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. 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

3. 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

4. 0.5 k
57.7 k
RC = 6 k
7.46 k
RE = 0.6 k + vb -

5. STEPS
OUTPUT SIDE
Get the equivalent resistance at the output side, ROUT
Get the vo equation where vo = -  ib ROUT
INPUT SIDE
Calculate Rib = vb / ib : KVL at loop (extra step)
Calculate Ri
Get vb in terms of vs – eg: using voltage divider.
Go back to vo equation and calculate the voltage gain

6. 1. Rout = RC = 6 k
2. Equation of vo : vo = -  ib RC = ib
3. Calculate Rib  using KVL: ib r + ie RE - vb = 0
but ie = (1+ ) ib = 121 ib
so: ib [ 121(0.6) ] = vb  Rib = k
4. Calculate Ri  RTH||Rib = k
5. vb in terms of vs  use voltage divider:
vb = [ Ri / ( Ri + Rs )] * vs = vs
vs = vb

7. AV = vo / vs = - 8.86 so: vs = 1.0149 vb 6. Go back to equation of vo
vo = ib = [ vb / Rib ] = -720 vb / = vb
vo / vs = vb / vb
vo / vs =
AV = vo / vs =

8. Current Gain Output side: io = vo / RC = vo / 6
RS = 0.5 k vs
RC = 6k
Ri = k
Output side: io = vo / RC = vo / 6
Input side: ii = vs / (RS + Ri ) = vS / 33.53
Current gain
= io / ii
= vo (33.53) = * 5.588 =
vs (6)

9. 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

10. 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

11. 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 = 200 V
Bypass capacitor
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 10 k
VTH = 5 V

12. 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 = 1.8 / ( ) = mA
IC = βIB = 0.84 mA
Calculate the small-signal parameters
r = 3.87 k , ro = 238 k and gm = 32.3 mA/V

13. Follow the steps 1. Rout = ro || RC = 2.278 k
gmvbe
RTH = 10 k vS vO
RC = 2.3 k
3.74 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 in terms of vs
vbe = vs since connected in parallel

14. 6. Go back to equation of vo vo / vs = -73.58 vbe / vbe
gmvbe
RTH = 10 k vS vO
RC = 2.3 k
3.74 k
vbe
238 k
so: vbe = vs
6. Go back to equation of vo
vo / vs = vbe / vbe
vo / vs =
AV = vo / vs =

15. Current Gain Output side: io = vo / RC = vo / 2.3vs
RC = 2.3 k
Ri = 2.79 k
Output side: io = vo / RC = vo / 2.3
Input side: ii = vs / Ri = vS / ( 2.79)
Current gain
= io / ii
= vo (2.79) = * 1.213 =
vs (2.3)