Download EE101: BJT basics - Department of Electrical Engineering, Indian

EE101: BJT basics
M. B. Patil
[email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Base
pnp transistor
Collector
Emitter
n
p
n
Collector
Base
npn transistor
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Base
pnp transistor
Collector
Emitter
n
p
n
Collector
Base
npn transistor
* Bipolar: both electrons and holes contribute to conduction
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Base
pnp transistor
Collector
Emitter
n
p
n
Collector
Base
npn transistor
* Bipolar: both electrons and holes contribute to conduction
* Junction: device includes two p-n junctions (as opposed to a “point-contact”
transistor, the first transistor)
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Collector
Base
pnp transistor
Emitter
n
p
n
Collector
Base
npn transistor
* Bipolar: both electrons and holes contribute to conduction
* Junction: device includes two p-n junctions (as opposed to a “point-contact”
transistor, the first transistor)
* Transistor: “transfer resistor”
When Bell Labs had an informal contest to name their new invention, one engineer pointed
out that it acts like a resistor, but a resistor where the voltage is transferred across the
device to control the resulting current.
(http://amasci.com/amateur/trshort.html)
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Collector
Base
pnp transistor
Emitter
n
p
n
Collector
Base
npn transistor
* Bipolar: both electrons and holes contribute to conduction
* Junction: device includes two p-n junctions (as opposed to a “point-contact”
transistor, the first transistor)
* Transistor: “transfer resistor”
When Bell Labs had an informal contest to name their new invention, one engineer pointed
out that it acts like a resistor, but a resistor where the voltage is transferred across the
device to control the resulting current.
(http://amasci.com/amateur/trshort.html)
* invented in 1947 by Shockley, Bardeen, and Brattain at Bell Laboratories.
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Collector
Emitter
Base
pnp transistor
n
p
n
Collector
Base
npn transistor
* Bipolar: both electrons and holes contribute to conduction
* Junction: device includes two p-n junctions (as opposed to a “point-contact”
transistor, the first transistor)
* Transistor: “transfer resistor”
When Bell Labs had an informal contest to name their new invention, one engineer pointed
out that it acts like a resistor, but a resistor where the voltage is transferred across the
device to control the resulting current.
(http://amasci.com/amateur/trshort.html)
* invented in 1947 by Shockley, Bardeen, and Brattain at Bell Laboratories.
* “A BJT is two diodes connected back-to-back.”
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Emitter
p
n
p
Collector
Emitter
Base
pnp transistor
n
p
n
Collector
Base
npn transistor
* Bipolar: both electrons and holes contribute to conduction
* Junction: device includes two p-n junctions (as opposed to a “point-contact”
transistor, the first transistor)
* Transistor: “transfer resistor”
When Bell Labs had an informal contest to name their new invention, one engineer pointed
out that it acts like a resistor, but a resistor where the voltage is transferred across the
device to control the resulting current.
(http://amasci.com/amateur/trshort.html)
* invented in 1947 by Shockley, Bardeen, and Brattain at Bell Laboratories.
* “A BJT is two diodes connected back-to-back.”
WRONG! Let us see why.
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Consider a pnp BJT in the following circuit:
E
R1
5V
1 k I1
p n p
B
I3
C
I 2 R2 1 k
10 V
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Consider a pnp BJT in the following circuit:
E
R1
1 k I1
C
p n p
I 2 R2 1 k
B
I3
5V
10 V
If the transistor is replaced with two diodes connected back-to-back, we get,
E
R1
5V
1 k I1
C
D1
B D2
I3
I 2 R2 1 k
10 V
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Consider a pnp BJT in the following circuit:
E
R1
1 k I1
C
p n p
I 2 R2 1 k
B
I3
5V
10 V
If the transistor is replaced with two diodes connected back-to-back, we get,
E
R1
5V
1 k I1
C
D1
B D2
I 2 R2 1 k
I3
10 V
Assuming Von = 0.7 V for D1, we get
5 V − 0.7 V
I1 =
= 4.3 mA,
R1
I2 = 0 (since D2 is reverse biased), and
I3 ≈ I1 = 4.3 mA.
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Using a more accurate equivalent circuit for the BJT, we obtain,
E
R1
5V
1 k I1
p n p
B
C
E
I 2 R2 1 k
R1
I3
5V
10 V
1 k I1
α I1
B
C
I2 R2 1 k
I3
10 V
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Using a more accurate equivalent circuit for the BJT, we obtain,
E
R1
5V
1 k I1
p n p
B
C
E
I 2 R2 1 k
R1
I3
5V
10 V
1 k I1
α I1
B
C
I2 R2 1 k
I3
10 V
We now get,
5 V − 0.7 V
I1 =
= 4.3 mA (as before),
R1
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Using a more accurate equivalent circuit for the BJT, we obtain,
E
R1
5V
1 k I1
p n p
B
C
E
I 2 R2 1 k
R1
I3
1 k I1
5V
10 V
α I1
B
C
I2 R2 1 k
I3
10 V
We now get,
5 V − 0.7 V
I1 =
= 4.3 mA (as before),
R1
I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Using a more accurate equivalent circuit for the BJT, we obtain,
E
R1
5V
1 k I1
p n p
B
C
E
I 2 R2 1 k
R1
I3
1 k I1
5V
10 V
α I1
B
C
I2 R2 1 k
I3
10 V
We now get,
5 V − 0.7 V
I1 =
= 4.3 mA (as before),
R1
I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and
I3 = I1 − I2 = (1 − α) I1 ≈ 0 A.
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Using a more accurate equivalent circuit for the BJT, we obtain,
E
R1
5V
1 k I1
p n p
B
C
E
I 2 R2 1 k
R1
I3
1 k I1
5V
α I1
B
C
I2 R2 1 k
I3
10 V
10 V
We now get,
5 V − 0.7 V
I1 =
= 4.3 mA (as before),
R1
I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and
I3 = I1 − I2 = (1 − α) I1 ≈ 0 A.
The values of I2 and I3 are dramatically different than the ones obtained earlier.
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
Using a more accurate equivalent circuit for the BJT, we obtain,
E
R1
5V
1 k I1
p n p
B
C
E
I 2 R2 1 k
R1
I3
1 k I1
5V
α I1
B
C
I2 R2 1 k
I3
10 V
10 V
We now get,
5 V − 0.7 V
I1 =
= 4.3 mA (as before),
R1
I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and
I3 = I1 − I2 = (1 − α) I1 ≈ 0 A.
The values of I2 and I3 are dramatically different than the ones obtained earlier.
Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although
it does have two p-n junctions).
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
What is wrong with the two-diode model of a BJT?
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
What is wrong with the two-diode model of a BJT?
* When we replace a BJT with two diodes, we assume that there is no interaction
between the two diodes, which may be expected if they are “far apart.”
Emitter
p
p
n
Collector
Base
Emitter
Collector
D1
Base
D2
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
What is wrong with the two-diode model of a BJT?
* When we replace a BJT with two diodes, we assume that there is no interaction
between the two diodes, which may be expected if they are “far apart.”
Emitter
p
p
n
Collector
Base
Emitter
Collector
D1
Base
D2
* However, in a BJT, exactly the opposite is true. For a higher performance, the
base region is made as short as possible (subject to certain constraints), and the
two diodes therefore cannot be treated as independent devices.
Emitter
p
n
p
Collector
Base
M. B. Patil, IIT Bombay
Bipolar Junction Transistors
What is wrong with the two-diode model of a BJT?
* When we replace a BJT with two diodes, we assume that there is no interaction
between the two diodes, which may be expected if they are “far apart.”
Emitter
p
p
n
Collector
Base
Emitter
Collector
D1
Base
D2
* However, in a BJT, exactly the opposite is true. For a higher performance, the
base region is made as short as possible (subject to certain constraints), and the
two diodes therefore cannot be treated as independent devices.
Emitter
p
n
p
Collector
Base
* Later, we will look at the “Ebers-Moll model” of a BJT, which is a fairly
accurate representation of the transistor action.
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
IC
C
E
IC
IE
IB
B
C
E
IE
n
p
IC
C
E
IC
IE
IB
IB
B
n
B
B
C
IB
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
IC
C
E
IC
IE
IB
B
C
E
IE
n
p
IC
C
E
IC
IE
IB
IB
B
n
B
B
C
IB
* In the active mode of a BJT, the B-E junction is under forward bias, and the
B-C junction is under reverse bias.
- For a pnp transistor, VEB > 0 V , and VCB < 0 V .
- For an npn transistor, VBE > 0 V , and VBC < 0 V .
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
IC
C
E
IC
IE
IB
B
C
E
IE
n
p
IC
C
E
IC
IE
IB
IB
B
n
B
B
C
IB
* In the active mode of a BJT, the B-E junction is under forward bias, and the
B-C junction is under reverse bias.
- For a pnp transistor, VEB > 0 V , and VCB < 0 V .
- For an npn transistor, VBE > 0 V , and VBC < 0 V .
* Since the B-E junction is under forward bias, the voltage (magnitude) is typically
0.6 to 0.75 V .
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
IC
C
E
IC
IE
IB
B
C
E
IE
n
p
IC
C
E
IC
IE
IB
IB
B
n
B
B
C
IB
* In the active mode of a BJT, the B-E junction is under forward bias, and the
B-C junction is under reverse bias.
- For a pnp transistor, VEB > 0 V , and VCB < 0 V .
- For an npn transistor, VBE > 0 V , and VBC < 0 V .
* Since the B-E junction is under forward bias, the voltage (magnitude) is typically
0.6 to 0.75 V .
* The B-C voltage can be several Volts (or even hundreds of Volts), and is limited
by the breakdown voltage of the B-C junction.
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
IC
C
E
IC
IE
IB
B
C
E
IE
n
p
IC
C
E
IC
IE
IB
IB
B
n
B
B
C
IB
* In the active mode of a BJT, the B-E junction is under forward bias, and the
B-C junction is under reverse bias.
- For a pnp transistor, VEB > 0 V , and VCB < 0 V .
- For an npn transistor, VBE > 0 V , and VBC < 0 V .
* Since the B-E junction is under forward bias, the voltage (magnitude) is typically
0.6 to 0.75 V .
* The B-C voltage can be several Volts (or even hundreds of Volts), and is limited
by the breakdown voltage of the B-C junction.
* The symbol for a BJT includes an arrow for the emitter terminal, its direction
indicating the current direction when the transistor is in active mode.
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
IC
C
E
IC
IE
IB
B
C
E
IE
n
p
IC
C
E
IC
IE
IB
IB
B
n
B
B
C
IB
* In the active mode of a BJT, the B-E junction is under forward bias, and the
B-C junction is under reverse bias.
- For a pnp transistor, VEB > 0 V , and VCB < 0 V .
- For an npn transistor, VBE > 0 V , and VBC < 0 V .
* Since the B-E junction is under forward bias, the voltage (magnitude) is typically
0.6 to 0.75 V .
* The B-C voltage can be several Volts (or even hundreds of Volts), and is limited
by the breakdown voltage of the B-C junction.
* The symbol for a BJT includes an arrow for the emitter terminal, its direction
indicating the current direction when the transistor is in active mode.
* Analog circuits, including amplifiers, are generally designed to ensure that the
BJTs are operating in the active mode.
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
C
E
IE
n
p
B
B
α IE
E
IE
C
IC
B
IB
n
C
IC
E
IC
IE
IB
IB
B
B
α IE
E
IE
C
IB
C
IC
IB
B
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
C
E
IE
n
p
B
B
α IE
E
IE
C
IC
B
n
C
IC
E
IC
IE
IB
IB
B
B
α IE
E
IE
IB
C
IB
C
IC
IB
B
* In the active mode, IC = α IE , α ≈ 1 (slightly less than 1).
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
C
E
IE
n
p
B
B
α IE
E
IE
C
IC
B
n
C
IC
E
IC
IE
IB
IB
B
B
α IE
E
IE
IB
C
IB
C
IC
IB
B
* In the active mode, IC = α IE , α ≈ 1 (slightly less than 1).
* IB = IE − IC = IE (1 − α) .
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
C
E
IE
n
p
B
B
α IE
E
IE
C
IC
B
n
C
IC
E
IC
IE
IB
IB
B
B
α IE
E
IE
IB
C
IB
C
IC
IB
B
* In the active mode, IC = α IE , α ≈ 1 (slightly less than 1).
* IB = IE − IC = IE (1 − α) .
* The ratio IC /IB is defined as the current gain β of the transistor.
IC
α
β=
=
.
IB
1−α
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
C
E
IE
n
p
B
B
α IE
E
IE
C
IC
B
n
C
IC
E
IC
IE
IB
IB
B
B
α IE
E
IE
IB
C
IB
C
IC
IB
B
* In the active mode, IC = α IE , α ≈ 1 (slightly less than 1).
* IB = IE − IC = IE (1 − α) .
* The ratio IC /IB is defined as the current gain β of the transistor.
IC
α
β=
=
.
IB
1−α
* β is a function of IC and temperature. However, we will generally treat it as a
constant, a useful approximation to simplify things and still get a good insight.
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
E
IE
n
p
B
α IE
E
IE
C
IC
B
IB
n
C
IC
E
IC
IE
IB
IB
B
β=
C
B
B
α IE
E
IE
C
IB
C
IC
IB
B
IC
α
=
IB
1−α
α
β
0.9
9
0.95
19
0.99
99
0.995
199
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
E
IE
n
p
B
α IE
E
IE
α
β
0.9
9
0.95
19
0.99
99
0.995
199
C
IC
E
IC
IE
B
B
α IE
E
IC
B
IC
α
=
IB
1−α
C
n
IB
IB
B
β=
C
IE
C
IB
C
IC
IB
IB
B
* β is a sensitive function of α.
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
E
IE
n
p
B
α IE
E
IE
α
β
0.9
9
0.95
19
0.99
99
0.995
199
C
IC
E
IC
IE
B
B
α IE
E
IC
B
IC
α
=
IB
1−α
C
n
IB
IB
B
β=
C
IE
C
IB
C
IC
IB
IB
B
* β is a sensitive function of α.
* Transistors are generally designed to get a high value of β
(typically 100 to 250, but can be as high as 2000 for
“super-β” transistors).
M. B. Patil, IIT Bombay
BJT in active mode
E
IE
p
n
p
C
IC
E
IC
IE
IB
E
IE
n
p
B
α IE
E
IE
α
β
0.9
9
0.95
19
0.99
99
0.995
199
C
IC
E
IC
IE
B
B
α IE
E
IC
B
IC
α
=
IB
1−α
C
n
IB
IB
B
β=
C
IE
C
IB
C
IC
IB
IB
B
* β is a sensitive function of α.
* Transistors are generally designed to get a high value of β
(typically 100 to 250, but can be as high as 2000 for
“super-β” transistors).
* A large β ⇒ IB IC or IE when the transistor is in the
active mode.
M. B. Patil, IIT Bombay
A simple BJT circuit
1k
RC
C
100 k B
RB
2V
VBB
VCC
β = 100 10 V
E
A simple BJT circuit
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
RC
n
2V
VBB
100 k p
RB
β = 100
n
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
M. B. Patil, IIT Bombay
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB .
M. B. Patil, IIT Bombay
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 13 µA.
RB
100 k
M. B. Patil, IIT Bombay
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 13 µA.
RB
100 k
IC = β × IB = 100 × 13 µA = 1.3 mA.
M. B. Patil, IIT Bombay
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 13 µA.
RB
100 k
IC = β × IB = 100 × 13 µA = 1.3 mA.
VC = VCC − IC RC = 10 V − 1.3 mA × 1 k = 8.7 V .
M. B. Patil, IIT Bombay
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 13 µA.
RB
100 k
IC = β × IB = 100 × 13 µA = 1.3 mA.
VC = VCC − IC RC = 10 V − 1.3 mA × 1 k = 8.7 V .
Let us check whether our assumption of active mode is correct. We need to check
whether the B-C junction is under reverse bias.
M. B. Patil, IIT Bombay
A simple BJT circuit
10 V VCC
10 V VCC
1k
RC
C
100 k B
RB
2V
VBB
1k
VCC
β = 100 10 V
E
1k
RC
α IE
n
2V
VBB
100 k p
RB
β = 100
n
RC
IC
2V
VBB
100 k IB
RB
IE
Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 13 µA.
RB
100 k
IC = β × IB = 100 × 13 µA = 1.3 mA.
VC = VCC − IC RC = 10 V − 1.3 mA × 1 k = 8.7 V .
Let us check whether our assumption of active mode is correct. We need to check
whether the B-C junction is under reverse bias.
VBC = VB − VC = 0.7 V − 8.7 V = −8.0 V ,
i.e., the B-C junction is indeed under reverse bias.
M. B. Patil, IIT Bombay
A simple BJT circuit (continued)
10 V VCC
1k
IC
2V
VBB
10 k
p
RB I B
RC
n
β = 100
n
What happens if RB is changed from 100 k to 10 k?
M. B. Patil, IIT Bombay
A simple BJT circuit (continued)
10 V VCC
1k
IC
2V
VBB
10 k
p
RB I B
RC
n
β = 100
n
What happens if RB is changed from 100 k to 10 k?
Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and
IC = β IB .
M. B. Patil, IIT Bombay
A simple BJT circuit (continued)
10 V VCC
1k
IC
2V
VBB
10 k
p
RB I B
RC
n
β = 100
n
What happens if RB is changed from 100 k to 10 k?
Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and
IC = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 130 µA.
RB
10 k
IC = β × IB = 100 × 130 µA = 13 mA.
VC = VCC − IC RC = 10 V − 13 mA × 1 k = −3 V .
M. B. Patil, IIT Bombay
A simple BJT circuit (continued)
10 V VCC
1k
IC
2V
VBB
10 k
p
RB I B
RC
n
β = 100
n
What happens if RB is changed from 100 k to 10 k?
Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and
IC = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 130 µA.
RB
10 k
IC = β × IB = 100 × 130 µA = 13 mA.
VC = VCC − IC RC = 10 V − 13 mA × 1 k = −3 V .
VBC = VB − VC = 0.7 V − (−3) V = 3.7 V ,
M. B. Patil, IIT Bombay
A simple BJT circuit (continued)
10 V VCC
1k
IC
2V
VBB
10 k
p
RB I B
RC
n
β = 100
n
What happens if RB is changed from 100 k to 10 k?
Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and
IC = β IB .
VBB − VBE
2 V − 0.7 V
IB =
=
= 130 µA.
RB
10 k
IC = β × IB = 100 × 130 µA = 13 mA.
VC = VCC − IC RC = 10 V − 13 mA × 1 k = −3 V .
VBC = VB − VC = 0.7 V − (−3) V = 3.7 V ,
VBC is not only positive, it is huge!
The BJT cannot be in the active mode, and we need to take another look at the
circuit.
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
B
p
IC
C
E
IC
IE
IB
IB
B
C
E
C
IE
B
IB
αF IE
IC
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
p
IC
C
E
IC
IE
IB
B
C
E
C
IE
IB
B
B
IB
αF IE
IC
Reverse active mode: B−E in r. b., B−C in f. b.
IE
E
p
n
B
p
IC
αR (−IC )
C
E
IC
IE
IB
IB
B
C
(−IC )
E
IE
C
IC
B
IB
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
p
IC
C
E
IC
IE
IB
B
C
E
C
IE
IB
B
B
IB
αF IE
IC
Reverse active mode: B−E in r. b., B−C in f. b.
IE
E
p
n
B
p
IC
αR (−IC )
C
E
IC
IE
IB
IB
B
C
(−IC )
E
IE
C
IC
B
IB
In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the
terminals with their original names.)
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
p
IC
C
E
IC
IE
IB
B
C
E
C
IE
IB
B
B
IB
αF IE
IC
Reverse active mode: B−E in r. b., B−C in f. b.
IE
E
p
n
B
p
IC
αR (−IC )
C
E
IC
IE
IB
IB
B
C
(−IC )
E
IE
C
IC
B
IB
In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the
terminals with their original names.)
The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different.
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
p
IC
C
E
IC
IE
IB
B
C
E
C
IE
IB
B
B
IB
αF IE
IC
Reverse active mode: B−E in r. b., B−C in f. b.
IE
E
p
n
B
p
IC
αR (−IC )
C
E
IC
IE
IB
C
(−IC )
E
IE
IB
B
C
IC
B
IB
In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the
terminals with their original names.)
The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different.
Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5.
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
p
IC
C
E
IC
IE
IB
B
C
E
C
IE
IB
B
B
IB
αF IE
IC
Reverse active mode: B−E in r. b., B−C in f. b.
IE
E
p
n
B
p
IC
αR (−IC )
C
E
IC
IE
IB
C
(−IC )
E
IE
IB
B
C
IC
B
IB
In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the
terminals with their original names.)
The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different.
Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5.
The corresponding current gains (βF and βR ) differ significantly, since β = α/(1 − α).
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
Active mode ("forward" active mode): B−E in f. b., B−C in r. b.
IE
E
p
n
p
IC
C
E
IC
IE
IB
B
C
E
C
IE
IB
B
B
IB
αF IE
IC
Reverse active mode: B−E in r. b., B−C in f. b.
IE
E
p
n
B
p
IC
αR (−IC )
C
E
IC
IE
IB
C
(−IC )
E
IE
IB
B
C
IC
B
IB
In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the
terminals with their original names.)
The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different.
Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5.
The corresponding current gains (βF and βR ) differ significantly, since β = α/(1 − α).
In amplifiers, the BJT is biased in the forward active mode (simply called the “active
mode”) in order to make use of the higher value of β in that mode.
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
The Ebers-Moll model combines the forward and reverse operations of a BJT in a single
comprehensive model.
IE
E
p
n
B
IC
p
IB
E
p
E
IC
IE
IB
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
p
I′C
IB
B
IC
n
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
The Ebers-Moll model combines the forward and reverse operations of a BJT in a single
comprehensive model.
IE
E
p
n
B
IC
p
IB
E
p
E
IC
IE
IB
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
p
I′C
IB
B
IC
n
The currents IE0 and IC0 are given by the Shockley diode equation:
»
„
«
–
»
„
«
–
VEB
VCB
− 1 , IC0 = ICS exp
−1 .
IE0 = IES exp
VT
VT
M. B. Patil, IIT Bombay
Ebers-Moll model for a pnp transistor
The Ebers-Moll model combines the forward and reverse operations of a BJT in a single
comprehensive model.
IE
p
E
n
B
IC
p
IB
E
p
E
IC
IE
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
p
I′C
IB
B
B
IC
n
The currents IE0 and IC0 are given by the Shockley diode equation:
»
„
«
–
»
„
«
–
VEB
VCB
− 1 , IC0 = ICS exp
−1 .
IE0 = IES exp
VT
VT
Mode
B-E
B-C
Forward active
forward
reverse
IE0 IC0
Reverse active
reverse
forward
IC0 IE0
Saturation
forward
forward
IE0 and IC0 are comparable.
Cut-off
reverse
reverse
IE0 and IC0 are negliglbe.
M. B. Patil, IIT Bombay
Ebers-Moll model
pnp transistor
IE
E
p
n
IC
p
IB
B
E
p
E
IC
IE
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
B
p
I′C = ICS [exp(VCB /VT ) − 1]
I′C
IB
B
IC
I′E = IES [exp(VEB /VT ) − 1]
n
npn transistor
IE
E
n
IC
p
n
IB
E
B
E
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
IB
D2
C
αR I′C
n
I′C = ICS [exp(VBC /VT ) − 1]
I′C
IB
B
IC
I′E = IES [exp(VBE /VT ) − 1]
p
M. B. Patil, IIT Bombay
Ebers-Moll model
pnp transistor
IE
E
p
n
IC
p
IB
B
E
p
E
IC
IE
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
B
p
I′C = ICS [exp(VCB /VT ) − 1]
I′C
IB
B
IC
I′E = IES [exp(VEB /VT ) − 1]
n
npn transistor
IE
E
n
IC
p
n
IB
E
B
E
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
IB
D2
C
αR I′C
n
I′C = ICS [exp(VBC /VT ) − 1]
I′C
IB
B
IC
I′E = IES [exp(VBE /VT ) − 1]
p
For an npn transistor, the same model holds with current directions and voltage
polarities suitably changed.
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
IB
I′E = IES [exp(VBE /VT ) − 1]
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
D2
C
αR I′C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
A BJT is a three-terminal device, and its I -V chatacteristics can therefore be
represented in several different ways. The IC versus VCE characteristics are very useful
in amplifiers.
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
IB
I′E = IES [exp(VBE /VT ) − 1]
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
D2
IC
n
C
αR I′C
I′C
IB
B
p
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
A BJT is a three-terminal device, and its I -V chatacteristics can therefore be
represented in several different ways. The IC versus VCE characteristics are very useful
in amplifiers.
To start with, we consider a single point, IB = 10 µA, VCE = 5 V .
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
IB
I′E = IES [exp(VBE /VT ) − 1]
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
D2
IC
n
C
αR I′C
I′C
IB
B
p
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
A BJT is a three-terminal device, and its I -V chatacteristics can therefore be
represented in several different ways. The IC versus VCE characteristics are very useful
in amplifiers.
To start with, we consider a single point, IB = 10 µA, VCE = 5 V .
There are several ways to assign VBE and VCB so that they satisfy the constraint:
VCB + VBE = (VC − VB ) + (VB − VE ) = VCE = 5 V .
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
IB
I′E = IES [exp(VBE /VT ) − 1]
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
D2
IC
n
C
αR I′C
I′C
IB
B
p
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
A BJT is a three-terminal device, and its I -V chatacteristics can therefore be
represented in several different ways. The IC versus VCE characteristics are very useful
in amplifiers.
To start with, we consider a single point, IB = 10 µA, VCE = 5 V .
There are several ways to assign VBE and VCB so that they satisfy the constraint:
VCB + VBE = (VC − VB ) + (VB − VE ) = VCE = 5 V .
Let us consider some of these possibilities.
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
IB
αF I′E
I′E
C
I′E = IES [exp(VBE /VT ) − 1]
C
D1
IE
D2
C
αR I′C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
1V B
C
n
6V
p
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
1V B
C
n
D1 and D2 are both off, and we cannot satisfy the
condition, IB = 10 µA, since all currents are much
smaller than 10 µA.
6V
p
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
1V B
p
6V
C
n
D1 and D2 are both off, and we cannot satisfy the
condition, IB = 10 µA, since all currents are much
smaller than 10 µA.
⇒ This possibility (and similarly others with both
junctions reverse biased) is ruled out.
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
IB
I′E = IES [exp(VBE /VT ) − 1]
C
D1
n IE
IC
IE
B
αF I′E
I′E
C
D2
C
αR I′C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
C
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
D1
n IE
IC
IE
B
αF I′E
I′E
C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
6V B
C
n
1V
p
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
C
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
D1
n IE
IC
IE
B
αF I′E
I′E
C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
6V B
C
n
D1 and D2 are both conducting; however, the forward
bias for the B-E junction is impossibly large.
1V
p
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
C
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
D1
n IE
IC
IE
B
αF I′E
I′E
C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
6V B
1V
C
n
D1 and D2 are both conducting; however, the forward
bias for the B-E junction is impossibly large.
⇒ This possibility is also ruled out.
p
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
IB
αF I′E
I′E
C
I′E = IES [exp(VBE /VT ) − 1]
C
D1
IE
D2
C
αR I′C
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
0.7 V B
C
n
4.3 V
p
M. B. Patil, IIT Bombay
IC -VCE characteristics
IE
n
E
IC
p
n
IB
E
B
E
n
IC
IE
B
αF I′E
I′E
C
C
D1
IE
D2
C
αR I′C
IB
I′E = IES [exp(VBE /VT ) − 1]
I′C
IB
B
p
IC
n
I′C = ICS [exp(VBC /VT ) − 1]
αF = 0.99, ISE = 1 × 10−14 A
αR = 0.50, ISC = 2 × 10−14 A
Constraints: IB = 10 µA, VCE = 5 V .
5V
E
IC
n IE
IB
0.7 V B
p
4.3 V
C
n
D1 is on, D2 is off. This is a realistic possibility. Since
the B-C junction is under reverse bias, IC0 and αR IC0 are
much smaller than IE0 , and therefore the lower branches
in the Ebers-Moll model can be dropped (see next
slide).
M. B. Patil, IIT Bombay
IC -VCE characteristics
5V
E
IC
n IE
0.7 V B
IB
C
n
4.3 V
p
n IE
αF I′E
I′E
E
D1
C
IC
IB
B
n
p
(The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are
VBE = 0.656 V and VCB = 4.344 V .)
The BJT is in the active mode, and therefore
αF
IB = 99 × 10 µA = 0.99 mA.
IC = β IB =
1 − αF
IC -VCE characteristics
5V
IC
n IE
0.7 V B
IB
C
n
1
4.3 V
p
E
n IE
αF I′E
I′E
D1
C
IC
IB
B
p
IC (mA)
E
n
0
0
1
2
3
VCE (V)
4
5
(The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are
VBE = 0.656 V and VCB = 4.344 V .)
The BJT is in the active mode, and therefore
αF
IB = 99 × 10 µA = 0.99 mA.
IC = β IB =
1 − αF
IC -VCE characteristics
5V
IC
n IE
0.7 V B
IB
C
n
1
4.3 V
p
E
n IE
αF I′E
I′E
D1
C
IC
IB
B
p
IC (mA)
E
n
0
0
1
2
3
VCE (V)
4
5
(The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are
VBE = 0.656 V and VCB = 4.344 V .)
The BJT is in the active mode, and therefore
αF
IB = 99 × 10 µA = 0.99 mA.
IC = β IB =
1 − αF
If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and
once again, we find that IC = β IB = 0.99 mA.
IC -VCE characteristics
5V
IC
n IE
0.7 V B
IB
C
n
1
4.3 V
p
E
n IE
αF I′E
I′E
D1
C
IC
IB
B
p
IC (mA)
E
n
0
0
1
2
3
VCE (V)
4
5
(The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are
VBE = 0.656 V and VCB = 4.344 V .)
The BJT is in the active mode, and therefore
αF
IB = 99 × 10 µA = 0.99 mA.
IC = β IB =
1 − αF
If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and
once again, we find that IC = β IB = 0.99 mA.
Thus, the plot of IC versus VCE is simply a horizontal line.
IC -VCE characteristics
5V
IC
0.7 V B
IB
C
n
p
E
n IE
αF I′E
I′E
D1
C
IC
IB
B
p
1
1
4.3 V
IC (mA)
n IE
IC (mA)
E
n
0
0
1
2
3
VCE (V)
4
5
0
0
1
2
3
VCE (V)
4
5
(The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are
VBE = 0.656 V and VCB = 4.344 V .)
The BJT is in the active mode, and therefore
αF
IB = 99 × 10 µA = 0.99 mA.
IC = β IB =
1 − αF
If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and
once again, we find that IC = β IB = 0.99 mA.
Thus, the plot of IC versus VCE is simply a horizontal line.
M. B. Patil, IIT Bombay
IC -VCE characteristics
5V
IC
0.7 V B
IB
C
n
p
E
n IE
αF I′E
I′E
D1
C
IC
IB
B
p
1
1
4.3 V
IC (mA)
n IE
IC (mA)
E
n
0
0
1
2
3
VCE (V)
4
5
0
0
1
2
3
VCE (V)
4
5
(The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are
VBE = 0.656 V and VCB = 4.344 V .)
The BJT is in the active mode, and therefore
αF
IB = 99 × 10 µA = 0.99 mA.
IC = β IB =
1 − αF
If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and
once again, we find that IC = β IB = 0.99 mA.
Thus, the plot of IC versus VCE is simply a horizontal line.
However, as VCE → 0 V , things change (see next slide).
M. B. Patil, IIT Bombay
IC -VCE characteristics
0.7 V
E
IC
n IE
0.7 V B
IB
C
n
0V
p
When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V .
IC -VCE characteristics
0.7 V
E
IC
n IE
0.7 V B
IB
C
n
0V
p
0.3 V
E
IC
n IE
IB
0.7 V B
C
n
0.4 V
p
When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V .
As VCE is reduced further, the B-C junction gets forward biased. For example, with
VCE = 0.3 V , we may have a voltage distribution shown in the figure.
(The numbers are only representative; the actual VBE and VBC values can be obtained
by solving the E-M equations.)
IC -VCE characteristics
0.7 V
E
IC
n IE
0.7 V B
IB
C
0V
E
p
αF I′E
I′E
n
C
D1
n IE
D2
IC
n
0.3 V
E
IC
n IE
IB
0.7 V B
0.4 V
C
n
αR I′C
I′C
IB
B
p
p
When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V .
As VCE is reduced further, the B-C junction gets forward biased. For example, with
VCE = 0.3 V , we may have a voltage distribution shown in the figure.
(The numbers are only representative; the actual VBE and VBC values can be obtained
by solving the E-M equations.)
Now, the component IC0 in the E-M model becomes significant, IC = αF IE0 − IC0
reduces, and IC becomes smaller than βIB .
IC -VCE characteristics
saturation
0.7 V
IC
n IE
0.7 V B
IB
C
0V
E
p
1
C
D1
n IE
D2
0.3 V
E
IC
n IE
IB
0.7 V B
p
0.4 V
C
n
linear
αF I′E
I′E
n
αR I′C
n
I′C
IB
B
IC
IC (mA)
E
p
0
0
1
2
3
VCE (V)
4
5
When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V .
As VCE is reduced further, the B-C junction gets forward biased. For example, with
VCE = 0.3 V , we may have a voltage distribution shown in the figure.
(The numbers are only representative; the actual VBE and VBC values can be obtained
by solving the E-M equations.)
Now, the component IC0 in the E-M model becomes significant, IC = αF IE0 − IC0
reduces, and IC becomes smaller than βIB .
The region where IC < βIB is called the “saturation region.”
M. B. Patil, IIT Bombay
IC -VCE characteristics
saturation
linear
IC (mA)
2
IB = 20 µA
1
0
IB = 10 µA
0
1
2
3
VCE (V)
4
5
If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear
region. Apart from that, there is no qualitative change in the IC − VCE plot.
IC -VCE characteristics
saturation
linear
IC (mA)
2
IB = 20 µA
1
0
IB = 10 µA
0
1
2
3
VCE (V)
4
5
If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear
region. Apart from that, there is no qualitative change in the IC − VCE plot.
Clearly, the IC − VCE behaviour of a BJT is not represented by a single curve but by a
family of curves, known as the “IC − VCE characteristics.”
IC -VCE characteristics
saturation
saturation
linear
5
IB = 20 µA
1
50 µA
4
IC (mA)
IC (mA)
2
linear
IB = 10 µA
40 µA
3
30 µA
2
20 µA
1
0
0
1
2
3
VCE (V)
4
5
0
0
IB = 10 µA
1
2
3
VCE (V)
4
5
If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear
region. Apart from that, there is no qualitative change in the IC − VCE plot.
Clearly, the IC − VCE behaviour of a BJT is not represented by a single curve but by a
family of curves, known as the “IC − VCE characteristics.”
M. B. Patil, IIT Bombay
IC -VCE characteristics
saturation
saturation
linear
5
IB = 20 µA
1
50 µA
4
IC (mA)
IC (mA)
2
linear
IB = 10 µA
40 µA
3
30 µA
2
20 µA
1
0
0
1
2
3
VCE (V)
4
5
0
0
IB = 10 µA
1
2
3
VCE (V)
4
5
If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear
region. Apart from that, there is no qualitative change in the IC − VCE plot.
Clearly, the IC − VCE behaviour of a BJT is not represented by a single curve but by a
family of curves, known as the “IC − VCE characteristics.”
The IE − VCB and IC − VBE characteristics of a BJT are also useful in understanding
BJT circuits.
M. B. Patil, IIT Bombay
A simple BJT circuit (revisited)
10 V VCC
1k
IC
2V
VBB
p
RC
n
β = 100
RB I B
IE
n
We are now in a position to explain what happens when RB is decreased from 100 k
to 10 k in the above circuit.
A simple BJT circuit (revisited)
saturation
linear
15
10 V VCC
IC
2V
VBB
p
RC
n
β = 100
RB I B
IE
130 µA (RB = 10 k)
IC (mA)
1k
10
5
13 µA (RB = 100 k)
n
0
0
2
4
6
VCE (V)
8
10
We are now in a position to explain what happens when RB is decreased from 100 k
to 10 k in the above circuit.
VBB − 0.7 V
Let us plot IC − VCE curves for IB ≈
for the two values of RB .
RB
A simple BJT circuit (revisited)
saturation
linear
15
10 V VCC
IC
2V
VBB
p
RC
n
β = 100
RB I B
IE
130 µA (RB = 10 k)
IC (mA)
1k
10
load line
5
13 µA (RB = 100 k)
n
0
0
2
4
6
VCE (V)
8
10
We are now in a position to explain what happens when RB is decreased from 100 k
to 10 k in the above circuit.
VBB − 0.7 V
Let us plot IC − VCE curves for IB ≈
for the two values of RB .
RB
In addition to the BJT IC − VCE curve, the circuit variables must also satisfy the
constraint, VCC = VCE + IC RC , a straight line in the IC − VCE plane.
M. B. Patil, IIT Bombay
A simple BJT circuit (revisited)
saturation
linear
15
10 V VCC
IC
2V
VBB
p
RC
n
β = 100
RB I B
IE
130 µA (RB = 10 k)
IC (mA)
1k
10
load line
5
13 µA (RB = 100 k)
n
0
0
2
4
6
VCE (V)
8
10
We are now in a position to explain what happens when RB is decreased from 100 k
to 10 k in the above circuit.
VBB − 0.7 V
Let us plot IC − VCE curves for IB ≈
for the two values of RB .
RB
In addition to the BJT IC − VCE curve, the circuit variables must also satisfy the
constraint, VCC = VCE + IC RC , a straight line in the IC − VCE plane.
The intersection of the load line and the BJT characteristics gives the solution for the
circuit. For RB = 10 k, note that the BJT operates in the saturation region, leading
to VCE ≈ 0.2 V , and IC = 9.8 mA.
M. B. Patil, IIT Bombay