DC and Small Signal Components
• In realistic and useful electronic circuit, the input can be decomposed into two separate components:
 The DC component, V
I
 The small signal component, v i
(t)
• The DC component signal is not a function of time (as a constant e.g. V
I
=12V)
• The small-signal component v i signal is an AC signal
(t) is a function of time. This
• This signal v i
(t) is referred as the small-signal component because its magnitude is generally small for all time t

DC and Small Signal Components

DC and Small Signal Components
• The purpose of DC analysis is to establish a Q-point (quiescent point) or DC operating point . The goal is to set the Q-point such that it does not go into saturation or cutoff when an ac signal is applied
• If the Q-point is in active region , the transistor can operate as an amplifier .
• The purpose of AC analysis is to obtain the gain (ratio of output voltage and input voltage)
• To obtain
– Current and Voltage Gain, A i
and V i
– Input Impedance, Z i
– Output Impedance, Z o
by adopting simple circuit model or transistor equivalent model.

AC Analysis – AC Equivalent Circuit
■
AC analysis;
 Short all capacitors
 Open circuit all DC Supplies and ground them

AC Analysis – AC Equivalent Circuit
• Transistor can be redrawn for small-signal AC analysis.
• BJT is replaced with an equivalent circuit.
• Transistor equivalent model will be introduced
 The Hybrid Equivalent π Model

AC Analysis – Problem Solving Techniques
1. Analyze circuit with only dc sources to find Q point (DC analysis).
2. Replace the dc voltage sources with open-circuit and ground it
3. Replace the coupling and bypass capacitors with short-circuit
4. Replace transistor with its equivalent circuit, which is the smallsignal model or hybrid π model.
5. Analyze the small-signal equivalent circuit

AC Analysis
Equivalent Circuit for Common Emitter
• Note that for the capacitor has been shorted and the symbol used for the current is in small capital to indicate that AC signal is applied.

Hybrid π Model
βi b g m r



I
CQ
V
T

V
T
I
CQ g m r
 g m v be



 i b g m
= transconductance
V
T
= temperature equivalent voltage = 26mV at room temp

Common Emitter with Bypass Capacitor (C
3
)
Original Circuit After shorting the all the capacitors

Common Emitter with Bypass Capacitor (C
3
)
After replacing transistor with Hybrid π model, we get
Z ib
= r
π
Z i
= R
1
//R
2
//Z ib Z o
= R
C

Common Emitter with Bypass Capacitor (C
3
)
βi b
Z
A vs
A vs



( g m
R
C
(
 r



) R
C
R
B because g m
)( r


 r
 r


R
B
) or

Important Parameters (R
E
Bypass)
Z i
Z o

 r

R
C
A v
A v
A vs
 v o v i v o

(
  i b
) R
C
 v o v i
 v o v
S
 v i
(
 
) R
C

( i b
) r
 r

 v o v i v x v
S i
A vs
 v o v
S
Since Z i

(
 
 r

; r
 v i

) R
C
Z i
Z
 i
R
B
( v s
) x
Z i
Z
 i
R
B
A vs

(
 r



) R
C
R
B

Hybrid π Model (R
E
Un-bypass) (Example)
• Determine V
Th
, R
Th
, I
B
, I
C
, and
V
CE
• Draw an equivalent AC Circuit
• Calculate r 
, Z i
, Z o
, A vs
and A i
.

Z i
Z ib
Hybrid π Model (R
E
Un-bypass) (Answer)
Z o r

Z ib


 g m r

Z i

R
1
A vs
 v o v s g m

I
C , V
T

26 mV
V
T

( 1

R
2


) R
E
Z ib r



( 1


R
C

) R
E
Z o

(
Z i
R
C
Z
 i
R
S
)

Important Parameters (R
E
Un-bypass) r

Z ib


 g m r

Z i

R
1

( 1

R
2
Z ib g m

) R
E

I
V
T
C
Z o
, V
T

26 mV

R
C
A v
 v o v i v o
   i b
R
C
A v
 v o v i v i

A v
 r

 i b
Z ib


( 1


R
C

 i b
( r

) R
E

( 1
 
) R
E
)
A vs
 v o v s
 r
 v i



( 1


R
C

(
Z i
) R
E
Z
 i
R
S
(
Z i
) v s
Z
 i
R
S
)