The answer may be wrong!
Problem 1: Qualitative BJT Questions
Complete the table below by indicating whether the noted change in a BJT device parameter
increases, decreases, or has no effect on the listed performance parameters. Provide brief,
qualitative justifications (without equations or formulas) for your answers.
Change
Effect on αT
Increase base width, WB
Decrease
Increase minority-carrier lifetime within the base,
Increase
τB
Increase base doping, NB
Increase emitter doping, NE
Decrease
Slight
Increase
Justification:
(1) An increase in WB results in an increase in the quasi-neutral base width W. The longer W is,
the more likely that the injected minority carriers will recombine before they reach the collectorjunction depletion region to be swept into the collector, so the base transport factor αT decreases.
(2) An increase inτB results in an increase in the average distance that a minority carrier diffuses
before it is annihilated, i.e. the minority-carrier diffusion length LB. By increasing LB relative to
W, the probability that a minority carrier will “survive” long enough to reach the collectorjunction depletion region increases, so αT increases.
(3) Because the minority-carrier mobility decreases with increasing NB, the minority-carrier
diffusivity DB decreases (due to the Einstein relationship); therefore, the average distance that a
minority carrier diffuses before it is annihilated (i.e., LB) due to recombination decreases. Thus,
the probability that a minority carrier will recombine before reaching the collector-junction
depletion region increases, so αT decreases.
We also noted that an increase in base doping NB results in an decrease of W (W ~
sqrt(LnNd/Nd)). However, this dependence is much weaker than the former one.
(4) An increase in NE will result in a slightly higher built-in potential of the emitter junction and
hence a very slightly larger width of the emitter-junction depletion region. As a result, W will
decrease very slightly, so that a minority carrier will slightly less likely to recombine; so αT
increases.
Problem 2: BJT Current Components and Performance Parameters
Consider a Si npn BJT with emitter area A = 10-7 cm2, maintained at room temperature (T =
300K), with parameters for each of the regions as shown in the table below:
Parameter
Emitter
Base
Collector
-3
18
16
Dopant concentration (cm )
10 (n-type) 10 (p-type) 1015 (n-type)
1.0
0.5
2.0
Width (µm)
Minority-carrier lifetime (s)
10-6
10-6
10-5
Note that the emitter region and collector region are each short, so that
(
I Cp
)
DE
cosh(WE ' / LE ) qVBE / kT
pE 0
e
− 1 and
LE
sinh(WE ' / LE )
D
cosh(WC ' / LC ) qVBC / kT
= −qA C pC 0
(e
− 1)
LC
sinh(WC ' / LC )
I Ep = qA
The base region is short, so that
I En = qA
[
cosh(W / LB )
sinh(W / LB )
(e qVBE / kT − 1) − sinh(W1 / LB ) e qVBC / kT − 1
I Cn
[
1
sinh(W / LB )
(W / LB ) qVBC / kT
(e qVBE / kT − 1) − cosh
e
−1
sinh(W / LB )
DB
nB 0
LB
D
= qA B n B 0
LB
(
)]
(
)]
and
Suppose that the BJT is biased at the edge of saturation: the emitter junction is forward biased
such that exp(qVBE/kT) = 1010, and the base-collector junction is zero biased (i.e. VBC = 0V).
Ignore band-gap narrowing in the emitter. Calculate:
a) quasi-neutral emitter width W'E , quasi-neutral base width W, and quasi-neutral collector
width W'C
b) IEp, IEn, ICp, and ICn
c) IE, IC, and IB
d) γ, αT, βdc
An:
Given an npn BJT biased as follows:
kT
V BE =
ln[10 10 ] = ( 60 mV )(10 ) = 0 . 60 V
q
V BC = 0V
a) First we calculate the built-in potential for each junction:
NB
kT
10 16
V bi , BE = 0 . 56 +
ln[
] = 0 . 56 + . 026 ⋅ ln( 10 ) = 0 . 92 V
q
ni
10
(this is not precise)
N N
kT
10 16 10 15
V bi , BC =
ln[ B C ] = .026 ⋅ ln(
) = 0 .66 V
q
ni
10 10
Then we can obtain the widths of the depletion regions on the n-side and p-side of each junction:
ε N
1
x n, BE = [2 Si B (
)(Vbi, BE − V BE )]1 / 2 = 2.03 ×10 −7 cm = 2.03nm
q NE NB + NE
x p , BE = [2
x p , BC = [2
ε Si N E
1
(
)(Vbi, BE − V BE )]1 / 2 = 2.03 ×10 −5 cm = 0.2µm
q NB NB + NE
ε Si N C
q NB
(
1
)(Vbi , BC − V BC )]1 / 2 = 88.5 ×10 −7 cm = 0.089µm
N B + NC
x n, BC = [2
ε Si N B
(
q NC
1
)(Vbi, BC − V BC )]1 / 2 = 0.885 ×10 − 4 cm = 0.89µm
N B + NC
Therefore,
W
'
W
'
W
'
E
= W E − x n , BE = 1µ m − 2 . 03 nm ≈ W E = 1µ m
B
= W B − x p , BE − x p , BC = 0 .5 µ m − 0 . 2 µ m − 0 . 089 µ m ≈ 0 . 21 µ m
C
= W C − x n , BC = 2 µ m − 0 . 89 µ m ≈ 1 . 11 µ m
b) We need to determine the minority-carrier diffusivity and diffusion length for each of the
quasi-neutral regions of the BJT, in order to calculate the various current components:
µ E = 150 cm
2
⇒ DE = µE
kT
= 150 cm 2 V
q
D Eτ
⇒ LE =
µ B = 1248 cm
⇒ DB = µB
⇒ LB =
/V ⋅ s
=
E
3 . 9 ⋅ 10
−6
s − 1 ⋅ 0 . 026 V = 3 . 9 cm
cm
2
2
/s
= 1 . 97 × 10 − 3 cm = 20 µ m
/V ⋅ s
2
kT
= 1248 cm 2 V
q
D Bτ
−1
B
−1
s − 1 ⋅ 0 . 026 V = 32 . 3 cm
32 . 3 ⋅ 10 − 6 cm
=
2
= 5 . 68 × 10
−3
2
/s
cm = 60 µ m
µ C = 458 cm 2 / V ⋅ s
⇒ DC = µ C
kT
= 458 cm 2 V
q
D Cτ C =
⇒ LC =
−1
s − 1 ⋅ 0 . 026 V = 11 . 8 cm 2 / s
11 . 8 ⋅ 10 − 5 cm 2 = 0 . 011 cm = 110 µ m
The excess minority-carrier concentrations at the edges of the depletion regions are
nB 0 (e
qV
BE
/ kT
p E 0 (e
qV
BE
/ kT
10 20
− 1) ≅
10 10 = 10 14 cm
17
10
10 20
− 1) ≅
10 10 = 10 12 cm
18
10
−3
−3
p C 0 (e
I Ep = qA
qV
BC
10
− 1) =
10
/ kT
20
15
( e 0 − 1 ) = 0 cm
(
−3
)
DE
cosh(WE ' / LE ) qVBE / kT
pE 0
e
−1
LE
sinh(WE ' / LE )
⇒ I Ep = 1.6 ⋅ 10 −19 C (10 −7 cm 2 )
⇒ I Ep = 6 . 2 × 10
I En = qA
DB
nB 0
LB
− 10
cosh(W / LB )
sinh(W / LB )
⇒ I En = 1.6 ⋅ 10 −19 C (10 −7 cm 2 )
I Cp = −qA
−6
(
(
)]
(e qVBE / kT − 1) − sinh(W1 / LB ) e qVBC / kT − 1
32.3cm 2 / s 4 −3 cosh(0.21 ⋅ 10 −4 cm / 60 ⋅ 10 −4 cm) 10
10 cm [
10 − 1 − 0]
60 ⋅ 10 − 4 cm
sinh(0.21 ⋅ 10 − 4 cm / 60 ⋅ 10 − 4 cm)
(
A = 2 .5 µ A
DC
cosh(WC ' / LC ) qVBC / kT
pC 0
(e
− 1)
LC
sinh(WC ' / LC )
I Cp = − qA
DC
cosh( W C ' / L C ) 0
(0 )
( e − 1)
LC
sinh( W C ' / L C )
⇒ I Cp = 0 A
I Cn = qA
DB
nB0
LB
[
)
A = 0 . 62 nA
[
⇒ I En = 2 . 5 × 10
3.9cm 2 / s 2 −3 cosh(1 ⋅ 10 −4 cm / 20 ⋅ 10 −4 cm) 10
10 cm
10 − 1
20 ⋅ 10 − 4 cm
sinh(1 ⋅ 10 − 4 cm / 20 ⋅ 10 − 4 cm)
1
sinh(W / LB )
(
)]
(W / LB ) qVBC / kT
(e qVBE / kT − 1) − cosh
e
−1
sinh(W / LB )
)
⇒ I Cn = 1.6 ⋅ 10 −19 C (10 −7 cm 2 )
⇒ I Cn = 2 . 5 × 10
−6
(
A = 2 .5 µ A
c)
I E = I En + I Ep = 2 . 5 µ A + 0 . 62 nA ≅ 2 . 5 µ A
I C = I Cn + I Cp = 2 . 5 µ A + 0 A = 2 . 5 µ A
I B = I E − I C = 0 . 64 nA
(Including all significant digits!!)
d)
γ ≡
I En
≈1
I En + I Ep
αT ≡
β dc ≡
I Cn
≈1
I En
IC
≈ 3882
IB
Problem 3: BJT − Deviations from the Ideal
Consider the npn Si BJT in Problem 2, with the emitter junction forward biased such that
exp(qVBE/kT) = 1010. Calculate
a) Early voltage, VA.
Note: You should use the value of CJC (units F/cm2) for VBC = 0V.
b) punchthrough voltage (VCE = VCB + VBE for which the quasi-neutral base width W = 0)
c) Estimate the breakdown voltage VCEO using the formula
VCEO =
)
32.3cm 2 / s 4 −3
1
10 cm [
1010 − 1 − 0]
−3
−4
−3
6 ⋅ 10 cm
sinh(0.21 ⋅ 10 cm / 6 ⋅ 10 cm)
VCBO
β dc + 1
where VCBO is the reverse breakdown voltage of the base-collector pn junction. Assume
that the critical electric field for avalanche breakdown of the base-collector junction is
5×105 V/cm.
d) Based of your answers in parts (b) and (c), will punchthrough be observed in the commonemitter output characteristics (IC vs. VCE plot) before breakdown?
An:
a) The Early voltage (VA) is given by the following equation:
VA =
qN BW
C JC
where CJC has units of Farads per cm2:
C JC =
ε Si
W dep , BC
=
ε Si
x p , BC + x n , BC
(10 −12 F / cm )
=
= 1 .1 × 10 − 8 F / cm 2
−4
( 0 .089 + 0 .89 ) × 10 cm
Thus,
V
A
− 19
( 1 . 6 × 10
=
C )( 10 16 cm − 3 )( 0 . 2 × 10
1 . 1 × 10 − 8 F / cm 2
− 4
cm )
= 3 . 1V
b) Mirroring Exercise 11.8 in Pierret for an npn BJT, we arrive at the following equation for VBC
at punchthrough:
V
BC
⇒ V
⇒ V
= V
BC
BC
bi , BC
−
(W
B
2 ε si
(
q
N
= 0 . 66 V −
− x
p , BE
N
B
(N
C
C
)2
+ N
B
)
)
( 0 . 5 ⋅ 10 − 4 cm − 0 . 2 ⋅ 10 − 4 cm ) 2
2 × 1 . 0 ⋅ 10 − 12 F / cm
10 15 cm − 3
(
1 . 6 ⋅ 10 − 19 C
10 16 cm − 3 ( 1 . 1 ⋅ 10 16 cm
= − 6 .8V
Therefore,
V PT ≡ V CE , PT = V CB + V BE = 6 . 8V + 0 . 6V = 7 . 4V
c) The breakdown voltage (VCE0) is determined as follows:
First, calculate the reverse breakdown voltage of the base-collector junction (VCB0), considering
that it breaks down when the peak electric field
ε max =
qN
ε Si
B
x p , BC = [ 2
q
(
NC NB
ε Si N C + N B
εmax=5×105 V/cm:
)( V bi , BC + V CB 0 )] 1 / 2
− 3
)
)
⇒ VCB 0 =
⇒ V CB 0 =
ε Si ( N C + N B )
2 qN C N B
(ε max ) 2 − Vbi , BC
1 ⋅ 10 −12 F / cm (1 . 1 ⋅ 10 16 cm − 3 )
( 5 ⋅ 10 5 V / cm ) 2 − 0 .66 V
2 × 1 .6 ⋅ 10 −19 C × 10 15 cm − 3 × 10 16 cm − 3
⇒ V CB 0 ≈ 897 V
Therefore,
VCE 0 =
VCB 0
897V
=
≈ 14.4V
β dc + 1
3882 + 1
d) Comparing the answers obtained in parts (b) and (c), we see that the punchthrough voltage
(VPT) is smaller (in magnitude) than the (common-emitter) breakdown voltage (VCE0). Therefore,
punchthrough will be observed before breakdown in the common-emitter output characteristics of
this BJT.