Semiconductor physics Winter term 2012/2013 Prof. Dr. M. S. Brandt Exercise 13, to be discussed in the week from 28. 1. to 1. 2.

The PNP Transistor as an example of a bipolar transistor A silicon p-n-p transistor has impurity concentrations of 5 · 1018 cm−3 , 2 · 1017 cm−3 and 1 · 1016 cm−3 in the emitter, base and collector respectively. The base width is 1.0 µm, and the device cross-sectional area is 0.2 mm2 . The emitter-base junction is forward biased to 0.5 V and the base-collector junction is reverse biased to 5 V. a) Calculate the neutral base width. From the last Problem set we know the built in voltage of a diode to be Vbi =

kT ND NA ln . q n2i

(1)

If we solve the poisson equation inside the depletion region and integrate the result once we arrive at the following fields qNA (x + WDP ) r 0 qND (WDN − x) E(x) = − r 0 E(x) = −

for − WDP ≤ x ≤ 0

(2)

for 0 ≤ x ≤ WDN

(3)

where WDP and WDN are the widths in the n and the p region. If we integrate those equations again we find the potential across the n and the p region to be 2 qNA WDP 2r 0 2 qND WDN −Vn = . 2r 0

Vp =

1

(4) (5)

If we use NA WDP = Nd WDN and Vbi = Vp − Vn we find s 2r 0 Vbi ND WDP = qNA (NA + ND ) s 2r 0 Vbi NA WDN = . qND (NA + ND )

(6) (7)

An applied voltage in forward direction reduces the built in voltage and an applied voltage in the reverse direction adds to the built-in voltage. To calculate the width in the case of a bias voltage we have to replace Vbi by Vbi − Vbias in the formulas above. All that is left is determining the values in these formulas. To calculate Vbi we need to calculate Eg

n2i = NC NV e− kT

(8)

with 1 NC = √ 2



1 NV = √ 2



m∗e kT π¯ h2

 32

m∗h kT π¯ h2

 32

(9) .

(10)

The effective masses of electrons and holes in silicon are m∗e = 1.08m0 and m∗h = 0.81m0 respectively. With this we find NC = 2.82 · 1019 cm−3 NV = 1.81 · 1019 cm−3 n2i = 1.69 · 1020 cm−6

(11) (12) (13)

for a temperature T = 300K. The built in Voltage Vbi is calculated to be 0.9384 V in the emitter-base junction and 0.795V in the base-collector junction. With this we find the width of the depletion region of the emitter-base junction in the base to be 5.2 · 10−2 µm and the width of the base-collector junction in the base to be 4.3 · 10−2 µm. So the total neutral base width is 1µm − 5.2 · 10−2 µm − 4.3 · 10−2 µm = 0.905µm. b) Determine the minority carrier concentration at the emitter-base junction. From the last problem set we know, that the hole concentration at the edge of the depletion layer is qVEB kT

pn (0) = pn0 e n2 pn0 = i . ND 2

(14) (15)

If we insert the values for n2i which we have already calculated, we find the hole concentration to be 2.14 · 1011 cm−3 . 2

The diffusion constants of minority carriers in the emitter, the base and the collector are 52 cms , 2 2 40 cms and 115 cms , respectively and the corresponding lifetimes are 10−8 s, 10−7 s and 10−6 s. c) How large is the hole current from the emitter to the base and from the base to the collector? To calculate the currents we first need the minority carrier distribution in the emitter, base and collector. The minority carrier distribution is found by using the continuity equation ∂ρ + ∇j = 0 ∂t

(16)

and adjusting it to the minority carrier densities Dp

∂ 2 pn pn − pn0 = 0. − ∂x2 τp

(17)

The solution to this equation is − Lx

x

pn (x) = pn + C1 e Lp + C2 e with Lp =

p

(18)

p Dp τp . In the reverse biased case the boundary conditions are pn (0) = pn0 e pn (W ) = 0.

qVEB kT

(19) (20)

With these boundary conditions we can find the distribution in the limit of W/Lp  1 to be x pn (x) = pn (0)(1 − ). (21) W With the boundary conditions qVEB kT

(22)

q|VCB | − kT

(23)

nE (x = −xe ) = ne0 e nC (x = xc ) = nc0 e

the same calculation also leads to the distribution of electrons in the emitter nE (x) = nE0 + nE0 (e 3

qVEB kT

− 1)e

x+xe Le

x ≤ −xE

(24)

and the collector nc (x) = nC0 + nC0 (e−

q|VCB | kT

c − x−x L

− 1)e

C

.

(25)

See Figure 2 for a plot.

In steady state, the current into each area due to minoriy charge carriers has to be equal to the diffusion of this minority charge right at the boundary to the depletion region. To Find the hole current into the base, we therefore calculate   qADp pn0 qVEB ∂pn ' e kT = 3.03 · 10−5 A. (26) IEp = A −qDp ∂x x=0 W The neutral base width p of 0.905µm is much smaller than the diffusion length of the holes in the base (Lp = Dp τp ≈ 20µm), so we assume that no holes are lost in the base and set ICp = IEp . This also leads to IBB = 0 since this current is needed to resupply electrons which recombined with holes in the base. d) Calculate the electron current from the emitter to the base and from the base to the collector? Similar to problem c) we calculate the diffusion current at the boundary to the depletion region. !  qADE nE0  qVEB ∂nE kT ' e − 1 = 1.91 · 10−7 A (27) IEn = A −qDe ∂x x=−xE LE !  ∂nC qADC nC0  −q|VCB | kT ICn = A −qDc ' − 1 = 5.8 · 10−14 A. (28) e ∂x x=xC LC e) Use your results obtained in problem c) and d) to calculate the total currents into the emitter, the base and the collector.

IE = IEp + IEn = 3.05 · 10−5 A IC = ICp + ICn = 3.03 · 10−5 A IB = IEn + IBB − ICn = 1.9 · 10−7 A

(29) (30) (31)

f) Calculate the current gain if the transistor is used in a common emitter configuration. The current gain is defined by β= 4

IC ≈ 160 IB

(32)

g) How can the current gain be improved?

– Lower the base width so that even more holes reach the collector. – Increase the difference in doping between the emitter and the base, so that more holes are injected in the base. – Reduce the electron emitter-base-current e.g. by a heterojunction. A good introduction to bipolar transistors can be found in Sze, ”Semiconductor Devices: Physics and Technology”.

Figure 1: Currents in a PNP Transistor. Hole currents are blue, electron currents black. From Sze, ”Semiconductor Devices: Physics and Technology”

Figure 2: Minority charge carrier concentration in the emitter, the base and the collector. From Sze, ”Semiconductor Devices: Physics and Technology”

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Figure 3: PNP Transistor in common emitter configuration. From Sze, ”Semiconductor Devices: Physics and Technology” Name e-mail Tel.Nr. Office Prof. Dr. M. S. Brandt [email protected] 289-12758 S301 Patrick Altmann [email protected] 289-11527 S310 Benedikt Stoib [email protected] 289-11525 S304 Konrad Klein [email protected] 289-11565 S310 Florian Hrubesch [email protected] 289-11380 C201 The exercises take place on Tue 12:20 - 13:30 (WSI S101), Wed 15:00-17:00 (ZNN 0th floor seminar room), Thu 08:30-10:00 (WSI S101) and Thu 15:45-17:15 (WSI S101).

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