Transistors are semiconductor devices with three terminals - collector, base, and emitter. There are two main types, NPN and PNP, which differ in the doping of the semiconductor regions. Bipolar transistors can operate as amplifiers, switches, or oscillators depending on the biasing conditions. The common emitter configuration provides voltage gain but inverts the phase, while common base and common collector configurations do not invert phase. Transistor gain is determined by factors like current gain (beta) and varies based on operating point and temperature.

1. Transistors
Transfer Resistor
Chapter 9

2. Bipolar Transistors
Two PN junctions joined together
Two types available – NPN and PNP
The regions (from top to bottom) are called the collector (C), the
base (B), and the emitter (E)
Base
Collector
Emitter

3. Operation
 Begin by reverse biasing the CB junction
 Here we are showing an NPN transistor
as an example
 Now we apply a small forward bias on
the emitter-base junction
 Electrons are pushed into the base,
which then quickly flow to the collector
 The result is a large emitter-collector
electron current (conventional current is
C-E) which is maintained by a small E-B
voltage
 Some of the electrons pushed into the
base by the forward bias E-B voltage
end up depleting holes in that junction
 This would eventually destroy the
junction if we didn’t replenish the holes
 The electrons that might do this are
drawn off as a base current

4. Currents

5. Conventional View

6. Origin of the names
the Emitter 'emits' the electrons which
pass through the device
the Collector 'collects' them again once
they've passed through the Base
...and the Base?...

7. Original Manufacture

8. Base Thickness
 The thickness of the unmodified Base region has
to be just right.
 Too thin, and the Base would essentially vanish. The
Emitter and Collector would then form a continuous
piece of semiconductor, so current would flow
between them whatever the base potential.
 Too thick, and electrons entering the Base from the
Emitter wouldn't notice the Collector as it would be
too far away. So then, the current would all be
between the Emitter and the Base, and there'd be no
Emitter-Collector current.

9. Amplification Properties
The C-B voltage junction operates near
breakdown.
 This ensures that a small E-B voltage causes
avalanche
 Large current through the device

10. Common Base NPN

11. Common Emitter NPN

12. Common Collector NPN
How does IC vary with VCE for various IB?
Note that both dc sources are variable
Set VBB to establish a certain IB

13. Collector Characteristic Curve
 If VCC = 0, then IC = 0 and VCE = 0
 As VCC ↑ both VCE and IC ↑
 When VCE  0.7 V, base-collector
becomes reverse-biased and IC
reaches full value (IC = bIB)
 IC ~ constant as VCE ↑. There is a
slight increase of IC due to the
widening of the depletion zone
(BC) giving fewer holes for
recombinations with e¯ in base.
 Since IC = bIB, different base
currents produce different IC
plateaus.

14. NPN Characteristic Curves

15. PNP Characteristic Curves

16. Load Line
For a constant load, stepping IB gives different currents (IC) predicted by
where the load line crosses the characteristic curve. IC = bIBworks so long as
the load line intersects on the plateau region of the curve.
Slope of
the load
line is 1/RL

17. Saturation and Cut-off
Note that the load line intersects the 75 mA curve below the
plateau region. This is saturation and IC = bIB doesn’t work
in this region.
Cut-off

18. Example
We adjust the base current to 200 mA and note
that this transistor has a b = 100
 Then IC = bIB = 100(200 X 10-6A) = 20 mA
Notice that we can use Kirchhoff’s voltage law
around the right side of the circuit
 VCE = VCC – ICRC = 10 V – (20 mA)(220 W)
= 10 V – 4.4 V = 5.6 V

19. Example
Now adjust IB to 300 mA
 Now we get IC = 30 mA
 And VCE = 10 V – (30 mA)(220 W) = 3.4 V
Finally, adjust IB = 400 mA
 IB = 40 mA and VCE = 1.2 V

20. Plot the load line
VCE IC
5.6 V 20 mA
3.4 V 30 mA
1.2 V 40 mA

21. Gain as a function of IC
As temperature increases, the gain increases
for all current values.

22. Operating Limits
There will be a limit on the dissipated power
 PD(max) = VCEIC
 VCE and IC were the parameters plotted on the
characteristic curve.
 If there is a voltage limit (VCE(max)), then you can
compute the IC that results
 If there is a current limit (IC(max)), then you can compute
the VCE that results

23. Example
 Assume PD(max) = 0.5 W
VCE(max) = 20 V
IC(max) = 50 mA
PD(max) VCE IC
0.5 W 5 V 100 mA
10 50
15 33
20 25

24. Operating Range
Operating
Range

25. Voltage Amplifiers
Common Base PNP
Now we have added an ac source
The biasing of the junctions are:
BE is forward biased by VBB - thus a small resistance
BC is reverse biased by VCC – and a large resistance
Since IB is small, IC  IE

26. Equivalent ac Circuit
gain
voltage

 V
in
out
A
V
V
rE = internal ac emitter
resistance
IE = Vin/rE (Ohm’s Law)
Vout = ICRC  IERC
E
C
E
E
C
E
V
r
R
r
I
R
I
A 
 Recall the name – transfer resistor

27. Current Gains
 Common Base
 a = IC/IE < 1
 Common Emitter
 b = IC/IB
b
a
1
1
1
I
I
1
I
I
I
I
I
Law
Current
s
Kirchhoff'
From
C
B
C
E
B
C
E






b
b
a
b
a
b
ab
a
b
b
b
a








1
)
1
(
1
1

28. Example
If b = 50, then a = 50/51 = 0.98
 Recall a < 1
Rearranging,
b = a + ab
b(1-a) = a
b = a/(1-a)

29. Transistors as Switches

30. The operating points
We can control the base current using VBB (we
don’t actually use a physical switch). The circuit
then acts as a high speed switch.

31. Details
In Cut-off
 All currents are zero and VCE = VCC
In Saturation
 IB big enough to produce IC(sat)  bIB
Using Kirchhoff’s Voltage Law through the
ground loop
 VCC = VCE(sat) + IC(sat)RC
 but VCE(sat) is very small (few tenths), so
 IC(sat) VCC/RC

32. Example
a) What is VCE when Vin = 0 V?
Ans. VCE = VCC = 10 V
b) What minimum value of IB is
required to saturate the transistor if
b = 200? Take VCE(sat) = 0 V
IC(sat)  VCC/RC = 10 V/1000 W
= 10 mA
Then, IB = IC(sat)/b = 10 mA/200 = 0.05mA

33. Example
LED
If a square wave is input for VBB,
then the LED will be on when the
input is high, and off when the
input is low.

34. Transistors with ac Input
Assume that b is such that
IC varies between 20 and 40
mA. The transistor is
constantly changing curves
along the load line.

35. Pt. A corresponds to the positive peak. Pt. B
corresponds to the negative peak. This graph shows
ideal operation.

36. Distortion
The location of the point Q (size of the dc
source on input) may cause an operating
point to lie outside of the active range.
Driven to saturation
Driven into Cutoff

37. Base Biasing
 It is usually not necessary to provide two
sources for biasing the transistor.
The red arrows follow the base-emitter
part of the circuit, which contains the
resistor RB. The voltage drop across RB
is VCC – VBE (Kirchhoff’s Voltage
Law). The base current is then…
C
BE
CC
R
V
V 

B
I and IC = bIB

38. Base Biasing
 Use Kirchhoff’s Voltage Law on the black
arrowed loop of the circuit
VCC = ICRC + VCE
So, VCE = VCC – ICRC
VCE = VCC – bIBRC
 Disadvantge
 b occurs in the equation for both VCE and IC
 But b varies – thus so do VCE and IC
 This shifts the Q-point (b-dpendent)

39. Example
 Let RC = 560 W @ 25 °C b = 100
RB = 100 kW @ 75 °C b = 150
VCC = +12 V
mA
11.3
A)
(100)(113
I
I B
C 

 m
b
V
5.67
)
A)(560
(100)(113
-
V
12
R
I
V
V C
B
CC
CE




W
m
b
@ 75 °C
IB is the same
IC = 16.95 mA
VCE = 2.51 V
IC increases by 50%
VCE decreases by 56%
A
113
100,000
V
0.7
-
V
12
I
C
25
@
B m
W





B
BE
CC
R
V
V

40. Transistor Amplifiers
Amplification
 The process of increasing the strength of a
signal.
 The result of controlling a relatively large
quantity of current (output) with a small
quantity of current (input).
Amplifier
 Device use to increase the current, voltage, or
power of the input signal without appreciably
altering the essential quality.

41. Class A
Entire input waveform is faithfully
reproduced.
Transistor spends its entire time in the
active mode
 Never reaches either cutoff or saturation.
 Drive the transistor exactly halfway between
cutoff and saturation.
 Transistor is always on – always dissipating
power – can be quite inefficient

42. Class A

43. Class B
No DC bias voltage
 The transistor spends half its time in active
mode and the other half in cutoff

44. Push-pull Pair
Transistor Q1 "pushes" (drives the output voltage in a positive direction with
respect to ground), while transistor Q2 "pulls" the output voltage (in a negative
direction, toward 0 volts with respect to ground).
Individually, each of these transistors is operating in class B mode, active only for
one-half of the input waveform cycle. Together, however, they function as a team to
produce an output waveform identical in shape to the input waveform.

45. Class AB
Between Class A (100% operation) and
Class B (50% operation).

46. Class C
IC flows for less than half then cycle. Usually get
more gain in Class B and C, but more distortion

47. Common Emitter Transistor Amplifier
Notice that VBB forward biases the emitter-base junction and dc
current flows through the circuit at all times
The class of the amplifier is determined by VBB with respect to the
input signal.
Signal that adds to VBB causes transistor current to increase
Signal that subtracts from VBB causes transistor current to decrease

48. Details
 At positive peak of input, VBB is adding to the
input
 Resistance in the transistor is reduced
 Current in the circuit increases
 Larger current means more voltage drop across
RC (VRC = IRC)
 Larger voltage drop across RC leaves less
voltage to be dropped across the transistor
 We take the output VCE – as input increases, VCE
decreases.

49. More details
As the input goes to the negative peak
 Transistor resistance increases
 Less current flows
 Less voltage is dropped across RC
 More voltage can be dropped across C-E
The result is a phase reversal
 Feature of the common emitter amplifier
The closer VBB is to VCC, the larger the
transistor current.

50. PNP Common Emitter Amplifier

51. NPN Common Base Transistor
Amplifier
Signal that adds to VBB causes transistor current to increase
Signal that subtracts from VBB causes transistor current to decrease
• At positive peak of input, VBB is adding to the input
• Resistance in the transistor is reduced
• Current in the circuit increases
• Larger current means more voltage drop across RC (VRC = IRC)
• Collector current increases
• No phase reversal

52. PNP Common Base Amplifier

53. NPN Common Collector Transistor
Amplifier
Also called an Emitter Follower circuit – output on emitter is almost a replica of the
input
Input is across the C-B junction – this is reversed biased and the impedance is high
Output is across the B-E junction – this is forward biased and the impedance is low.
Current gain is high but voltage gain is low.

54. PNP Common Collector Transistor
Amplifier

55. Gain Factors
E
C
I
I

a Usually given for common base amplifier
B
C
I
I

b Usually given for common emitter amplifier
B
E
I
I

 Usually given for common collector amplifier

56. Gamma
 Recall from Kirchhoff’s Current Law
 IB + IC = IE

b
1
I
I
I
I
1
I
B
E
B
C
B





a

a
a
a

a
a
a
a
b
-
1
1
-
1
-
1
LCD
-
1
1
-
1
since
And






Ex. For b = 100 a = b/(1+b) = 0.99
 = 1 + b = 101

57. Bringing it Together
Type Common
Base
Common
Emitter
Common
Collector
Relation
between
input/output
phase
0° 180° 0°
Voltage Gain High Medium Low
Current Gain Low (a) Medium (b) High ()
Power Gain Low High Medium
Input Z Low Medium High
Output Z High Medium Low

58. Hybrid Parameters
Condition
hi Input resistance Output shorted
hr Voltage feedback ratio Input open
hf Forward current gain Output shorted
ho Output conductance Input open
Second subscript indicates common base (b), common emitter
(e), or common collector (c)

59. Hybrid Parameters
= b
= Slope of curve

60. Hybrid Parameters
hie = VB/IB Ohm’s Law
hie =input impedance
hre = VB/VC

61. Hybrid Parameters
hfe = IC/IB
Equivalent of b
hoe = IC/VC

62. Various Forms
Common
Emitter (e)
Common
Base (b)
Common
Collector (c)
hi (ohms) VB/IB VE/IB VB/IB
hr (unitless) VB/VC VE/VC VB/VE
hf (unitless) IC/IB IC/IE IE/IB
ho (watts) ICVC ICVC IEVE

63. Pin-outs
No standard – look at the spec sheet or the case

64. Loudness
 When the energy (intensity) of the sound
increases by a factor of 10, the loudness
increases by 1 bel
 Named for A. G. Bell
 One bel is a large unit and we use 1/10th bel, or
decibels
 When the energy (intensity) of the sound
increases by a factor of 10, the loudness
increases by 10 dB

65. Decibel Scale
For intensities
 L = 10 log(I/Io)
For energies
 L = 10 log(E/Eo)
For amplitudes
 L = 20 log(A/Ao)

66. Threshold of Hearing
 The Io or Eo or Ao refers to the intensity, energy, or
amplitude of the sound wave for the threshold of
hearing
 Io = 10-12 W/m2
 Loudness levels always compared to threshold
 Relative measure

67. Common Loud Sounds
160
Jet engine - close up
150
Snare drums played hard at 6 inches away
Trumpet peaks at 5 inches away
140 Rock singer screaming in microphone (lips on mic)
130
Pneumatic (jack) hammer Cymbal crash
Planes on airport runway 120 Threshold of pain - Piccolo strongly played
Fender guitar amplifier, full volume at 10 inches away
Power tools 110
Subway (not the sandwich shop) 100
Flute in players right ear - Violin in players left
ear

68. Common Quieter Sounds
90
Heavy truck traffic
Chamber music 80
Typical home stereo listening level
Acoustic guitar, played with finger at 1 foot away
Average factory
70
Busy street Small orchestra
60 Conversational speech at 1 foot away
Average office noise 50
Quiet conversation 40
Quiet office 30
Quiet living room 20
10 Quiet recording studio
0 Threshold of hearing for healthy youths

69. The Math
l1 = 10 log(I1/Io)
l2 = 10 log(I2/Io)
l2 – l1 = Dl = 10(log I2 – log Io – log I1 + log Io)
= 10(log I2 – log I1)
l2 – l1 = Dl = 10 log(I2/I1)
Threshold of Hearing when I = Io l = 0 dB
Threshold of Pain when I  1012 Io l = 120 dB

70. Example
 A loudspeaker produces loudness rated at 90 dB
(l1) at a distance of 4 ft (d1). How far can the
sound travel (d2) and still give a loudness at the
listener’s ear of 40 dB (l2 - conversation at 3 ft.)?
Sound follows the inverse square law I1/I2 = d2
2/d1
2
Dl = 50 dB = 10 log(I2/I1)
log(I2/I1) = 5 which means I2/I1 = 105
If d1 = 4 ft, then d2
2 = (I1/I2) d1
2 = 105 (4 ft)2
d2 = 1260 ft (about ¼ mile)

71. Common Emitter Current Gain
For the -3 dB point
 Dl = 3 dB = 10 log (I1/I2)
 I1/I2 = 2 = P1/P2
 so 3 dB below initial level mean half the power
Frequency
hfe
0 dB
-3 dB

72. Why do Frequency limits occur?
 It takes a certain time for e- to travel from emitter
to collector (transit time)
 If frequency is too high, applied current varies
too rapidly
 Electrons may be unable to dislodge rapidly
enough to move from E to C before current
surges in the other direction.
Making the base thinner reduces transit time and
improves frequency response

73. Interelement Capacitance
 As reverse bias increases on the C-B junction,
the depletion zone increases and C decreases
(C = eA/d and d increasing).
 As emitter current increases, C increases (d
decreasing).
 If capacitance changes, so does capacitive
reactance
 Increasing C decreases XC
C
f
2
1
XC



74. Feedback
Small base current provides a path back to
input
 If the feedback voltage aids the input voltage,
then it is positive (regenerative) feedback
 If the feedback is too large, the amplifier will
oscillate

75. Superheterodyne Receiver