# 08-BJT AC

```BJT –
AC Analysis
DC and Small Signal Components
• In realistic and useful electronic circuit, the input can be
decomposed into two separate components:
 The DC component, VI
 The small signal component, vi(t)
• The DC component signal is not a function of time (as a
constant e.g. VI=12V)
• The small-signal component vi(t) is a function of time. This
signal is an AC signal
• This signal vi(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, Ai and Vi
– Input Impedance, Zi
– Output Impedance, Zo
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
gm 
βib
r 
I CQ
VT
 VT
I CQ
g m r  
g m vbe   ib
gm = transconductance
VT = temperature equivalent voltage = 26mV at room temp
Common Emitter with Bypass Capacitor (C3)
Original Circuit
After shorting the all the capacitors
Common Emitter with Bypass Capacitor (C3)
After replacing transistor with Hybrid π model, we get
Zib = rπ
Zi = R1//R2//Zib
Zo = RC
Common Emitter with Bypass Capacitor (C3)
βib
Avs  ( g m RC )(
Avs 
r
)
r  RB
(   ) RC
r  RB
because
gm 

r
Z
or
Important Parameters (RE Bypass)
Z i  r
Z o  RC
Av 
vo
vi
vo  (   ib ) RC
vi  (ib ) r
Av 
vo
(   ) RC

vi
r
Avs 
vo
v
v
 o x i
vS
vi
vS
vi 
Avs 
Zi
(v s )
Z i  RB
vo
(   ) RC
Zi

x
vS
r
Z i  RB
Since Z i  r ;
Avs 
(   ) RC
r  RB
Hybrid π Model (RE Un-bypass) (Example)
• Determine VTh, RTh, IB, IC, and
VCE
• Draw an equivalent AC Circuit
• Calculate r, Zi, Zo, Avs and Ai.
Hybrid π Model (RE Un-bypass) (Answer)
Zi
Zib
Zo
r 

gm
gm 
IC
VT
, VT  26mV
Z ib  r  (1   ) RE
Z i  R1 R 2 Z ib
Avs 
Z o  RC
vo
  RC
Zi

(
)
vs
r  (1   ) RE Z i  RS
Important Parameters (RE Un-bypass)
r 

gm
IC
gm 
, VT  26mV
VT
Z ib  r  (1   ) RE
Z i  R1 R 2 Z ib
Av 
vo
vi
Z o  RC
vo   ib RC
vi  ib Z ib  ib (r  (1   ) RE )
Av 
vo
 RC
 Av 
vi
r  (1   ) RE
vi  (
Avs 
Zi
)v s
Z i  RS
vo
 RC
Zi

(
)
vs r  (1   ) RE Z i  RS
To be continued
```