# 1 A common-emitter amplifier is displayed above. The input VIN and

```VCC=12 V
HW #5
R1=47kΩ
Common Emitter Amp
AV = -RC/RE voltage gain
ZIN ~ β RE input impedance
ZOUT ~RC
RC=4.7kΩ
VC
C
VB
B ΔVCE = 0.7V
E
VOUT
VIN
R2=4.7kΩ
RE=470Ω
GND
#1 A common-emitter amplifier is displayed above. The input VIN and output VOUT share a common ground
at the emitter. The transistor base is biased to a voltage VB to allow the input signal VIN to always lie above
0.7V . Use the voltage divider theorem to determine VB in the circuit above (47kΩ and 4.7kΩ chain). What
is the maximum amplitude signal |V0 | that can be amplified such that VIN ! 0.7V ? What is the gain of
the amplifier AV ? What is the input impedance ZIN and output impedance ZOUT? (β~100)
|
VIN
VB = (4.9/4.9+49)12V=1.1V
|V0 | &lt;= 0.25V
|V0 | (amplitude)
VB
AV =-RC/RE = 10
V=0.7V
V
V=0.0V ground
ZIN = 100 x 4700 = 470000Ω
ZOUT = RC =4.7kΩ
10 (2Vo+0.7V) &lt; = 12V or 2Vo=1.2V-0.7= 0.5V
Vo&lt;= 0.25V
#2. Assume we use a C=1uF capacitor on the input of our amplifier, and there is 4 nF of stray capacitance
in the circuit. Determine the fLO and fHI cutoffs for this transistor amplifier.
flo = (2πZinCin)-1= ( 6.28x47000Ωx0.000001F)-1=3.9Hz
fhi = (2πZoutCout)-1= ( 6.28x4700Ωx4.0e-9F)-1=8470Hz
#3. On slide 5 we see the display of the collector current IC of a transistor versus the collector-emitter
voltage VCE. At about VCE =5V the transistor is operating in its linear range. Graph IC vs IB and determine
the β of this transistor. ( IC = β IB ).
#4. Derive the feedback equation
AV ! =
where
AV
!
1 &quot; ! AV
AV is the open loop gain and β is
the feedback fraction.
Av =
(
)
Vout
Vout
Vout
Av#
Vin
!!
!=
=
!!!!!!!
Vin
Vin &quot; #Vin 1 &quot; # Vout
1 &quot; # Av#
Vin
(
)
```