit to find R TH and V TH. R TH = 103kΩ & V TH = 1.6V 2 nd : Apply KVL around B-E loop to find equation for I BQ. Then, find I CQ and I EQ. IBQ = 3.92μA ICQ = 0.471mA & IEQ = 0.474mA 3 rd : Apply KVL around C-E loop to find equation for V CEQ. V CEQ = 3.8V 4 th : Sketch dc load line and indicate the Q- points. Find I C(max) at y-axis and V CE(cutoff) = V CC =V + -V - =18V at x-axis.