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Question 1. FIGURE 1 shows a transistor connected in common-emitter mode. Using the Ebers-Moll equations for an npn transistor, estimate the value of the base emitter voltage, VBE, required to make
(i) the collector-emitter voltage, VCE, 5 volts
(ii) the collector current 9.95 mA.

Determine VBE
The equation to find out IC is
IC = αFIEBS[exp(e/KT(VBE))-1]-ICBS[exp(-e/KT(VCE - VBE)-1]
Substituting the 0.98 known values in this
VCE = 5V
IC = 9.95mA
IEBS = ICBS =1×10-13
αF = 0.98
KT/e = 25mV
9.95×10-3 = 0.98×10-13[exp(VBE/(26×10-3 ))-1]-10-13[exp((-1)/(26×10-3 )(5-VBE))-1]
(9.95×10-3)/10-13 =0.98exp(VBE/(26×10-3 )) - 0.98 -exp((-5)/(26×10-3 )+VBE/(26×10-3 ))+1
9.95×1010 = 0.02+0.98exp(VBE/(26×10-3))-exp((-5)/(26×10-3) )+VBE/(26×10-3 ))
9.95×1010 = exp(VBE/(26×10-3 ))[0.98-exp((-5)/(26×10-3 ))]
9.95×1010 =exp(VBE/(26×10-3 ))×0.98
(9.95×1010)/0.98 = =exp(VBE/(26×10-3 ))
11.006×1010=exp(VBE/(26×10-3 ))
VBE/(26×10-3 ) = log(11.006×1010)
VBE = 286.168×10-3 =0.286V

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Question 2. An npn transistor has the doping concentrations as given in TABLE A, where NDE and NDC are the emitter and collector donor concentrations and NAB the acceptor concentration of the base.

(a) Justify the assumption made in FIGURE 2(a) that the majority carrier densities nE, nB and nC can be regarded as being the same value as the doping densities NDE, NAB and NDC, respectively.

Holes diffuse from emitter to collector and drift is negligible in the base region. The majority carrier densities are therefore considered as equal to doping concentrations.

(b) Determine the values of the unbiased minority carrier densities pE0, nB0 and PC0,

PE0 = 2X1023

nB0 = 5X1020

Pco= 4x1022

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Question 3. FIGURE 2(b) represents the carrier densities in the three neutral regions of the transistor under active conditions. The base-emitter junction is forward-biased by 0.6 volts and the base collector junction is reverse-biased by 1 volt.

The base-region is sufficiently thin (WB << LN, where LN is the electron diffusion length) as to allow us to assume a linear minority gradient dnB (x)/dx

(a) Calculate the values of pE(0), nB(0) and pc(0).

Pn(0) = pno (eVA/KT)

=2 X10 23 X 26mV

52 X 1020

nB(0) = 5 X10 20 X 26mV

=130 X 1017

Pc(0)=4 X10 22 X 26mV

=104 X 1018

(b) Estimate the collector current density if the width of the base is 1.1 μm.

= (qDnnpbo/Wb).e(qvbe/kt)

7.33 X 1013

(c) Calculate the minimum cross-sectional area of the base if the transistor is to be capable of carrying 1 ampere.

Answer: The minimum cross sectional area
I = neµEA
A = 0.67mm2

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Question 4. Determine the most suitable transistor from TABLE B for use in the circuit of FIGURE 3. You may assume in all cases that the transistor is biased with VBE = 0.6 V.

Answer: From the table the most suitable will be BCY88 as this transistor is having the VCEO value lesser than the supply voltage.

VBE = 0.6V

Let us determine the quiscent values

 Transistors VCEQ=24-ic(RC+RE) BC107 -42V negative BC109 -42V negative BCY88 5.0V positive

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