1. Recall Last Lecture Introduction to BJT Amplifier
Small signal or AC equivalent circuit parameters
Have to calculate the DC collector current by performing DC analysis first
Common Emitter-Emitter Grounded

2. STEPS OUTPUT SIDE Get the equivalent resistance at the output side, RO
Get the vo equation where vo = - gm vbeRO
INPUT SIDE
Calculate Ri
Get vbe in terms of vi

3. TYPE 2: Emitter terminal connected with RE – normally ro =  in this type
New parameter: input resistance seen from the base, Rib = vb / ib
VCC = 5 V
RC = 5.6 k
250 k
75 k
0.5 k
RE = 0.6 k
β = 120
VBE = 0.7V
VA = 
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 57.7 k
VTH = V

4. Perform DC analysis to obtain the value of IC
BE loop: 57.7 IB IE – = 0
IE = 121 IB
57.7 IB (121IB) – = 0
IB = / ( ) = mA
IC = βIB = mA
Calculate the small-signal parameters
r = 7.46 k , ro =  and gm = mA/V

5. 0.5 k
RTH =
57.7 k
RC = 6 k
7.46 k
RE = 0.6 k + vi -
gmvbe
vbe

6. STEPS OUTPUT SIDE Get the equivalent resistance at the output side, Ro
Get the vo equation where vo = - gmvbeRo
INPUT SIDE
Calculate Rib = r + (1+ ) RE
Calculate Ri = Rib // RTH
vbe in terms of vi

7. 1. Rout = RC = 6 k
2. Equation of vo : vo = - gmvbeRC = vbe
3. Calculate Rib  Rib = = r + (1+ ) RE = [ 121(0.6) ] = k
4. Calculate Ri  RTH // Rib = k
5. vbe in terms of vi  KVL
vi=vbe+ R E vbe rπ +gmvbe =vbe vbe =10.72 vbe

8. Equation of vo : vo = - gmvbeRC = - 96.36 vbe
vi=vbe+ R E vbe rπ +gmvbe =vbe vbe =10.72 vbe
5. Av vi = vo  open circuit voltage
Av vi = vbe = vi =
Av =  open circuit voltage gain
10.72

9. To find new voltage gain, vo/vs with input signal voltage source, vs
RS = 0.5 kΩ vS vo
Ri = 33.53 k
6 k
To find new voltage gain, vo/vs with input signal voltage source, vs
6. vi in terms of vs  use voltage divider:
vi = [ Ri / ( Ri + Rs )] * vs = vs
7. vo = Avvi  because there is no load resistor
vo = ( vs)
vo/vs = -8.86

10. Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k
VCC = 10 V
RC = 2.3 k
20 k
5 k
β = 125
VBE = 0.7V
VA = 
Bypass capacitor
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 10 k
VTH = 5 V

11. Perform DC analysis to obtain the value of IC
β = 125
VBE = 0.7V
VA = 
Perform DC analysis to obtain the value of IC
BE loop: 10 IB IE – 5 = 0
IE = 126 IB
10 IB (126 IB) – 5 = 0
IB = 4.3 / ( ) = mA
IC = βIB = 0.84 mA
Calculate the small-signal parameters
r = 3.87 k , ro =  k and gm = 32.3 mA/V

12. r = 3.87 k , ro =  k and gm = 32.3 mA/V
1. Rout = RC = 2.3 k
2. Equation of vo : vo = - gmvbeRC = vbe
3. Calculate Rib  Rib = = r + (1+ ) RE = [ 126(5) ] = k
4. Calculate Ri  RTH // Rib = 9.8 k
5. vbe in terms of vi  KVL
vi=vbe+ R E vbe rπ +gmvbe =vbe vbe = vbe

13. Equation of vo : vo = - gmvbeRC = - 74.29 vbe
vi= vbe
5. Av vi = vo  open circuit voltage
Av vi = vbe = vi =
Av =  open circuit voltage gain
163.79

14. To find new voltage gain, vo/vs with input signal voltage source, vs
2.3 k
9.8 k
To find new voltage gain, vo/vs with input signal voltage source, vs
6. vi in terms of vs  in parallel
vi = vs
7. vo = Avvi  because there is no load resistor
vo = (vs)
vo/vs =

15. TYPE 3: With Emitter Bypass Capacitor, CE
Circuit with Emitter Bypass Capacitor
There may be times when the emitter resistor must be large for the purpose of DC design, but degrades the small-signal gain too severely.
An emitter bypass capacitor can be used to effectively create a short circuit path during AC analysis hence avoiding the effect RE

16. CE becomes a short circuit path – bypass RE; hence similar to Type 1
gmvbe
RTH vS vO RC
vbe
CE becomes a short circuit path – bypass RE; hence similar to Type 1

17. Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k
VCC = 10 V
RC = 2.3 k
20 k
5 k
β = 125
VBE = 0.7V
VA = 
Bypass capacitor
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 10 k
VTH = 5 V

18. Follow the steps 1. Ro = RC = 2.3 k
gmvbe
RTH = 10 k vS vO
RC = 2.3 k
3.87 k
vbe
Follow the steps
1. Ro = RC = 2.3 k
2. Equation of vo : vo = - (RC ) gmvbe= vbe
3. Calculate Ri  RTH||r = 2.79 k
4. vbe = vi

19. Equation of vo : vo = - 74.29 vbe
vbe = vi
5. Avvi = vo  open circuit voltage
Av vi = vbe = vi
Av =  open circuit voltage gain

20. To find new voltage gain, vo/vs with input signal voltage source, vs
2.278 k
Ri = 2.79 k
To find new voltage gain, vo/vs with input signal voltage source, vs
6. vi = vs  in parallel
7. vo = Avvi
vo = (vs)
vo/vs =
CONFIGURATION
GAIN
With RE
With bypass capacitor, CE

21. Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k
EXAMPLE 2 – with ro
VCC = 10 V
RC = 2.3 k
20 k
5 k
β = 125
VBE = 0.7V
VA = 200 V
Bypass capacitor
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 10 k
VTH = 5 V

22. Perform DC analysis to obtain the value of IC
β = 125
VBE = 0.7V
VA = 200 V
Perform DC analysis to obtain the value of IC
BE loop: 10 IB IE – 5 = 0
IE = 126 IB
10 IB (126 IB) – 5 = 0
IB = 4.3 / ( ) = mA
IC = βIB = 0.84 mA
Calculate the small-signal parameters
r = 3.87 k , ro = 238 k and gm = 32.3 mA/V

23. Follow the steps 1. Rout = ro || RC = 2.278 k
gmvbe
RTH = 10 k vS vO
RC = 2.3 k
3.87 k
vbe
238 k
Follow the steps
1. Rout = ro || RC = k
2. Equation of vo : vo = - ( ro || RC ) gmvbe= vbe
3. Calculate Ri  RTH||r = 2.79 k
4. vbe = vi

24. Equation of vo : vo = - ( ro || RC ) gmvbe= -73.58 vbe
vbe = vi
5. Avvi = vo  open circuit voltage
Av vi = vbe = vi
Av =  open circuit voltage gain

25. To find new voltage gain, vo/vs with input signal voltage source, vs
2.278 k
2.79 k
To find new voltage gain, vo/vs with input signal voltage source, vs
6. vi = vs  in parallel
7. vo = Avvi
vo = (vs)
vo/vs =