Sze problems 5.3, 5.4 and 5.5
Figure 1: Overview of a bipolar pnp transistor.
1 Introduction
In this project for the TUG course "Semiconductor Physics" I want to show you how to solve
the practice question "Sze problems 5.3, 5.4 and 5.5" in the bipolar transistor section at the
course website.
The problem is about a bipolar pnp transistor in forward active mode (pn junction forward
biased and np junction reverse biased).
The following quantities are given:
• Emitter - base voltage (forward biased): Veb
• Base - collector voltage (reverse biased): Vbc
• Acceptor density in emitter: NA,e
• Donor density in base: ND
• Acceptor density in collector: NA,c
• Cross section of the transistor: A
• Length of the base: Lb
• Diusion constants of minority carriers: Dn,e , Dp , Dn,c
• Lifetimes of minority carriers: τn,e , τp , τn,c
The following quantities are assumed/known:
• Intrinsic carrier density of Silicon: ni ≈ 1.5 ∗ 1010 cm−3
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Sze problems 5.3, 5.4 and 5.5
• Relative permittivity for Silicon: r,Si = 11.9
• Temperature: T = 300K
Physical constants needed:
• vacuum permittivity: 0
• elementary charge: e
• Boltzmann constant: kB
2 Question 1
The rst question is to calculate the neutral base width.
To do this we need to remember the formula for the depletion width for a diode:
s
W =
2(ND + NA )(Vbi − V )
eND NA
(1)
We have 2 diodes (a pn and a np junction) in the transistor, so there are 2 dierent depletion
widths (one from emitter to base - Web and one from base to collector - Wbc ).
We want to know the "neutral base width" - that's the base length minus the depleted parts
of the base. To calculate this we need the length of the depletion regions only in the base.
That's given by equations 2 and 3 (compare with pn-junction section of course).
Wb e =
NA,e Web
NA,e + ND
(2)
Wb c =
NA,c Wbc
NA,c + ND
(3)
We can now use equation 1 to replace Web respectively Wbc in equation 2 and 3 (use the
correct NA ).
But there is 1 more quantity needed, the "built in voltage" (Vbi ) which occurs in Web and
Wbc . A quick look in the lecture notes provides us with the formula:
Vbi =
kB T
NA ND
ln
e
n2i
2
(4)
Sze problems 5.3, 5.4 and 5.5
Figure 2: Minority charge carrier concentration in a bipolar pnp transistor in forward active mode.
Of course we have 2 dierent built in voltages for the 2 junctions and both are needed in
equation 2 and 3. There also are 2 dierent bias voltages for each junction (V).
We now have all quantities needed to calculate the "neutral base width" - we just need to
subtract Wbe and Wbc from the given base width Lb .
3 Question 2
Calculate the minority carrier concentration at the emitter-base junction.
The minority carrier concentration in the emitter opposite to the junction equals the base
equilibrium value of the p-doped semiconductor (ne,0 ). If all dopants are fully inoized, this
quantity is given by:
ne,0 =
n2i
NA,e
(5)
This density rises exponentially and at the start of the depletion width in the emitter (x1 )
it's:
eVeb
ne,x1 = ne,0 ∗ e kB T
At the base side the minority charge carriers are holes and the equilibrium value is:
3
(6)
Sze problems 5.3, 5.4 and 5.5
pb,0 =
n2i
ND
(7)
The density of the holes in the base goes from a value below pb,0 (at the border of the basecollector depletion region - x3 ) up to a value above pb,0 (at the border of the emitter-base
depl. reg. - x2 ) in a linear way (because in a bipolar transistor, the base length is always
short compared to the diusion length to avoid recombination).
eVeb
pb,x2 = pb,0 ∗ e kB T > pb,0
pb,x3 = pb,0 ∗ e
eVbc
kB T
< pb,0
(8a)
(8b)
At the collector, the equilibrium value of the minority carriers is:
nc,0 =
n2i
NA,c
(9)
And it rises to:
eVbc
nc,x4 = nc,0 ∗ e kB T
(10)
Figure 2 shows the described situation (compare with lecture notes).
4 Question 3
Find the current components Ien , Iep , Icn , Icp and Ibb .
These currents can be calculated through diusion of the minority charge carriers. The diusion constant (D) is given for the emitter, base and collector, but we have to calculate dn/dx
for each region.
If you missed them in the lessons, here are the basic diusion equations for the diusion
currents:
dp
dx
dn
jdi,n = |e| Dn
dx
jdi,p = − |e| Dp
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(11a)
(11b)
Sze problems 5.3, 5.4 and 5.5
For the neutral base width we have a linear rise of minority carriers from collector side to the
emitter side (base width is short compared to the diusion length of the holes). So we just
need to divide the dierence of the minority carrier density by the neutral base length to get
dn/dx. With that we can calculate Iep (hole current injected in base), which is - as long as
base width is short - about the same as Icp (hole current collected from base). Because of this
Ibb is zero (Ibb = Icp − Iep ).
Iep = A ∗ jdi,ep = eADn,e
ne,x2 − ne,x3
≈ Icp
Le
(12)
Now we consider the electron ow from collector to base (Icn ) and the electron ow from
base to emitter (Ien ). In a semiconductor, a minority carrier decays in 1 diusion length.
We can calculate the diusion length in all regions, because the carrier lifetimes and diusion
constants are given (see equation 15).
√
L=
(13)
τD
To estimate dn/dx in the emitter and collector regions, we use the minority carrier density
increase from the equilibrium value per diusion length.
In the emitter this is:
Ien = A ∗ jdi,en = eADn,e
ne,x1 − ne,0
Le
(14)
Icn = A ∗ jdi,cn = eADn,c
nc,0 − nc,x4
Lc
(15)
and in the collector:
5 Question 4
Finally we can use the results from question 3 to calculate the emitter (Ie ), base (Ib ) and
collector current (Ic ).
Ie = Ien + Iep
(16a)
Ic = Icn + Icp
(16b)
Ib = Ien − Icn + Ibb
(16c)
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Sze problems 5.3, 5.4 and 5.5
Now we can calculate the transistor eciency (γ ), the base transport factor (αT ) and the
common base and common emitter current gains (α0 , β0 ):
γ=
αT =
Ie
Iep
Icp
Iep
α0 = γαT
α0
β0 =
1 − α0
(17a)
(17b)
(17c)
(17d)
To increase the transistor eciency the doping of the emitter needs to be increased (compared
to the base doping). To improve the base transport factor, the base width needs to be
reduced.
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