1. EKT104 ANALOG ELECTRONIC CIRCUITS [LITAR ELEKTRONIK ANALOG] BASIC BJT AMPLIFIER (PART II)
DR NIK ADILAH HANIN BINTI ZAHRI

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
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. VOLTAGE GAIN WITH AND WITHOUT BYPASS CAPACITOR
RS=0.5k
R1=56k
R2=12.2k
RE=
0.4k
RC=2k vs vO CC
VCC=10V CE
Compare the Voltage gain value with and without Bypass Capacitor of the following circuit.

8. VOLTAGE GAIN WITH AND WITHOUT BYPASS CAPACITOR
Voltage gain Measurement Without Bypass Capacitor
RS=0.5k
R1=56k
R2=12.2k
RE=
0.4k
RC=2k vs vO CC
VCC=10V CE

9. VOLTAGE GAIN WITH AND WITHOUT BYPASS CAPACITOR
Voltage gain Measurement With Bypass Capacitor
RS=0.5k
R1=56k
R2=12.2k
RE=
0.4k
RC=2k vs vO CC
VCC=10V CE

10. STABILITY OF VOLTAGE GAIN
Stability : Measure of how well an amplifier maintains its design values over changes
Bypassing external RE does produce maximum voltage gain, however there is stability problem because ac voltage depends internal ac emitter resistance, re
Where re depends on IE and on temperature

11. COMMON-EMITTER AMPLIFIER WITH EMITTER BYPASS CAPACITOR
Swamping Method
Method to minimize effect of re without reducing the voltage gain to minimum value
RE is partially bypassed so that reasonable gain can be achieved and the effect of re on the gain can be greatly eliminated
RE is formed with two separate emitter resistor, where RE2 is bypassed and RE1 is not
Common-emitter amplifier with emitter bypass capacitor

12. 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

13. COMMON-EMITTER AMPLIFIER WITH EMITTER BYPASS CAPACITOR
Example 1: Determine the Q-point (VBE=0.7V, β=150, VA=∞) and DC & AC Load Line. Then plot the graph.

14. DC LOAD LINE
SOLUTION...
KVL on C-E loop

15. DC & AC LOAD LINES
FULL SOLUTION

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

17. DC LOAD LINE
To determine dc Q-point, KVL around B-E loop

18. AC LOAD LINE
Small signal hybrid-π equivalent circuit

19. DC & AC LOAD LINES
FULL SOLUTION

20. MAXIMUM SYMMETRICAL SWING
Symmetrical sinusoidal signal applied to the input of an amplifier produces an output of 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

21. 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

22. MAXIMUM SYMMETRICAL SWING
Example 3
Determine the maximum symmetrical swing in the output voltage of the following circuit (same as Example 2).

23. MAXIMUM SYMMETRICAL SWING
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:

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

25. 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