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BIPOLAR JUNCTION TRANSISTORS AT LOWFREQUENCIES:

PRINCIPLES AND SOLVED PROBLEMS

The three inventors of the transistor: William Shockley, (seated), John Bardeen (left) and Walter Brattain (right) in 1948;

the three inventors shared the Nobel prize in 1956.(Courtesy of Bell Labs, Lucent Technologies.)

S. O. KasapDepartment of Electrical Engineering

University of SaskatchewanCANADA

An e-BookletOptimized for laser printing

1991-2001 S.O. Kasap

Bipolar Junction Transistors at Low Frequencies ( S.O. Kasap, 1991-2002: v.1.1) 2An e-Booklet

BIPOLAR JUNCTION TRANSISTORS AT LOWFREQUENCIES:

PRINCIPLES AND SOLVED PROBLEMS

"If current trends endure, future computers will consist of a single chip.No one will have the foggiest idea what is on it. Somewhere in thebasement of Intel or its successor will be a huge computer file withchip's listing. The last electrical engineer will sit nearby, handcuffed tothe disk drive in a scene out of Ben Hur. That engineer will beextremely well paid, and his or her every demand will be immediatelysatisfied. That engineer will be last keeper of the secret of theuniverse: E = IR."

Robert LuckySpectrum, May 1998 Issue, p.21

1. DC Characteristics of an npn BJT

An npn bipolar junction transistor (BJT) connected in the common base configuration is shown in Figure1-1. In the normal and active mode of operation, the base-emitter junction is forward biased and base-collector junction is reverse biased. Electrons from the emitter become injected into the base. Injectedelectrons in the base are minority carriers. These minority carriers (electrons) diffuse across the base andwhen they reach the collector junction depletion region they become drifted by the electric field in thisregion.

n+ p n

np(0)

IE

x

E B C

VCB

VEB

IC

IB

Input

np(x)

A

Output

A

pn(0)

pn(x')

SCL SCL

pn(x)

x'

Minority carrier concentration profiles in the emitter, base andcollector of an npn BJT. Depletion regions are marked as SCL(space charge layer). is the electric field in the collectorjunction SCL.

Figure 1-1


The minority carrier concentration profiles in the emitter, base and collector are shown in Figure 1-1. The minority carrier concentrations just outside the depletion regions, marked as SCL (space chargelayer), are determined by the law of the junction. For the minority carrier concentration at x = 0 in the basejust outside the SCL,

n neV

kTp poEB( ) exp0 =

where VEB is the forward bias voltage across the emitter-base junction and the other symbols have theirusual meanings. The emitter current IE(electron) due to electron diffusion in the base is determined by theminority carrier concentration profile at x = 0. If De is the electron diffusion coefficient in the base, then

I AeDdn

dx

eAD n

W

eAD n

W N

eV

kTE ep

x

e p

B

e i

B a

EB(electron) =

≈ =

=

±( )

exp0

20

or, I IeV

kTE soeEB

(electron) =

exp ; I

eAD n

N Wsoee i

a B=

2

where npo = ni2/Na and Na is the acceptor concentration in the base.

If the emitter width is much longer than the hole diffusion length Lh in the emitter, then the holecomponent of the emitter current is,

I IeV

kTE sohEB

(hole) =

exp ; I

eAD n

N Lsohh i

d h

2

Assuming that the recombination component in the emitter current is negligible (not entirely true),then the total emitter current IE = IE(electron) + IE(hole). The emitter injection efficiency γ is defined as thefraction of the emitter current due to minority carriers injected from the emitter into the base,

γ =+

=+

I

I I

I

I IE

E E

soe

soe soh

(electron)

(electron) (hole)

The collector current is due to electrons reaching the base-collector SCL and IC = αBIE(electron) where

αB is the base transport factor that accounts for some of the injected electrons recombining in the base. Ifthe hole component IE(hole) (minority carriers injected into the emitter) of the emitter current is negligible,then αB represents IC/IE and is the common base (CB) current gain (current transfer ratio). αB is given by,

α ττB

t

e

= − = −1 1Transit (diffusion) time across base

Minority carrier recombination time in base

where the base transit time τt is the time it takes for the minority carriers to diffuse across the neutral base

region. If W′B is the width of the neutral base region, then the diffusion time τt is,

τ tB

e

W

D= ′2

2

In practice, the emitter current is not totally due to electron injection into the base and γ is not

exactly unity. The effective CB current gain α is then given by

α = γαB.

The common emitter current gain, β = IC/IB, is given by


β αα

=−1

1.1. Example: Characteristics of an npn Si BJTConsider an idealized silicon npn bipolar transistor with the properties listed below in Table 1-1. The baseregion has a relatively uniform doping. The emitter and collector donor concentrations are mean values.The cross sectional area is 100 µm × 100 µm. The transistor is biased to operate in the normal activemode. The base-emitter forward bias voltage is 0.6 V and the reverse bias base-collector voltage is 18 V.

Table 1-1

Properties of an npn bipolar transistor.

Emitterwidth

Emitterdoping

Holelifetime inemitter

Base width Base doping Electronlifetime inbase

Collectordoping

10 µm 2 × 1018 cm-3 10 ns 4 µm 1 × 1016 cm-3 400 ns 1 × 1016 cm-3

a Calculate the depletion layer width extending from the collector into the base and also from theemitter into the base. What is the width of the neutral base region?

b Calculate α and hence β for this transistor assuming unity emitter injection efficiency. How do αand β change with VCB?

c What is the emitter injection efficiency and what are α and β taking into account the emitterinjection efficiency is not unity?

d What are the emitter, collector and base currents?

e What is the collector current when VCB = 19 V but VEB = 0.6 V? What is the incremental collectoroutput resistance, defined as ∆VCB/∆ΙC?


IE

E B C

VCB

VEB

IC

IB

Input

A

Output

A

WBCp WBCn

WEB

Electrondiffusion Drift

Holes

WB

W B

n+ p n

An npn transistor operated in the normal mode, in the active region, in thecommon base (CB) configuration (Notation: W = width, = electric field).

Figure 1-2

Solution

a Figure 1-2 shows the principle of operation of the npn BJT and also defines various devicecharacteristics such as the depletion widths and the base widths. With VBC >> Vo (built-in voltage) thereverse bias across the base-collector junction Vr = VBC + Vo ≈ VBC. Thus, the depletion layer WBC at thebase-collector junction is given by

WN N V

eN NBCo r a d BC

a d

= +

21 2

ε ε ( )/

i.e. WBC =2(8.854 10 F m )(11.9)(10 10 m )(18 V)

(1.6 10 C)(10 m )(10 m )

-12

-1 22 22

-3

19 22

-3 22

-3

1/2× +

×

−

i.e. WBC = 2.18 × 10-6 m or 2.18 µm

Only a portion of WBC is in the base side. Suppose that WBCp and WBCn are the depletion widths inthe base and collector sides of the SCL respectively. Since the total charge on the p and n-sides of the SCLmust be the same

NaWBCp = NdWBCn

and since WBC = WBCp + WBCn

we can find WBCp,

WN

N NWBCp

d

a dBC=

+=

+10

10 102 18

16

16 16 ( . )µm = 1.09


Since Nd(E) >> Na(B), the depletion layer width WEB is almost totally in the p-side (in the base).With forward bias, VEB = 0.6 V across the emitter-base junction, WEB is given by

WV V

eNEBo r o EB

a

= −

21 2

ε ε ( )/

We first need to calculate the built-in voltage Vo between the emitter and base,

VkT

e

N N

noa d

i

=

= × ××

ln ( . ) ln( )( )

( . )2

18 16

10 20 02592 10 1 10

1 5 10V

i.e. Vo = 0.830 V

Then, WV V

eNEBo r o EB

a

= −

21 2

ε ε ( )/

i.e. WEB =2(8.854 10 F m )(11.9)(0.830 ± 0.6 V)

(1.6 10 C)(10 m )

-12

-1

19 22

-3

1/2×

×

−

or WEB = 1.74 × 10-7 m or 0.174 µm

Notice that due to the forward bias across the EB junction, WEB is an order of magnitude smallerthan WBCp.

If WB is the base width between emitter and collector metallurgical junctions, then the width W′B ofthe neutral region in the base between the borders of the depletion regions is given by,

W′B = WB − WBCp − WEB

so that W′B = 4 µm − 1.09 µm − 0.174 µm = 2.74 µm

b The electron drift mobility µe in the base is determined by the dopant (acceptor) concentration here.

For Na = 1 × 1016 cm-3, µe = 1250 cm2 V-1 s-1 and the diffusion coefficient De from the Einsteinrelationship is,

De = kTµe/e = (0.02585 V)(1250 × 10-4 m2 V-1 s-1) = 3.23 × 10-3 m2 s-1

The electron diffusion length Le in the base is

Le = (Deτe)1/2 = [(3.23 × 10-3 m2 s-1)(400 × 10-9 s)]1/2

i.e. Le = 36.0 × 10-6 m (= 36.0 µm)

In order to calculate α we first need to find the transit (diffusion) time τt through the base

τ tB

e

W

D= ′ = ×

×

−

− −

2 6 2

322 74 10

2 3 23 10( . )

( . )m

m s2 1 = 1.161 × 10-9 s

If we assume unity injection (γ = 1), then α = αB, base transport factor,

α α ττ

= = − = −Bt

e

1 1Transit (diffusion) time across base


i.e. α = 1 − (1.161 ns)/(400 ns) = 0.99710


The current gain β is

β αα

=−

=−

=1

0 99711 09971

343.

.

c The hole drift mobility µh in the emitter is determined by the dopant (donor) concentration here.

For Nd = 2 × 1018 cm-3, µh ≈ 100 cm2 V-1 s-1 and the diffusion coefficient Dh from the Einstein relationshipis,

Dh = kTµh/e = (0.02585 V)(100 × 10-4 m2 V-1 s-1) = 2.59 × 10-4 m2 s-1

The hole diffusion length Lh in the emitter is

Lh = (Dhτh)1/2 = [(2.59 × 10-4 m2 s-1)(10 × 10-9 s)]1/2

Lh = 1.61 × 10-6 (= 1.61 µm)

Thus the hole diffusion length is much shorter than the emitter width.

The emitter current is given by electron diffusion in the base and hole diffusion in the emitter sothat

IE = IE(electron) + IE(hole)

where for electrons diffusing in the base,

I IeV

kTE soeEB

(electron) =

exp ; I

eAD n

N Wsoee i

a B

=2

and holes diffusing in the emitter,

I IeV

kTE sohEB

(hole) =

exp ; I

eAD n

N Lsohh i

d h

2

where we used Lh instead of WE because WE >> Lh (emitter width is 10 µm and the hole diffusion length is

1.61 µm).

Substituting the values we find,

Isoe = × × × ×× ×

− − − − −

− −( . )( )( . )( . )

)( . 1 601 10 1 10 3 23 10 1 5 10

1 10 274 10

19 8 2 3 2 1 16 2

22 6

C m m s m( m m)

3

3

i.e. Isoe = 4.267 × 10-14 A or 42.67 fA

and Isoh = × × × ×× ×

− − − − −

− −( . )( )( . )( . )

)( . 1 601 10 1 10 2 59 10 1 5 10

2 10 161 10

19 8 2 4 2 1 16 2

24 6

C m m s m( m m)

3

3

i.e. Isoh = 2.93 × 10-17 A or 0.0293 fA

The emitter injection efficiency is the fraction of the emitter current that is due to minority carriersinjected into the base; i.e., those that diffuse across the base towards the collector.

γ =+

=+

I

I I

I

I IE

E E

soe

soe soh

(electron)

(electron) (hole)


i.e. γ = ×× + ×

−

− −4 267 10

4 267 10 2 93 10

14

14 17

.. .

= 0.99931

The current gains, taking into account the emitter injection efficiency, are

α = γαB = (0.99931)(0.9971) = 0.99641

and β = α/(1−α)= 0.99641/(1−0.99641) = 278

d The emitter current with VEB = 0.6 V is

IE = (Isoe + Isoh)exp(eVEB/kT)

IE = (4.267 × 10-14 A + 2.93 × 10-17 A)exp(0.6/0.2585)

IE = 5.13 × 10-4 A or 0.513 mA

The collector current is determined by those minority carriers in the base that reach the collectorjunction. Only γIE of IE is injected into the base as minority carriers and only a fraction αB make it to thecollector,

IC = αBγIE = αIE = (0.99641)(0.513 mA) = 0.511 mA

The base current is given by,

IB = IC/β = (0.511 mA)/278 = 1.83 × 10-3 mA = 1.83 µA

Table 1-2

Characteristics of BJT with VCB = 18 and 19 V.

VCB W′B α β IE IC

18 V 2.74 µm 0.99641 278 0.513 mA 0.511 mA

19 V 2.71 µm 0.99649 283 0.517 mA 0.515 mA

e Suppose that we increase VCB to 19 V and repeat all the calculations above. We then find resultstabulated in Table 1-2. We can calculate the small signal collector incremental resistance from,

rV

IcCB

C

= = −−

∆∆

19 180 515 0 511

. .

V VmA mA

= 250 kΩ

We can also calculate the BJT characteristics using the hyperbolic expressions given in Problem6.9, p. 504, in Principles of Electrical Engineering Materials and Devices, S.O. Kasap (McGraw-Hill).

The base transport factor αB is given by,

α µµB

B

e

W

L≈

=

=sech sechmm

2 7436 0

0 99711. .

.

which is, within calculation errors, almost identical to the simplified theory.

The emitter current is given by


I IeV

kTE soeEB

(electron) =

exp ; I

eAD n

N L

W

Lsoee i

a e

B

e

=

2

coth

Isoe = × × × ×

× ×

− − − − −

− −

( . )( )( . )( . )

)( tanh.

1 601 10 1 10 3 23 10 1 5 10

1 10 36 102 7436

19 8 2 3 2 1 16 2

22 6

C m m s m

( m m)m

m

3

3 µµ

i.e. Isoe = 4.260 × 10-14 A or 42.60 fA

which, for all practical purposes, is identical to the simplified theory. NOTE: We have ignored any bandgap narrowing in the emitter due to heavy doping in this region.

1.2. Example: Characteristics of a pnp Si BJT

Consider an idealized Si pnp bipolar transistor with the properties listed below in Table 1-3. The baseregion has a relatively uniform doping. The emitter and collector donor concentrations are mean values.The effective cross-sectional area is 0.01 mm2 (a square area of side 100 µm).

Table 1-3

Properties of an pnp bipolar transistor.

Emitterwidth

Emitterdoping

Electronlifetime inemitter

Base width(junction-junction)

Base doping Holelifetime inbase

Collectordoping

1 µm 1 × 1018 cm-3 ~25 ns 2 µm 1 × 1016 cm-3 ~500 ns 1 × 1015 cm-3

a Calculate current gains α and β for this transistor in the absence of any applied bias voltages, buttaking into account the emitter injection efficiency.

b Calculate α and β for this transistor when VEB = 0.6 V and VBC = 6 V such that the emitter-basejunction is forward biased and the base-collector junction is reverse biased (normal active mode ofoperation). What is your conclusion?

c Suppose that the transistor is biased, for example, in the common emitter configuration, with a dcbase current IB of 10 µA. What are the collector and emitter currents? What is the emitter-base voltage?

Solution

Given a Si transistor, we have ni = 1.5 × 1016 m-3, and the following dopant concentrations, NE = 1 × 1018

cm-3 in the emitter; NB = 1 × 1016 cm-3 in the base; NC = 1 × 1015 cm-3 in the collector. The minority carrier

lifetimes in the base and emitter respectively are τB = 10-6 s (holes in the base) and τE = 500 × 10-9 s(electrons in the emitter).

a No bias voltages

The built-in voltage across the base-collector (BC) junction is


VkT

e

N N

nOBCB C

i

=

= × ××

ln ( . ) ln( )( )

( . )2

16 15

10 20 02591 10 1 10

1 5 10V

i.e. VOBC = 0.634 V

The built-in voltage across the emitter-base (EB) is

VkT

e

N N

nOEBE B

i

=

= × ××

ln ( . ) ln( )( )

( . )2

18 16

10 20 02591 10 1 10

1 5 10V

i.e. VOEB = 0.812 V

The depletion layer WBC at the base-collector junction is given by;

WN N V

eN NBCo r B C OBC

B C

≈

+21 2

ε ε ( )/

i.e. WBC =2(8.854 10 F m )(11.9)(10 10 m )(0.634 V)

(1.6 10 C)(10 m )(10 m )

-12

-1 22 21

-3

19 22

-3 21

-3

1/2× +

×

−

giving, WBC = 0.958 × 10-6 m or 0.958 µm

Only a portion of WBC is in the base side (n-side) which is given by

WN

N NWBCn

C

B CBC=

+=

+10

10 100 958

15

16 15 ( . )µm = 0.0871 µm

Since NE >> NB, the depletion layer width WEB is almost totally in the base. With no bias, WEB isgiven by

WV

eNEBo r OEB

B

=

21 2

ε ε/

i.e. WEB =2(8.854 10 F m )(11.9)(0.812 V)

(1.6 10 C)(10 m )

-12

-1

19 22

-3

1/2×

×

−

giving WEB = 0.329 µm

If WB is the base width between the emitter and collector metallurgical junctions, then the widthW′B of the neutral base region between the depletion regions is given by,

W′B = WB − WBCn − WEB

so that W′B = 2 − 0.0871 − 0.329 µm = 1.584 µm

Electron and hole drift mobilities in the base and emitter are determined by the dopant

concentrations. For NB = 1 × 1016 cm-3, µh = 400 cm2 V-1 s-1 and the minority carrier diffusion coefficientDB in the base, from the Einstein relationship, is

DB = kTµh/e = (0.02585 V)(400 × 10-4 m2 V-1 s-1) = 1.034 × 10-3 m2 s-1

The minority carrier (hole) diffusion length LB in the base is

LB = (DBτB)1/2 = [(1.034 × 10-3 m2 s-1)(500 × 10-9 s)]1/2


i.e. LB = 22.74 × 10-6 m (= 22.74 µm)

In order to calculate α, first we need to find the transit (diffusion) time τt through the base

τ tB

B

W

D= ′ = ×

×

−

− −

2 6 2

321 584 10

2 1034 10( . )

( . )m

m s2 1 = 1.213 × 10-9 s or 1.213 ns

The base transport factor αB is then

α ττB

t

B

= − = −1 1Transit (diffusion) time across base


i.e. αB = 1 − (1.213 ns)/(500 ns) = 0.99757

The electron drift mobility µe in the emitter is determined by the dopant (acceptor) concentration

here. For NE = 1 × 1018 cm-3, µe ≈ 250 cm2 V-1 s-1 and the minority carrier diffusion coefficient DE in theemitter, from the Einstein relationship is,

DE = kTµe/e = (0.02585 V)(250 × 10-4 m2 V-1 s-1) = 6.46 × 10-4 m2 s-1

The minority carrier (electron) diffusion length LE in the emitter is

LE = (DEτE)1/2 = [(6.46 × 10-4 m2 s-1)(25 × 10-9 s)]1/2

i.e. LE = 4.02 × 10-6 (= 4.02 µm)

Thus the minority carrier diffusion length LE is "much" longer than the emitter width WE (= 1 µm)and we have to use the short diode equation for the minority carrier diffusion current in the emitter.

The emitter current is given by minority carrier diffusions in the base and emitter (holes diffusingin the base and electrons diffusing in the emitter) so that

IE = IE(hole) + IE(electron)

where for holes diffusing in the base,

I IeV

kTE sohEB

(hole) =

exp ; I

eAD n

N WsohB i

B B

=′

2

and electrons diffusing in the emitter,

I IeV

kTE soeEB

(electron) =

exp ; I

eAD n

N WsoeE i

E E

=2

where we used WE instead of Lh because Lh > WE (emitter width is 1 µm and electron diffusion length is

4.02 µm).

The emitter injection efficiency is the fraction of the emitter current that is due to minority carriersinjected into the base; those that diffuse across the base towards the collector.

γ =+

=+

I

I I

I

I IE

E E

soh

soh soe

(hole)

(hole) (electron)


or γ =+

=

+

′

=+ ′

1

1

1

1

1

12

2

I

IeAD n

N WeAD n

N W

D N W

D N Wsoe

soh

E i

E E

B i

B B

E B B

B E E

i.e. γ =+ × × ×

× × ×

− − − −

− − − −

1

16 46 10 1 10 1 584 10

1 03 10 1 10 1 10

4 2 1 22 6

3 2 1 24 6

( . ) )( . ( . ) )(

m s ( m m) m s ( m m)

3

3

thus, γ = 0.99016

The emitter-to-collector current gain, taking into account the emitter injection efficiency, is

α = γαB = (0.99016)(0.99757) = 0.9878

and

β = α/(1−α) = 0.9878/(1−0.9878) = 81

b With bias voltages

When the transistor has bias voltages applied, the depletion region widths change, which alters the widthof the neutral base region. The built-in voltages stay the same.

The depletion layer WBC at the base-collector junction with a reverse bias is given by,

WN N V V

eN NBCo r B C OBC BC

B C

= + +

21 2

ε ε ( )( )/

i.e. WBC =2(8.854 10 F m )(11.9)(10 10 m )(0.634 V + 6 V)

(1.6 10 C)(10 m )(10 m )

-12

-1 22 21

-3

19 22

-3 21

-3

1/2× +

×

−

WBC = 3.098 µm

Only a portion of WBC is in the base side (n-side) which is given by

WN

N NWBCn

C

B CBC=

+=

+10

10 103 098

15

16 15 ( . )µm = 0.282 µm

Since NE >> NB, the depletion layer width WEB is almost totally in the base. With forward bias, WEBis given by

WV V

eNEBOEB EB

B

= −

××

−

21 2

ε( )/

=2(8.854 10 F m )(11.9)(0.812 ± 0.6 V)

(1.6 10 C)(10 m )

-12

-1

19 22

-3

1/2

or WEB = 0.168 µm

Thus, W′B = WB − WBCn − WEB

i.e. W′B = 2 µm − 0.282 µm − 0.168 µm = 1.55 µm

New transit (diffusion) time τt through the base


τ tB

B

W

D= ′ = ×

×

−

− −

2 6 2

321 55 10

2 1034 10( . )

( . )mm s2 1 = 1.162 × 10-9 s or 1.16 ns

New base transport factor αB is then

αB = 1 − (τ t /τB) = 1 − (1.16 ns)/(500 ns) = 0.99768

New emitter injection efficiency is

γ =+ ′ =

+ × × ×× × ×

− − − −

− − − −

1

1

1

16 46 10 1 10 1 55 10

1 03 10 1 10 1 10

4 2 1 22 6

3 2 1 24 6

D N W

D N WE B B

B E E

( . ) )( . ( . ) )(

m s ( m m) m s ( m m)

3

3

thus, γ = 0.99039

New emitter-to-collector current gain is

α = γαB = (0.99039)(0.99768) = 0.9881

and β = α/(1−α) = 0.9881/(1−0.9881) = 83.0

The change in the current gain β is only ~2.5%.

c BJT biased with IB = 10 µA

If the base current is 10 µA, then the collector current is given by

IC = βIB ≈ 83(0.01 mA) = 0.83 mA

The emitter current due to holes diffusing in the base is

IE(hole) = IC + IB = 0.83 mA + 0.01 mA = 0.84 mA

This is the hole component of the emitter current, and the total emitter current is

IE = IE(hole)/γ = (0.84 mA)/(0.99039) = 0.848 mA

We can find the VEB as follows. The emitter current due to hole diffusion in the base is

I IeV

kTE sohEB

(hole) =

exp ; I

eAD n

N WsohB i

B B

=′

2

where Isoh = × × × ×× ×

− − − − −

− −( . )( . )( . )( . )

)( . 1 601 10 0 01 10 1 03 10 1 5 10

1 10 155 10

19 6 2 3 2 1 16 2

22 6

C m m s m( m m)

3

3

i.e. Isoh = 2.40 × 10-14 A

Then VkT

e

I

IEBE

soh

=

= ×

×

−

−ln ( . ) ln. .

(hole) VAA

0 025850 84 102 40 10

3

14

i.e. VEB = 0.628 V

Note: The fact that we used β for VEB = 0.6 V does not really matter because β does not changesignificantly with bias (in this transistor) and the emitter current is in the logarithm of the expression forVEB. Thus, when IB = 10 µA, VEB = 0.63 V.


2. Nonuniform Base Doping in the BJT

x

1018

1016

1020

n+ np

E B C0 2 µm

1014

EB

E B1 CB2

(a) Typical net dopant concentration profilein an npn bipolar transistor.

(b) The non-uniform acceptor doping leadsto a built-in field EB in the base.

As a consequence of the fabrication process, the base is non-uniformly doped.

WB

Net Dopant Concentration (cm-3)

Acceptor ion h+

Figure 2-1

Due to the diffusion processes used in the fabrication process, the base region of a BJT normally has anon-uniform dopant concentration as shown in Figure 2-1 for an npn BJT. There is a greater acceptorconcentration in the base at the emitter side, B1, than at the collector side B2. This leads to an initial netdiffusion of holes from B1 to B2 which exposes negatively charged acceptor ions around B1 andaccumulates excess holes around B2. An internal field EB therefore builds up until an equilibrium isreached and no further holes can diffuse from B1 to B2 because the built-in field EB prevents furtherdiffusion just as in the case of the pn junction under open circuit conditions. Suppose that the net acceptorconcentration, Na(x), in the base can be approximated by

Na(x) = Noexp(−x/b)

from x = 0 to x = WB as shown in Figure 2-1, where x in this equation is measured from around B1towards B2, and b is a parameter that characterizes the nonuniform doping profile. The parameter bdepends on the fabrication process. Suppose that the hole concentration approximately follows the dopingconcentration, that is p(x) ≈ Na(x). Then, under open circuit conditions, the net current in the base due toholes is zero, that is

J ep eD

dp

dxhole h h= − =µ EB 0

We can now substitute p(x) = Na(x) = Noexp(−x/b) and use the Einstein relation Dh/µh = kT/e tofind,

EB = kT

eb

This is the built-in field in the base due to non-uniform doping. When this npn transistor isoperating under normal and active conditions, the electrons injected from the emitter into the base are


drifted by the built-in field to the collector. This effectively shortens the transit time across the base. Thus,the built-in field speeds up the transit of electrons through the base and improves the gain and thefrequency response. If the drift time due to EB is shorter than the diffusion time, then the minority carriertransit time across the base is given by,

τ

µtBW≈

e BE

2.1. Example: Nonuniform base doping and transit timea Consider a BJT with a nonuniform base doping as in Figure 2-1. Suppose that the base dopantconcentration in Figure 2-1 at B1, x =0, is 1017 cm-3 and at B2, x = WB, Na = 1015 cm-3. Calculate b and thebuilt-in field EB.

b Taking WB ≈ 2 µm, Calculate drift transit time of electrons across the base due to the built-in field.How does this compare with the transit time due to diffusion (use a geometric average acceptorconcentration to find µe)? What is your conclusion?

Solution

a Given two dopant concentrations at two locations and two unknowns, b and No, we have

1017 = Noexp(−0/b) and 1015 = Noexp(−2/b)

Thus No = 1017 cm-3 and,

b = −××

2

1 101 10

15

17ln

= 0.43 µm

The built-in field is

EB = = ×

×

−

−kT

eb

258 104 3 10

3

5

.

Vm

= 600 V m-1

b The geometric mean doping concentration is 1016 cm-3 and at this doping level µe ≈ 1300 cm2 V-1

s-1. Taking WB ≈ 2 µm, drift time of electrons across the base is

τ

µtBW

(drift)

-6

-4 2 -1 -1 -1

2 10 m(1300 10 m V s )(600 V m )

≈ = ××e BE

( )= 2.59 × 10-10 s or 0.259 ns

The diffusion time across the base is

τ µtB

e

B

e

W

D

WkT

e

(diffusion)

-6

2 -1 -1 -4

2 10 m2(1300 m V s 10 0 V)

= =

= ×× ×

2 2 2

2 2 02585( )

.

= 5.95 × 10-10 s or 0.595 ns

Thus, the drift time is about half the diffusion time. We would expect the base transport to be morethan doubled with such a nonuniform base doping.

3. Bandgap Narrowing and Emitter Injection Efficiency

Heavy doping in semiconductors leads to what is called bandgap narrowing, which is an effectivenarrowing of the bandgap Eg [H.P.D. Lanyon and R.A. Tuft, "Bandgap Narrowing in Heavily Doped


Silicon”, IEEE Trans. Electron Devices, ED-26, 1014 (1979)]. The model is actually quite complicated.It is a direct consequence of the electrostatic screening action of a large number of majority carriers onminority carriers. If ∆Eg is the reduction in the energy gap then for an n-type semiconductor, according toLanyon and Tuft,

∆En

g( ) ./

in meV =

22 5

1018

1 2

where n (in cm-3) is the concentration of majority carriers which is equal to the dopant concentration if theyare all ionized (for example, at room temperature). The new effective intrinsic concentration nieff due to thereduced bandgap is given by

n N NE E

kTieff c vg g2 = −

−

exp( )∆

i.e. n nE

kTieff ig2 2=

exp∆

where ni is the intrinsic concentration, that in the absence of emitter bandgap narrowing.

The equilibrium electron and hole concentrations, nno and pno respectively, then obey

n p nno no ieff= 2

where nno = Nd since all donors would be ionized at room temperature.

3.1. Example: Emitter injection efficiency

Consider a Si npn bipolar transistor with narrow emitter and base regions. The emitter region WE is ofthickness 1 µm and has a donor doping of 1019 cm-3. The width WB of the base is 1 µm and has anacceptor doping of 1017 cm-3.

a Obtain an expression for the emitter injection efficiency taking into account the emitter bandgapnarrowing effect above.

b Calculate the emitter injection efficiency with and without the emitter bandgap narrowing.

c Calculate the common emitter current gain β with and without the emitter bandgap narrowing effect

given perfect base transport factor (αB = 1).

Solution

a Assuming thin base and emitter regions ,WB << Le (electron diffusion length in base) and WE <<Lh (hole diffusion length in the emitter), we can use the short diode equations for the electron and holecomponents of the emitter current. Suppose that WE and WB are the widths of the neutral regions (outsidethe depletion regions). Then,

γ =+

=+

I

I I

I

I IE

E E

soe

soe soh

(electron)

(electron) (hole)

or γ =+

=

+

=+

1

1

1

1

1

12

2

2

2

I

IeAD n

N W

eAD n

N W

D N W n

D N W n

soh

soe

h ieff

d E

e i

a B

h a B ieff

e d E i

(emitter)

(emitter)

(base)

(base)

(emitter) (base)

(base) (emitter)


i.e. γµµ

=+

1

1 h a B

e d E

gN W

N W

E

kT(emitter) (base)

(base) (emitter)

exp∆

b Given Nd(emitter) ≈ 1 × 1019 cm-3 we can find the hole (minority carrier) drift mobility from the µh vs.

dopant concentration graph, which gives µh(emitter) ≈ 50 cm2 V-1 s-1. Further, in the emitter, n = Nd(emitter) =1019 cm-3 so that,

∆En

g( ) . ./ /

in meV =

=

22 510

22 5101018

1 2 19

18

1 2

= 71.1 meV

In the base Na(base) ≈ 1 × 1017 cm-3 and the minority carrier drift mobility µe(base) ≈ 900 cm2 V-1 s-1.

Emitter injection efficiency without bandgap narrowing is

γµµ

=+

=+

− − −

− − −

1

1

1

150 10 1900 10 1

2 1 1 17 3

2 1 1 19 3h a B

e d E

N W

N W(emitter) (base)

(base) (emitter)

cm V s cm mcm V s cm m

( )( )( )( )( )( )

µµ

γ = 0.99945

Emitter injection efficiency γ′ with bandgap narrowing is

′ =+

=+

γµµ

1

1

1

150 10 1900 10 1

0 07110 02585

17

19h a B

e d E

gN W

N W

E

kT(emitter) (base)

(base) (emitter)

exp( )( )( )( )( )( )

exp..

∆

γ′ = 0.99138

c Assuming perfect base transport, i.e. the base transport factor αB = 1, the corresponding commonemitter current gain without bandgap narrowing is

β γγ

=+

=+10 99945

1 099945.

.= 1800

and with bandgap narrowing this is,

′ = ′+ ′

=+

β γγ1

0 991381 099138

..

= 115

Clearly the emitter bandgap narrowing has a substantial effect on the transistor current gain.


4. Small Signal Low Frequency Equivalent Circuit of a BJT

np(0)

EIB + ib x

E

B

C

VCC

VBB

RC

vbe(t)np(x) n′

p(0)

IC

IC + ic

IE + ie

Input

Output

vce(t)Electron

diffusion

An npn transistor operated in the active region in the common emitteramplifier configuration. The applied signal vbe modulates the dc voltageacross the BE junction and hence modulates the injected electronconcentration up and down about the dc value np(0). The solid line showsnp(x) when only the dc bias VBB is present. The dashed lines show how np(x)is modulated up by a positive small signal vbe superimposed on VBB. Thecollector current is in the opposite direction to electron diffusion.

Figure 4-1

The npn bipolar transistor in the CE (common emitter) amplifier configuration is shown in Figure 4-1. Theinput circuit has a dc bias VBB to forward bias the base-emitter BE junction and the output circuit has a dcvoltage VCC (larger than VBE) to reverse bias the base-collector BC junction through a collector resistor RC.The actual reverse bias voltage across the BC junction is VCE − VBE where VCE is,

VCE = VCC − ICRC (1)

An input signal in the form of a small ac signal vbe is applied in series with the bias voltage VBE andmodulates the forward voltage across the BE junction about its dc value, VBE. The varying voltage acrossthe BE modulates np(0) up and down about its dc value which leads to a varying emitter current and henceto an almost identically varying collector current in the output circuit. The variation in the collector currentis converted to an output voltage signal by the collector resistance RC.

Since the BE junction is forward biased the relationship between IE and VBE is exponential,

I IeV

kTE EOBE=

exp (2)

where IEO is a constant. The collector current is approximately the same as the emitter current, IC ≈ IE.When the input voltage vbe(t) increases, so does VBE (= VBB + vbe) and hence the collector current IC. Note


however that when IC increases, VCE, according to Equation (1), actually decreases so that the voltagechanges at the output are 180° out of phase with the input voltage changes.

We can differentiate Equation (2) to relate small variations in IE and VBE as in the presence of smallsignals superimposed on dc values. For small signals, we have vbe = δVBE, ib = δIB, ie = δIE, ic = δIC.

Inasmuch as IC = βIB, we have δIC = β δIB so that ic = β ib. Since α ≈ 1, ie ≈ ic.

What is advantage of the CE circuit over the common base (CB) configuration? First, the inputcurrent is the base current which is about a factor of β smaller than the emitter current. The ac input

resistance of the CE circuit is therefore a factor of β higher than that of the CB circuit. This means that theamplifier does not load the ac source; the input resistance of the amplifier is much greater than the internal(or output) resistance of the ac source at the input. The small signal input resistance rbe is

rv

i

V

I

V

I

kT

eI Ibebe

b

BE

B

BE

E E C

= = ≈ = ≈( )

δδ

β δδ

β β25mA

where we differentiated Equation (2).

The output ac signal vce develops across CE and is tapped out through a capacitor. As IC increases,VCE decreases as apparent in Equation (1). This means vce and ic are 180° out of phase.

vce = δVce = −RCδIC = −RCic

The voltage amplification is

Av

v

R i

r i

R

r

R IV

ce

be

C c

be b

C

be

C C= = − = − ≈ − ( )β mA25

which is the same as that in the CB configuration. However, in the CE configuration the output to inputcurrent ratio, ic/ib = β whereas this is almost unity in the CB configuration. Consequently, the CEconfiguration provides a greater power amplification, which is the second advantage of the CE circuit.

The input signal vbe gives rise to an output current ic. This input voltage to output currentconversion is defined into a parameter called the mutual conductance, or transconductance, gm.

gmc

be

E

BE

E

e

i

v

I

V

I

r= ≈ = ( ) =δ

δmA25

1

The voltage amplification of the CE amplifier is then

AV = −gmRC.

We generally find it convenient to use a small signal equivalent circuit for the low-frequencybehavior of a BJT in the CE configuration. Between the base and emitter, the applied ac source voltage vssees only an input resistance of rbe as shown in Figure 4-2. To underline the importance of the inputresistance, the output (or the internal) resistance of the ac source is also shown. In the output circuit thereis a voltage controlled current source ic which generates a current of gmvbe. The current ic passes throughthe load (or collector) resistance RC across which the voltage signal develops. As we are only interested inac signals, the batteries are taken as a short circuit path for the ac current which means that the internalresistances of the batteries are taken as zero. This model, of course, is only valid under normal and activeoperating conditions and small signals about dc values, and at low frequencies.

The bipolar transistor general dc current equation, IC = βIB, where β ≈ τh/τt is a material dependent

constant, implies that the ac small signal collector current is δIC = β δIB or ic = βib. Thus the CE dc and ac


small signal current gains are the same. This is a reasonable approximation in the low frequency range,typically at frequencies below 1/τh. It is useful to have a relationship between β, gm and rbe. Using theequations above,

β = gmrbe

In transistor data books, the dc current gain, IC/IB, is denoted as hFE, whereas the small signal accurrent gain, ic/ib, is denoted as hfe. Except at high frequencies, hfe is approximately equal to hFE.

ib

vs

rbe

B

E E

C

vcevbe

Rs

S

S

vinic=gmvbe RC

Small signal equivalent circuitAC source Load

Low frequency small signal simplified equivalent circuit of thebipolar transistor in the CE configuration with a load resistor RCin the collector circuit.

Figure 4-2

4.1. Example: A CE amplifier using a pnp Si BJT

Consider an idealized Si pnp bipolar transistor with a β = 100 used in a simple CE amplifier circuit.

Assume that β does not change with the collector current (or bias voltages). The transistor is biased with a

dc base current IB of 10 µA.

a Calculate the small signal equivalent circuit parameters rbe and gm of this transistor in the commonemitter configuration.

b What should be the collector circuit resistance RC if a small signal voltage gain of 200 is required?What should be the battery voltage VCC ?

What is the voltage gain if RC is as calculated in b but the output resistance Rs of the ac supply is 50 Ω?

c If the ac source output voltage, without being connected to any device, is 1 mV peak-to-peak,calculate the input and output power and power gain of this CE amplifier circuit?

Solution

a If the base current is 10 µA, then the collector current is given by

IC = βIB ≈ 100(0.01 mA) = 1 mA

The base-emitter small signal resistance is given by


rkT

eI IbeC C

≈ ≈ =β β25 25 1001( )

( )mA

= 2500 Ω

The mutual conductance (transconductance) is

gm

C CeI

kT

I≈ ≈ =(mA)25

125

= 0.04 A/V

b The magnitude of the voltage gain is given by,

AV = gmRC

Thus a gain of 200 requires,

R

A

VCV

m

≈ =g

2000 04. /A

= 5000 Ω

The dc voltage across RC is ICRC = (5000 Ω)(1 × 10-3 A) = 5 V. The battery voltage has to besubstantially larger than this to allow the required reverse bias to appear across the base-collector junction.

c When the ac source is connected to the B and E terminals, the input resistance rbe of the BJT loadsthe ac source so that vbe across BE is,

v vr

r Rbe sbe

be s

=+( )

The actual voltage gain, or the effective gain, is

Av

v

R v

vR

r

r RVce

s

m C be

sm C

be

be s( )effective = = =

+( )g

g

i.e. AV( ) ( )effective =+( )

2002500

2500 50= 196

The loading effect reduces the gain of the amplifier. To diminish the loading of the ac source, weneed to increase rbe, i.e. reduce the collector current, but that also reduces the gain. So to keep the gain thesame, we need to reduce IC and increase RC. However, RC cannot be increased indefinitely because RCitself is loaded by the input of the next stage and, in addition, there is an incremental resistance between thecollector and emitter terminals (usually ~ 100 kΩ) that shunts RC.

d Given vs(p-p) = 1 mV, vs(rms) = 1 mV/(2√2) = 0.3536 mV.

v vr

r Rbe sbe

be s( ) ( ) ( . rms rms mV)=

+( ) =+( )

0 35362500

2500 50= 0.3466 mV

The input power into the BE terminals is

P v iv

rin be bbe

be

= = =×( )−

(rms) (rms)(rms) V2 3 2

0 3466 10

2500

.

Ω = 4.80 × 10-11 W

The output power is

P v iv

Rin ce cce

C

= = =× ×( )−

(rms) (rms)(rms) V2 3 2

200 0 3466 10

5000

.

Ω = 9.61 × 10-7 W

The power amplification is


AP

PPout

in

= = ××

−

−9 61 104 80 10

7

11

..

W W

= 2.0 × 104

Note that the power amplification of the CE BJT is,

Ai v

i vAP

c ce

b beV= = β

so that, AP = (100)(200) = 20000

NOTATION

A area; cross sectional area; amplificationAV, AP voltage amplification, power amplificationB base terminal of a BJTBC base-collectorBE base-emitterBJT bipolar junction transistorC capacitance (F), collector terminal of a BJTCB common baseCE common emitterD diffusion coefficient (m2 s-1)E emitter terminal of a BJTEg band gap energyE electric field (V m-1)e electronic charge (1.60218 × 10-19 C)e (subscript) electron, e.g. µe = electron drift

mobilityEB emitter baseEHP electron hole pair

gm mutual transconductance (A/V)

h (subscript) hole, e.g. µh = hole drift mobilityhFE dc current gainhfe small signal ac current gainHF high frequencyI electric current (A)IB, IC, IE base, collector and emitter currents in a BJTJ current density (A m-2)k Boltzmann constant (k = R/NA = 1.3807 × 10-23

J K-1)

Le , Lh electron and hole diffusion lengths

ln, lp lengths of the neutral n and p regions outsidedepletion region in a pn junction

Na, Nd acceptor and donor concentrations (m-3)ni intrinsic concentration (m-3)nno, ppo equilibrium majority carrier concentrationsnpo, pno equilibrium minority carrier concentrationsnp(0) minority carrier concentration in the base just

outside the emitter-base SCLrbe small signal input resistance (Ω)T temperature (K)SCL space charge layerV voltageVo built-in voltagevbe small signal at base with respect to the emittervce small signal at collector with respect to the

emitterW widthα gain or current transfer ratio from emitter to

collector of a BJTβ current gain IC/IB of a BJTγ emitter injection efficiencyεo permittivity of free space or absolute permittivity

(8.8542×10-12 C V-1 m-1 or F m-1)εr relative permittivity or dielectric constantµh , µe hole drift mobility, electron drift mobility (m2

V-1 s-1)

τ t minority carrier diffusion time across the base

USEFUL DEFINITIONS

Acceptor atoms are dopants that have one less valency than the host atom. They therefore accept electrons from thevalence band (VB) and thereby create holes in the VB which leads to p > n and hence to a p-type semiconductor.

Bipolar junction transistor (BJT) is a transistor whose normal operation is based on the injection of minority carriersfrom the emitter into the base region and their diffusion to the collector where they give rise to a collector current.The voltage between the base and the emitter controls the collector current; this is the transistor action.

Collector junction is the metallurgical junction between the base and the collector of a bipolar transistor.


Depletion layer (or space charge layer, SCL) is a region around the metallurgical junction where recombination ofelectrons and holes has depleted this region of its large number of equilibrium majority carriers.

Donor atoms are dopants in the semiconductor that have a valency one more than the host atom. They therefore donateelectrons to the conduction band (CB) and thereby create electrons in the CB which leads to n > p and hence to an n-type semiconductor (n is the electron concentration in the CB and p is the hole concentration in the valence band(VB)).

Drift mobility is the drift velocity per unit applied field. If µd is the mobility then the defining equation is vd=µdE wherevd is the drift velocity and E is the electric field.

Drift velocity is the average velocity, over all the conduction electrons in the conductor, in the direction of an appliedelectrical force (F = −eE for electrons). In the absence of an applied field, all the electrons are moving aroundrandomly and the average velocity, over all the electrons, in any direction is zero. With an applied field, Ex, there is anet velocity per electron, vdx, in the opposite direction to the field where vdx depends on Ex via vdx=µdEx where µd isthe drift mobility.

Einstein relation relates the diffusion coefficient D of a given species of charge carriers to their drift mobility µ via D/µ =kT/q where q is the charge of the carrier, k is the Boltzmann constant and T is the temperature.

Emitter of a bipolar transistor is one of the two similarly doped (e.g. both n-type) regions that surrounds the oppositelydoped (p-type) base and injects minority carriers into the base when the emitter-base junction is forward biased.

Emitter injection efficiency is the fraction of the emitter current due to minority carriers injected from the emitter intothe base.

Emitter junction is the metallurgical junction between the emitter and the base of a bipolar junction transistor.

Forward bias is the application of an external voltage to a pn junction such that the positive terminal is connected to the p-side and negative to the n-side. The applied voltage reduces the built-in potential.

Hole (h+) is a missing electron in an electronic state that is in the valence band. Intuitively, it is a missing electron in abond between two neighboring atoms in the semiconductor crystal. The region around this “ruptured” bond is a netpositive charge of +e. It can drift in an applied field because an electron in a neighboring bond can tunnel into thisvacant site and thereby cause the positively charged bond-vacancy to become displaced, shifted. Thus holes contributeto electrical conduction in semiconductors as well. In a full valence band there is no net contribution to the current.There are equal number of electrons (e.g. at b and b') with opposite momenta. If there is an empty state, hole, at b atthe top of the valence band (VB) then the electron at b' contributes to the current. The reason that the presence of ahole makes conduction possible is the fact that the momenta of all the VB electrons are canceled except that at b’.Thus, we can consider the net result of the motions of all the electrons in the VB just by examining the behavior ofthe missing electron at b’ and assigning to it a positive charge +e and an effective mass mh*.

Law of the junction relates the injected minority carrier concentration just outside the depletion layer to the appliedvoltage. For holes in the n-side, it is pn(0) = pnoexp[eV/(kT)] where pn(0) is the hole concentration just outside thedepletion layer.

Minority carrier diffusion length (L) is the mean distance a minority carrier diffuses before recombination, L = √[Dτ]where D is the diffusion coefficient and τ is the minority carrier lifetime.

Minority carriers are electrons in a p-type and holes in an n-type semiconductor.

Reverse bias is the application of an external voltage to a pn junction such that the positive terminal is connected to the n-side and negative to the p-side. The applied voltage increases the built-in potential and hence the internal field in thespace charge layer.

Semiconductor is a nonmetallic element (e.g. Si or Ge) that contains both electrons and holes as charge carriers in contrastto an enormous number of electrons only as in metals. A hole is essentially a "half-broken" covalent bond which hasa missing electron and therefore behaves effectively as if positively charged. Under the action of an applied field thehole can move by accepting an electron from a neighboring bond thereby passing on the "hole". Electron and holeconcentrations in a semiconductor are generally many orders of magnitude less than those in metals thus leading tomuch smaller conductivities.

Small signal equivalent circuit of a transistor replaces the transistor with an equivalent circuit that consists ofresistances, capacitances and dependent sources (current or voltage). The equivalent circuit represents the transistorbehavior under small signal ac conditions. The batteries are replaced with short circuits (or their internal resistances).Small signals imply small variations about dc values.


Transistor is a three terminal solid state device in which a current flowing between two electrodes is controlled by thevoltage between the third and one of the other terminals or by a current flowing into the third terminal.

All material in this publication is copyrighted.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, ortransmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise,

without the prior written permission of the author.Permission is granted to individuals for downloading this document from the author’s website or his

CD-ROM for self-study only. Permission is hereby granted to instructors to use this publication as a class-handout if the author’s McGraw-Hill textbook Principles of Electronic Materials and Devices, SecondEdition, has been adopted as a requisite course textbook. The permission is valid only while the book

remains in adoption.

SPECIAL CUSTOM PUBLISHED e-BOOKLET S.O. Kasap, 1990-2002The author reserves all rights

Last Updated: 27 November 2001 (v.1.1)

First published in Web-Materials(Established 1996)

http://Materials.Usask.Ca

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