1. Transistor Modelling
A model is a combination of circuit elements, properly chosen, that best approximates the actual behaviour of a semiconductor device under specific operating condition.
There are three models:
re model
hybrid  model
hybrid equivalent model.

2. Before discussing model, we must make preparation to the discussion.
The superposition theorem is applicable and the the investigation of the dc conditions can be totally separated from the ac response.
An suitable Q-point has been chosen. Then the dc levels can be ignored in the
ac analysis network.

3. The coupling capacitor C1, C2 and bypass capacitor C3 are chosen to have a very small reactance. Therefore, each is replaced by a short circuit.
This also results in the “shorting out” of dc biasing resistor RE .
Those important parameters, such as Zi , Zo , Vi , Vo , Ii and Io should be kept unchanged.

4. The input voltage Vi is defined between base and ground.
The input current Ii is defined as the base current of the transistor.
The input impedance Zi is defined from base to ground.
The output voltage Vo is defined between collector and ground.

5. The output current Io is defined as the current through the load resistor RC.
The output impedance Zo is defined from collector to ground.
By introducing a common ground, R1
and R2 will be in parallel.
RC will appear from collector to emitter.

6. The transistor equivalent circuit also consists of resistors and independent controlled source.
The circuit analysis techniques such as superposition, Thevenin theorem can be applied to determine the desired quantities.
There are important quantities indicating amplification effects:

7. Voltage gain: Vo / Vi
Current gain: Io / Ii
Briefly, steps to obtain ac equivalent circuit:
dc sources to zero
capacitors to short circuits
Removing elements bypassed by short circuits and rearranging network

8. Figure: Transistor circuit for ac analysis

9. The re Model for CB
As shown in the figure, it is the common- base BJT circuit.
Now, re model is introduced.
On the input port, there is a resistor, re .
where re = 26mV / IE
(Eq. 5.1)
The subscript e of re indicates that it is the dc level of IE that determines the ac level of the resistance.

10. The re Model for CB where re = 26mV / IE (Eq. 5.1)
Figure: re model for common-base circuit

11. On the output port, there is a controlled current source, denoted by a diamond shape.
The Ic is controlled by the level of Ie .
The input impedance Zi is re .
The output impedance Zo  .
In general, for common-base configuration, the Zi is relatively small and Zo quite high.

12. Ai = Io / Ii = (-Ic)/(Ie) = (- Ie )/(Ie)  -1
Voltage gain:
Vo = - Io RL = - (- Ic) RL =Ie RL
Vi = Ii Zi = Ie Zi =Ie re
so that
Av = Vo / Vi = (Ie RL)/(Ie re) = ( RL)/ re  RL / re
Current gain:
Ai = Io / Ii = (-Ic)/(Ie) = (- Ie )/(Ie)  -1
For common-base configuration, the Vo and
Vi are in phase.

13. The re Model for CE
As shown in the figure, it is the common- emitter BJT circuit.
Now, replace the transistor with re model
On the input side, there exists re .
Zi = Vi / Ii = Vbe /Ib = (Ie re)/ Ib = [( +1)Ib re]/ Ib
= ( +1) re   re
( if  is large enough)
Typically, Zi is at moderate level.
BJT AC Analysis

14. Ideally, the output impedance Zo  .
On the output side, the controlled- current source is connected between collector and emitter, Isource=  Ib.
Ideally, the output impedance Zo  .
Or, Zo = ro , as shown in the figure. At this time, Ic  Ib .
Normally, voltage gain Av and current gain Ai are high.
For common-collector configuration, this model is also applicable.
BJT AC Analysis

15. CE Fixed Bias Circuit
As shown in the figure, it is the common- emitter fixed-bias configuration.
The input signal Vi is applied to the base and the output Vo is off the collector.
The input current Ii is not the base current and the Io is the collector current.
For small-signal analysis, VCC is replaced with ground.
BJT AC Analysis

16. So the RB is in parallel with the input port and RC output port.
Those dc blocking capacitors C1 and C2 are replaced with short circuits.
So the RB is in parallel with the input port and RC output port.
The parameters of Zi , Zo , Ii and Io should be in the same places as original network.
Finally, substitute the re model for the transistor.
BJT AC Analysis

17. The  is read from the datasheet or measured with testing instrument.
In practice,
The  is read from the datasheet or measured with testing instrument.
The re is determined from dc analysis. The ro is obtained from the datasheet.
In the small-signal analysis, we assume that , ro and re have been determined in advance.
BJT AC Analysis

18. From the figure, it is obvious that Zi = RB|| re
Input impedance Zi :
From the figure, it is obvious that
Zi = RB|| re
For the majority of situation, RB is greater than  re by more than a factor of 10.
So we get
Zi   re
BJT AC Analysis

19. From the definition of Zo , we get Zo = RC ||ro If ro  10RC , then
Output impedance Zo:
From the definition of Zo , we get
Zo = RC ||ro
If ro  10RC , then
Zo RC
Voltage gain Av:
From the figure, we get
Vo =( - Ib )(RC ||ro)
BJT AC Analysis

20. Vo =( - Ib )(RC ||ro) =( - Vi /( re))(RC ||ro)
And
Ib =Vi /( re) So
Vo =( - Ib )(RC ||ro) =( - Vi /( re))(RC ||ro)
=( - Vi / re)(RC ||ro) Finally,
Av = Vo / Vi
= - (RC ||ro) / re
BJT AC Analysis

21. Phase relationship:
The negative sign in Av ,reveals that a 180 phase shift occurs between the input and output signals.
This also means that a single-stage of amplifier of this type is not enough.
The magnitude of output signal is larger than that of input signal. But the frequencies of them should be the same.
BJT AC Analysis

22. Figure: re model for CE fixed-bias circuit
BJT AC Analysis

23. Figure: phase shift of input & output
BJT AC Analysis

24. Example 5.4  12V  0.7V  24.04A  VCC VBE
As shown in the figure, it is the common- emitter fixed-bias configuration. Determine: re , Zi , Zo, Av with ro = 50k .
Solution:
From dc analysis, we get
 12V  0.7V  24.04A
 VCC VBE I B R
470k B
BJT AC Analysis

25. re = (100)(10.71  ) = 1.071 k IE= (β+1) IB = (100+1)  24.04A
= 2.428mA
re = (26mV)/IE
= 
= (26mV)/ 2.428mA
Then, ac analysis
re = (100)(10.71  ) = k
Zi = RB|| re = (470 k ) || (1.071 k )
= k
BJT AC Analysis

26. Zo = RC || ro = (3 k ) || (50 k ) = 2.83k
Av = - (RC ||ro) / re = k / 
= - 264
From the Av ,we can see that the output signal has been amplified but out of
phase with the input signal. 
BJT AC Analysis

27. Figure: Example 5.4
BJT AC Analysis

28. Voltage Divider Bias
As shown in the figure, it is the voltage divider bias configuration.
Substituting re equivalent circuit, note that:
RE is absent due to the low impedance of the bypass capacitor CE .
When VCC is set to zero, one end of R1
and RC are connected to ground.
R1 and R2 remain part of the input circuit while RC is part of output circuit.
BJT AC Analysis

29. Some parameter of the equivalent circuit: Input impedance Zi :
Zi = R1|| R2|| re
Output impedance Zo:
Zo = RC ||ro
If ro  10RC , then
Zo RC
Voltage gain Av:
BJT AC Analysis

30. Vo =( - Ib )(RC ||ro) =[- Vi /( re)] (RC ||ro)
So,
Av = Vo / Vi = - (RC ||ro) / re
If ro  10RC , then
Av = Vo / Vi
Phase relationship:
 - RC / re
180 phase shift occurs between the input and output signals.
BJT AC Analysis

31. Figure: Voltage divider bias & its equivalent circuit
BJT AC Analysis

32. Example 5.5
As shown in the figure, it is the voltage divider bias configuration. Determine: re , Zi , Zo, Av with ro = 50k .
Solution:
dc analysis, testing RE >10 R2 ,
901.5k > 108.2k 135k > 82k (satisfied)
BJT AC Analysis

33. Using the approximate approach (Sec. 4.5, p155), we obtain:
R2 R1  R2
 8.2k  22V  2.81V
V  V B CC
56k  8.2k
VE = VB - VBE = 2.81V-0.7V = 2.11V
IE = VE / RE = 2.11V/1.5k = 1.41mA
re = 26mV / IE
= 
= 26mV / 1.41mA
BJT AC Analysis

34. ac analysis, Zi = R1|| R2|| re = 56k || 8.2k || (90)(18.44 )
Zo = RC ||ro
= 6.8k || 50k = 5.99k
Av= - (RC ||ro) / re
= - 324
= k /18.44  
BJT AC Analysis

35. Figure: Example 5.5
BJT AC Analysis

36. CE Emitter Bias
As shown in the figure, it is the emitter bias configuration, without CE .
Substituting re equivalent circuit, note that:
The resistance ro is ignored for simplicity.
First, let us obtain Zb .
Vi =Ibre+ Ie RE =Ibre+(+1) Ib RE
So, Zb = Vi / Ib = re+(+1) RE
BJT AC Analysis

37. At input port, Zi = RB ||Zb At output port, Zo = RC
Vo = -Io RC = - Ib RC = - (Vi / Zb) RC
So, Av = Vo / Vi = - RC / Zb
Phase relationship:
180 phase shift occurs between the output and input signals.
BJT AC Analysis

38. Figure: Emitter bias & its equivalent circuit
BJT AC Analysis

39. Example 5.6 VCC VBE  20V  0.7V  35.89A
As shown in the figure, it is the emitter bias configuration. Determine:
re , Zi , Zo, Av.
Solution:
dc analysis,
VCC VBE
R  (  1)R
I  B
B E
 20V  0.7V  35.89A
470k  (120  1)  0.56k
BJT AC Analysis

40. IE= (β+1) IB = (120+1)  35.89A =4.34mA re = (26mV)/IE
= 5.99 
= (26mV)/ 4.34mA
Then, ac analysis
Zb = re+(+1) RE
=1205.99 + 121560
= k
BJT AC Analysis

41. Zi = RB|| Zb = (470 k ) || (68.48k ) = 59.77 k Zo = RC = 2.2 k
Av = - RC / Zb
= -3.86 
= -120  2.2k / 68.48k
For emitter bias with C , see Ex. 5.7. E
For emitter bias + voltage divider, see Ex & 5.9.
BJT AC Analysis

42. Figure: Example 5.6
BJT AC Analysis

43. Emitter-Follower
As shown in the figure, it is the emitter- follower configuration.
Actually, it is a common-collector network.
The output is always slightly less than the input, but this is good for practical use.
Also the Vo is in phase with Vi and this accounts for the name of “emitter-follower”.
BJT AC Analysis

44. This is the reason why it is used for impedance matching purpose.
The emitter-follower configuration has a high input impedance and a low output impedance.
This is the reason why it is used for impedance matching purpose.
This can give a weak load to the previous stage and a strong output ability to the next stage.
BJT AC Analysis

45. Substituting re equivalent circuit, note that:
The resistance ro is ignored because for most applications a good approximation for the actual results can still be obtained.
First, the same as before, Zb is obtained:
Zb = re+(+1) RE
Then,
Zi = RB ||Zb
BJT AC Analysis

46. Ie = (+1) Ib = (+1) (Vi /Zb )
At output port,
Ie = (+1) Ib = (+1) (Vi /Zb )
So,
(  1)V
Ie   i
re  (  1)RE
If  is sufficiently large, we get Vi
I  e
r  R
e E
This means that is Ie generated by Vi .
BJT AC Analysis

47. Also, Vo is the potential drop across RE .
So we construct a network from the viewpoint of output port.
So, by setting the Vi to zero, we get
Zo = RE ||re
Furthermore, from this network it is obvious that RE
V  V o
R  r i
E e
BJT AC Analysis

48. Since, RE is usually much greater than re , Av  1 Phase relationship:
This leads to V RE
A  o  v
V R  r
i E e
Since, RE is usually much greater than re ,
Av  1
Phase relationship:
Vo and Vi are in phase.
BJT AC Analysis

49. Figure: Emitter-follower & its equivalent circuit
BJT AC Analysis

50. Example 5.10 VCC VBE  20V  0.7V  20.43A
As shown in the figure, it is the emitter- follower configuration. Determine:
re , Zi , Zo, Av.
Solution:
dc analysis,
VCC VBE
R  (  1)R
I  B
B E
 20V  0.7V  20.43A
220k  (100  1)  3.3k
BJT AC Analysis

51. IE= (β+1) IB = (100+1)  20.43A =2.063mA re = (26mV)/IE
= 12.6 
= (26mV)/ 2.063mA
Then, ac analysis
Zb = re+(+1) RE
=10012.6 + 1013.3k
= k
BJT AC Analysis

52. Zi = RB|| Zb = (220 k ) || (334.56k )
Zo = RE ||re = 3.3k ||12.6  =  V RE
A  o 
 3.3k  v
V R  r
3.3k  12.6 i
E e 
For some variations of emitter follower configuration, see Fig & 5.55
BJT AC Analysis

53. Figure: Example 5.10
BJT AC Analysis

54. Common-base Configuration
As shown in the figure, it is the common- base configuration.
It has a relatively low Zi and high Zo and a current gain less than 1.
However, the Av can be quite large.
Substituting re equivalent circuit into the network, note that:
BJT AC Analysis

55. So that Av = Vo / Vi =  RC / re
The resistance ro is typically in the M and can be ignored in parallel with RC.
Zi = RE || re Zo= RC
First,
Then,
And Vo = - Io RC
With Vi = Ie re ,
= Ic RC =  Ie RC
So that Av = Vo / Vi =  RC / re
BJT AC Analysis

56. At last, assuming RE >> re , we get Ii = Ie And
Io = - Ie = - Ii
So that Ai = Io / Ii = -  -1
Phase relationship:
Vo and Vi are in phase in common- base configuration.
BJT AC Analysis

57. Figure: Common-base & its equivalent circuit
BJT AC Analysis

58. Example 5.11  2V  0.7V  1.3mA  VEE VBE
As shown in the figure, it is the common- base configuration. Determine:
re , Zi , Zo, Av and Ai.
Solution:
dc analysis,
 2V  0.7V  1.3mA
 VEE VBE I E R
1k E
re = (26mV)/IE = (26mV)/ 1.3mA
= 20 
BJT AC Analysis

59. Then, ac analysis Zi = RE || re = 1 k || 20 = 19.61 Zo= RC = 5 k
Av= RC / re
=0.985 k / 20 = 245
Ai = - =  -1 
BJT AC Analysis

60. Figure: Example 5.11
BJT AC Analysis

61. Summary of Chapter 5
The re model is applied to the analysis of transistor configurations:
 Common-emitter:
Fixed bias; Voltage divider; Emitter bias
 Common-collector: emitter-follower
 Common-base.
The calculation of those parameters, like Vi ,
Vo , Ii , Io, Zi , Zo, Av and Ai ,in the circuits.
BJT AC Analysis