1. Common-Collector (Emitter-Follower) Amplifier

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

3. β = 100
VBE = 0.7V
VA = 80 V

4. 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

5. vb β = 100 VBE = 0.7 V VA = 80 V RS = 0.5 k 3.28 k RTH = 25 k vSvO
3.28 k k vb
RE = 2 k
RS = 0.5 k
Output at emitter terminal

6. Redraw the small signal equivalent circuit so that all signal grounds connected together.x vS vb
RS = 0.5 k
RTH = 25 k
3.28 k x x vO
RE = 2 k x

7. RTH = 25 k vS vO
3.28 k k vb
RE = 2 k
RS = 0.5 k x

8. 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

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

10. 2. At node x, use KCL and get io in terms of ib
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

11. 4. Find vb in terms of ib using supermesh: vb = ibr + io(Rout)
3.28 k
RS = 0.5 k x vO io vb
RTH = 25 k vS
Rout
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

12. 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 =

13. Current Gain Output side: io = vo / RE = vo / 2
RS = 0.5 k vs
RE = 2 k
Ri = k
Output side: io = vo / RE = vo / 2
Input side: iI = vs / (RS + Ri ) = vS / 22.74
Current gain
= io / iI
= vo (22.74) = * = 10.94
vs (2)

14. The voltage gain of common collector is less than 1.
Hence, the common collector doesn’t do any voltage amplification.
For that reason, this circuit is sometimes called a voltage follower.
But, as you can see, this circuit does have great potential as a current amplifier.

15. Output Resistance of a Common-Collector

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

18. vx The output resistance, 1. vbe in terms of vx 3.28 k + Vx - + Vx -
Independent voltage source turned off
In parallel
In parallel
The output resistance,
1. vbe in terms of vx
3.28 k + Vx - + Vx -
vbe = vx
vbe = vx
0.49 k
1.96 k

19. r + 0.49 k + Vx - 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

20. 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

21. + 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

22. The circuit above is a common collector configuration
The circuit above is a common collector configuration. Given  = 135 and VBEon = 0.66 V and VA = 
Calculate the DC collector current, IC
Draw the AC equivalent circuit

23. ii iL