1. TRANSISTOR AMPLIFIER CONFIGURATION -BJT Common-Emitter Amplifier-
By:
Syahrul Ashikin Azmi
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. Introduction
3 basic single-transistor amplifier configuration that can be formed are:
Common-emitter (C-E configuration)
Common collector / emitter follower (C-C configuration)
Common base (C-B configuration)
Each configuration has its own advantages in form of:
Input impedance
Output impedance
Current / voltage amplification

4. Basic common-emitter circuit
Dc voltage
-> power the amplifier
Voltage divider biasing
-> set Q-point
Coupling capacitor ->
dc isolation between amplifier
and signal source
Emitter at ground
-> common emitter

5. 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.

6. 1st: Perform DC analysis
The circuit can be analyzed by forming a Thevenin equivalent circuit.
CC acts as an open circuit to dc.

7. Thevenin circuit analysis
We know that,
Thevenin resistance, RTH is:
Thevenin voltage, VTH is:
Apply KVL around B-E loop;
The collector current, ICQ is then:

8. Cont. Thevenin circuit analysis
Apply KVL to collector-emitter loop;
Thus, Q-point of the amplifier circuit is the coordinate between ICQ and VCEQ.

9. 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

10. 2nd: Perform AC analysis -small-signal equivalent circuit-
Inside the transistor

11. Small-signal hybrid-π parameters
Control voltage, Vπ
Output voltage, Vo
Input resistance, Ri
Small-signal input resistance, rπ
Transconductance, gm
Small-signal output resistance, ro

12. Small-signal hybrid-π parameters
Output resistance, Ro
Voltage gain, Av

13. Example 1
Given VCC=12V,RS=0.5kΩ, R1=93.7kΩ, R2=6.3kΩ, RC=6kΩ, β=100, VBE(on)=0.7V and VA=100V.
Determine small-signal voltage gain, input resistance and output resistance of the circuit.

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

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

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

17. Cont Solution Example 1
O/p resistance, Ro -> by setting independent source Vs = 0 -->no excitation to input portion, Vπ=0, so gmVπ=0 (open cct).

18. Common-emitter circuit with emitter resistor
Why we need to add emitter resistor, RE in the circuit design?
Without RE, when β increases or decreases -> ICQ and VCEQ also vary, thus Q-point will be shifted and makes the circuit unstable.
By adding RE, 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.

19. Common-emitter circuit with emitter resistor

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

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

22. Ac analysis -small-signal equivalent circuit-

23. 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:

24. Small-signal hybrid-π parameters
By voltage divider, we get relate Vin and Vs:
Small-signal voltage gain is then:
If Ri>>RS and if (1+β)RE >> rπ, voltage gain is:
Exact value
Approximate value

25. Example 2
Given VCC=10V, R1=56kΩ, R2=12.2kΩ, RC=2kΩ, RE=0.4kΩ, RS=0.5kΩ, VBE(on)=0.7V, β=100 and VA=∞.
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, AV.

26. 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.

27. Example 3
A simple transistor circuit biased with both +ve and –ve dc voltages is shown in figure below. Given β=100 and VBE(on)=0.7V. Calculate IEQ, ICQ and VCEQ.

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

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

30. 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

31. Solution Example 4
1st: Sketch Thevenin equivalent circuit to find RTH and VTH.
RTH = 103kΩ & VTH = 1.6V
2nd: Apply KVL around B-E loop to find equation for IBQ. Then, find ICQ and IEQ.
IBQ = 3.92μA
ICQ = 0.471mA & IEQ = 0.474mA
3rd: Apply KVL around C-E loop to find equation for VCEQ.
VCEQ = 3.8V
4th: Sketch dc load line and indicate the Q-points. Find IC(max) at y-axis and VCE(cutoff)= VCC=V+-V-=18V at x-axis.

32. C-E Amplifier with Emitter Bypass Capacitor
CE provides a short circuit to ground for the ac signals

33. Cont.. By include RE, it provide stability of Q-point.
If RE is too high +++> small-signal voltage gain will be reduced severely. (see Av equation)
Thus, RE is split to RE1 & RE2 and the second resistor is bypassed with “emitter bypass capacitor”. CE provides a short circuit to ground for ac signal.
So, only RE1 is a part of ac equivalent circuit.
For dc stability: RE=RE1+RE2
For ac gain stability: RE=RE1 since CE will short RE2 to ground.

34. Example 6
Given β=100, VBE=0.7V and VA=100V. Determine: (a) small-signal voltage gain (b) input resistance seen by the signal source, Rin and the output resistance looking back into the output terminal, Ro.

35. AC Load Line Analysis
Dc load line -> a way of visualizing r/ship between Q-point and transistor characteristic.
When capacitor included in cct, a new effective load line  ac load line exist.
Ac load line -> visualizing r/ship between small-signal response and transistor characteristic.
Ac operating region is on ac load line.

36. Ac load line cont..

37. Ac load line cont.. For Dc load line:
Apply KVL around collector-emitter loop,
But
Substitute and rearrange both equations:
If β>>1, then we can approximate
Dc load line equation

38. Cont.. For ac analysis, apply KVL around collector-emitter loop,
Assume ic ≈ ie,
The slope is given by:
The slope of ac load differ from dc load line  RE2 is not included in the equivalent circuit. Small-signal C-E voltage and collector current response are functions of resistor RC and RE1.

39. Dc and ac load lines for CE circuit

40. AC load line cont..
+ IC Q
ICQ
VCEQ
+ VCE

41. Maximum symmetrical swing
When symmetrical sinusoidal signal applied to i/p of amplifier, symmetrical sinusoidal signal generated at o/p.
Use ac load line to determine the maximum output symmetrical swing.
If output exceed limit, a portion of o/p signal will be clipped and signal distortion occur.