Bipolar Junction Transistors: Part 1

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

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor npn transistor

M. B. Patil, IIT Bombay * 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 of this device. * “A BJT is two diodes connected back-to-back.” WRONG! Let us see why.

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor npn transistor

* Bipolar: both electrons and holes contribute to conduction

M. B. Patil, IIT Bombay * 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 of this device. * “A BJT is two diodes connected back-to-back.” WRONG! Let us see why.

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor 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 * 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 of this device. * “A BJT is two diodes connected back-to-back.” WRONG! Let us see why.

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor npn transistor

M. B. Patil, IIT Bombay * BJT is still used extensively, and anyone interested in electronics must have at least a working knowledge of this device. * “A BJT is two diodes connected back-to-back.” WRONG! Let us see why.

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

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

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor npn transistor

M. B. Patil, IIT Bombay WRONG! Let us see why.

Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor npn transistor

M. B. Patil, IIT Bombay Bipolar Junction Transistors

Emitterp n p Collector Emittern p n Collector

Base Base pnp transistor npn transistor

M. B. Patil, IIT Bombay If the transistor is replaced with two diodes connected back-to-back, we get

R1 EC R2

I1 D1 D2 I 1kB 2 1k

5 V I3 10 V

Assuming Von = 0.7 V for D1, we get 5 V 0.7 V I1 = − = 4.3 mA, R1 I = 0 (since D2 is reverse biased), and I I = 4.3 mA. 2 3 ≈ 1

Bipolar Junction Transistors

Consider a pnp BJT in the following circuit:

R1 E C R2 p n p 1 k I1 I2 1 k B I 5 V 3 10 V

M. B. Patil, IIT Bombay Assuming Von = 0.7 V for D1, we get 5 V 0.7 V I1 = − = 4.3 mA, R1 I = 0 (since D2 is reverse biased), and I I = 4.3 mA. 2 3 ≈ 1

Bipolar Junction Transistors

Consider a pnp BJT in the following circuit:

R1 E C R2 p n p 1 k I1 I2 1 k B I 5 V 3 10 V

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

R1 EC R2

I1 D1 D2 I 1kB 2 1k

5 V I3 10 V

M. B. Patil, IIT Bombay Bipolar Junction Transistors

Consider a pnp BJT in the following circuit:

R1 E C R2 p n p 1 k I1 I2 1 k B I 5 V 3 10 V

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

R1 EC R2

I1 D1 D2 I 1kB 2 1k

5 V I3 10 V

Assuming Von = 0.7 V for D1, we get 5 V 0.7 V I1 = − = 4.3 mA, R1 I = 0 (since D2 is reverse biased), and I I = 4.3 mA. 2 3 ≈ 1 M. B. Patil, IIT Bombay We now get, 5 V 0.7 V I1 = − = 4.3 mA (as before), R1 I = αI 4.3 mA (since α 1 for a typical BJT), and 2 1 ≈ ≈ I = I I = (1 α) I 0 A. 3 1 − 2 − 1 ≈ The values of I and I are dramatically diﬀerent than the ones obtained earlier, viz., I 0, I 4.3 mA. 2 3 2 ≈ 3 ≈ Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although it does have two p-n junctions).

Bipolar Junction Transistors

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

R1 E C R2 R1 ECα I1 R2 p n p 1 k I1 I2 1 k 1 k I1 I2 1 k B B I I 5 V 3 10 V 5 V 3 10 V

M. B. Patil, IIT Bombay I = αI 4.3 mA (since α 1 for a typical BJT), and 2 1 ≈ ≈ I = I I = (1 α) I 0 A. 3 1 − 2 − 1 ≈ The values of I and I are dramatically diﬀerent than the ones obtained earlier, viz., I 0, I 4.3 mA. 2 3 2 ≈ 3 ≈ Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although it does have two p-n junctions).

Bipolar Junction Transistors

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

R1 E C R2 R1 ECα I1 R2 p n p 1 k I1 I2 1 k 1 k I1 I2 1 k B B I I 5 V 3 10 V 5 V 3 10 V

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

M. B. Patil, IIT Bombay I = I I = (1 α) I 0 A. 3 1 − 2 − 1 ≈ The values of I and I are dramatically diﬀerent than the ones obtained earlier, viz., I 0, I 4.3 mA. 2 3 2 ≈ 3 ≈ Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although it does have two p-n junctions).

Bipolar Junction Transistors

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

R1 E C R2 R1 ECα I1 R2 p n p 1 k I1 I2 1 k 1 k I1 I2 1 k B B I I 5 V 3 10 V 5 V 3 10 V

We now get, 5 V 0.7 V I1 = − = 4.3 mA (as before), R1 I = αI 4.3 mA (since α 1 for a typical BJT), and 2 1 ≈ ≈

M. B. Patil, IIT Bombay The values of I and I are dramatically diﬀerent than the ones obtained earlier, viz., I 0, I 4.3 mA. 2 3 2 ≈ 3 ≈ Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although it does have two p-n junctions).

Bipolar Junction Transistors

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

R1 E C R2 R1 ECα I1 R2 p n p 1 k I1 I2 1 k 1 k I1 I2 1 k B B I I 5 V 3 10 V 5 V 3 10 V

We now get, 5 V 0.7 V I1 = − = 4.3 mA (as before), R1 I = αI 4.3 mA (since α 1 for a typical BJT), and 2 1 ≈ ≈ I = I I = (1 α) I 0 A. 3 1 − 2 − 1 ≈

M. B. Patil, IIT Bombay Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although it does have two p-n junctions).

Bipolar Junction Transistors

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

R1 E C R2 R1 ECα I1 R2 p n p 1 k I1 I2 1 k 1 k I1 I2 1 k B B I I 5 V 3 10 V 5 V 3 10 V

We now get, 5 V 0.7 V I1 = − = 4.3 mA (as before), R1 I = αI 4.3 mA (since α 1 for a typical BJT), and 2 1 ≈ ≈ I = I I = (1 α) I 0 A. 3 1 − 2 − 1 ≈ The values of I and I are dramatically diﬀerent than the ones obtained earlier, viz., I 0, I 4.3 mA. 2 3 2 ≈ 3 ≈

M. B. Patil, IIT Bombay Bipolar Junction Transistors

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

R1 E C R2 R1 ECα I1 R2 p n p 1 k I1 I2 1 k 1 k I1 I2 1 k B B I I 5 V 3 10 V 5 V 3 10 V

M. B. Patil, IIT Bombay * 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 n p 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, and the two diodes 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.

Bipolar Junction Transistors

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

M. B. Patil, IIT Bombay * 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.

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.

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 n p Collector

Base

Emitter Collector

D1 Base D2

M. B. Patil, IIT Bombay * Later, we will look at the “Ebers-Moll model” of a BJT, which is a fairly accurate representation of the transistor action.

Bipolar Junction Transistors

Emitter p n p 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, and the two diodes cannot be treated as independent devices.

Emitter p n p Collector

Base

M. B. Patil, IIT Bombay Bipolar Junction Transistors

Emitter p n p 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, and the two diodes 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 * 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

M. B. Patil, IIT Bombay * 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

* 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 * 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

* In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias.

M. B. Patil, IIT Bombay * 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

* In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias.

M. B. Patil, IIT Bombay * Analog circuits, including ampliﬁers, are generally designed to ensure that the BJTs are operating in the active mode.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

* In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias.

M. B. Patil, IIT Bombay BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

* In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias.

M. B. Patil, IIT Bombay * 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. I α β = C = . 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

M. B. Patil, IIT Bombay * IB = IE IC = IE (1 α). − − * The ratio IC /IB is deﬁned as the current gain β of the transistor. I α β = C = . 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

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

M. B. Patil, IIT Bombay * The ratio IC /IB is deﬁned as the current gain β of the transistor. I α β = C = . 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B 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 * 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.

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B 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. I α β = C = . IB 1 α −

M. B. Patil, IIT Bombay BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B 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. I α β = C = . 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 * β increases substantially as α 1. → α β * Transistors are generally designed to get a high value of β (typically 100 to 250, but can be as high as 2000 for 0.9 9 “super-β” transistors). 0.95 19 * A large β IB IC or IE when the transistor is in the 0.99 99 active mode.⇒ 0.995 199

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

I α β = C = IB 1 α −

M. B. Patil, IIT Bombay * 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.⇒

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

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

M. B. Patil, IIT Bombay * 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.⇒

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

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

M. B. Patil, IIT Bombay * A large β IB IC or IE when the transistor is in the active mode.⇒

BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

I α β = C = IB 1 α − * β increases substantially as α 1. → α β * Transistors are generally designed to get a high value of β (typically 100 to 250, but can be as high as 2000 for 0.9 9 “super-β” transistors). 0.95 19 0.99 99 0.995 199

M. B. Patil, IIT Bombay BJT in active mode

E p n p C E C E n p n C E C IE IC IE IC IE IC IE IC

IB I IB B I B B B B B

α IE α IE E C E C IE IC IE IC I IB B B B

M. B. Patil, IIT Bombay 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.

10 V VCC

10 V VCC 1 k RC

IC 1 k RC

n 100k p 100k αIE

VBB RB VBB RB IB 2 V n 2 V

A simple BJT circuit

1 k RC

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

VBC = VB VC = 0.7 V 8.7 V = 8.0 V , − − − i.e., the B-C junction is indeed under reverse bias.

10 V VCC

1 k RC

100k αIE

VBB RB IB 2 V

A simple BJT circuit

10 V VCC

1 k RC 1 k RC

C 10 V V n 100k CC 100k p β = 100 R B VBB R B B n E 2 V VBB 2 V

VBC = VB VC = 0.7 V 8.7 V = 8.0 V , − − − i.e., the B-C junction is indeed under reverse bias.

A simple BJT circuit

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

M. B. Patil, IIT Bombay 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.

A simple BJT circuit

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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

M. B. Patil, IIT Bombay 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.

A simple BJT circuit

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V 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 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.

A simple BJT circuit

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V 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 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.

A simple BJT circuit

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V 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 VBC = VB VC = 0.7 V 8.7 V = 8.0 V , − − − i.e., the B-C junction is indeed under reverse bias.

A simple BJT circuit

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V 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 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 100k 100k p 100k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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 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 IC = β IB = 100 130 µA = 13 mA. RB 10 k → × × 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. →

A simple BJT circuit: continued

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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

M. B. Patil, IIT Bombay VBB VBE 2 V 0.7 V IB = − = − = 130 µA IC = β IB = 100 130 µA = 13 mA. RB 10 k → × × 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. →

A simple BJT circuit: continued

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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 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. →

A simple BJT circuit: continued

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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

M. B. Patil, IIT Bombay 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. →

A simple BJT circuit: continued

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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 IC = β IB = 100 130 µA = 13 mA. RB 10 k → × ×

M. B. Patil, IIT Bombay 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. →

A simple BJT circuit: continued

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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 IC = β IB = 100 130 µA = 13 mA. RB 10 k → × × VC = VCC IC RC = 10 V 13 mA 1 k = 3 V − − × −

M. B. Patil, IIT Bombay 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. →

A simple BJT circuit: continued

10 V VCC

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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

10 V VCC 1 k RC

IC 1 k RC 1 k RC

C 10 V n VCC 10 k 10 k p 10 k αIE β = 100 R B VBB R VBB R IB B B n B E 2 V 2 V VBB 2 V IE

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 IC = β IB = 100 130 µA = 13 mA. RB 10 k → × × 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 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.

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

Ebers-Moll model for a pnp transistor

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

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

M. B. Patil, IIT Bombay 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.

Ebers-Moll model for a pnp transistor

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

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

M. B. Patil, IIT Bombay 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.

Ebers-Moll model for a pnp transistor

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

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

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

M. B. Patil, IIT Bombay Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5.

Ebers-Moll model for a pnp transistor

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

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

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 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.

Ebers-Moll model for a pnp transistor

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

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

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 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.

Ebers-Moll model for a pnp transistor

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

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

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.

α IE E p n p C E C E C IE IC IE IC IE IC

IB IB IB B B B

Reverse active mode: B-E in r.b. B-C in f.b.

IC αR ( IC) − E p n p C E C E − C IE IC IE IC IE IC

IB IB IB B B B

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.

use of the higher value of β in that mode. M. B. Patil, IIT Bombay 0 0 The currents IE and IC are given by the Shockley diode equation: 0 VEB 0 VCB IE = IES exp − 1 , IC = ICS exp − 1 . VT VT

Mode B-E B-C 0 0 Forward active forward reverse IE IC 0 0 Reverse active reverse forward IC IE 0 0 Saturation forward forward IE and IC are comparable. 0 0 Cut-oﬀ reverse reverse IE and IC are negliglbe.

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 p n p C IE′ αFIE′ IE IC IE D1 IC IB E C B (p) D2 (p) E C αRIC′ IE IC IB IC′

IB (n) B B

M. B. Patil, IIT Bombay Mode B-E B-C 0 0 Forward active forward reverse IE IC 0 0 Reverse active reverse forward IC IE 0 0 Saturation forward forward IE and IC are comparable. 0 0 Cut-oﬀ reverse reverse IE and IC are negliglbe.

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 p n p C IE′ αFIE′ IE IC IE D1 IC IB E C B (p) D2 (p) E C αRIC′ IE IC IB IC′

IB (n) B B

0 0 The currents IE and IC are given by the Shockley diode equation: 0 VEB 0 VCB IE = IES exp − 1 , IC = ICS exp − 1 . 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.

E p n p C IE′ αFIE′ IE IC IE D1 IC IB E C B (p) D2 (p) E C αRIC′ IE IC IB IC′

IB (n) B B

0 0 The currents IE and IC are given by the Shockley diode equation: 0 VEB 0 VCB IE = IES exp − 1 , IC = ICS exp − 1 . VT VT

M. B. Patil, IIT Bombay STOP

Ebers-Moll model

pnp transistor

E p n p C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VEB/VT) 1] IB − E C B D2 (p) (p) IC′ = ICS [exp(VCB/VT) 1] E C − IE IC αRIC′ IB IC′ IB (n) B B

npn transistor

E n p n C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VBE/VT) 1] IB − E C B D2 (n) (n) IC′ = ICS [exp(VBC/VT) 1] E C − IE IC αRIC′ IB IC′ IB (p) B B

M. B. Patil, IIT Bombay Ebers-Moll model

pnp transistor

E p n p C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VEB/VT) 1] IB − E C B D2 (p) (p) IC′ = ICS [exp(VCB/VT) 1] E C − IE IC αRIC′ IB IC′ IB (n) B B STOP npn transistor

E n p n C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VBE/VT) 1] IB − E C B D2 (n) (n) IC′ = ICS [exp(VBC/VT) 1] E C − IE IC αRIC′ IB IC′ IB (p) B B

M. B. Patil, IIT Bombay pnp transistor

E p n p C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VEB/VT) 1] IB − E C B D2 (p) (p) IC′ = ICS [exp(VCB/VT) 1] E C − α I′ IE IC R C I′ IB C IC = αF IE = βF IB IB (n) B B

npn transistor

E n p n C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VBE/VT) 1] IB − E C B D2 (n) (n) IC′ = ICS [exp(VBC/VT) 1] E C − α I′ IE IC R C I′ IB C IC = αF IE = βF IB IB (p) B B

Ebers-Moll model in active mode

M. B. Patil, IIT Bombay Ebers-Moll model in active mode

pnp transistor

E p n p C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VEB/VT) 1] IB − E C B D2 (p) (p) IC′ = ICS [exp(VCB/VT) 1] E C − α I′ IE IC R C I′ IB C IC = αF IE = βF IB IB (n) B B

npn transistor

E n p n C IE′ αFIE′ IE IC IE D1 IC IE′ = IES [exp(VBE/VT) 1] IB − E C B D2 (n) (n) IC′ = ICS [exp(VBC/VT) 1] E C − α I′ IE IC R C I′ IB C IC = αF IE = βF IB IB (p) B B

M. B. Patil, IIT Bombay * 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.

BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

M. B. Patil, IIT Bombay - 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.

BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

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

M. B. Patil, IIT Bombay - 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.

BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

* 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 - 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.

BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

* 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 * 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.

BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

* 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 * The IC -VCE characteristics for diﬀerent IB values are useful in understanding ampliﬁer biasing.

BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

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

M. B. Patil, IIT Bombay BJT I -V characteristics

IC VCB p n B VCE IB n

IE VBE

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

* The IC -VCE characteristics for diﬀerent IB values are useful in understanding ampliﬁer biasing.

M. B. Patil, IIT Bombay 1.0

0.5 VBE (Volts) VCE 0.0 IE′ αFIE′ −0.5 VBC (Volts) IE D1 IC E C −1.0 (n) D2 (n) −1.5 20 αRIC′ IB IC′ (p) B IC′ (µA) IB0 10 10 µA

0 IE′ = IES [exp(VBE/VT) 1] − 1.2 I′ = I [exp(V /V ) 1] sat lin C CS BC T − IC = αF IE = βF IB in active mode 0.8

0.4 * linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB IC (mA) * saturation region: B-E under forward bias, B-C under forward bias,IC < βFIB 0 IE′ (mA)

VCE:0 0.5 1 1.5 2

BJT I -V characteristics

VCE E C

IE IC

IB B

IB0 10 µA

αF αF = 0.99 βF = = 99 → 1 αF α− α = 0.5 β = R = 1 R → R 1 α − R 14 IES = 1 10− A × 14 I = 2 10− A CS ×

M. B. Patil, IIT Bombay 1.0

0.5 VBE (Volts) 0.0

−0.5 VBC (Volts)

−1.0

−1.5 20

IC′ (µA) 10

0 1.2 sat lin

0.8

0.4 * linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB IC (mA) * saturation region: B-E under forward bias, B-C under forward bias,IC < βFIB 0 IE′ (mA)

VCE:0 0.5 1 1.5 2

BJT I -V characteristics

VCE

IE′ αFIE′ VCE E C IE D1 IC E C IE IC (n) D2 (n)

IB αRIC′ B IB IC′

IB0 (p) B 10 µA IB0 10 µA αF αF = 0.99 βF = = 99 → 1 αF α−R α = 0.5 β = = 1 IE′ = IES [exp(VBE/VT) 1] R → R 1 α − − R 14 IC′ = ICS [exp(VBC/VT) 1] I = 1 10− A − ES × 14 I = α I = β I in active mode I = 2 10− A C F E F B CS ×

M. B. Patil, IIT Bombay * 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

BJT I -V characteristics 1.0

0.5 VBE (Volts) VCE 0.0 IE′ αFIE′ VCE −0.5 VBC (Volts) E C IE D1 IC E C −1.0 IE IC (n) D2 (n) −1.5

IB 20 αRIC′ B IB IC′

VCE:0 0.5 1 1.5 2

M. B. Patil, IIT Bombay BJT I -V characteristics 1.0

0.5 VBE (Volts) VCE 0.0 IE′ αFIE′ VCE −0.5 VBC (Volts) E C IE D1 IC E C −1.0 IE IC (n) D2 (n) −1.5

IB 20 αRIC′ B IB IC′

IB0 (p) B I′ (µA) 10 µA C IB0 10 10 µA αF αF = 0.99 βF = = 99 → 1 αF α− 0 R I′ = IES [exp(VBE/VT) 1] αR = 0.5 βR = = 1 E − → 1 αR 1.2 − lin 14 IC′ = ICS [exp(VBC/VT) 1] sat I = 1 10− A − ES × 14 IC = αF IE = βF IB in active mode 0.8 ICS = 2 10− A × 0.4 * linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB IC (mA) * saturation region: B-E under forward bias, B-C under forward bias,IC < βFIB 0 IE′ (mA)

VCE:0 0.5 1 1.5 2 M. B. Patil, IIT Bombay 20 µA 20 µA IB = 20 µA

BJT I -V characteristics

1.0

VCE 0.5 VBE (Volts) IE′ αFIE′ VCE 0.0 IE D1 IC E C V (Volts) E BC

IE IC (n) D2 C −0.5 (n) IB α I′ R C ′ −1.0 B IB IC 10 µA (p) B −1.5 10 µA sat lin 2 α α = 0.99 β = F = 99 F → F 1 α F α−R α = 0.5 β = = 1 IE′ = IES [exp(VBE/VT) 1] R → R 1 α − − R 14 IC′ = ICS [exp(VBC/VT) 1] I = 1 10− A − 1 ES × 14 I = α I = β I in active mode IB = 10 µA I = 2 10− A C F E F B CS ×

* linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB IC (mA) 0 * saturation region: B-E under forward bias, B-C under forward bias,IC < βFIB 0 0.5 1 1.5 2 VCE (Volts)

M. B. Patil, IIT Bombay BJT I -V characteristics

1.0

VCE 0.5 VBE (Volts) IE′ αFIE′ VCE 0.0 IE D1 IC E C V (Volts) E BC

IE IC (n) D2 C −0.5 (n) IB α I′ R C ′ −1.0 B IB IC 10 µA (p) B 20 µA −1.5 10 µA sat lin 20 µA 2 α I = 20 µA α = 0.99 β = F = 99 B F → F 1 α F α−R α = 0.5 β = = 1 IE′ = IES [exp(VBE/VT) 1] R → R 1 α − − R 14 IC′ = ICS [exp(VBC/VT) 1] I = 1 10− A − 1 ES × 14 I = α I = β I in active mode IB = 10 µA I = 2 10− A C F E F B CS ×

* linear region: B-E under forward bias, B-C under reverse bias, IC = βFIB IC (mA) 0 * saturation region: B-E under forward bias, B-C under forward bias,IC < βFIB 0 0.5 1 1.5 2 VCE (Volts)

M. B. Patil, IIT Bombay 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. ≈

saturation

linear 15

IB =130 µA (RB =10 k) 10

(mA) load line C I 5

IB =13 µA (RB =100 k) 0 0 2 4 6 8 10

VCE (V)

A simple BJT circuit (revisited)

10 V VCC

1 k RC

β = 100 IC n p V BB RB IB n 2 V IE

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

M. B. Patil, IIT Bombay 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. ≈

saturation

linear 15

IB =130 µA (RB =10 k) 10

(mA) load line C I 5

IB =13 µA (RB =100 k) 0 0 2 4 6 8 10

VCE (V)

A simple BJT circuit (revisited)

10 V VCC

1 k RC

β = 100 IC n p V BB RB IB n 2 V IE

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

load line

A simple BJT circuit (revisited)

saturation

linear 15 10 V VCC

IB =130 µA (RB =10 k) 1 k RC 10 β = 100 IC

n (mA)

p C I 5 V BB RB IB n 2 V IE IB =13 µA (RB =100 k) 0 0 2 4 6 8 10

VCE (V)

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

M. B. Patil, IIT Bombay 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. ≈

load line

A simple BJT circuit (revisited)

saturation

linear 15 10 V VCC

IB =130 µA (RB =10 k) 1 k RC 10 β = 100 IC

n (mA)

p C I 5 V BB RB IB n 2 V IE IB =13 µA (RB =100 k) 0 0 2 4 6 8 10

VCE (V)

A simple BJT circuit (revisited)

saturation

linear 15 10 V VCC

IB =130 µA (RB =10 k) 1 k RC 10 β = 100 IC

n (mA) load line

p C I 5 V BB RB IB n 2 V IE IB =13 µA (RB =100 k) 0 0 2 4 6 8 10

VCE (V)

M. B. Patil, IIT Bombay A simple BJT circuit (revisited)

saturation

linear 15 10 V VCC

IB =130 µA (RB =10 k) 1 k RC 10 β = 100 IC

n (mA) load line

p C I 5 V BB RB IB n 2 V IE IB =13 µA (RB =100 k) 0 0 2 4 6 8 10

VCE (V)

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 4.3 V VEB VEE + IE RE = 0 IE RE = 5 0.7 RE = = 2.15 k. − → − → 2 mA

VBC + IC RC VCC = 0 IC RC = VCC VBC . − → − 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

M. B. Patil, IIT Bombay 4.3 V IE RE = 5 0.7 RE = = 2.15 k. → − → 2 mA

VBC + IC RC VCC = 0 IC RC = VCC VBC . − → − 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

VEB VEE + IE RE = 0 −

M. B. Patil, IIT Bombay 4.3 V RE = = 2.15 k. → 2 mA

VBC + IC RC VCC = 0 IC RC = VCC VBC . − → − 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

VEB VEE + IE RE = 0 IE RE = 5 0.7 − → −

M. B. Patil, IIT Bombay VBC + IC RC VCC = 0 IC RC = VCC VBC . − → − 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

4.3 V VEB VEE + IE RE = 0 IE RE = 5 0.7 RE = = 2.15 k. − → − → 2 mA

M. B. Patil, IIT Bombay IC RC = VCC VBC . → − 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

4.3 V VEB VEE + IE RE = 0 IE RE = 5 0.7 RE = = 2.15 k. − → − → 2 mA

VBC + IC RC VCC = 0 −

M. B. Patil, IIT Bombay 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

4.3 V VEB VEE + IE RE = 0 IE RE = 5 0.7 RE = = 2.15 k. − → − → 2 mA

VBC + IC RC VCC = 0 IC RC = VCC VBC . − → −

M. B. Patil, IIT Bombay 4 V IE RC 5 1 RC = = 2 k. → ≈ − → 2 mA

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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

4.3 V VEB VEE + IE RE = 0 IE RE = 5 0.7 RE = = 2.15 k. − → − → 2 mA

VBC + IC RC VCC = 0 IC RC = VCC VBC . − → −

Since α 1, IC IE ≈ ≈

M. B. Patil, IIT Bombay 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). ≈

IE E C IC

B RE RC 5V 5V

VEE VCC

4.3 V VEB VEE + IE RE = 0 IE RE = 5 0.7 RE = = 2.15 k. − → − → 2 mA

VBC + IC RC VCC = 0 IC RC = VCC VBC . − → − 4 V Since α 1, IC IE IE RC 5 1 RC = = 2 k. ≈ ≈ → ≈ − → 2 mA

M. B. Patil, IIT Bombay