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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
Current gain
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
vb
CE becomes a short circuit path –
bypass RE; hence similar to Type 1
β = 125
VBE = 0.7V
VA = 200 V
IC = 0.84 mA
VCC = 5 V
20 k
RC = 2.3 k
0 k
20 k
RE =
5k
Bypass
capacitor
β = 125
VBE = 0.7V
VA = 200 V
IC = 0.84 mA
3.87
k
10 k
RC = 2.3 k
vbe
238 k
Short-circuited
(bypass) by the
capacitor CE
r =3.87 k , ro = 238 k and gm = 32.3 mA/V
3.87
k
10 k
RC = 2.3 k
vbe
238 k
Follow the steps
1. Rout = ro || RC = 2.278 k
2. Equation of vo : vo = - ( ro || RC ) gmvbe= -73.58 vbe
3. Calculate Ri  RTH||r = 2.79 k
4. vbe in terms of vs
vbe = vs since connected in parallel
3.87
k
10 k
RC = 2.3 k
vbe
238 k
so: vbe = vs
6. Go back to equation of vo
vo / vs = -73.58 vbe / vbe
vo / vs = - 73.58
AV = vo / vs = - 73.58
Current Gain
is
iout
3.87
k
10 k
RC = 2.3 k
vbe
238 k


Output side: iout = vo / RC = vo / 2.3
Input side: is = vs / (RS + Ri ) = vS / (0 + 2.79)
Current gain
= iout / is
= vo (2.79) = -73.58 * 1.213
vs (2.3)
= - 89.25
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 get the voltage gain

Supermesh

Supermesh is defined as the combination of two
meshes which have current source on their
boundary
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
7. Get vb in terms of vs using voltage divider
vb =
Vb
22.24
vs
22.24 + 0.5
vb = 0.978vs
vs = 1.02249 vb
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 ( vb) = 0.9837 vb
201.24
vo / vs = 0.9837 vb / 1.02249 vb
vo / vs = 0.9621
AV = vo / vs = 0.9621
Current Gain
Vb


Output side: iout = vo / RE = vo / 2
Input side: is = vs / RS +Ri = vS / 22.74
Current gain
= iout / is
= vo (22.74) = 0.9621* 11.37 = 10.94
vs (2)
Output Resistance of
a Common-Collector
Steps



Turn off independent sources
Place test voltage supply, VX having current
IX at the output
Hence, RO = VX / IX
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
-
-
2. Use nodal analysis
- Vx + gmvbe 3.77
1.96
k
vbe = - 0.87 vx and gm = 30.5 mA/V
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