Semiconductor Devices

Bipolar Junction Transistors: Part 2

M. B. Patilmbpatil@ee.iitb.ac.in

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

Bipolar junction transistors: Ebers-Moll model

* We have considered a BJT in the active mode (B-E junction under forward bias, B-C junction underreverse bias) and obtained α.

The BJT can now be replaced with its equivalent circuit.

IB= (1− α) IEIB= (1− α) IE

IE IC IE IC

IC=αIEIEIC=αIEIE

VEB VBCVBE VCB

B (base)

(emitter) (collector)

B (base)

* A generalised model valid in all modes can be obtained by removing the conditions of a forward biasacross the E -B junction and a reverse bias across the C -B junction → Ebers-Moll model.

IB= (1− α) IEIB= (1− α) IE

IE IC IE IC

IC=αIEIEIC=αIEIE

VEB VBCVBE VCB

B (base)

IB= (1− α) IEIB= (1− α) IE

IE IC IE IC

IC=αIEIEIC=αIEIE

VEB VBCVBE VCB

B (base)

IB= (1− α) IEIB= (1− α) IE

IE IC IE IC

IC=αIEIEIC=αIEIE

VEB VBCVBE VCB

B (base)

Outline of derivation for a pnp BJT

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

forward activebias is not shown.Variation of W with

* Boundary conditions:

∆n(xE ) = n0E

)− 1

]∆n(−∞) = 0

∆p(0) = p0B

)− 1

]∆p(W ) = p0B

)− 1

]∆n(xC ) = n0C

)− 1

]∆n(∞) = 0

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

∆n(xE ) = n0E

)− 1

]∆n(−∞) = 0

∆p(0) = p0B

)− 1

]∆p(W ) = p0B

)− 1

]∆n(xC ) = n0C

)− 1

]∆n(∞) = 0

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

∆n(xE ) = n0E

)− 1

]∆n(−∞) = 0

∆p(0) = p0B

)− 1

]∆p(W ) = p0B

)− 1

]∆n(xC ) = n0C

)− 1

]∆n(∞) = 0

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

∆n(xE ) = n0E

)− 1

]∆n(−∞) = 0

∆p(0) = p0B

)− 1

]∆p(W ) = p0B

)− 1

∆n(xC ) = n0C

)− 1

]∆n(∞) = 0

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

∆n(xE ) = n0E

)− 1

]∆n(−∞) = 0

∆p(0) = p0B

)− 1

]∆p(W ) = p0B

)− 1

]∆n(xC ) = n0C

)− 1

]∆n(∞) = 0

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

* Solve the minority-carrier continuity

equations in the neutral emitter, base, and

collector regions.

* From the solutions, obtain the following

currents.

InE (xE ) = qADnEdn

dx(xE ).

IpB(0) = −qADpBdp

dx(0).

IpB(W ) = −qADpBdp

dx(W ).

InC (xC ) = qADnCdn

dx(xC ).

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

collector regions.

currents.

InE (xE ) = qADnEdn

dx(xE ).

IpB(0) = −qADpBdp

dx(0).

dx(W ).

InC (xC ) = qADnCdn

dx(xC ).

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

collector regions.

currents.

InE (xE ) = qADnEdn

dx(xE ).

IpB(0) = −qADpBdp

dx(0).

dx(W ).

InC (xC ) = qADnCdn

dx(xC ).

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

* Obtain the terminal currents, ignoring G-R

in the depletion regions.

IE = InE (xE ) + IpB(0).

IC = InC (xC ) + IpB(W ).

IB = IE − IC .

Ebers-Moll model:

xCW0xE

n(x) p(x) n(x)

cutoff

reverse active

saturation

* Obtain the terminal currents, ignoring G-R

in the depletion regions.

IE = InE (xE ) + IpB(0).

IC = InC (xC ) + IpB(W ).

IB = IE − IC .

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

* Current directions are assigned such that IC , IE , IB are all positive if the BJT operates in the active mode.

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

* Current directions are assigned such that IC , IE , IB are all positive if the BJT operates in the active mode.

Ebers-Moll model

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

* The Ebers-Moll model can be interpreted as two transistors connected in parallel, each acting in the active mode.

* The forward transistor is represented by the E -B diode and the corresponding dependent source (the upper branches),and the reverse transistor by the C -B diode and the corresponding dependent source (the lower branches).

Ebers-Moll model

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

Ebers-Moll model

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

Ebers-Moll model

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

* The model has four parameters: IES , ICS , αF , αR (F for forward, R for reverse) which can be related to the geometry(base width, device area) and material parameters (doping densities, diffusion coefficients, lifetimes) of the transistor.

* With the assumptions we have made, αF IES = αR ICS .

Ebers-Moll model

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

* The model has four parameters: IES , ICS , αF , αR (F for forward, R for reverse) which can be related to the geometry

(base width, device area) and material parameters (doping densities, diffusion coefficients, lifetimes) of the transistor.1

1R.F. Pierret, Semiconductor Device Fundamentals. New Delhi: Pearson Education, 1996.M. B. Patil, IIT Bombay

Ebers-Moll model

npn transistor

pnp transistor

I′E = IES

)− 1

I′C = ICS

)− 1

I′E = IES

)− 1

I′C = ICS

)− 1

ICIEIC

αRI′C

αFI′E

VEB VBC

VBE VCB

* The model has four parameters: IES , ICS , αF , αR (F for forward, R for reverse) which can be related to the geometry

(base width, device area) and material parameters (doping densities, diffusion coefficients, lifetimes) of the transistor.1

1R.F. Pierret, Semiconductor Device Fundamentals. New Delhi: Pearson Education, 1996.M. B. Patil, IIT Bombay

Ebers-Moll model

* Assumptions made:

- Low-level injection

- Uniform doping densities, non-degenerate carrier statistics

- One-dimensional device, with the emitter and collector regions much longer than

the respective minority carrier diffusion lengths

- No generation/recombination in the depletion regions

- Constant width (W ) of the neutral base region, independent of bias voltages

* In practice, the above assumptions do not hold, e.g., as we have seen, the dopingdensities are not uniform.

Furthermore, several details about the device such as lifetimes, mobilities, and basewidth, are not known.

* The Ebers-Moll model can still be used as a “phenomenological” description of the deviceif model parameters are suitably extracted using measured data.

* More advanced BJT models are available (e.g., the SPICE model) and are used for circuitsimulation.

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

Ebers-Moll model

* Assumptions made:

* More advanced BJT models are available (e.g., the SPICE model2) and are used forcircuit simulation.

2P. Antognetti and G. Massobrio, Semiconductor Device Modeling with SPICE. New York: McGraw-Hill, 1988.M. B. Patil, IIT Bombay

Ebers-Moll model: special cases

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

* IE = 0, i.e., emitter open-circuited.

IC = −I ′C + αF I′E

= −I ′C + αF (IE + αR I′C )

= −I ′C (1− αFαR) + αF IE .

When the C -B junction is under reverse bias, I ′C ≈−ICS , and with IE = 0, we get

IC ≡ ICBO = ICS (1− αFαR).

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

= −I ′C + αF (IE + αR I′C )

= −I ′C (1− αFαR) + αF IE .

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

= −I ′C + αF (IE + αR I′C )

= −I ′C (1− αFαR) + αF IE .

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

= −I ′C + αF (IE + αR I′C )

= −I ′C (1− αFαR) + αF IE .

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

* IB = 0, i.e., base open-circuited.

IC = −I ′C (1− αFαR) + αF IE = −I ′C (1− αFαR) + αF (IC + IB)

=−I ′C (1− αFαR)

1− αF+

(1− αF )IB =

−I ′C (1− αFαR)

1− αF+ βF IB .

When the C -B junction is under reverse bias, I ′C ≈−ICS , and with IB = 0, we get

IC ≡ ICEO =ICS (1− αFαR)

(1− αF )=

(1− αF ),

which is much larger than ICBO since αF is close to 1.

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

=−I ′C (1− αFαR)

1− αF+

(1− αF )IB =

1− αF+ βF IB .

(1− αF )=

(1− αF ),

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

=−I ′C (1− αFαR)

1− αF+

(1− αF )IB =

1− αF+ βF IB .

(1− αF )=

(1− αF ),

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

=−I ′C (1− αFαR)

1− αF+

(1− αF )IB =

1− αF+ βF IB .

(1− αF )=

(1− αF ),

npn transistor

I′E = IES

)− 1

I′C = ICS

)− 1

αRI′C

αFI′E

E E CC

VBE VCB

=−I ′C (1− αFαR)

1− αF+

(1− αF )IB =

1− αF+ βF IB .

(1− αF )=

(1− αF ),

BJT I -V description

pnp transistor npn transistor

VEB VBCVBE VCB

* Unlike the diode (where there is only one current and one voltage), the BJT has

three currents and three voltages.

* The current-voltage relationship is described in the form of a “family” of curves,

with a current selected as the y variable, a voltage as the x variable, and a third

variable as a quantity to be held constant for each I -V curve.

* Two descriptions, which are related to the “common-base” and “common-emitter”

configurations, are commonly used.

VEB VBCVBE VCB

Common-base configuration

αRI′C

αFI′E

IES = 1× 10−14 A

ICS= 2× 10−14 A

αF= 0.99

αR= 0.5

0.2mA0mA

VBE (volts) VCB (volts)

IE= 1mA

0 0.7 −1 0 1 2 3 4 5

* VCB > 0 V:

C-B junction is reverse biased, I ′C ≈−ICS , which is negligibly small.

IE ≈ I ′E , IC ≈αF IE → for a given IE , IC is a constant.

On the input side, the IC versus VBE curve (for a positive VCB value) is like a diode I -V relationship.

* VCB < 0 V:

C-B junction is forward biased, but I ′C becomes substantial only when VCB ≈−0.5 V.

Beyond this point, IC drops sharply since IC =αF I′E − I ′C → saturation mode.

αRI′C

αFI′E

IES = 1× 10−14 A

ICS= 2× 10−14 A

αF= 0.99

αR= 0.5

0.2mA0mA

IE= 1mA

0 0.7 −1 0 1 2 3 4 5

* VCB > 0 V:

* VCB < 0 V:

αRI′C

αFI′E

IES = 1× 10−14 A

ICS= 2× 10−14 A

αF= 0.99

αR= 0.5

0.2mA0mA

IE= 1mA

0 0.7 −1 0 1 2 3 4 5

* VCB > 0 V:

* VCB < 0 V:

αRI′C

αFI′E

IES = 1× 10−14 A

ICS= 2× 10−14 A

αF= 0.99

αR= 0.5

0.2mA0mA

IE= 1mA

0 0.7 −1 0 1 2 3 4 5

* VCB > 0 V:

* VCB < 0 V:

αRI′C

αFI′E

IES = 1× 10−14 A

ICS= 2× 10−14 A

αF= 0.99

αR= 0.5

0.2mA0mA

IE= 1mA

0 0.7 −1 0 1 2 3 4 5

* VCB > 0 V:

* VCB < 0 V:

αRI′C

αFI′E

IES = 1× 10−14 A

ICS= 2× 10−14 A

αF= 0.99

αR= 0.5

0.2mA0mA

IE= 1mA

0 0.7 −1 0 1 2 3 4 5

* VCB > 0 V:

* VCB < 0 V:

Common-emitter configuration

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

* In the active region (where IC is constant for a given IB), the B-C junction is reverse biased.

→ I ′C ≈ 0→ IC =αF IE =βIB , irrespective of VCE .

* When VBC becomes greater than about 0.4 V, I ′C becomes significant, and IC =αF I′E − I ′C decreases → IC < βIB .

* In the active region (e.g., VCE = 1 V), VBE is nearly constant (∼ 0.65 V).

* In the saturtion region, VCE is 0.2 V or smaller. This is generally true for all low-power BJTs.

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

* Comparison of IB versus VBE for VCE = 0 V and VCE = 1 V:

- VCE = 1 V (active region):

VBC = VBE − VCE ≈ 0.7− 1 = −0.3 V → I ′C ≈ 0 → IB = I ′E − αF I′E ≈ (1− αF ) IESe

VBE/VT .

- VCE = 0 V (saturation region):

VBC = VBE − VCE = VBE → I ′C is comparable to I ′E .

→ IB = (1− αF )I ′E + (1− αR)I ′C → IB -VBE curve shifts left.

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBC = VBE − VCE ≈ 0.7− 1 = −0.3 V

→ I ′C ≈ 0 → IB = I ′E − αF I′E ≈ (1− αF ) IESe

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBC = VBE − VCE ≈ 0.7− 1 = −0.3 V → I ′C ≈ 0

→ IB = I ′E − αF I′E ≈ (1− αF ) IESe

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBC = VBE − VCE ≈ 0.7− 1 = −0.3 V → I ′C ≈ 0 → IB = I ′E − αF I′E

≈ (1− αF ) IESeVBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

VBC = VBE − VCE = VBE

→ I ′C is comparable to I ′E .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

→ IB = (1− αF )I ′E + (1− αR)I ′C

→ IB -VBE curve shifts left.

VBE (volts)

I B(µ

IES = 1× 10−14 A

ICS = 2× 10−14 A

αF= 0.99

αR= 0.5

I′CIB

αRI′C

αFI′E

2µA0µA

IB = 10µA

VCE (volts)

VBC=0.4 V

VBC=0VVBC=−1V

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0 1 2 3 4 5

VBE/VT .

BJT: second-order effects

* The Ebers-Moll model does remarkably well in capturing the basic transistor action.

* For a higher accuracy in circuit simulation, second-order effects need to be considered.

* We will consider

- base width modulation

- breakdown phenomena

* We will consider

Base width modulation

C-Bdepletionregion

E-Bdepletionregion

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

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

We have assumed so far that the width of the neutral base region

(in the active mode) is independent of VBE and VBC .

This is a reasonable assumption for the following reasons.

- The B-E junction voltge is nearly constant, say 0.6 to 0.75 V

for a silicon BJT, and the variation of the B-E depletion

width is negliglble.

- Since VCB – the reverse bias across the B-C junction – can

vary substantially, the B-C depletion width can also change

significantly.

However, the change occurs mostly on the collector side since

Na(B) � Nd (C).

C-Bdepletionregion

E-Bdepletionregion

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

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

significantly.

Na(B) � Nd (C).

C-Bdepletionregion

E-Bdepletionregion

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

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

significantly.

Na(B) � Nd (C).

C-Bdepletionregion

E-Bdepletionregion

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

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

significantly.

Na(B) � Nd (C).

VCB=VCB1

VCB=VCB2

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

IB8µA

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

- As VCB ↑, the B-C depletion region expands, W ↓

→ IC ∝dn

dx(W ) ↑

- This “base width modulation” effect (also called the “Early

effect”) gives rise to a finite slope of the IC -VCE curves in the

active region.

(VCE ↑ → VCB (= VCE − VBE ) ↑ → W ↓ → IC ↑)

VCB=VCB1

VCB=VCB2

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

IB8µA

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

→ IC ∝dn

dx(W ) ↑

active region.

VCB=VCB1

VCB=VCB2

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

IB8µA

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

→ IC ∝dn

dx(W ) ↑

active region.

VCB=VCB1

VCB=VCB2

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

IB8µA

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

→ IC ∝dn

dx(W ) ↑

active region.

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

VCE−VA

* When the active region parts of the IC -VCE curves are extended backwards, they intersect

the VCE axis approximately at the same point, −VA.

* VA is called the Early voltage.

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

VCE−VA

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

VCE−VA

VCE (volts)

IB= 10µA

VsatCE

0 1 2 3 4 5

VCE−VA

Breakdown phenomena

* We have seen that a pn junction diode cannot withstand arbitrarily large reverse voltages,

it breaks down at some point.

* Similarly, if the reverse bias across the B-C junction of a BJT is increased, it breaks down

at some point, i.e., the collector current becomes very large.

* We will look at two breakdown mechanisms:

- punchthrough

- avalanche breakdown

Breakdown phenomena

- punchthrough

Breakdown phenomena

- punchthrough

Punchthrough

VCE=10V

VCE=5V

E-Bdepletionregion

C-Bdepletionregionn p n

VCE=30V

VCE=40V

E-B and C-Bdepletion regionsmerged together

* As the reverse bias VCB is increased,

the C-B depletion region expands

→ the neutral base region shrinks.

* At some point, the E-B and C-B

depletion regions consume the entire

base region. This condition is called

punchthrough.

(The band bending in the emitter

region is due to non-uniform doping in

the simulated structure.)

Punchthrough

VCE=10V

VCE=5V

E-Bdepletionregion

VCE=30V

VCE=40V

punchthrough.

Punchthrough

VCE=10V

VCE=5V

E-Bdepletionregion

VCE=30V

VCE=40V

punchthrough.

Punchthrough

VCE=10V

VCE=5V

E-Bdepletionregion

VCE=30V

VCE=40V

punchthrough.

Punchthrough

VCE=10V

VCE=5V

E-Bdepletionregion

VCE=30V

VCE=40V

punchthrough.

Punchthrough

VCE=30V

VCE=40VVCE=10V

VCE=5V

E-Bdepletionregion

* Prior to punchthrough, an increase in

the C-B reverse bias only affects the

bands in the base and collector

regions, leaving the E-B barrier (for

electron flow) unchanged.

* After punchthrough, any further

increase in VCB lowers the E-B

potential barrier. The number of

electrons injected from the emitter

increases dramatically. They get swept

away toward the collector, resulting in

a large collector current.

Punchthrough

VCE=30V

VCE=40VVCE=10V

VCE=5V

E-Bdepletionregion

Punchthrough

VCE=30V

VCE=40VVCE=10V

VCE=5V

E-Bdepletionregion

Avalanche breakdown

E pn n

* Avalanche multiplication because of impact ionisation can take place in a semiconductor

if the electric field is high (∼ critical field Ec ).

* In a BJT operating in the active mode, the C-B junction is under reverse bias. If the

reverse voltage is sufficiently large, avalanche breakdown can take place.

Avalanche breakdown

E pn n

Avalanche breakdown

E pn n

Avalanche breakdown

E pn n

VR (volts)

m=5m=4

050 1000

* The avalanche multiplication process is characterised by a multiplication factor M.

* Let I0 = current through the high-field region without multiplication

I = current through the high-field region with multiplication

Then, M =I

* Empirically, it is observed that M =1

)m , where 3 < m < 6 (depending on the semiconductor),

VR is the reverse bias, and V BR is the breakdown voltage.

Avalanche breakdown

E pn n

VR (volts)

m=5m=4

050 1000

Then, M =I

Avalanche breakdown

E pn n

VR (volts)

m=5m=4

050 1000

Then, M =I

Avalanche breakdown

E pn n

VR (volts)

m=5m=4

050 1000

Then, M =I

VR is the reverse bias, and V BR is the breakdown voltage.M. B. Patil, IIT Bombay

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

* Collector current without multiplication is αF I′E − I ′C .

* Collector current with multiplication is M(αF I′E − I ′C

), i.e.,

IC = M(αF I′E + ICS

)∵ I ′C ≈ −ICS

= M[αF

(IE + αR I

)+ ICS

]= M αF IE + M ICS (1− αFαR)

= M αF IE + M ICBO .

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

), i.e.,

)∵ I ′C ≈ −ICS

= M[αF

(IE + αR I

)+ ICS

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

), i.e.,

)∵ I ′C ≈ −ICS

= M[αF

(IE + αR I

)+ ICS

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

IC = M αF IE + M ICBO .

* Emitter open:

IC = MICBO = ICBO ×1

V BRBC

Breakdown voltage: As VR → V BRBC , IC →∞, and therefore

the breakdown voltage with the emitter open (denoted by

VCBO) is simply VCBO =V BRBC .

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Emitter open:

V BRBC

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Emitter open:

V BRBC

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Emitter open:

V BRBC

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Emitter open:

V BRBC

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

IC = M αF (IC + IB) + M ICBO

= M αF IC + M ICBO

→ IC (1−MαF ) = M ICBO → IC =M ICBO

1−MαF.

Breakdown condition: MαF → 1 or M → 1

V BRBC

αF=βF + 1

→ VR ≡ VCEO =V BRBC

(βF + 1)1/m≈ VCBO

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open:

= M αF IC + M ICBO

1−MαF.

V BRBC

αF=βF + 1

β1/mF

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open: VCEO =V BRBC

β1/mF

As an example, for βF = 200 and m = 4, VCEO =VCBO/3.8, which is

significantly smaller than VCBO .

Avalanche breakdown

IE IEIC

I′CαRI

αFI′E

E Cpn n

* Base open: VCEO =V BRBC

β1/mF

As an example, for βF = 200 and m = 4, VCEO =VCBO/3.8, which is

significantly smaller than VCBO .

IE= 1mA

∼ 0mA

without impact ionisation

with impact ionisation

VCB (volts)

0 50 100 150 200 250

IB = 10µA

∼ 0µA2µA

VCE (volts)

0 50 100

Significance of VCBO and VCEO :

* When IE = 0, the breakdown voltage (VCB) is given by VCBO .

In the above example, it is ∼ 230 V.

* When IB = 0, the breakdown voltage (VCE ) is given by VCEO .

In the above example, it is ∼ 90 V, which is significantly

lower, as we would expect.

* Note that the slope of the IC -VCE curves in the linear region

without impact ionisation is because of base width

modulation.

IE= 1mA

∼ 0mA

VCB (volts)

0 50 100 150 200 250

IB = 10µA

∼ 0µA2µA

VCE (volts)

0 50 100

modulation.

IE= 1mA

∼ 0mA

VCB (volts)

0 50 100 150 200 250

IB = 10µA

∼ 0µA2µA

VCE (volts)

0 50 100

modulation.

IE= 1mA

∼ 0mA

VCB (volts)

0 50 100 150 200 250

IB = 10µA

∼ 0µA2µA

VCE (volts)

0 50 100

modulation.

IE= 1mA

∼ 0mA

VCB (volts)

0 50 100 150 200 250

IB = 10µA

∼ 0µA2µA

VCE (volts)

0 50 100

modulation.

IE= 1mA

∼ 0mA

VCB (volts)

0 50 100 150 200 250

IB = 10µA

∼ 0µA2µA

VCE (volts)

0 50 100

modulation.

IB = 10µA

∼ 0µA2µA

IE= 1mA

∼ 0mA

VCE (volts)

VCB (volts)

with impact ionisationpn n

0 50 100 150 200 250

0 50 100

IC = M αF IE + M ICBO

=MαF IB

1−MαF+

1−MαF, with M =

V BRBC

* Common-base characteristics: For the same VCB , i.e., the

same multiplication factor, the increase in IC due to

multiplication is larger for larger IE .

* Common-emitter characteristics: For the same VCE (∼ VCB),

i.e., the same multiplication factor, the increase in IC due to

multiplication is larger for larger IB .

IB = 10µA

∼ 0µA2µA

IE= 1mA

∼ 0mA

VCE (volts)

VCB (volts)

0 50 100 150 200 250

0 50 100

=MαF IB

1−MαF+

1−MαF, with M =

V BRBC

)m* Common-base characteristics: For the same VCB , i.e., the

IB = 10µA

∼ 0µA2µA

IE= 1mA

∼ 0mA

VCE (volts)

VCB (volts)

0 50 100 150 200 250

0 50 100

=MαF IB

1−MαF+

1−MαF, with M =

V BRBC

)m* Common-base characteristics: For the same VCB , i.e., the

VCEO < VCBO : qualitative exaplnation

E Cpn n

* VCEO is sbustantially smaller than VCBO although, in both cases, the

breakdown is related to the same junction (the C-B junction). Why?

* With the emitter open, the breakdown process is really no different than an

isolated C-B junction.

* With the base open, the situation is different.

E Cpn n

Consider an electron undergoing impact ionisation with the base

* A hole generated by impact ionisation experiences an electric

field pointing toward the base, and it enters the neutral base

region.

* Since the base is open, the hole gets injected to the emitter

* The electron and hole components at the B-E junction are

related by

γ =InE

InE + IpE

→ InE

1− γ .

→ injection of one hole into the emitter region causes

injection ofγ

1− γ electrons from the emitter into the base

region.

→ multiplication of carriers is enhanced → lower breakdown

voltage [Pierret].

region.

related by

γ =InE

InE + IpE

→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE

→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE

→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE

→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE

→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

region.

related by

γ =InE

InE + IpE→ InE

1− γ .

injection ofγ

region.

voltage [Pierret].

A typical discrete transistor: 2N3904 (npn)

* VCEO = 40 V, VCBO = 60 V, VEBO = 6 V:

We have already seen why VCEO is smaller than VCBO .

VEBO , the E -B breakdown voltage is substantially lower because of the larger doping

density in the emitter region.

x (µm)0 1.5 3

(representative plot)1020

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

In the active or saturation modes, the E-B junction is under forward bias, and a low VEBO

is not a concern.

* Maximum collector current (continuous) ImaxC : 200 mA (DC).

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

is not a concern.

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

is not a concern.

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

is not a concern.

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

is not a concern.

x (µm)0 1.5 3

Dopingdensity(cm

Nd (E)Nd (C)

Na (B)

is not a concern.

* Maximum power dissipation PD = 625 mW:

The power dissipated by a BJT (as heat) is

P =VBE IB + VCE IC .

In the active mode, IC =βIB is much larger than IB .

→ P ≈VCE IC .

In the common-emitter output characteristics (IC -VCE ), the

constraint P =PD is therefore a hyperbola.

In designing a BJT amplifier, the DC bias values are subject

to the constraints due to ImaxC , VCEO , and PD .

→ P ≈VCE IC .

* DC current gain (βF ) = 100 to 300 at IC = 10 mA, VCE = 1 V:

- A range of values for βF is specified because of device-to-device

variation in the doping profiles and geometric parameters

(especially the base width).

- Since βF varies in practice with bias conditions, the specification

includes the bias values.

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

Ebers-moll model (active mode):

IC = αF IESeVBE/VT , IB = (1− αF ) IESe

VBE/VT .

* At lower values of VBE , the diffusion component of the E -B diode

current becomes small, and recombination in the emitter-depletion

region, which adds to the base current, cannot be neglected any

more. This causes IB to be larger than that predicted by the above

equation.

* At high values of VBE (large IC ),

- The voltage drop across the base resistance rb becomes

significant.

- The minority carrier concentration in the base becomes

comparable to the majority carrier concentration (high-level

injection)

As a result, βF =IC

IBis constant only for a range of IC values.

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

(substrate)

Collector

BaseEmitter

IC, IB (log scale)

logβF

IC2IC1 log IC

logβF

VBE/VT .

equation.

significant.

injection)

0.1 1 10010IC (mA)

VCE= 1V

(Note: βF varies from device to device.)

* V satCE = 0.2 V at IC = 10 mA, V sat

CE = 0.3 V at IC = 50 mA.

VsatCE

* Output conductance hoe = 1 to 40 µf at IC = 1 mA, VCE = 10 V:

The slope of the IC versus VCE curve at a constant IB is defined as the output

conductance hoe .

slope= hoe

IB= constant

From hoe , we can get an idea of the Early voltage VA of the device. For example, with

hoe = 10µf, IC1 = 1 mA, VCE1 = 10 V, we get

VA + VCE1= hoe → VA =

1× 10−3

10× 10−6− 10 = 90 V.

conductance hoe .

slope= hoe

IB= constant

1× 10−3

10× 10−6− 10 = 90 V.

conductance hoe .

slope= hoe

IB= constant

1× 10−3

10× 10−6− 10 = 90 V.

Semiconductor Devices - IIT Bombay

Transcript of Semiconductor Devices - IIT Bombay

Semiconductor Devices - IIT Bombay

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