1. Recall Lecture 13 Biasing of BJT Applications 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

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:
and

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. Let’s assume that Model 2 is used
IDQ
VDQ = V rd id
DC equivalent
AC equivalent

10. AC equivalent circuit –
Small-Signal Hybrid-π Equivalent
ib OR

11. 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

12. Hence from the equation of the AC parameters, we HAVE to perform DC analysis first in order to calculate them.

13. 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

14. Voltage Gain, AV = vo / vs Current Gain, Ai = iout / is
CALCULATION OF GAIN
Voltage Gain, AV = vo / vs
Current Gain, Ai = iout / is

15. Small-Signal Voltage Gain: Av = Vo / Vs
ib

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 = -  ib ROUT
INPUT SIDE
Calculate Rib using KVL where Rib = vb / ib
Calculate Ri
Get vb in terms of vs – eg: using voltage divider.
Go back to vo equation and replace where necessary

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 VCC = 12 V RC = 6 k

22. Follow the steps 1. Rout = ro || RC = 5.677 kvb
Follow the steps
1. Rout = ro || RC = k
2. Equation of vo : vo = - ( ro || RC )  ib= ib
3. Calculate Rib  using KVL: ib r - vb = 0  hence Rib = r
4. Calculate Ri  RTH||r = 1.87 k
5. vb in terms of vs  use voltage divider:
vb = [ Ri / ( Ri + Rs )] * vs = vs

23. so: vb = 0.789 vs replace in equation from step 1
6. Go back to equation of vo
vo = ib
but ib = vb / Rib (from step 3)
vo = [0.789 vs] / 2.74
vo = vs
AV = vo / vs =
bring VS over

24. TYPE 2: Emitter terminal connected with RE – normally ro =  in this type
β = 120
VBE = 0.7V
VA = 
VCC = 5 V
RC = 5.6 k
250 k
75 k
0.5 k
RE = 0.6 k

25. 0.5 k
57.7 k
RC = 6 k
7.46 k
RE = 0.6 k vb

26. 2. Equation of vo : vo = - RC  ib= - 720 ibvb
1. Rout = RC = 6 k
2. Equation of vo : vo = - RC  ib= ib
3. Calculate Rib  using KVL: ib r + ie RE - vb = 0
but ie = (1+ ) ib = 121 ib
so: ib [ 121(0.6) ] = vb  Rib = k
4. Calculate Ri  RTH||Rib = k
5. vb in terms of vs  use voltage divider:
vb = [ Ri / ( Ri + Rs )] * vs = vs

27. 6. Go back to equation of vo vo = - 720 ib = - 720 [ vb / Rib ]
so: vb = vs
6. Go back to equation of vo
vo = ib = [ vb / Rib ]
vo = [ vs / ]
vo = vs
AV = vo / vs =
bring VS over

28. 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

29. vb
CE becomes a short circuit path – bypass RE; hence similar to Type 1

30. β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA Bypass capacitor VCC = 5 V
RC = 2.3 k
20 k
0 k
RE = 5k
Bypass capacitor

31. Short-circuited (bypass) by the capacitor CE
β = 125
VBE = 0.7V
VA = 200 V
IC = 0.84 mA
0 k
10 k
RC = 2.3 k
238 k
3.87 k vb
Short-circuited (bypass) by the capacitor CE

32. Follow the steps 1. Rout = ro || RC = 2.278 kvb
Follow the steps
1. Rout = ro || RC = k
2. Equation of vo : vo = - ( ro || RC )  ib= ib
3. Calculate Rib  using KVL: ib r - vb = 0  hence Rib = r = 3.87 k
4. Calculate Ri  RTH||r = 2.79 k
5. vb in terms of vs  use voltage divider:
vb = [ Ri / ( Ri + Rs )] * vs = vs since RS = 0 k

33. 6. Go back to equation of vo vo = - 284.75 ib = - 284.75 [ vb / Rib ]
so: v b = vs
6. Go back to equation of vo
vo = ib = [ vb / Rib ]
vo = (vs) / 3.87
vo = vs
AV = vo / vs =
bring VS over