1. ANALOG ELECTRONIC CIRCUITS 1
EKT 204
Basic BJT Amplifiers (Part 2)

2. Basic Common-Emitter Amplifier
The basic common-emitter circuit used in previous analysis causes a serious defect :
If BJT with VBE=0.7 V is used, IB=9.5 μA & IC=0.95 mA
But, if new BJT with VBE=0.6 V is used, IB=26 μA & BJT goes into saturation; which is not acceptable  Previous circuit is not practical
So, the emitter resistor is included: Q-point is stabilized against variations in β, as will the voltage gain, AV
Assumptions
CC acts as a short circuit
Early voltage = ∞ ==> ro neglected due to open circuit

3. Common-Emitter Amplifier with Emitter Resistor
inside transistor
CE amplifier with emitter resistor
Small-signal equivalent circuit
(with current gain parameter, β)

4. Common-Emitter Amplifier with Emitter Resistor
ac output voltage
Input voltage loop
Input resistance, Rib
Input resistance to amplifier, Ri
Voltage divider equation of Vin to Vs
Remember: Assume VA is infinite,  ro is neglected

5. Common-Emitter Amplifier with Emitter Resistor
Cont..
So, small-signal voltage gain, AV
If Ri >> Rs and (1 + β)RE >> rπ
Remember: Assume VA is infinite,  ro is neglected

6. Common-Emitter Amplifier with Emitter Bypass CapacitorRS R1 R2 RE RC vs vO CC
VCC CE
Emitter bypass capacitor, CE provides a short circuit to ground for the ac signals
Emitter bypass capacitor is used to short out a portion or all of emitter resistance by the ac signal. Hence no RE appear in the hybrid-π equivalent circuit Vo Vs RC RS r ro
R1|| R2
gmV
Small-signal hybrid-π equivalent circuit

7. DC & AC LOAD LINE ANALYSIS
DC load line
Visualized the relationship between Q-point & transistor characteristics
AC load line
Visualized the relationship between small-signal response & transistor characteristics
Occurs when capacitors added in transistor circuit

8. Common Emitter Amplifier with emitter bypass capacitor
Example 1
Common-emitter amplifier with emitter bypass capacitor

9. DC Load Line
Solution...
KVL on C-E loop

10. AC Load Line
Solution...
KVL on C-E loop
AC equivalent circuit

11. DC & AC Load Lines
Full solution

12. AC LOAD LINE ANALYSIS
Determine the dc and ac load line. VBE=0.7V, β=150, VA=∞
Example 2

13. DC Load Line
To determine dc Q-point, KVL around B-E loop

14. AC Load Line
Small signal hybrid-π equivalent circuit

15. DC & AC Load lines
Full solution

16. Maximum Symmetrical Swing
When symmetrical sinusoidal signal applied to the input of an amplifier, the output generated is also a symmetrical sinusoidal signal
AC load line is used to determine maximum output symmetrical swing
If output is out of limit, portion of the output signal will be clipped & signal distortion will occur

17. Maximum Symmetrical Swing
Steps to design a BJT amplifier for maximum symmetrical swing:
Write DC load line equation (relates of ICQ & VCEQ)
Write AC load line equation (relates ic, vce ; vce = - icReq, Req = effective ac resistance in C-E circuit)
Generally, ic = ICQ – IC(min), where IC(min) = 0 or some other specified min collector current
Generally, vce = VCEQ – VCE(min), where VCE(min) is some specified min C-E voltage
Combination of the above equations produce optimum ICQ & VCEQ values to obtain maximum symmetrical swing in output signal

18. Maximum Symmetrical Swing
Example 3
Determine the maximum symmetrical swing in the output voltage of the circuit given in Example 2.
Solution:
From the dc & ac load line, the maximum negative swing in the Ic is from mA to zero (ICQ). So, the maximum possible peak-to-peak ac collector current:
The max. symmetrical peak-to-peak output voltage:
Maximum instantaneous collector current:

19. Self-Reading
Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007
Chapter 6: Basic BJT Amplifiers
Page: ,

20. Exercise
Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007
Exercise 6.5, 6.6, 6.7,6.9
Exercise 6.10 , 6.11