Semiconductor

1. CHAPTER 3: BIPOLAR
JUNCTION TRANSISTOR

2.  Two main categories of transistors:
◦ bipolar junction transistors (BJTs) and
◦ field effect transistors (FETs).
 Transistors have 3 terminals where the application of
current (BJT) or voltage (FET) to the input terminal
increases the amount of charge in the active region.
 The physics of "transistor action" is quite different for the
BJT and FET.
 In analog circuits, transistors are used in amplifiers and
linear regulated power supplies.
 In digital circuits they function as electrical switches,
including logic gates, random access memory (RAM),
and microprocessors.

3.  A bipolar transistor
essentially consists of a pair of
PN Junction diodes that are
joined back-to-back.
 There are therefore two kinds
of BJT, the NPN and PNP
varieties.
 The three layers of the
sandwich are conventionally
called the Collector, Base,
and Emitter.

4.  Three terminals:
◦ Base (B): very thin and lightly doped central region (little
recombination).
◦ Emitter (E) and collector (C) are two outer regions
sandwiching B.
 Normal operation (linear or active region):
◦ B-E junction forward biased; B-C junction reverse biased.
◦ The emitter emits (injects) majority charge into base region
and because the base very thin, most will ultimately reach
the collector.
◦ The emitter is highly doped while the collector is lightly
doped.
◦ The collector is usually at higher voltage than the emitter.

8.  Active:
◦ Most importance mode, e.g. for amplifier operation.
◦ The region where current curves are practically flat.
 Saturation:
◦ Barrier potential of the junctions cancel each other out
causing a virtual short.
◦ Ideal transistor behaves like a closed switch.
 Cutoff:
◦ Current reduced to zero
◦ Ideal transistor behaves like an open switch.

9.  There are three different currents in transistors :
collector current I C , base current I B and emitter current I E
.
 Since the emitter is the source of the electrons, it has
the largest of the three currents.
 The collector current is slightly lower than the emitter
current, while the base current is usually very small.
 The relationship is expressed as follows:
IE = IB + IC

10. DC Gain
 The relationship between the collector current
and the base current is expressed in the
following formula
β = I / I
C B
 Current gain(H FE )
 The ratio between collector current and emitter
current is referred as α (alpha) and is shown in
the following relationship :
 α = I /I
C E

11. Voltages in Transistor Circuit
 Emitter voltage,VE
 Base voltage,VB
 Collector voltage,VC RC
 Collector emitter C
RB
voltage,VCE B VCC
VCE
 Base emitter VBE VC
voltage,VBE E
VB
 Collector Base VE
voltage,VCB
 DC supply voltage,VCC

12. I-V Characteristics
I - V characteristics are used to explain the operation of
transistors.
Input I-V characteristics Output I-V characteristics curves

13. Load Line
 A load line is a line
drawn over the
collector curves to
show each and
every possible Lo
operating point of a dl
ine
a transistor.

14.  The point at which
the base current Q
intersects the load
line is referred to as
the operating point
or the quiescent
point.

17. Fixed Base Bias Circuit (Simple Bias
Circuit) and Formula
• It consists only of a fixed bias
resistor and load resistor. VCC = VRB + VBE
VRB = VCC - VBE
IB = VCC -VBE
RB RC IC
RB
IB
VCC
VCE IC = βIB
VBE VCC = VCE + VRC
IE
VCE = VCC - ICRC

19. Potential Divider Bias Circuit
It offers the best resilience against
changes in temperature and device
characteristics and by far more stabilized.
R1 RC R1 and R2 form a potential divider, which
IC
will fix the base potential of the transistor.
IB The current is usually set at 10 times
VCC greater than the base current required by
VCE the transistor
VBE
The base emitter voltage drop is
R2 RE IE approximated as 0.7volt. There will also
be a voltage drop across the emitter
resistor, RE, this is generally set to about
10% of the supply voltage.
If IC increases, then so will IE.

20. Formula
VCC = IR1 + IR2 IE = I B + I C
VCE = VCC –VRC - VRE
VC = VCE + VRE
VB = R2 VCC
R1 + R2
VRE = VB - VBE
IE = VRE/ RE
IB can be ignored if the value is too small

21.  One factor can affect the IB is AC input signal.
 AC input signal must not too big so that can control the
IB.

22. IC(mA)
Q point is at IB = 60µA, Ic =
100µA
10 6mA, Vc = 8V.
80µA
8
60µA
If IB is between 40µA and
6
80µA, while Ic is between 4mA
4
40µA
and 8mA
20µA
2
So Vc is between 4V and 12V.
0V
VC (V)
4 5 8 10 12 15 20 Vc is output voltage for the
amplifier circuit, Vo.

23.  If input signal is too big, It will affect the IB.
 IB will become big, followed by Ic and Vc.
 Ic and Vc will over the saturation point and will
cut the signal.
 When this happen, there is a distortion to the
output signal.
 If there is a distortion to the output signal, the
amplifier is not efficient.

24. From figure above, Load resistance,
 rL : R3 // R6
Will have a new AC loadline (new value of cut off
and saturation point).
 Ic(sat)new = ICQ + (VCQ/RL)
 Vc(sat)new = VCQ + ICQ . RL

25. IC(mA)
Garis Beban a.u
3.5
3
2.5 Titik-Q
2
1.5 Garis Beban a.t
1
0.5
5 10 15 20 25 30
VC (V)
Rajah 6.36

26.  Av = Vo/Vi = rL / re
Where re = 25mV/ IE
AC loadline exists when the load for AC voltage
output is different with the load for DC voltage
output.

27. IC(mA)
3.5
Garis Beban a.u
3
2.5 Titik-Q
2
1.5 Garis Beban a.t
1
0.5
0 3 5 10 15 20 25 30 VC (V)
VCQ 27V
15Vp 12Vp VC(alih)
Bahagian 1 :
VC(ALIH)(AU) - VCQ = 27V – 15V
= 12Vp
24Vp-p
Bahagian 2 :
VCQ – 0 = 15V – 0 = 15Vp
30Vp-p

28.  Is output voltage that oscillated within the
operation point symmetrically and no distortion.

29. Can be determine from equation Av = rL / re
Use the Av to determine maximum input voltage, Vi.
 Av = Vo/Vi, so Vi = Vo/Av.
It can define as undistorted maximum input
signal because it has undistorted maximum
output signal.

32.  Frequency response is characterized by the
response magnitude (measure in dB) versus
frequency.
 From frequency response graph above :
 Y – axis (shows response in decibel (dB))
 X – axis (shows frequency (Hz))
 Important parameters of frequency response
curve :
◦ Maximum voltage gain, Av(max).
◦ Maximum voltage gain, Av (max) in dB
◦ Cut off frequency (-3dB)
◦ Frequency bandwidth.

33.  Maximum voltage gain , Av(max) =
maximum ratio of output voltage, Vo over
input voltage, Vi.
 Av in dB = 20 log Av(max)
 Cut off frequency – or corner frequency is
the frequency either above or below
power output of a circuit.

34.  fc1 – lower 3dB cut off frequency
 fc2 – upper 3dB cut off frequency
 Frequency Bandwidth – range of
frequency between fc1 and fc2.

35. Use Av(max)dB instead of Av(max)
because to scale a large measurement
down to a much smaller and more
useable range.
Eg : Vo = 100, Vi = 1
 Av(max) = Vo/Vi = 100/1 = 100
 In dB, Av(max)dB = 20 log Av(max)
 = 20 log 100 = 40.

36.  This frequency is where a device will quit
working or operating in an efficient manner
resulting in it shutting down or being cut off.
 Refer to the 3dB point since a fall of 3dB
corresponds approximately to half power of
the output circuit.
 There is lower cut off frequency and upper
cut off frequency.

37. a band of a given width can carry the
same amount of information.
So to relate bandwidth to the
performance of amplifier,
◦ wide bandwidth, the amplifier is more
efficient.
◦ BW = fc2 – fc1

39.  fB = 1 / (2π RCB)
where :
R = Rs + Rin ; Rin = R1 ║R2 ║hie
 CB = value of the Base coupling capacitor,
C1

40.  fC = 1 / [2π (RC + RL) CC]
 RC +RL = sum of the resistance in the
collector circuit
 CC = value of the Base coupling capacitor,
C2

41.  Rout = RE ║ ( re + Rin / hfe ); re = VT / IE
 Rin = R1 ║ R2 ║ Rs
 fE = 1 / (2π Rout CE )
CE = value of the Base coupling capacitor,C3

42.  There are 4 classes of amplifier.
◦ Class A
◦ Class B
◦ Class AB
◦ Class C

43. Class A Amplifier
 The most common and simplest form of power
amplifier that uses the switching transistor in the
standard common emitter circuit configuration as
seen previously.
 The transistor is always biased "ON" so that it
conducts during one complete cycle of the input
signal waveform producing minimum distortion and
maximum amplitude to the output.

44.  Is the ideal operating mode, because there can be
no crossover or switch-off distortion to the output
waveform even during the negative half of the
cycle.
 Class A power amplifier output stages may use a
single power transistor or pairs of transistors
connected together to share the high load current.

46.  Class-B amplifiers only amplify half of the input
wave cycle, thus creating a large amount of
distortion, but their efficiency is greatly improved
and is much better than class A.
 Class B has a maximum theoretical efficiency of
π/4. (i.e. 78.5%) This is because the amplifying
element is switched off altogether half of the time,
and so cannot dissipate power.

47.  A single class-B element is rarely found in
practice, though it has been used for driving
the loudspeaker in the early IBM Personal
Computers with beeps, and it can be used in RF
power amplifier where the distortion levels are
less important. However, class C is more
commonly used for this.

49.  A practical circuit using class-B elements is
the push–pull stage, such as the very simplified
complementary pair arrangement .
 Complementary or quasi-complementary devices
are each used for amplifying the opposite halves
of the input signal, which is then recombined at
the output.

52.  This arrangement gives excellent efficiency, but
can suffer from the drawback that there is a small
mismatch in the cross-over region – at the "joins"
between the two halves of the signal, as one
output device has to take over supplying power
exactly as the other finishes. This is
called crossover distortion

53.  In class-AB operation, each device operates the
same way as in class B over half the waveform,
but also conducts a small amount on the other
half.
 As a result, the region where both devices
simultaneously are nearly off (the "dead zone") is
reduced. The result is that when the waveforms
from the two devices are combined, the crossover
is greatly minimised or eliminated altogether.

55.  Collector current flows for less than one half cycle
of the input signal.
 By reverse biasing the emitter-base junction,
which sets the dc operating point below cut off
and allows only the portion of the input signal that
overcomes the reverse bias to cause collector
current flow.

57.  Class A – amps sound the best, cost the most,
and are the least practical. They waste power and
return very clean signals.
 Class B - operated amplifier is used extensively
for audio amplifiers that require high-power
outputs. It is also used as the driver- and power-
amplifier stages of transmitters.

58.  Class C - operated amplifier is used as a radio-
frequency amplifier in transmitters.
 Class AB - operated amplifier is commonly used
as a push-pull amplifier to overcome a side effect
of class B operation called crossover distortion

59.  Bipolar transistor amplifiers must be properly
biased to operate correctly.
 There are :
 - base biased with emitter feedback
technique.
 - biased voltage divider technique

60. Vin

61. Fig 1 Fig 2
Circuit in Fig 1 is Circuit in Fig 2 is stable
unstable

62.  Adding resistance (RE) to the emitter of transistor.
 RE improves performance by adding negative feedback.
 IC begins to increase as temperature rises.
 The increase in IC increases IE and as a result VE rises.
 VBE (or VB - VE) is fixed, but the rise in VE reduces VBE. The
overall effect of reducing VBE is to reduce IC which in turn
makes the circuit stable. Negative feedback.

63.  R E greatly reduces error (variation in gain and
consequent distortion).
 Emitter resistor RE also solves other problems
such as temperature instability and distortion.

64.  The voltage divider is
formed using external
resistors R1 and R2.
 The voltage across R2
forward biases the emitter
junction.
 By proper selection of
resistors R1 and R2, the
operating point of the
transistor can be made
independent of β.

65.  RTH = R1//R2
 RTH = R1R2
 R1 + R2
 VTH = R2 x VCC
 R1 + R2
 IE = (β+1)IB

66. KVL from input loop :
 - VTH + IBRTH + VBE + IERE = 0
 - VTH + IBRTH + VBE + (β+1)IBRE = 0
 - VTH + IB(RTH + (β+1)RE) + VBE = 0
 IB = VTH – VBE
(RTH + (β+1)RE)

67. KVL from output loop:
- VCC – ICRC – VCE – IERE = 0
 IC = βIB
 IE = (β+1)IB

68. Application
Transistors are commonly used for :
1. Amplifier circuit 2. Switching circuit