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

2. Chapter 5 basic bjt amplifiers (AC ANALYSIS)

3. 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 time-varying 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.

4. 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

5. EXAMPLE
iC , iB and iE,
vCE and vBE
Sum of both ac and dc components

6. Graphical Analysis and ac Equivalent Circuit
From the concept of small signal, all the time-varying signals are superimposed on dc values. Then:

7. PERFORMING DC and AC analysis
DC ANALYSIS
AC ANALYSIS
Turn off DC SUPPLY = short circuit
Turn off AC SUPPLY = short circuit

8. DO YOU STILL REMEMBER?

9. IDQ
VDQ = V
+ - rd id
DC equivalent
AC equivalent

10. CALCULATE DC CURRENT, ID CALCULATE AC CURRENT, id
DC ANALYSIS
AC ANALYSIS
DIODE = MODEL 1 ,2 OR 3
CALCULATE rd
DIODE = RESISTOR, rd
CALCULATE DC CURRENT, ID
CALCULATE AC CURRENT, id

11. What about bjt?

12. AC equivalent circuit – Small-Signal Hybrid-π EquivalentOR

13. 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 = small signal transistor output resistance
VA is normally equals to , hence, if that is the case, rO =   open circuit
ro = VA / ICQ
Hence from the equation of the AC parameters, we HAVE to perform DC analysis first in order to calculate them.

14. 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= k and ro = 38.5 k

15. Voltage Gain, AV = vo / vs Current Gain, Ai = io / is
CALCULATION OF GAIN
Voltage Gain, AV = vo / vs
Current Gain, Ai = io / is

16. Common-Emitter Amplifier

17. 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

18. STEPS
OUTPUT SIDE
Get the equivalent resistance at the output side, ROUT
Get the vo equation where vo = - gm vbeROUT
INPUT SIDE
Calculate Ri
Get vbe in terms of vs – eg: using voltage divider.
Go back to vo equation and calculate the voltage gain

19. Emitter Grounded β = 100 VBE = 0.7V VA = 100 V
VCC = 12 V
RC = 6 k
93.7 k
6.3 k
0.5 k
β = 100
VBE = 0.7V
VA = 100 V
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 5.9 k
VTH = V

20. Perform DC analysis to obtain the value of IC
BE loop: 5.9IB – = 0
IB =
IC = βIB = mA
Calculate the small-signal parameters
r = 2.74 k , ro = k and gm = 36.5 mA/V

21. Emitter Grounded β = 100 VBE = 0.7V VA = 100 V vo RC RTH vS
off - becomes short circuit
off - becomes short circuit
CC becomes short circuit during AC vS
RTH RC vo

22. Follow the steps 1. Rout = ro || RC = 5.677 k
vbe
gmvbe
RTH
RC = 6 k
RS = 0.5 k vS vO
5.9 k
2.74 k
105.37 k
Follow the steps
1. Rout = ro || RC = k
2. Equation of vo : vo = - ( ro || RC ) gmvbe= ( 5.677) vbe = vbe
3. Calculate Ri  RTH||r = 1.87 k
4. vbe in terms of vs  use voltage divider:
vbe = [ Ri / ( Ri + Rs )] * vs = vs
so vs = vbe

23. 5. Go back to equation of vo and calculate the gain
vbe
gmvbe
RTH
RC = 6 k
RS = 0.5 k vS vO
5.9 k
2.74 k
105.37 k
so: vs = vbe
5. Go back to equation of vo and calculate the gain
vo / vs = vbe / vbe
vo / vs = /
vo / vs =
AV = vo / vs =

24. Current Gain Output side: io = vo / RC = vo / 6
vbe
gmvbe
RTH
RC = 6 k
RS = 0.5 k vS vO
5.9 k
2.74 k
105.37 k is io
Output side: io = vo / RC = vo / 6
Input side: is = vs / (RS + Ri ) = vS / 2.37 RS Ri is vS
Current gain
= iout / is
= vo (2.37) = * 0.395 =
vs (6)

25. RECALL CHAPTER 1 RS vS Ri RS vS
Rload

26. Example
β = 139
VBE = V
VA = 

27. Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 4 k
VTH = 0.7 V
β = 139
VBE = V
VA = 
Perform DC analysis to obtain the value of IC
BE loop: 4 IB – 0.7 = 0
IB = 0.008
IC = βIB = mA
Calculate the small-signal parameters
r = 3.25 k , ro =  and gm = mA/V

28. Follow the steps 1. Rout = RC || RL = 0.3 || 100 = 0.3 k
vbe
0.3 k
100 k
0.5 k
4 k
3.25 k V1
gmvbe RC RL
Follow the steps
1. Rout = RC || RL = 0.3 || 100 = 0.3 k
2. Equation of vo : vo = - (RC || RL ) gmvbe= ( 42.77) vbe = vbe
3. Calculate Ri  RTH||r = 4 || 3.25 = k
4. vbe in terms of vs  use voltage divider:
vbe = [ Ri / ( Ri + Rs )] * v1 = v1
so v1 = vbe

29. 5. Go back to equation of vo and calculate the gain
vbe
0.3 k
100 k
0.5 k
4 k
3.25 k V1
gmvbe RL RC
so: v1 = vbe
5. Go back to equation of vo and calculate the gain
vo / v1 = vbe / vbe
vo / v1 = / 1.279
vo / v1 = -10
AV = vo / v1 = - 10

30. Output side: io = vo / 100 = vo / 100
RS = 0.5 k v1
100k
Ri = 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 * =
v1 (100)