Download Recall Last Lecture

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
TYPE 2: Emitter terminal connected with RE
– normally ro =  in this type
New parameter: input resistance seen from the base, Rib = vb / ib
β = 120
VBE = 0.7V
VA = 
VCC = 5 V
RC = 5.6 k
250 k
Voltage Divider biasing:
Change to Thevenin
0.5 k
Equivalent
RTH = 57.7 k
VTH = 1.154 V
75 k
RE = 0.6 k

Perform DC analysis to obtain the value of IC
BE loop:

57.7 IB + 0.7 + 0.6 IE – 1.154 = 0
IE = 121 IB
57.7 IB + 0.7 + 0.6 (121IB) – 1.154 = 0
IB = 0.454 / (72.6 + 57.7) = 0.00348 mA
IC = βIB = 0.4176 mA
Calculate the small-signal parameters
r = 7.46 k , ro =  and gm = 16.06 mA/V
0.5 k
+
7.46 k
vb
57.7 k
RC = 6 k
RE = 0.6 k
-
STEPS
OUTPUT SIDE
1.
Get the equivalent resistance at the output side, ROUT
2.
Get the vo equation where vo = -  ib ROUT
INPUT SIDE
3.
Calculate Rib = vb / ib : KVL at loop (extra step)
4.
Calculate Ri
5.
Get vb in terms of vs – eg: using voltage divider.
6.
Go back to vo equation and calculate the voltage gain
1. Rout = RC = 6 k
2. Equation of vo : vo = -  ib RC = - 720 ib
3. Calculate Rib  using KVL: ib r + ie RE - vb = 0
but ie = (1+ ) ib = 121 ib
so: ib [ 121(0.6) + 7.46 ] = vb  Rib = 80.06 k
4. Calculate Ri  RTH||Rib = 33.53 k
5. vb in terms of vs  use voltage divider:
vb = [ Ri / ( Ri + Rs )] * vs = 0.9853 vs
vs = 1.0149 vb
so: vs = 1.0149 vb
6. Go back to equation of vo
vo = - 720 ib = - 720 [ vb / Rib ] = -720 vb / 80.06 = - 8.993 vb
vo / vs = - 8.993 vb / 1.0149 vb
vo / vs = - 8.86
AV = vo / vs = - 8.86
Current Gain
RS = 0.5 k
vs


Ri = 33.53 k
RC = 6k
Output side: io = vo / RC = vo / 6
Input side: ii = vs / (RS + Ri ) = vS / 33.53
Current gain
= io / ii
= vo (33.53) = -8.86 * 5.588
vs (6)
= - 49.5
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
vO
vS
RTH
vbe
gmvbe
CE becomes a short circuit path –
bypass RE; hence similar to Type 1
RC
VCC = 10 V
β = 125
VBE = 0.7V
VA = 200 V
20 k
RC = 2.3 k
20 k
5 k
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 10 k
VTH = 5 V
Bypass
capacitor
β = 125
VBE = 0.7V
VA = 200 V

Perform DC analysis to obtain the value of IC
BE loop:

10 IB + 0.7 + 5 IE – 5 = 0
IE = 126 IB
10 IB + 0.7 + 5 (126 IB) – 5 = 0
IB = 1.8 / (10 + 630) = 0.00672mA
IC = βIB = 0.84 mA
Calculate the small-signal parameters
r = 3.87 k , ro = 238 k and gm = 32.3 mA/V
3.74 k
vS
RTH = vbe
10 k
vO
gmvbe
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
RC = 2.3 k
238 k
3.74 k
RTH = vbe
10 k
vS
vO
gmvbe
RC = 2.3 k
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
vs


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) = -73.58 * 1.213
vs (2.3)
= - 89.25
RC = 2.3 k