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
Common Emitter – with RE
Voltage gain

2. 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

3. vb
CE becomes a short circuit path – bypass RE; hence similar to Type 1

4. β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA Bypass capacitor VCC = 5 V
RC = 2.3 k
20 k
0 k
RE = 5k
Bypass capacitor

5. Short-circuited (bypass) by the capacitor CE
β = 125
VBE = 0.7V
VA = 200 V
IC = 0.84 mA
10 k
RC = 2.3 k
238 k
3.87 k
vbe
Short-circuited (bypass) by the capacitor CE
r =3.87 k , ro = 238 k and gm = 32.3 mA/V

6. Follow the steps 1. Rout = ro || RC = 2.278 k
vbe
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

7. 6. Go back to equation of vo vo / vs = -73.58 vbe / vbe
RC = 2.3 k
238 k
3.87 k
vbe
so: vbe = vs
6. Go back to equation of vo
vo / vs = vbe / vbe
vo / vs =
AV = vo / vs =

8. Common-Collector (Emitter-Follower) Amplifier

9. Remember that for Common Collector Amplifier,
the output is measured at the emitter terminal.
the gain is a positive value

10. β = 100
VBE = 0.7V
VA = 80

11. Perform DC analysis to obtain the value of IC
BE loop: 25IB IE – 2.5 = 0
25IB (1+ β)IB = 2.5
IC = βIB = mA
Calculate the small-signal parameters
r = 3.28 k and ro = k

12. Output at emitter terminal
β = 100
VBE = 0.7V
VA = 80
Small-signal equivalent circuit
of the emitter-follower amplifier
Output at emitter terminal

13. Redraw the small signal equivalent circuit so that all signal grounds connected together.x Vb x
RTH = 25 k
RS = 0.5 k
r = 3.28 k
RE = 2 k
ro = k Vb

14. STEPS
OUTPUT SIDE
Get the equivalent resistance at the output side, ROUT
At node x, use KCL and get io in terms of ib where io = ib +  ib
Get the vo equation where vo = io ROUT
INPUT SIDE
Find vb in terms of ib using supermesh
Calculate Rib – input resistance seen from base: Rib = vb / ib
Calculate Ri
Get vb in terms of vs.
Go back to vo equation and get the voltage gain

15. Supermesh
Supermesh is defined as the combination of two meshes which have current source on their boundary

16. 2. At node x, use KCL and get io in terms of ib vb
1. Get the equivalent resistance at the output terminal, ROUT  ROUT = ro ||RE = 1.96 k
2. At node x, use KCL and get io in terms of ib
 io = ib+ib = ( 1+ )ib = 101 ib
3. Get the vo equation where vo = io ROUT
 vo = Rout ( 1+ )ib = ib

17. 4. Find vb in terms of ib using supermesh: vb = ibr + io(Rout) x vb
4. Find vb in terms of ib using supermesh:
vb = ibr + io(Rout)
vb = ib (r +101 (1.96)) = ib
5. Calculate Rib Rib = vb / ib  k
6. Calculate Ri  Ri = RTH||Rib = k

18. Small-Signal Voltage Gain
7. Get vb in terms of vs using voltage divider
8. Go back to vo equation and replace where necessary
vb = vs
vb = 0.978vs Vb
vs = vb
vo = ib but ib = vb / Rib
vo = (vb / Rib) = ( vb) = vb
201.24
AV = vo / vs =
vo / vs = vb / vb
vo / vs =

19. Output Resistance of a Common-Collector

20. Steps Turn off independent sources
Place test voltage supply, VX having current IX at the output
Hence, RO = VX / IX

21. vx The output resistance, 1. vbe in terms of vx + Vx - + Vx - vx
vbe = vx
0.49 k
1.96 k

22. r + RS + Vx - 2. Use nodal analysis
3.77 k
1.96 k
2. Use nodal analysis
vbe = vx and gm = 30.5 mA/V
- Vx gmvbe Vx Ix = 0
Vx – Vx Vx + Ix = 0
Vx – Vx Vx + Ix = 0
Vx + Ix = 0
Ix = Vx
= Vx Ix
The output resistance, k

23. Output Resistance
The input signal source is short circuited and assume it is an ideal source so RS = 0
The output resistance,
1. vbe in terms of vx + Vx - + Vx -
vbe = - vx
1.96 k

24. + Vx - r 2. Use nodal analysis - Vx + gmvbe - Vx + Ix = 0 3.28 1.96
1.96 k r
2. Use nodal analysis
- Vx gmvbe Vx Ix = 0
Vx – 30.5 Vx Vx + Ix = 0
Vx – 30.5 Vx Vx + Ix = 0
Vx + Ix = 0
Ix = Vx
= Vx Ix
The output resistance, k