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Bipolar Junction Transistors Emitter p n p Collector Emitter n - - PowerPoint PPT Presentation

Feb 21, 2024 294 likes •1.26k views

Bipolar Junction Transistors Emitter p n p Collector Emitter n p n Collector Base Base pnp transistor npn transistor M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Collector Emitter n p n Collector Base

slide-1
SLIDE 1

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

  • M. B. Patil, IIT Bombay
slide-2
SLIDE 2

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* Bipolar: both electrons and holes contribute to conduction

  • M. B. Patil, IIT Bombay
slide-3
SLIDE 3

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* 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
slide-4
SLIDE 4

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* 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
slide-5
SLIDE 5

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* 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
slide-6
SLIDE 6

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* 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. * BJT is still used extensively, and anyone interested in electronics must have at least a working knowledge

  • f this device.
  • M. B. Patil, IIT Bombay
slide-7
SLIDE 7

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* 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. * BJT is still used extensively, and anyone interested in electronics must have at least a working knowledge

  • f this device.

* “A BJT is two diodes connected back-to-back.”

  • M. B. Patil, IIT Bombay
slide-8
SLIDE 8

Bipolar Junction Transistors

Base Emitter Collector Base Emitter Collector npn transistor pnp transistor

p n p n p n

* 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. * BJT is still used extensively, and anyone interested in electronics must have at least a working knowledge

  • f this device.

* “A BJT is two diodes connected back-to-back.” WRONG! Let us see why.

  • M. B. Patil, IIT Bombay
slide-9
SLIDE 9

Bipolar Junction Transistors

Consider a pnp BJT in the following circuit:

p n p B C E 5 V 10 V

I3 I1 I2 1 k 1 k R1 R2

  • M. B. Patil, IIT Bombay
slide-10
SLIDE 10

Bipolar Junction Transistors

Consider a pnp BJT in the following circuit:

p n p B C E 5 V 10 V

I3 I1 I2 1 k 1 k R1 R2

If the transistor is replaced with two diodes connected back-to-back, we get

D1 D2 B E C 5 V 10 V

I3 I2 I1 1 k 1 k R2 R1

  • M. B. Patil, IIT Bombay
slide-11
SLIDE 11

Bipolar Junction Transistors

Consider a pnp BJT in the following circuit:

p n p B C E 5 V 10 V

I3 I1 I2 1 k 1 k R1 R2

If the transistor is replaced with two diodes connected back-to-back, we get

D1 D2 B E C 5 V 10 V

I3 I2 I1 1 k 1 k R2 R1

Assuming Von = 0.7 V for D1, we get I1 = 5 V − 0.7 V R1 = 4.3 mA, I2 = 0 (since D2 is reverse biased), and I3 ≈ I1 = 4.3 mA.

  • M. B. Patil, IIT Bombay
slide-12
SLIDE 12

Bipolar Junction Transistors

Using a more realistic equivalent circuit for the BJT, we obtain,

10 V 10 V B 5 V E C p n p B C E 5 V

R1 1 k I3 I2 I1 α I1 I3 I1 I2 1 k 1 k 1 k R2 R1 R2

  • M. B. Patil, IIT Bombay
slide-13
SLIDE 13

Bipolar Junction Transistors

Using a more realistic equivalent circuit for the BJT, we obtain,

10 V 10 V B 5 V E C p n p B C E 5 V

R1 1 k I3 I2 I1 α I1 I3 I1 I2 1 k 1 k 1 k R2 R1 R2

We now get, I1 = 5 V − 0.7 V R1 = 4.3 mA (as before),

  • M. B. Patil, IIT Bombay
slide-14
SLIDE 14

Bipolar Junction Transistors

Using a more realistic equivalent circuit for the BJT, we obtain,

10 V 10 V B 5 V E C p n p B C E 5 V

R1 1 k I3 I2 I1 α I1 I3 I1 I2 1 k 1 k 1 k R2 R1 R2

We now get, I1 = 5 V − 0.7 V R1 = 4.3 mA (as before), I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and

  • M. B. Patil, IIT Bombay
slide-15
SLIDE 15

Bipolar Junction Transistors

Using a more realistic equivalent circuit for the BJT, we obtain,

10 V 10 V B 5 V E C p n p B C E 5 V

R1 1 k I3 I2 I1 α I1 I3 I1 I2 1 k 1 k 1 k R2 R1 R2

We now get, I1 = 5 V − 0.7 V R1 = 4.3 mA (as before), 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
slide-16
SLIDE 16

Bipolar Junction Transistors

Using a more realistic equivalent circuit for the BJT, we obtain,

10 V 10 V B 5 V E C p n p B C E 5 V

R1 1 k I3 I2 I1 α I1 I3 I1 I2 1 k 1 k 1 k R2 R1 R2

We now get, I1 = 5 V − 0.7 V R1 = 4.3 mA (as before), 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, viz., I2 ≈ 0, I3 ≈ 4.3 mA.

  • M. B. Patil, IIT Bombay
slide-17
SLIDE 17

Bipolar Junction Transistors

Using a more realistic equivalent circuit for the BJT, we obtain,

10 V 10 V B 5 V E C p n p B C E 5 V

R1 1 k I3 I2 I1 α I1 I3 I1 I2 1 k 1 k 1 k R2 R1 R2

We now get, I1 = 5 V − 0.7 V R1 = 4.3 mA (as before), 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, viz., I2 ≈ 0, I3 ≈ 4.3 mA. 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
slide-18
SLIDE 18

Bipolar Junction Transistors

What is wrong with the two-diode model of a BJT?

  • M. B. Patil, IIT Bombay
slide-19
SLIDE 19

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.”

p D1 n p Base Emitter Base Collector Collector D2 Emitter

  • M. B. Patil, IIT Bombay
slide-20
SLIDE 20

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.”

p D1 n p Base Emitter Base Collector Collector D2 Emitter

* However, in a BJT, exactly the opposite is true. For a higher performance, the base region is made as short as possible, and the two diodes cannot be treated as independent devices.

p n p Emitter Collector Base

  • M. B. Patil, IIT Bombay
slide-21
SLIDE 21

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.”

p D1 n p Base Emitter Base Collector Collector D2 Emitter

* However, in a BJT, exactly the opposite is true. For a higher performance, the base region is made as short as possible, and the two diodes cannot be treated as independent devices.

p n p Emitter 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
slide-22
SLIDE 22

BJT in active mode

n p p n p n B E B C B E C B E C E C

IC IC IC IE IC IE IE IB IE IB IB IB

  • M. B. Patil, IIT Bombay
slide-23
SLIDE 23

BJT in active mode

n p p n p n B E B C B E C B E C E C

IC IC IC IE IC IE IE IB IE IB IB 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
slide-24
SLIDE 24

BJT in active mode

n p p n p n B E B C B E C B E C E C

IC IC IC IE IC IE IE IB IE IB IB 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
slide-25
SLIDE 25

BJT in active mode

n p p n p n B E B C B E C B E C E C

IC IC IC IE IC IE IE IB IE IB IB 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

  • f the B-C junction.
  • M. B. Patil, IIT Bombay
slide-26
SLIDE 26

BJT in active mode

n p p n p n B E B C B E C B E C E C

IC IC IC IE IC IE IE IB IE IB IB 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

  • f 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
slide-27
SLIDE 27

BJT in active mode

n p p n p n B E B C B E C B E C E C

IC IC IC IE IC IE IE IB IE IB IB 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

  • f 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
slide-28
SLIDE 28

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

  • M. B. Patil, IIT Bombay
slide-29
SLIDE 29

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

* In the active mode, IC = α IE , α ≈ 1 (slightly less than 1).

  • M. B. Patil, IIT Bombay
slide-30
SLIDE 30

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

* In the active mode, IC = α IE , α ≈ 1 (slightly less than 1). * IB = IE − IC = IE (1 − α) .

  • M. B. Patil, IIT Bombay
slide-31
SLIDE 31

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

* 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. β = IC IB = α 1 − α .

  • M. B. Patil, IIT Bombay
slide-32
SLIDE 32

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

* 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. β = IC IB = α 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
slide-33
SLIDE 33

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

β = IC IB = α 1 − α

  • M. B. Patil, IIT Bombay
slide-34
SLIDE 34

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

β = IC IB = α 1 − α α β 0.9 9 0.95 19 0.99 99 0.995 199

  • M. B. Patil, IIT Bombay
slide-35
SLIDE 35

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

β = IC IB = α 1 − α α β 0.9 9 0.95 19 0.99 99 0.995 199 * β increases substantially as α → 1.

  • M. B. Patil, IIT Bombay
slide-36
SLIDE 36

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

β = IC IB = α 1 − α α β 0.9 9 0.95 19 0.99 99 0.995 199 * β increases substantially as α → 1. * 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
slide-37
SLIDE 37

BJT in active mode

E B C E B C E C B E B C E B C n p p n p n B E C

IC IC IE IE IB IB α IE α IE IC IC IC IE IC IE IE IB IE IB IB IB

β = IC IB = α 1 − α α β 0.9 9 0.95 19 0.99 99 0.995 199 * β increases substantially as α → 1. * 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
slide-38
SLIDE 38

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

slide-39
SLIDE 39

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k

slide-40
SLIDE 40

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

  • M. B. Patil, IIT Bombay
slide-41
SLIDE 41

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB.

  • M. B. Patil, IIT Bombay
slide-42
SLIDE 42

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB. IB = VBB − VBE RB = 2 V − 0.7 V 100 k = 13 µA.

  • M. B. Patil, IIT Bombay
slide-43
SLIDE 43

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB. IB = VBB − VBE RB = 2 V − 0.7 V 100 k = 13 µA. IC = β × IB = 100 × 13 µA = 1.3 mA.

  • M. B. Patil, IIT Bombay
slide-44
SLIDE 44

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB. IB = VBB − VBE RB = 2 V − 0.7 V 100 k = 13 µA. 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
slide-45
SLIDE 45

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB. IB = VBB − VBE RB = 2 V − 0.7 V 100 k = 13 µA. 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
slide-46
SLIDE 46

A simple BJT circuit

RB RC VCC VBB B C E 10 V 2 V 1 k 100 k β = 100

p n n

RB RC VCC VBB 10 V 2 V 1 k 100 k IC IB IE αIE RB RC VCC VBB 10 V 2 V 1 k 100 k

Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB. IB = VBB − VBE RB = 2 V − 0.7 V 100 k = 13 µA. 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
slide-47
SLIDE 47

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

What happens if RB is changed from 100 k to 10 k?

  • M. B. Patil, IIT Bombay
slide-48
SLIDE 48

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

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
slide-49
SLIDE 49

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

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. IB = VBB − VBE RB = 2 V − 0.7 V 10 k = 130 µA

  • M. B. Patil, IIT Bombay
slide-50
SLIDE 50

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

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. IB = VBB − VBE RB = 2 V − 0.7 V 10 k = 130 µA → IC = β × IB = 100 × 130 µA = 13 mA.

  • M. B. Patil, IIT Bombay
slide-51
SLIDE 51

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

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. IB = VBB − VBE RB = 2 V − 0.7 V 10 k = 130 µA → 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
slide-52
SLIDE 52

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

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. IB = VBB − VBE RB = 2 V − 0.7 V 10 k = 130 µA → 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
slide-53
SLIDE 53

A simple BJT circuit: continued

p n n

IC IB IE αIE RB RB RB RC RC RC VCC VBB VCC VBB VCC VBB B C E 10 V 2 V 10 V 2 V 10 V 2 V 1 k 10 k 1 k 10 k 1 k 10 k β = 100

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. IB = VBB − VBE RB = 2 V − 0.7 V 10 k = 130 µA → 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
slide-54
SLIDE 54

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

slide-55
SLIDE 55

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

E B C E B C n p p B E C

IC IE IB −IC αR (−IC) IC IC IE IE IB IB Reverse active mode: B-E in r.b. B-C in f.b.

  • M. B. Patil, IIT Bombay
slide-56
SLIDE 56

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

E B C E B C n p p B E C

IC IE IB −IC αR (−IC) IC IC IE IE IB IB Reverse active mode: B-E in r.b. B-C in f.b.

In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their

  • riginal names.)
  • M. B. Patil, IIT Bombay
slide-57
SLIDE 57

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

E B C E B C n p p B E C

IC IE IB −IC αR (−IC) IC IC IE IE IB IB Reverse active mode: B-E in r.b. B-C in f.b.

In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their

  • riginal names.)

The two α’s, αF (forward α) and αR (reverse α) are generally quite different.

  • M. B. Patil, IIT Bombay
slide-58
SLIDE 58

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

E B C E B C n p p B E C

IC IE IB −IC αR (−IC) IC IC IE IE IB IB Reverse active mode: B-E in r.b. B-C in f.b.

In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their

  • riginal 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
slide-59
SLIDE 59

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

E B C E B C n p p B E C

IC IE IB −IC αR (−IC) IC IC IE IE IB IB Reverse active mode: B-E in r.b. B-C in f.b.

In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their

  • riginal 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
slide-60
SLIDE 60

Ebers-Moll model for a pnp transistor

E B C E B C n p p B E C

IC IE IB α IE IC IC IE IE IB IB Active mode (“forward” active mode): B-E in f.b. B-C in r.b.

E B C E B C n p p B E C

IC IE IB −IC αR (−IC) IC IC IE IE IB IB Reverse active mode: B-E in r.b. B-C in f.b.

In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their

  • riginal 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
slide-61
SLIDE 61

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. E B C E B C B E C (p) (n) (p) n p p IC IE IC IC IB IE IB I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1

  • M. B. Patil, IIT Bombay
slide-62
SLIDE 62

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. E B C E B C B E C (p) (n) (p) n p p IC IE IC IC IB IE IB I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1 The currents I ′

E and I ′ C are given by the Shockley diode equation:

I ′

E = IES

  • exp

VEB VT

  • − 1
  • ,

I ′

C = ICS

  • exp

VCB VT

  • − 1
  • .
  • M. B. Patil, IIT Bombay
slide-63
SLIDE 63

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. E B C E B C B E C (p) (n) (p) n p p IC IE IC IC IB IE IB I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1 The currents I ′

E and I ′ C are given by the Shockley diode equation:

I ′

E = IES

  • exp

VEB VT

  • − 1
  • ,

I ′

C = ICS

  • exp

VCB VT

  • − 1
  • .

Mode B-E B-C Forward active forward reverse I ′

E ≫ I ′ C

Reverse active reverse forward I ′

C ≫ I ′ E

Saturation forward forward I ′

E and I ′ C are comparable.

Cut-off reverse reverse I ′

E and I ′ C are negliglbe.

  • M. B. Patil, IIT Bombay
slide-64
SLIDE 64

Ebers-Moll model

E B C E B C E B C E B C (p) (n) (p) n p p E B C p n n E B C (n) (p) (n)

IC IC IC IE IC IC IE IC IE IB IB IE IB IB I′

C = ICS [exp(VBC/VT) − 1]

I′

E = IES [exp(VBE/VT) − 1]

I′

E = IES [exp(VEB/VT) − 1]

I′

C = ICS [exp(VCB/VT) − 1]

pnp transistor npn transistor I′

E

I′

E

I′

C

I′

C

αFI′

E

αFI′

E

αRI′

C

αRI′

C

IE IE IB IB D2 D2 D1 D1

  • M. B. Patil, IIT Bombay
slide-65
SLIDE 65

Ebers-Moll model

E B C E B C E B C E B C (p) (n) (p) n p p E B C p n n E B C (n) (p) (n)

IC IC IC IE IC IC IE IC IE IB IB IE IB IB I′

C = ICS [exp(VBC/VT) − 1]

I′

E = IES [exp(VBE/VT) − 1]

I′

E = IES [exp(VEB/VT) − 1]

I′

C = ICS [exp(VCB/VT) − 1]

pnp transistor npn transistor I′

E

I′

E

I′

C

I′

C

αFI′

E

αFI′

E

αRI′

C

αRI′

C

IE IE IB IB D2 D2 D1 D1

STOP

  • M. B. Patil, IIT Bombay
slide-66
SLIDE 66

Ebers-Moll model in active mode

  • M. B. Patil, IIT Bombay
slide-67
SLIDE 67

Ebers-Moll model in active mode

E B C E B C E B C E B C E E B C n p p (p) (n) (p) p n n B C (n) (p) (n)

IC = αF IE = βF IB IC = αF IE = βF IB IC IC IC IE IC IC IE IC I′

C = ICS [exp(VBC/VT) − 1]

I′

E = IES [exp(VBE/VT) − 1]

I′

E = IES [exp(VEB/VT) − 1]

I′

C = ICS [exp(VCB/VT) − 1]

pnp transistor npn transistor IB IE IB IE IB IB I′

E

I′

E

I′

C

I′

C

αFI′

E

αFI′

E

αRI′

C

αRI′

C

IE IE IB IB D2 D2 D1 D1

  • M. B. Patil, IIT Bombay
slide-68
SLIDE 68

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

  • M. B. Patil, IIT Bombay
slide-69
SLIDE 69

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

* Since BJT is a three-terminal device, its behaviour can be described in many different ways, e.g.,

  • M. B. Patil, IIT Bombay
slide-70
SLIDE 70

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

* Since BJT is a three-terminal device, its behaviour can be described in many different ways, e.g.,

  • IC versus VCB for different values of IE
  • M. B. Patil, IIT Bombay
slide-71
SLIDE 71

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

* Since BJT is a three-terminal device, its behaviour can be described in many different ways, e.g.,

  • IC versus VCB for different values of IE
  • IC versus VCE for different values of VBE
  • M. B. Patil, IIT Bombay
slide-72
SLIDE 72

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

* Since BJT is a three-terminal device, its behaviour can be described in many different ways, e.g.,

  • IC versus VCB for different values of IE
  • IC versus VCE for different values of VBE
  • IC versus VCE for different values of IB
  • M. B. Patil, IIT Bombay
slide-73
SLIDE 73

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

* Since BJT is a three-terminal device, its behaviour can be described in many different ways, e.g.,

  • IC versus VCB for different values of IE
  • IC versus VCE for different values of VBE
  • IC versus VCE for different values of IB

* The I-V relationship for a BJT is not a single curve but a “family” of curves or “characteristics.”

  • M. B. Patil, IIT Bombay
slide-74
SLIDE 74

BJT I-V characteristics

B E C n n p

IB VBE VCE VCB IC IE

* Since BJT is a three-terminal device, its behaviour can be described in many different ways, e.g.,

  • IC versus VCB for different values of IE
  • IC versus VCE for different values of VBE
  • IC versus VCE for different values of IB

* The I-V relationship for a BJT is not a single curve but a “family” of curves or “characteristics.” * The IC -VCE characteristics for different IB values are useful in understanding amplifier biasing.

  • M. B. Patil, IIT Bombay
slide-75
SLIDE 75

BJT I-V characteristics

B C E

IC IE IB0 10 µA VCE IB αF = 0.99 → βF = αF 1 − αF = 99 αR = 0.5 → βR = αR 1 − αR = 1 IES = 1 × 10−14 A ICS = 2 × 10−14 A

slide-76
SLIDE 76

BJT I-V characteristics

B C E

IC IE IB0 10 µA VCE IB αF = 0.99 → βF = αF 1 − αF = 99 αR = 0.5 → βR = αR 1 − αR = 1 IES = 1 × 10−14 A ICS = 2 × 10−14 A

B E C (p) (n) (n)

IC = αF IE = βF IB in active mode IC VCE IB0 10 µA I′

E = IES [exp(VBE/VT) − 1]

I′

C = ICS [exp(VBC/VT) − 1]

I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1

slide-77
SLIDE 77

BJT I-V characteristics

B C E

IC IE IB0 10 µA VCE IB αF = 0.99 → βF = αF 1 − αF = 99 αR = 0.5 → βR = αR 1 − αR = 1 IES = 1 × 10−14 A ICS = 2 × 10−14 A

B E C (p) (n) (n)

IC = αF IE = βF IB in active mode IC VCE IB0 10 µA I′

E = IES [exp(VBE/VT) − 1]

I′

C = ICS [exp(VBC/VT) − 1]

I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1

1.0 0.5 0.0 −0.5 −1.0 −1.5 10 20 0.5 1 1.5 2 1.2 0.8 0.4

lin sat VCE: VBC (Volts) VBE (Volts) I′

C (µA)

IC (mA) I′

E (mA)

slide-78
SLIDE 78

BJT I-V characteristics

B C E

IC IE IB0 10 µA VCE IB αF = 0.99 → βF = αF 1 − αF = 99 αR = 0.5 → βR = αR 1 − αR = 1 IES = 1 × 10−14 A ICS = 2 × 10−14 A

B E C (p) (n) (n)

IC = αF IE = βF IB in active mode IC VCE IB0 10 µA I′

E = IES [exp(VBE/VT) − 1]

I′

C = ICS [exp(VBC/VT) − 1]

I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1

1.0 0.5 0.0 −0.5 −1.0 −1.5 10 20 0.5 1 1.5 2 1.2 0.8 0.4

lin sat VCE: VBC (Volts) VBE (Volts) I′

C (µA)

IC (mA) I′

E (mA)

* linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB * saturation region: B-E under forward bias, B-C under forward bias, IC < βFIB

  • M. B. Patil, IIT Bombay
slide-79
SLIDE 79

BJT I-V characteristics

1 2 E B C B C E (p) (n) (n) 0.5 1 1.5 2 1.0 −1.0 0.5 0.0 −0.5 −1.5

10 µA 10 µA sat lin IC = αF IE = βF IB in active mode IC IE IC VCE * linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB * saturation region: B-E under forward bias, B-C under forward bias, IC < βFIB I′

E = IES [exp(VBE/VT) − 1]

I′

C = ICS [exp(VBC/VT) − 1]

IB αF = 0.99 → βF = αF 1 − αF = 99 αR = 0.5 → βR = αR 1 − αR = 1 IES = 1 × 10−14 A ICS = 2 × 10−14 A VCE (Volts) IB = 10 µA VCE I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1 IC (mA) VBE (Volts) VBC (Volts)

slide-80
SLIDE 80

BJT I-V characteristics

1 2 E B C B C E (p) (n) (n) 0.5 1 1.5 2 1.0 −1.0 0.5 0.0 −0.5 −1.5

10 µA 10 µA sat lin IC = αF IE = βF IB in active mode IC IE IC VCE * linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB * saturation region: B-E under forward bias, B-C under forward bias, IC < βFIB I′

E = IES [exp(VBE/VT) − 1]

I′

C = ICS [exp(VBC/VT) − 1]

IB αF = 0.99 → βF = αF 1 − αF = 99 αR = 0.5 → βR = αR 1 − αR = 1 IES = 1 × 10−14 A ICS = 2 × 10−14 A VCE (Volts) IB = 10 µA VCE I′

E

I′

C

αFI′

E

αRI′

C

IE IB D2 D1 IC (mA) VBE (Volts) VBC (Volts) IB = 20 µA 20 µA 20 µA

  • M. B. Patil, IIT Bombay
slide-81
SLIDE 81

A simple BJT circuit (revisited)

n n p

IB RB RC VCC VBB 10 V 2 V 1 k IC IE β = 100

We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit.

slide-82
SLIDE 82

A simple BJT circuit (revisited)

n n p

IB RB RC VCC VBB 10 V 2 V 1 k IC IE β = 100

We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. Let us plot IC − VCE curves for IB ≈ VBB − 0.7 V RB for the two values of RB.

slide-83
SLIDE 83

A simple BJT circuit (revisited)

n n p

IB RB RC VCC VBB 10 V 2 V 1 k IC IE β = 100

linear saturation 8 10 2 4 6 5 10 15

VCE (V) IC (mA) IB = 130 µA (RB = 10 k) IB = 13 µA (RB = 100 k)

We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. Let us plot IC − VCE curves for IB ≈ VBB − 0.7 V RB for the two values of RB.

slide-84
SLIDE 84

A simple BJT circuit (revisited)

n n p

IB RB RC VCC VBB 10 V 2 V 1 k IC IE β = 100

linear saturation 8 10 2 4 6 5 10 15

VCE (V) IC (mA) IB = 130 µA (RB = 10 k) IB = 13 µA (RB = 100 k)

We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. Let us plot IC − VCE curves for IB ≈ VBB − 0.7 V RB for the two values of 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.

slide-85
SLIDE 85

A simple BJT circuit (revisited)

n n p

IB RB RC VCC VBB 10 V 2 V 1 k IC IE β = 100

linear saturation 8 10 2 4 6 5 10 15

VCE (V) IC (mA) IB = 130 µA (RB = 10 k) IB = 13 µA (RB = 100 k)

load line

We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. Let us plot IC − VCE curves for IB ≈ VBB − 0.7 V RB for the two values of 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.

slide-86
SLIDE 86

A simple BJT circuit (revisited)

n n p

IB RB RC VCC VBB 10 V 2 V 1 k IC IE β = 100

linear saturation 8 10 2 4 6 5 10 15

VCE (V) IC (mA) IB = 130 µA (RB = 10 k) IB = 13 µA (RB = 100 k)

load line

We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. Let us plot IC − VCE curves for IB ≈ VBB − 0.7 V RB for the two values of 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
slide-87
SLIDE 87

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B

  • M. B. Patil, IIT Bombay
slide-88
SLIDE 88

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0

  • M. B. Patil, IIT Bombay
slide-89
SLIDE 89

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0 → IE RE = 5 − 0.7

  • M. B. Patil, IIT Bombay
slide-90
SLIDE 90

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0 → IE RE = 5 − 0.7 → RE = 4.3 V 2 mA = 2.15 k.

  • M. B. Patil, IIT Bombay
slide-91
SLIDE 91

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0 → IE RE = 5 − 0.7 → RE = 4.3 V 2 mA = 2.15 k. VBC + IC RC − VCC = 0

  • M. B. Patil, IIT Bombay
slide-92
SLIDE 92

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0 → IE RE = 5 − 0.7 → RE = 4.3 V 2 mA = 2.15 k. VBC + IC RC − VCC = 0 → IC RC = VCC − VBC .

  • M. B. Patil, IIT Bombay
slide-93
SLIDE 93

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0 → IE RE = 5 − 0.7 → RE = 4.3 V 2 mA = 2.15 k. VBC + IC RC − VCC = 0 → IC RC = VCC − VBC . Since α ≈ 1, IC ≈ IE

  • M. B. Patil, IIT Bombay
slide-94
SLIDE 94

BJT circuit example

Assuming the transistor to be operating in the active region, find RE and RC to obtain IE = 2 mA, and VBC = 1 V (α ≈ 1). RE VEE VCC RC 5 V 5 V IE IC C E B VEB − VEE + IE RE = 0 → IE RE = 5 − 0.7 → RE = 4.3 V 2 mA = 2.15 k. VBC + IC RC − VCC = 0 → IC RC = VCC − VBC . Since α ≈ 1, IC ≈ IE → IE RC ≈ 5 − 1 → RC = 4 V 2 mA = 2 k.

  • M. B. Patil, IIT Bombay