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Question 1. FIGURE 1 shows a transistor connected in commonemitter mode. Using the EbersMoll equations for an npn transistor, estimate the value of the base emitter voltage, VBE, required to make
(i) the collectoremitter voltage, VCE, 5 volts
(ii) the collector current 9.95 mA.
Answer:
Determine VBE
The equation to find out IC is
I_{C} = α_{F}I_{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} )(5V_{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
9.95×10^{10} = 0.02+0.98exp(V_{BE}/(26×10^{3}))exp((5)/(26×10^{3}) )+V_{BE}/(26×10^{3} ))
9.95×10^{10} = exp(V_{BE}/(26×10^{3} ))[0.98exp((5)/(26×10^{3} ))]
9.95×10^{10} =exp(V_{BE}/(26×10^{3} ))×0.98
(9.95×10^{10})/0.98 = =exp(V_{BE}/(26×10^{3} ))
11.006×10^{10}=exp(V_{BE}/(26×10^{3} ))
V_{BE}/(26×10^{3} ) = log(11.006×10^{10})
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 = 2X10^{23}
nB0 = 5X10^{20}
Pco= 4x10^{22}
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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 baseemitter junction is forwardbiased by 0.6 volts and the base collector junction is reversebiased by 1 volt.
The baseregion 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 10^{20}
nB(0) = 5 X10 ^{20} X 26mV
=130 X 10^{17}
Pc(0)=4 X10 ^{22} X 26mV
=104 X 10^{18}
(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 10^{13}
(c) Calculate the minimum crosssectional area of the base if the transistor is to be capable of carrying 1 ampere.
Answer: The minimum cross sectional area
I = neµEA
A = 0.67mm^{2}
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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
Transistors

VCEQ=24ic(RC+RE)


BC107

42V

negative

BC109

42V

negative

BCY88

5.0V

positive

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