Download BJT - AC Analysis (Common Emitter Part 1)

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
Derive input load
line, IB versus VBE
Formulas
KVL at BE
Loop
DC Analysis
npn
Derive output load
line, IC versus VCE
KVL at CE
Loop
pnp
Bipolar Junction
Transistor
Cutoff
Mode of operation
CHAPTER 4
Active
Saturation
What
happened
to each
junctions?
Fixed Bias
Biasing
Circuit
Collector
to Base
Feedback
Resistor
Voltage
Divider
Biasing
Circuit
Bipolar Junction
Transistor (Biasing)
CHAPTER 4 CONTINUE
DC Analysis
Voltage transfer characteristic, VO versus VI
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:
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?
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
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 = vo / vs
Current Gain = io / is
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, RO
2.
Get the vo equation where vo = - gm vbeRO
INPUT SIDE
3.
Calculate Ri
4.
Get vbe in terms of vi
Emitter Grounded
93.7 k
0.5 k
6.3 k
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 5.9 k
VTH = 0.756 V
VCC = 12 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
β = 100
VBE = 0.7V
VA = 100 V
off - becomes
short circuit
off becomes
short circuit
CC becomes short circuit
during AC
vo
vS
RTH
RC
vS
RS = 0.5 k
+
vi
RTH
5.9 k
2.74 k
vbe
-
vO
gmvbe
105.37
k
RC = 6 k
Follow the steps
1. Ro = 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. vbe = vi
Equation of vo : vo = - ( ro || RC ) gmvbe= - 36.5 ( 5.677) vbe = -207.21 vbe
vbe = vi
5. Av vi = vo  open circuit voltage
Avvi = -207.21 vbe = -207.21 vi
Av = -207.21  open circuit voltage gain
RS = 0.5kΩ
vS
5.677 k
Ri =
1.87
k
vo
To find new voltage gain, vo/vs with input signal voltage source,
vs
6.
vi in terms of vs  use voltage divider:
vi = [ Ri / ( Ri + Rs )] * vs = 0.789 vs
7.
vo = Avvi  because there is no load resistor
vo = -207.21 (0.789 vs)
vo/vs = -163.49
Example
β = 139
VBE = 0.668 V
VA = 
V1
RS
C1
0.5k
1n
R1
20k
RC
0.3k
C2
VCC
1n
3.5V
R2
5k
RL
100k
0
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 4 k
β = 139
VBE = 0.668 V
VA = 
VTH = 0.7 V

Perform DC analysis to obtain the value of IC
BE loop:

4 IB + 0.668 – 0.7 = 0
IB = 0.008
IC = βIB = 1.112 mA
Calculate the small-signal parameters
r = 3.25 k , ro =  and gm = 42.77 mA/V
V1
0.5
k
+
vi
4 k
vbe
3.25
k
gmvbe
-
RC
0.3 k
Follow the steps
1. Ro = RC = 0.3 k
2. Equation of vo : vo = - (RC ) gmvbe= - 0.3 ( 42.77) vbe = -12.831 vbe
3. Calculate Ri  RTH||r = 4 || 3.25 = 1.793 k
4. vbe = vi
Equation of vo : vo = - (RC ) gmvbe= - 0.3 ( 42.77) vbe = -12.831 vbe
vbe = vi
5. Avvi = vo  open circuit voltage
Avvi = - 12.831 vbe = -12.831 vi
Av = -12.831  open circuit voltage gain
0.5
k
v1
0.3 k
1.793
k
RL = 100 k
To find new voltage gain, vo/v1 now with signal voltage, vs and
RL
6. vi in terms of vs  use voltage divider:
vi = [ Ri / ( Ri + Rs )] * v1 = 0.782 v1
7.
vo = [ RL / ( RL + Ro )] * Avvi  this is we have load resistor RL
vo = 0.997 (-12.831 ) (0.782 v1)
vo/v1 = -10
CURRENT GAIN
0.5
k
ii
v1


0.3 k
1.793
k
Output side: io = vo / 100 = vo / 100
Input side: ii = v1 / (RS + Ri ) = v1 / 2.293
Current gain
= io / ii
= vo (2.293) = -10 * 0.02293
v1 (100)
= - 0.2293
io
RL = 100 k