BIPOLAR JUNCTION TRANSISTORS AT LOW
FREQUENCIES:
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. Kasap
Department of Electrical Engineering
University of Saskatchewan
CANADA
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BIPOLAR JUNCTION TRANSISTORS AT LOW
FREQUENCIES:
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 the
basement of Intel or its successor will be a huge computer file with
chip's listing. The last electrical engineer will sit nearby, handcuffed to
the disk drive in a scene out of Ben Hur. That engineer will be
extremely well paid, and his or her every demand will be immediately
satisfied. That engineer will be last keeper of the secret of the
universe: E = IR."
Robert Lucky
Spectrum, 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 Figure
1-1. In the normal and active mode of operation, the base-emitter junction is forward biased and basecollector junction is reverse biased. Electrons from the emitter become injected into the base. Injected
electrons in the base are minority carriers. These minority carriers (electrons) diffuse across the base and
when they reach the collector junction depletion region they become drifted by the electric field in this
region.
E
B
SCL
+
SCL
n
p
n
C
np(0)
IE
IC
np(x)
x'
pn(0)
A
pn(x')
Input
pn(x)
x
IB
VEB
A
Output
VCB
Minority carrier concentration profiles in the emitter, base and
collector of an npn BJT. Depletion regions are marked as SCL
(space charge layer). is the electric field in the collector
junction SCL.
Figure 1-1
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The minority carrier concentration profiles in the emitter, base and collector are shown in Figure 11. The minority carrier concentrations just outside the depletion regions, marked as SCL (space charge
layer), are determined by the law of the junction. For the minority carrier concentration at x = 0 in the base
just outside the SCL,
eV
n p (0) = n po exp EB 
 kT 
where VEB is the forward bias voltage across the emitter-base junction and the other symbols have their
usual meanings. The emitter current IE(electron) due to electron diffusion in the base is determined by the
minority carrier concentration profile at x = 0. If De is the electron diffusion coefficient in the base, then
eADe n p (0) eADe ni2
eV
 dn 
IE (electron) = ± AeDe  p  ≈
exp EB 
=
 kT 
WB
WB Na
 dx  x = 0
or,
IE (electron)
eADe ni2
eVEB 

= Isoe exp
; I =
 kT  soe
N a WB
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 L h in the emitter, then the hole
component of the emitter current is,
eV
eADh ni2
IE (hole) = Isoh exp EB  ; Isoh
 kT 
Nd Lh
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 the
fraction of the emitter current due to minority carriers injected from the emitter into the base,
γ =
IE (electron)
Isoe
=
IE (electron) + IE (hole) Isoe + Isoh
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. If
the 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,
αB = 1−
Transit (diffusion) time across base
τ
=1− t
Minority carrier recombination time in base
τe
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,
τt =
WB′ 2
2 De
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
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α
1−α
1.1. Example: Characteristics of an npn Si BJT
Consider an idealized silicon npn bipolar transistor with the properties listed below in Table 1-1. The base
region 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 active
mode. 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.
Emitter
width
Emitter
doping
Hole
lifetime in
emitter
Base width
Base doping
Electron
lifetime in
base
Collector
doping
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 the
emitter 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 α
and β change with VCB?
c
What is the emitter injection efficiency and what are α and β taking into account the emitter
injection 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 collector
output resistance, defined as ∆VCB/∆ΙC?
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WBCp
5
WBCn
WEB
B
E
C
WB
WB
IE
IC
Electron
diffusion
n+
A
Drift
p
n
IB
Output
A
Holes
Input
VEB
VCB
An npn transistor operated in the normal mode, in the active region, in the
common 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 device
characteristics such as the depletion widths and the base widths. With VBC >> V o (built-in voltage) the
reverse bias across the base-collector junction V r = VBC + V o ≈ V BC. Thus, the depletion layer W BC at the
base-collector junction is given by
1/ 2
WBC
2ε oε r ( Na + Nd )VBC 
=


eNa Nd

i.e.
WBC
 2(8.854 × 10 -12 F m -1 )(11.9)(10 22 + 10 22 m -3 )(18 V) 
=

(1.6 × 10 −19 C)(10 22 m -3 )(10 22 m -3 )


i.e.
WBC = 2.18 × 10-6 m or 2.18 µm





1/2
Only a portion of WBC is in the base side. Suppose that W BCp and W BCn are the depletion widths in
the base and collector sides of the SCL respectively. Since the total charge on the p and n-sides of the SCL
must be the same
N aW BCp = N dW BCn
and since
W BC = W BCp + W BCn
we can find WBCp,
WBCp =
Nd
1016
WBC = 16
(2.18 µm ) = 1.09
Na + Nd
10 + 1016
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Since Nd(E) >> N a(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
1/ 2
 2ε ε (V − VEB ) 
WEB =  o r o

eNa


We first need to calculate the built-in voltage Vo between the emitter and base,
Vo =
 (2 × 1018 )(1 × 1016 ) 
kT  Na Nd 
ln 2  = (0.0259 V)ln 
10 2

e  ni 
 (1.5 × 10 )

i.e.
Vo = 0.830 V
Then,
 2ε ε (V − VEB ) 
WEB =  o r o

eNa


i.e.
 2(8.854 × 10 -12 F m -1 )(11.9)(0.830 ± 0.6 V) 
WEB = 

(1.6 × 10 −19 C)(10 22 m -3 )


or
WEB = 1.74 × 10-7 m or 0.174 µm
1/ 2
1/2
Notice that due to the forward bias across the EB junction, WEB is an order of magnitude smaller
than W BCp.
If WB is the base width between emitter and collector metallurgical junctions, then the width W ′B of
the neutral region in the base between the borders of the depletion regions is given by,
W ′B = W B − W BCp − W EB
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 N a = 1 × 1016 cm-3, µe = 1250 cm2 V-1 s-1 and the diffusion coefficient De from the Einstein
relationship 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
τt =
WB′ 2
(2.74 × 10 −6 m )2
=
= 1.161 × 10-9 s
−3
2 −1
2 De 2(3.23 × 10 m s )
If we assume unity injection (γ = 1), then α = αB, base transport factor,
α = αB = 1−
i.e.
Transit (diffusion) time across base
τ
=1− t
Minority carrier recombination time in base
τe
α = 1 − (1.161 ns)/(400 ns) = 0.99710
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The current gain β is
β=
α
0.9971
=
= 343
1 − α 1 − 0.9971
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 relationship
is,
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 so
that
IE = IE(electron) + IE(hole)
where for electrons diffusing in the base,
eV
eADe ni2
IE (electron) = Isoe exp EB  ; Isoe =
 kT 
Na WB
and holes diffusing in the emitter,
eADh ni2
eV
IE (hole) = Isoh exp EB  ; Isoh
 kT 
Nd Lh
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 −19 C)(1 × 10 −8 m 2 )(3.23 × 10 −3 m 2 s −1 )(1.5 × 1016 m −3 )2
(1 × 10 22 m −3 )(2.74 × 10 −6 m)
i.e.
Isoe = 4.267 × 10-14 A or 42.67 fA
and
Isoh =
i.e.
Isoh = 2.93 × 10-17 A or 0.0293 fA
(1.601 × 10 −19 C)(1 × 10 −8 m 2 )(2.59 × 10 −4 m 2 s −1 )(1.5 × 1016 m −3 )2
(2 × 10 24 m −3 )(1.61 × 10 −6 m)
The emitter injection efficiency is the fraction of the emitter current that is due to minority carriers
injected into the base; i.e., those that diffuse across the base towards the collector.
γ =
IE (electron)
Isoe
=
IE (electron) + IE (hole) Isoe + Isoh
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4.267 × 10 −14
= 0.99931
4.267 × 10 −14 + 2.93 × 10 −17
The current gains, taking into account the emitter injection efficiency, are
α = γαB = (0.99931)(0.9971) = 0.99641
β = α/(1−α)= 0.99641/(1−0.99641) = 278
and
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 collector
junction. Only γIE of IE is injected into the base as minority carriers and only a fraction αB make it to the
collector,
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 results
tabulated in Table 1-2. We can calculate the small signal collector incremental resistance from,
rc =
∆VCB
19 V − 18 V
=
= 250 kΩ
∆IC
0.515 mA − 0.511 mA
We can also calculate the BJT characteristics using the hyperbolic expressions given in Problem
6.9, p. 504, in Principles of Electrical Engineering Materials and Devices, S.O. Kasap (McGraw-Hill).
The base transport factor αB is given by,
W 
 2.74 µm 
α B ≈ sech B  = sech
 = 0.99711
 36.0 µm 
 Le 
which is, within calculation errors, almost identical to the simplified theory.
The emitter current is given by
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W 
eV
eADe ni2
IE (electron) = Isoe exp EB  ; Isoe =
coth B 
 kT 
Na Le
 Le 
Isoe =
(1.601 × 10 −19 C)(1 × 10 −8 m 2 )(3.23 × 10 −3 m 2 s −1 )(1.5 × 1016 m −3 )2
 2.74 µm 
(1 × 10 22 m −3 )(36 × 10 −6 m) tanh

 36 µm 
Isoe = 4.260 × 10-14 A or 42.60 fA
i.e.
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 base
region 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.
Emitter
width
Emitter
doping
1 µm
1 × 1018 cm-3
Electron
lifetime in
emitter
~25 ns
Base width
(junctionjunction)
2 µm
Base doping
1 × 1016 cm-3
Hole
lifetime in
base
~500 ns
Collector
doping
1 × 1015 cm-3
a
Calculate current gains α and β for this transistor in the absence of any applied bias voltages, but
taking 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-base
junction is forward biased and the base-collector junction is reverse biased (normal active mode of
operation). What is your conclusion?
c
Suppose that the transistor is biased, for example, in the common emitter configuration, with a dc
base 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
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 (1 × 1016 )(1 × 1015 ) 
kT  N B NC 
ln 2  = (0.0259 V)ln 
10 2

e  ni 
 (1.5 × 10 )

VOBC = 0.634 V
The built-in voltage across the emitter-base (EB) is
VOEB =
i.e.
 (1 × 1018 )(1 × 1016 ) 
kT  N E N B 
ln 2  = (0.0259 V)ln 
10 2

e  ni 
 (1.5 × 10 )

VOEB = 0.812 V
The depletion layer WBC at the base-collector junction is given by;


≈

BC


W
2ε oε r ( N B + NC )VOBC 


eN B NC

1/ 2
1/2
 2(8.854 × 10 -12 F m -1 )(11.9)(10 22 + 10 21 m -3 )(0.634 V) 
=

(1.6 × 10 −19 C)(10 22 m -3 )(10 21 m -3 )


i.e.
WBC
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
WBCn
NC
1015
=
WBC = 16
(0.958 µm ) = 0.0871 µm
N B + NC
10 + 1015
Since NE >> NB, the depletion layer width WEB is almost totally in the base. With no bias, WEB is
given by
1/ 2
 2ε ε V 
WEB =  o r OEB 
 eN B 
1/2
i.e.
 2(8.854 × 10 -12 F m -1 )(11.9)(0.812 V) 
WEB = 

(1.6 × 10 −19 C)(10 22 m -3 )


giving
WEB = 0.329 µm
If W B is the base width between the emitter and collector metallurgical junctions, then the width
W′B of the neutral base region between the depletion regions is given by,
W ′B = W B − W BCn − W EB
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 coefficient
DB 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
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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
τt =
WB′ 2
(1.584 × 10 −6 m )2
= 1.213 × 10-9 s or 1.213 ns
=
2 DB 2(1.034 × 10 −3 m 2 s −1 )
The base transport factor αB is then
αB = 1−
i.e.
Transit (diffusion) time across base
τ
=1− t
Minority carrier recombination time in base
τB
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 the
emitter, 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 W E (= 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 diffusing
in the base and electrons diffusing in the emitter) so that
IE = IE(hole) + IE(electron)
where for holes diffusing in the base,
eV
eADB ni2
IE (hole) = Isoh exp EB  ; Isoh =
 kT 
N BWB′
and electrons diffusing in the emitter,
eV
eADE ni2
IE (electron) = Isoe exp EB  ; Isoe =
 kT 
N E WE
where we used WE instead of L h because L h > W E (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 carriers
injected into the base; those that diffuse across the base towards the collector.
γ =
IE (hole)
Isoh
=
IE (hole) + IE (electron) Isoh + Isoe
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or
γ =
1
=
I
1 + soe
Isoh
i.e.
γ =
1
(6.46 × 10 m s )(1 × 10 22 m −3 )(1.584 × 10 −6 m)
1+
(1.03 × 10 −3 m 2 s −1 )(1 × 10 24 m −3 )(1 × 10 −6 m)
thus,
γ = 0.99016
12
1
1
2 =
D
N W′
eADE ni
1+ E B B
DB N E WE
N W
1 + E E2
eADB ni
N BWB′
−4
2
−1
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 width
of 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,
1/ 2
WBC
2ε oε r ( N B + NC )(VOBC + VBC ) 
=


eN B NC

WBC
 2(8.854 × 10 -12 F m -1 )(11.9)(10 22 + 10 21 m -3 )(0.634 V + 6 V) 
=

(1.6 × 10 −19 C)(10 22 m -3 )(10 21 m -3 )







1/2
i.e.
WBC = 3.098 µm
Only a portion of WBC is in the base side (n-side) which is given by
WBCn =
NC
1015
WBC = 16
(3.098 µm ) = 0.282 µm
N B + NC
10 + 1015
Since NE >> NB, the depletion layer width WEB is almost totally in the base. With forward bias, WEB
is given by
1/ 2
 2ε (VOEB − VEB ) 
WEB = 

eN B


1/2
 2(8.854 × 10 -12 F m -1 )(11.9)(0.812 ± 0.6 V) 
=

(1.6 × 10 −19 C)(10 22 m -3 )


or
WEB = 0.168 µm
Thus,
W ′B = W B − W BCn − W EB
i.e.
W′B = 2 µm − 0.282 µm − 0.168 µm = 1.55 µm
New transit (diffusion) time τt through the base
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τt =
 S.O. Kasap, 1991-2002: v.1.1)
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13
WB′ 2
(1.55 × 10 −6 m )2
= 1.162 × 10-9 s or 1.16 ns
=
−3
2 −1
2 DB 2(1.034 × 10 m s )
New base transport factor αB is then
αB = 1 − (τt /τB) = 1 − (1.16 ns)/(500 ns) = 0.99768
New emitter injection efficiency is
γ =
thus,
1
1
=
−4
2 −1
D N W′
(6.46 × 10 m s )(1 × 10 22 m −3 )(1.55 × 10 −6 m)
1+ E B B 1+
DB N E WE
(1.03 × 10 −3 m 2 s −1 )(1 × 10 24 m −3 )(1 × 10 −6 m)
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
eV
IE (hole) = Isoh exp EB  ;
 kT 
Isoh =
eADB ni2
N BWB′
(1.601 × 10 −19 C)(0.01 × 10 −6 m 2 )(1.03 × 10 −3 m 2 s −1 )(1.5 × 1016 m −3 )2
(1 × 10 22 m −3 )(1.55 × 10 −6 m)
where
Isoh =
i.e.
Isoh = 2.40 × 10-14 A
Then
 0.84 × 10 −3 A 
kT   IE (hole) 

VEB =
ln
= (0.02585 V)ln

 e   Isoh 
 2.40 × 10 −14 A 
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 change
significantly with bias (in this transistor) and the emitter current is in the logarithm of the expression for
VEB. Thus, when IB = 10 µA, VEB = 0.63 V.
 S.O. Kasap, 1991-2002: v.1.1)
(
Bipolar Junction Transistors at Low Frequencies
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2.
14
Nonuniform Base Doping in the BJT
Net Dopant Concentration (cm-3)
n+
1020
p
n
WB
Acceptor ion
EB
h+
1018
1016
E
1014
x
E
B
B1
B2
C
C
0
2 µm
(a) Typical net dopant concentration profile
in an npn bipolar transistor.
(b) The non-uniform acceptor doping leads
to a built-in field EB in the base.
As a consequence of the fabrication process, the base is non-uniformly doped.
Figure 2-1
Due to the diffusion processes used in the fabrication process, the base region of a BJT normally has a
non-uniform dopant concentration as shown in Figure 2-1 for an npn BJT. There is a greater acceptor
concentration in the base at the emitter side, B 1 , than at the collector side B 2 . This leads to an initial net
diffusion of holes from B 1 to B 2 which exposes negatively charged acceptor ions around B 1 and
accumulates excess holes around B 2 . An internal field EB therefore builds up until an equilibrium is
reached and no further holes can diffuse from B 1 to B 2 because the built-in field EB prevents further
diffusion just as in the case of the pn junction under open circuit conditions. Suppose that the net acceptor
concentration, Na(x), in the base can be approximated by
N a(x) = N oexp(−x/b)
from x = 0 to x = W B as shown in Figure 2-1, where x in this equation is measured from around B 1
towards B 2 , and b is a parameter that characterizes the nonuniform doping profile. The parameter b
depends on the fabrication process. Suppose that the hole concentration approximately follows the doping
concentration, that is p(x) ≈ Na(x). Then, under open circuit conditions, the net current in the base due to
holes is zero, that is
Jhole = epµhEB − eDh
dp
=0
dx
We can now substitute p(x) = N a(x) = Noexp(−x/b) and use the Einstein relation Dh/µh = kT/e to
find,
EB =
kT
eb
This is the built-in field in the base due to non-uniform doping. When this npn transistor is
operating under normal and active conditions, the electrons injected from the emitter into the base are
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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 the
frequency response. If the drift time due to EB is shorter than the diffusion time, then the minority carrier
transit time across the base is given by,
τt ≈
WB
µe EB
2.1. Example: Nonuniform base doping and transit time
a
Consider a BJT with a nonuniform base doping as in Figure 2-1. Suppose that the base dopant
concentration in Figure 2-1 at B1, x =0, is 1017 cm-3 and at B2, x = WB, Na = 1015 cm-3. Calculate b and the
built-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 acceptor
concentration 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
= 0.43 µm
 1 × 1015 
ln

 1 × 1017 
The built-in field is
EB =
kT 258 × 10 −3 V
= 600 V m-1
=
eb 4.3 × 10 −5 m
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 t (drift) ≈
WB
(2 × 10 -6 m )
=
= 2.59 × 10-10 s or 0.259 ns
µe EB (1300 × 10 -4 m 2 V -1 s-1 )(600 V m -1 )
The diffusion time across the base is
τ t (diffusion) =
WB2
WB2
(2 × 10 -6 m )2
=
=
2 De 2 kTµe  2(1300 m 2 V -1 s-1 × 10 -4 × 0.02585 V)
 e 
= 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 more
than 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 effective
narrowing of the bandgap Eg [H.P.D. Lanyon and R.A. Tuft, "Bandgap Narrowing in Heavily Doped
Bipolar Junction Transistors at Low Frequencies
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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 on
minority carriers. If ∆Eg is the reduction in the energy gap then for an n-type semiconductor, according to
Lanyon and Tuft,
n
∆Eg (in meV) = 22.5 18 
 10 
1/ 2
where n (in cm-3) is the concentration of majority carriers which is equal to the dopant concentration if they
are all ionized (for example, at room temperature). The new effective intrinsic concentration nieff due to the
reduced bandgap is given by
 ( E − ∆Eg ) 
2
nieff
= Nc Nv exp − g

kT


 ∆E 
2
nieff
= ni2 exp g 
 kT 
i.e.
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
2
nno pno = nieff
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 W E is of
thickness 1 µm and has a donor doping of 1019 cm-3. The width W B of the base is 1 µm and has an
acceptor doping of 1017 cm-3.
a
Obtain an expression for the emitter injection efficiency taking into account the emitter bandgap
narrowing 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 ,W B << Le (electron diffusion length in base) and W E <<
L h (hole diffusion length in the emitter), we can use the short diode equations for the electron and hole
components of the emitter current. Suppose that WE and W B are the widths of the neutral regions (outside
the depletion regions). Then,
IE (electron)
Isoe
γ =
=
IE (electron) + IE (hole) Isoe + Isoh
or
γ =
1
=
I
1 + soh
Isoe
1
 eAD
n 
 N

 d (emitter)WE 
1+
 eADe (base)ni2 
 N

 a (base)WB 
2
h (emitter) ieff
=
1
2
Dh (emitter) Na (base)WB nieff
1+
De (base) Nd (emitter)WE ni2
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i.e.
γ =
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1
µ
N
W 
 ∆E 
1 +  h (emitter) a (base) B  exp g 
 kT 
 µe (base) Nd (emitter)WE 
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,
n
∆Eg (in meV) = 22.5 18 
 10 
1/ 2
 1019 
= 22.5 18 
 10 
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
γ =
1
µ
N
W 
1 +  h (emitter) a (base) B 
 µe (base) Nd (emitter)WE 
=
1
(50 cm V s )(1017 cm −3 )(1 µm )
1+
(900 cm 2 V −1 s −1 )(1019 cm −3 )(1 µm )
2
−1
−1
γ = 0.99945
Emitter injection efficiency γ′ with bandgap narrowing is
1
γ′=
µ
N
W 
 ∆E 
1 +  h (emitter) a (base) B  exp g 
 kT 
 µe (base) Nd (emitter)WE 
=
1
0.0711 
(50)(10 )(1)
1+
exp
19

0.02585 
(900)(10 )(1)
17
γ′ = 0.99138
c
Assuming perfect base transport, i.e. the base transport factor αB = 1, the corresponding common
emitter current gain without bandgap narrowing is
β=
γ
0.99945
=
= 1800
1 + γ 1 + 0.99945
and with bandgap narrowing this is,
β′ =
γ′
0.99138
=
= 115
1 + γ ′ 1 + 0.99138
Clearly the emitter bandgap narrowing has a substantial effect on the transistor current gain.
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4.
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18
Small Signal Low Frequency Equivalent Circuit of a BJT
Output
I C + ic
C
RC
E
IB + ib B
Input
vbe(t)
np(x)
VBB
IC
Electron
diffusion
x
vce(t)
VCC
n′p(0) np(0)
E
IE + ie
An npn transistor operated in the active region in the common emitter
amplifier configuration. The applied signal vbe modulates the dc voltage
across the BE junction and hence modulates the injected electron
concentration up and down about the dc value np(0). The solid line shows
np(x) when only the dc bias V BB is present. The dashed lines show how np(x)
is modulated up by a positive small signal vbe superimposed on VBB. The
collector 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. The
input circuit has a dc bias VBB to forward bias the base-emitter BE junction and the output circuit has a dc
voltage VCC (larger than VBE) to reverse bias the base-collector BC junction through a collector resistor R C .
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 and
modulates the forward voltage across the BE junction about its dc value, V BE. The varying voltage across
the BE modulates np(0) up and down about its dc value which leads to a varying emitter current and hence
to an almost identically varying collector current in the output circuit. The variation in the collector current
is 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,
eV
IE = IEO exp BE 
 kT 
(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 I C . Note
Bipolar Junction Transistors at Low Frequencies
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however that when IC increases, V CE, according to Equation (1), actually decreases so that the voltage
changes 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 small
signals superimposed on dc values. For small signals, we have v be = δVBE, ib = δIB, ie = δIE, ic = δ IC .
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 input
current 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 the
amplifier 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
rbe =
vbe δVBE
δV
βkT
25β
=
≈ β BE =
≈
ib
eIE
IC ( mA )
δI B
δIE
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 =
vce − RC ic − RC β
R I ( mA )
=
=
≈− C C
25
vbe
rbeib
rbe
which is the same as that in the CB configuration. However, in the CE configuration the output to input
current ratio, ic/ib = β whereas this is almost unity in the CB configuration. Consequently, the CE
configuration provides a greater power amplification, which is the second advantage of the CE circuit.
The input signal v be gives rise to an output current ic. This input voltage to output current
conversion is defined into a parameter called the mutual conductance, or transconductance, gm.
gm =
ic
δI
I ( mA ) 1
≈ E = E
=
vbe δVBE
25
re
The voltage amplification of the CE amplifier is then
A V = − g mR C.
We generally find it convenient to use a small signal equivalent circuit for the low-frequency
behavior of a BJT in the CE configuration. Between the base and emitter, the applied ac source voltage v s
sees only an input resistance of rbe as shown in Figure 4-2. To underline the importance of the input
resistance, the output (or the internal) resistance of the ac source is also shown. In the output circuit there
is a voltage controlled current source ic which generates a current of g m v be. The current ic passes through
the load (or collector) resistance RC across which the voltage signal develops. As we are only interested in
ac signals, the batteries are taken as a short circuit path for the ac current which means that the internal
resistances of the batteries are taken as zero. This model, of course, is only valid under normal and active
operating 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
Bipolar Junction Transistors at Low Frequencies
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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 the
equations above,
β = gmrbe
In transistor data books, the dc current gain, IC/IB, is denoted as hFE, whereas the small signal ac
current gain, ic/ib, is denoted as hfe. Except at high frequencies, hfe is approximately equal to hFE.
AC source
Small signal equivalent circuit
ib
S
B
vin
vbe rbe
S
E
Load
C
Rs
ic=gmvbe
RC vce
vs
E
Low frequency small signal simplified equivalent circuit of the
bipolar transistor in the CE configuration with a load resistor RC
in 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 common
emitter 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
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rbe ≈
 S.O. Kasap, 1991-2002: v.1.1)
(
21
βkT
25β
25(100)
≈
=
= 2500 Ω
eIC
IC ( mA )
1
The mutual conductance (transconductance) is
eIC IC (mA) 1
≈
=
= 0.04 A/V
kT
25
25
The magnitude of the voltage gain is given by,
gm ≈
b
A V = g mR C
Thus a gain of 200 requires,
RC ≈
AV
200
=
= 5000 Ω
gm 0.04 A / V
The dc voltage across R C is ICR C = (5000 Ω)(1 × 10-3 A) = 5 V. The battery voltage has to be
substantially 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 loads
the ac source so that vbe across BE is,
vbe = vs
rbe
(rbe + Rs )
The actual voltage gain, or the effective gain, is
AV ( effective ) =
i.e.
vce gm RC vbe
rbe
=
= gm RC
vs
vs
(rbe + Rs )
AV ( effective ) = (200)
2500
= 196
(2500 + 50)
The loading effect reduces the gain of the amplifier. To diminish the loading of the ac source,
need to increase rbe, i.e. reduce the collector current, but that also reduces the gain. So to keep the gain
same, we need to reduce IC and increase R C . However, R C cannot be increased indefinitely because
itself is loaded by the input of the next stage and, in addition, there is an incremental resistance between
collector 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.
vbe ( rms ) = vs ( rms )
rbe
2500
= (0.3536 mV)
= 0.3466 mV
(2500 + 50)
(rbe + Rs )
The input power into the BE terminals is
−3
vbe2 (rms) (0.3466 × 10 V)
=
=
= 4.80 × 10-11 W
2500 Ω
rbe
2
Pin = vbe (rms)ib (rms)
The output power is
v2
(200 × 0.3466 × 10 −3 V) = 9.61 × 10-7 W
= ce (rms) =
5000 Ω
RC
2
Pin = vce (rms)ic (rms)
The power amplification is
we
the
RC
the
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AP =
 S.O. Kasap, 1991-2002: v.1.1)
(
22
Pout
9.61 × 10 −7 W
=
= 2.0 × 104
−11
Pin 4.80 × 10 W
Note that the power amplification of the CE BJT is,
AP =
so that,
ic vce
= βAV
ib vbe
AP = (100)(200) = 20000
NOTATION
A
A V, A P
B
BC
BE
BJT
C
CB
CE
D
E
Eg
E
e
e
EB
EHP
gm
area; cross sectional area; amplification
voltage amplification, power amplification
base terminal of a BJT
base-collector
base-emitter
bipolar junction transistor
capacitance (F), collector terminal of a BJT
common base
common emitter
diffusion coefficient (m2 s-1)
emitter terminal of a BJT
band gap energy
electric field (V m-1)
electronic charge (1.60218 × 10-19 C)
(subscript) electron, e.g. µe = electron drift
mobility
emitter base
electron hole pair
mutual transconductance (A/V)
h
(subscript) hole, e.g. µh = hole drift mobility
hFE
dc current gain
h fe
small signal ac current gain
HF
high frequency
I
electric current (A)
IB, IC, IE base, collector and emitter currents in a BJT
J
current density (A m-2)
k
Boltzmann constant (k = R/NA = 1.3807 × 10-23
J K-1)
L e , Lh electron and hole diffusion lengths
ln, lp
lengths of the neutral n and p regions outside
depletion region in a pn junction
-3
N a, N d acceptor and donor concentrations (m )
-3
ni
intrinsic concentration (m )
nno, ppo equilibrium majority carrier concentrations
npo, pno equilibrium minority carrier concentrations
np(0)
minority carrier concentration in the base just
outside the emitter-base SCL
rbe
small signal input resistance (Ω)
T
temperature (K)
SCL
space charge layer
V
voltage
Vo
built-in voltage
vbe
small signal at base with respect to the emitter
vce
small signal at collector with respect to the
emitter
W
width
α
gain or current transfer ratio from emitter to
collector of a BJT
β
current gain IC/IB of a BJT
γ
emitter injection efficiency
εo
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 constant
µ h , µ e hole drift mobility, electron drift mobility (m2
V-1 s-1)
τt
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 the
valence 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 carriers
from 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.
Bipolar Junction Transistors at Low Frequencies
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Depletion layer (or space charge layer, SCL) is a region around the metallurgical junction where recombination of
electrons 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 donate
electrons to the conduction band (CB) and thereby create electrons in the CB which leads to n > p and hence to an ntype 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 µ d is the mobility then the defining equation is vd=µdE where
vd 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 applied
electrical force (F = −eE for electrons). In the absence of an applied field, all the electrons are moving around
randomly and the average velocity, over all the electrons, in any direction is zero. With an applied field, Ex, there is a
net velocity per electron, vdx, in the opposite direction to the field where vdx depends on E x via vdx=µdEx where µ d is
the 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 oppositely
doped (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 into
the 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 pside 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 a
bond between two neighboring atoms in the semiconductor crystal. The region around this “ruptured” bond is a net
positive charge of +e. It can drift in an applied field because an electron in a neighboring bond can tunnel into this
vacant site and thereby cause the positively charged bond-vacancy to become displaced, shifted. Thus holes contribute
to 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 at
the top of the valence band (VB) then the electron at b' contributes to the current. The reason that the presence of a
hole 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 of
the 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 applied
voltage. For holes in the n-side, it is pn(0) = pnoexp[eV/(kT)] where pn(0) is the hole concentration just outside the
depletion 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 nside and negative to the p-side. The applied voltage increases the built-in potential and hence the internal field in the
space charge layer.
Semiconductor is a nonmetallic element (e.g. Si or Ge) that contains both electrons and holes as charge carriers in contrast
to an enormous number of electrons only as in metals. A hole is essentially a "half-broken" covalent bond which has
a missing electron and therefore behaves effectively as if positively charged. Under the action of an applied field the
hole can move by accepting an electron from a neighboring bond thereby passing on the "hole". Electron and hole
concentrations in a semiconductor are generally many orders of magnitude less than those in metals thus leading to
much smaller conductivities.
Small signal equivalent circuit of a transistor replaces the transistor with an equivalent circuit that consists of
resistances, capacitances and dependent sources (current or voltage). The equivalent circuit represents the transistor
behavior under small signal ac conditions. The batteries are replaced with short circuits (or their internal resistances).
Small signals imply small variations about dc values.
Bipolar Junction Transistors at Low Frequencies
An e-Booklet
 S.O. Kasap, 1991-2002: v.1.1)
(
24
Transistor is a three terminal solid state device in which a current flowing between two electrodes is controlled by the
voltage between the third and one of the other terminals or by a current flowing into the third terminal.
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 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or
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 S.O. Kasap, 1990-2002
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Last Updated: 27 November 2001 (v.1.1)
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