Transcript Common-Collector (Emitter

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


Voltage gain
Current gain
Common-Collector (EmitterFollower) Amplifier

Remember that for Common Collector
Amplifier,


the output is measured at the emitter terminal.
the gain is a positive value
β = 100
VBE = 0.7V
VA = 80

Perform DC analysis to obtain the value of IC
BE loop:

25IB + 0.7 + 2IE – 2.5 = 0
25IB + 0.7 + 2(1+ β)IB = 2.5
IC = βIB = 0.793 mA
Calculate the small-signal parameters
r = 3.28 k and ro = 100.88 k
β = 100
VBE = 0.7V
VA = 80
Small-signal equivalent circuit
of the emitter-follower amplifier
Output at emitter
terminal

Redraw the small signal equivalent circuit so that all signal
grounds connected together.
x
Vb
r = 3.28
k
RS = 0.5 k
Vb
RTH = 25
k
x
ro =
100.88
k
RE = 2 k
STEPS
OUTPUT SIDE
1.
Get the equivalent resistance at the output side, ROUT
2.
At node x, use KCL and get io in terms of ib where io = ib +  ib
3.
Get the vo equation where vo = io ROUT
INPUT SIDE
4.
Find vb in terms of ib using supermesh
5.
Calculate Rib – input resistance seen from base: Rib = vb / ib
6.
Calculate Ri
7.
Get vb in terms of vs.
8.
Go back to vo equation and replace where necessary
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 = 197.96 ib
x
vb
4.
Find vb in terms of ib using supermesh:
vb = ibr + io(Rout)
vb = ib (r +101 (1.96)) = 201.24 ib
5.
Calculate Rib Rib = vb / ib  201.24 k
6.
Calculate Ri  Ri = RTH||Rib = 22.24 k
Small-Signal Voltage Gain
7. Get vb in terms of vs using voltage divider
Vb
vb =
22.24
vs
22.24 + 0.5
vb = 0.978vs
8. Go back to vo equation and replace where necessary
vo = 197.96 ib but ib = vb / Rib
vo = 197.96 (vb / Rib) = 197.96 ( 0.978 vs) = 0.9621 vs
201.24
AV = vo / vs = 0.9621
Output Resistance
Output Resistance
Assume that an imaginary
voltage source VX is
connected into the output
terminal and there is an
imaginary
current
IX
flowing through the circuit
The output resistance,
1. vbe in terms of vx
0.49
k
+
+
Vx
Vx
-
-
vbe = 1.96
k
3.28
3.28 + 0.49
vbe = - 0.87 vx
vx
r + RS
3.77
k
+
+
Vx
Vx
-
-
1.96
k
2. Use nodal analysis
- Vx + gmvbe 3.77
Vx + Ix = 0
- 0.2653 Vx – 26.535 Vx - 0.5102 Vx + Ix = 0
1.96
- 0.2653 Vx – 26.535 Vx - 0.5102 Vx + Ix = 0
- 27.3105 Vx + Ix = 0
Ix = 27.3105 Vx
1
=
Vx
27.3105
The output resistance,
Ix
0.0366 k
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
-
-
1.96
k
vbe = - vx
r
+
+
Vx
Vx
-
-
1.96
k
2. Use nodal analysis
- Vx + gmvbe -
3.28
Vx + Ix = 0
- 0.3049 Vx – 30.5 Vx - 0.5102 Vx + Ix = 0
1.96
- 0.3049 Vx – 30.5 Vx - 0.5102 Vx + Ix = 0
- 31.3151 Vx + Ix = 0
Ix = 31.3151 Vx
1
=
Vx
31.3151
The output resistance,
Ix
0.0319 k