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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.

Answer:

Determine VBE
The equation to find out IC is
I_C = α_FI_EBS[exp(e/KT(VBE))-1]-I_CBS[exp(-e/KT(V_CE - V_BE)-1]
Substituting the 0.98 known values in this
V_CE = 5V
IC = 9.95mA
I_EBS = I_CBS =1×10^-13
α_F = 0.98
KT/e = 25mV
9.95×10^-3 = 0.98×10^-13[exp(V_BE/(26×10^-3 ))-1]-10^-13[exp((-1)/(26×10^-3 )(5-V_BE))-1]
(9.95×10^-3)/10^-13 =0.98exp(V_BE/(26×10^-3 )) - 0.98 -exp((-5)/(26×10^-3 )+VBE/(26×10^-3 ))+1
9.95×10¹⁰ = 0.02+0.98exp(V_BE/(26×10^-3))-exp((-5)/(26×10^-3) )+V_BE/(26×10^-3 ))
9.95×10¹⁰ = exp(V_BE/(26×10^-3 ))[0.98-exp((-5)/(26×10^-3 ))]
9.95×10¹⁰ =exp(V_BE/(26×10^-3 ))×0.98
(9.95×10¹⁰)/0.98 = =exp(V_BE/(26×10^-3 ))
11.006×10¹⁰=exp(V_BE/(26×10^-3 ))
V_BE/(26×10^-3 ) = log(11.006×10¹⁰)
V_BE = 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 n_E, n_B and n_C can be regarded as being the same value as the doping densities N_DE, N_AB and N_DC, respectively.

Answer: Justifications

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 p_E0, n_B0 and P_C0,

Answer:

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 (W_B << L_N, where L_N is the electron diffusion length) as to allow us to assume a linear minority gradient dn_B (x)/dx

(a) Calculate the values of p_E(0), n_B(0) and p_c(0).

Answer: Determine

          Pn(0) = pno (eV_A/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.

Answer: Jn=eDndn/dx

= (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 V_BE = 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.

V_BE = 0.6V

Let us determine the quiscent values

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