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Transcript of ecm1106_n3
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
________________________________________________________________________________
1
THE BIPOLAR JUNCTION TRANSISTOR (BJT)
1. Bias Stability
2. Transistor Switches
________________________________________________________________________________
1. Bias Stability
For amplifying signals, DC bias was used to fix ICand VCEfor constant IBon the linear or active
part of the characteristics. The position of the operating point depends on the current gain .
, however, changes with temperatureandfrom transistor to transistor. The consequence is ashift in the operating point on the characteristic curve outside the active region of the
characteristics leading to distortion, making the circuit not suitable for amplification.
Example: Determine the values of ICand VCEfor the following circuits:
1k5286k
typ=100
+15V
(a)
1k5286k
max
=200
+15V
(b)
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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1.1. Voltage-Divider Bias Circuit
The circuit shown below provides DC bias that is almost independent of . The idea here is to
fix the emitter current IE(IC) through the emitter resistor REsuch that it is independent of .
In this case, IBis no longer fixed but will change to maintain the constant emitter (and collector)currents.
Resistors R1and R2form a voltage-divider to set the DC voltage level of the base terminal (to
ensure that the base-emitter junction is forward biased).
VCC
C1
R1
R2
RC
RE
vivo
C2
VCC
C1
R1
R2
RC
RE
vivo
C2
Aim: choose resistor values to set the operating point (i.e. VCEand IC) in the linear region of the
output characteristics.
1.1.1. Input Circuit
VCC
R1
R2 Ri
I1
I2
IB
VCC
R1
R2RE
I1
I2
IE
IB
Ri
VB
VCC
R1
R2 Ri
I1
I2
IB
VCC
R1
R2RE
I1
I2
IE
IB
Ri
VB
Resistance between base and ground: RiRE.
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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Provided Riis much larger than R2such that:
RE10R2
then IB
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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1.1.3. Self-Stabilising Network Action
IC increase
IE increase
VE=IERE increase
VBE=VB-VE decrease VB = constant
determined by
R1, R2 and VCC
IB decrease
VBE
IB
0
IC=IB decrease
IC increase
IE increase
VE=IERE increase
VBE=VB-VE decrease VB = constant
determined by
R1, R2 and VCC
IB decrease
VBE
IB
0 VBE
IB
0
IC=IB decrease
Any tendency to increase ICwill be met with a network reaction to decrease it.
1.1.4. Coupling Capacitors
The input capacitor C1serves two purposes:
1. Couples (superimposes) the AC input signal with the bias current.2. Blocks DC components of input signals (XC= ) that would affect the set DC bias
level.
The output capacitor C2is used to AC couple the output signal to a load and remove DC levels
produced by biasing stages.
The coupling capacitors should ideally provide zero reactance (XC=0, i.e. C) to AC signals.
Typical value for coupling capacitors are 1 10F.
1.2. Amplifier Design Considerations
RCis designed to drop about half of the supply voltage to provide a symmetrical voltage swing
of the output signal.
REcannot be unreasonably large because voltage across it limits range of V CE. VEis typically
set to be 1/4 to 1/0 of VCC.
The base resistor R2is chosen such that 10R2RE.
The value of R1is then determined from CC
21
2B VRR
R
V +=
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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Example: The following circuit is used to amplify the signals coming from a microphone.
(a) Determine the values of R1, R2, REand RCfor the operating point indicated.
(b) If an input AC signal with amplitude 0.5V and 2kHz frequency is applied, sketch the input,
output and emitter voltage waveforms.
(c) Estimate the voltage gain of this circuit.
15V
10F
R1
R2
RC
RE
vivo
0.7V
10F
15V
10F
R1
R2
RC
RE
vivo
0.7V
10F
VCE (V)0 1 2 3 4 5 6 7 8 9 10
-3
0
3
6
9
12
15
18
21
24
27
30
IB=0A
20A
40A
60A
80A
100A
120A
140A
IC(mA)
VCE (V)0 1 2 3 4 5 6 7 8 9 10
-3
0
3
6
9
12
15
18
21
24
27
30
IB=0A
20A
40A
60A
80A
100A
120A
140A
IC(mA)
0 1 2 3 4 5 6 7 8 9 10-3
0
3
6
9
12
15
18
21
24
27
30
IB=0A
20A
40A
60A
80A
100A
120A
140A
IC(mA)
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1.3. Signal Voltage Gain
Increasing IBdue to input signal and hence ICincreases the voltage drop ICRC. But since VC=
VCC-ICRC, the output collector voltage decreases.
Likewise, decreasing IB due to input signal and hence IC, decreases voltage drop ICRC. Butsince VC= VCC-ICRC, the output collector voltage increases.
Therefore there is 180o phase shift between base (input) and collector (output) signals
inverting amplifier.
R1
R2
RC
RE
vivo
+VCC
vB
vC
R1
R2
RC
RE
vivo
+VCC
vB
vC
Signal voltage gain of amplifier,B
C
i
ov
vbase,atamplitudevoltagesignalAC
vcollector,atamplitudevoltagesignalAC
v
vA ==
Because base-emitter junction is forward biased, then vBvE, and hence:
EE
CC
B
C
vRi
Ri
v
v
A ==
But and since iCiEthen:
E
Cv
R
RA
Example: Compare the estimated signal voltage gain in the previous example with the calculated
one using the above equation.
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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2. Transistor Switches
The second major application of transistors is electronic switching for computer and control
applications.
Typical transistor switching circuit is shown below, where it can be seen that the DC source isconnected to the collector (or output) side only (and not to the base).
VCC
RC
RB
vi
VCEorvo
IB
IC
VCC
RC
RB
vi
VCEorvo
IB
IC
When used as an electronic switch, the transistor switches between cutoffand saturation.
2.1. Cutoff
vi= 0V, IB= 0A and hence IC0mA.
The voltage drop across RCis ICRC= 0V resulting in:
VCEVCC
2.2. Saturation
Input signal vi forward biases B-E junction and
provides sufficient base current IB to increase IC to its
maximum (ICsat) such that most of the supply voltageVCCdrops across RC.
VCEis then very small (VCEsat0.1 - 0.3V) and can beneglected and replaced by short circuit. The collector
current (or value of RC to achieve saturation) is then
determined from:
C
CC
CsatR
VI =
RC
RB
0V
IC=0
+VCC
RC
+VCC
C
E
RC
RB
0V
IC=0
+VCC
RC
+VCC
C
E
RC
RB
vi
ICsat
+VCC
RC
+VCC
C
E
IB
ICsatRC
RB
vi
ICsat
+VCC
RC
+VCC
C
E
IB
ICsat
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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To ensure saturation:
> CsatBI
I (typically set IB2ICsat/)
The level of IBis set through applying Kirchhoff's voltage law around the input circuit:
B
BEiB
R
VvI
=
Example: For the following circuit:
(a) What is VCEwhen vi= 0?
(b) What is the minimum value of IB that will saturate the
transistor?
(c) Calculate maximum RBwhen vi= 5V.
Example: For the following circuit:
(a) Determine ICsat.
(b) Determine IBat saturation.
(c) Plot the output voltage waveform across the collector.
820R
68k
+5V
=125
vi
5V
t0
820R
68k
+5V
=125
vi
5V
t0
1k
RB
vi
+10V
=200
1k
RB
vi
+10V
=200
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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2.3. Inductive Load Drivers
Transistors are used to establish the necessary current to
energise (or drive) devices with coils such as relays and
motors.
When positive pulse is applied, the transistor is turned on
(saturation) to establish sufficient current to drive the
inductive load.
When the signal to the base is removed to turn off the
transistor (cutoff), the rapidly decreasing collector current
develops a large reverse voltage across the coil (VL =
LdI/dt).
This also induces a large positive voltage at the collector terminal which exceeds the breakdown
voltage of the transistor causing permanent damage to the transistor.
RB
+VCC
vi
t0
off
I
t0
Large voltage
spike, VL = LdI/dt
I
boomRB
+VCC
vi
t0
off
I
t0
Large voltage
spike, VL = LdI/dt
I
boom
Solution: add a diode in parallel with the inductive load diode forward biased during turn offto short out and dissipate the voltage spike and current.
RB
+VCC
vi
t0
Diode forward
biased during
turn off
ION
IOFF
RB
+VCC
vi
t0
Diode forward
biased during
turn off
ION
IOFF
RB
+VCC
vi
t0
on
I
RB
+VCC
vi
t0
on
I
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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)
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Example: A motor switch circuit is shown below. The motor is on when the control signal is 5V,
and off when the control signal is 0V.
(a) Determine the minimum IBrequired to turn on the motor (saturation).
(b) Determine the maximum value for RBto maintain saturation of the transistor.
(c) Sketch the input and output voltage waveforms for this circuit.(d) What protection you need to apply to prevent damage to the transistor during turn off?
RB
+5V
vi
t0
250mH
5R
=50
Motor
5V
RB
+5V
vi
t0
250mH
5R
=50
Motor
5V
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2.4. Logic Gates
Logic gates are the fundamental building blocks of digital integrated circuits and components.
They consist of transistors operating as switches in the saturation and cutoff regions.
For example, an AND gate implementation using transistors is shown.
10k
B
+5V
10k
A
3k3 C=A AND BA B C
0 0
0 1
1 0
1 1
Q1
Q210k
B
+5V
10k
A
3k3 C=A AND BA B C
0 0
0 1
1 0
1 1
Q1
Q2
If both inputs A and B have a turn-on voltage, then both transistors Q1 and Q2 are approximately
short circuits (saturation) providing a path from 5V supply to output terminal high or binary 1state.
If one or both transistors are off due to 0V at input terminal, an open circuit (cutoff) is placed in
series with the path from the 5V supply to output terminal 0V or binary 0 state.