Download AC ANALYSIS

Recall Last Lecture

Biasing of BJT

Three types of biasing




Fixed Bias Biasing Circuit
Biasing using Collector to Base Feedback Resistor
Voltage Divider Biasing Circuit
Applications of BJT

As digital logic gates


NOT
NOR
CHAPTER 5
BASIC BJT AMPLIFIERS
(AC ANALYSIS)
The Bipolar Linear Amplifier

Bipolar transistors have been traditionally used in linear amplifier
circuits because of their relatively high gain.

To use the circuit as an amplifier, the transistor needs to be biased
with a dc voltage at a quiescent point (Q-point) such that the transistor
is biased in the forward-active region.

If a time-varying signal is superimposed on the dc input voltage, the
output voltage will change along the transfer curve producing a timevarying output voltage.

If the time-varying output voltage is directly proportional to and larger
than the time-varying input voltage, then the circuit is a linear
amplifier.

The linear amplifier applies superposition
principle


Response – sum of responses of the circuit for
each input signals alone
So, for linear amplifier,


DC analysis is performed with AC source turns off or
set to zero
AC analysis is performed with DC source set to zero
EXAMPLE
 iC , iB and iE,

vCE and vBE
Sum of both
ac and dc
components
Graphical Analysis and ac Equivalent Circuit
 From the concept of small signal, all the time-varying
signals are superimposed on dc values. Then:
and
PERFORMING DC and AC
analysis
DC ANALYSIS
Turn off AC
SUPPLY = short
circuit
AC ANALYSIS
Turn off DC
SUPPLY = short
circuit
DO YOU STILL REMEMBER?
Let’s assume that
Model 2 is used
VDQ = V
IDQ
DC equivalent
rd
id
AC equivalent
DC ANALYSIS
DIODE = MODEL 1 ,2
OR 3
CALCULATE DC
CURRENT, ID
AC ANALYSIS
CALCULATE
rd
DIODE = RESISTOR,
rd
CALCULATE AC
CURRENT, id
WHAT ABOUT BJT?
AC equivalent circuit –
Small-Signal Hybrid-π Equivalent
OR
 ib
THE SMALL SIGNAL
PARAMETERS
The resistance rπ is called diffusion
resistance or B-E input resistance. It
is connected between Base and
Emitter terminals
The term gm is called a
transconductance
ro = VA / ICQ
rO = small signal transistor output
resistance
VA is normally equals to , hence, if
that is the case, rO =   open
circuit

Hence from the equation of the AC
parameters, we HAVE to perform DC
analysis first in order to calculate them.
EXAMPLE

The transistor parameter are  = 125 and
VA=200V. A value of gm = 200 mA/V is
desired. Determine the collector current, ICQ
and then find r and ro
ANSWERS: ICQ = 5.2 mA, r= 0.625 k and ro = 38.5 k
CALCULATION OF
GAIN
Voltage Gain, AV = vo / vs
Current Gain, Ai = iout / is

Small-Signal Voltage Gain: Av = Vo / Vs
 ib
Common-Emitter
Amplifier

Remember that for Common Emitter Amplifier,



the output is measured at the collector terminal.
the gain is a negative value
Three types of common emitter



Emitter grounded
With RE
With bypass capacitor CE
STEPS
OUTPUT SIDE
1.
Get the equivalent resistance at the output side, ROUT
2.
Get the vo equation where vo = - gm vbeROUT
INPUT SIDE
3.
Calculate Ri
4.
Get vbe in terms of vs – eg: using voltage divider.
5.
Go back to vo equation and replace where necessary
Emitter Grounded
VCC = 12 V
93.7 k
0.5 k
6.3 k
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 5.9 k
VTH = 0.756 V
RC = 6 k
β = 100
VBE = 0.7V
VA = 100 V

Perform DC analysis to obtain the value of IC
BE loop:

5.9IB + 0.7 – 0.756 = 0
IB = 0.00949
IC = βIB = 0.949 mA
Calculate the small-signal parameters
r = 2.74 k , ro = 105.37 k and gm = 36.5 mA/V
Emitter Grounded
VCC = 12 V
93.7 k
0.5 k
6.3 k
RC = 6 k
β = 100
VBE = 0.7V
VA = 100 V
vbe
Follow the steps
1. Rout = ro || RC = 5.677 k
2. Equation of vo : vo = - ( ro || RC ) gmvbe= - 36.5 ( 5.677) vbe = -207.21 vbe
3. Calculate Ri  RTH||r = 1.87 k
4. vb in terms of vs  use voltage divider:
vbe = [ Ri / ( Ri + Rs )] * vs = 0.789 vs
vbe
so: vb = 0.789 vs replace in equation from step 1
5. Go back to equation of vo
vo = -207.21 vbe
vo = - 207.21 [0.789 vs]
vo = -163.5 vs
AV = vo / vs = - 163.5
bring VS over
TYPE 2: Emitter terminal connected with RE
– normally ro =  in this type
New parameter: input resistance seen from the base, Rib = vb / ib
VCC = 5 V
β = 120
VBE = 0.7V
VA = 
250 k
RC = 5.6 k
0.5 k
75 k
RE = 0.6 k
7.46 k
0.5 k
RC = 6 k
vb
57.7 k
RE = 0.6 k
vb
1. Rout = RC = 6 k
2. Equation of vo : vo = - RC  ib= - 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
vb
so: vb = 0.9853 vs
6. Go back to equation of vo
vo = - 720 ib = - 720 [ vb / Rib ]
vo = - 720 [ 0.9853 vs / 80.06 ]
vo = - 8.86vs
bring VS over
AV = vo / vs = - 8.86
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 = -73.58 vs
bring VS over
AV = vo / vs = - 73.58