Download Recall Last Lecture

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
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, AV = vo / vs
Current Gain, Ai = 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, 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 calculate the voltage gain
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
RTH
5.9 k
2.74 k
vbe
vO
gmvbe
105.37
k
RC = 6 k
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. vbe in terms of vs  use voltage divider:
vbe = [ Ri / ( Ri + Rs )] * vs = 0.789 vs
so vs = 1.2674 vbe
vS
RS = 0.5 k
RTH
5.9 k
2.74 k
vbe
vO
gmvbe
105.37
k
RC = 6 k
so: vs = 1.2674 vbe
5. Go back to equation of vo and calculate the gain
vo / vs = -207.21 vbe / 1. 2674 vbe
vo / vs = - 207.21 / 1.2674
AV = vo / vs = - 163.5
vo / vs = -163.5
Current Gain
is
RS = 0.5 k
vS
RTH
5.9 k


io
gmvbe
vbe
105.37
k
Output side: io = vo / RC = vo / 6
Input side: is = vs / (RS + Ri ) = vS / 2.37
is
Current gain
= iout / is
RS
vS
2.74 k
vO
Ri
= vo (2.37) = -163.5 * 0.395
vs (6)
= - 64.6
RC = 6 k
RECALL CHAPTER 1
RS
vS
Ri
RS
vS
Rload
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
0.5 k
V1
3.25 k
4k
vbe
0.3 k
gmvbe
RC
RL
100 k
Follow the steps
1. Rout = RC || RL = 0.3 || 100 = 0.3 k
2. Equation of vo : vo = - (RC || RL ) gmvbe= - 0.3 ( 42.77) vbe = -12.831 vbe
3. Calculate Ri  RTH||r = 4 || 3.25 = 1.793 k
4. vbe in terms of vs  use voltage divider:
vbe = [ Ri / ( Ri + Rs )] * v1 = 0.782 v1
so v1 = 1.279 vbe
0.5 k
V1

3.25 k
4k
vbe
0.3 k
gmvbe
RC
RL
100 k
so: v1 = 1.279 vbe
5. Go back to equation of vo and calculate the gain

vo / v1 = -12.831 vbe / 1.279 vbe
vo / v1 = - 12.831 / 1.279
vo / v1 = -10
AV = vo / v1 = - 10
RS = 0.5 k
v1


Ri = 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
100k