A bipolar amplifier circuit shown below,exhibits the following characteristic
IC=Isexp\(\left ( \frac{V_{BE}}{2V_{T}} \right )\),VT=25 mV
If there is no early effect, then voltage gain of the amplifier for a bias current IC= 1 mA is __________
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Solution
IC=Isexp\(\left ( \frac{V_{BE}}{2V_{T}} \right )\)
Transconductance
gm=\(\frac{\sigma I_{c}}{\sigma V_{BE}}=\frac{I_{C}}{2V_{T}}\)
Output impedance
Rout= Rc(Since there is not early effect)
Equivalent circuit is
Voltage is
\(\left | \frac{V_{o}}{V_{in}} \right |=g_{m}R_{C}=\frac{(1)(1)}{(2)(0.025)}=20\)
For the circuit shown below, the transistor parameter are V2TN= 0.8V and k’n = 30µA/V2. If output voltage VO= 0.1V,when input voltage is Vi= 4.2 V, the required transistor width-to length ratio is __________
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Solution
VGS=4.2V, VDC= 0.1 V
VDS< VGS– VTN
Thus transistor is in non saturation region.
ID=\(\frac{5-0.1}{10k}=0.49mA\)
ID=\(\frac{K_{n}}{2}\frac{W}{L}\left \{ 2(V_{GS}-V_{TN}) \right \}V_{DS}-V_{DS}^{2}\)
⇒0.49=0.015\(\left ( \frac{W}{L} \right )\left \{ 2(4.2-0.8) \right \}(0.1)-(0.1)^{2}\)
⇒0.49=\(\left ( \frac{W}{L} \right )(0.67)\Rightarrow \frac{W}{L}=0.731\)
A system consisting of 2 plants connected by a transmission line and the load is located at plant 2. When 100 MW is transmitted from plant 1, a loss of 10 MW takes place in transmission line.
Determine the power received by the load when λ for the system in ₹25 MWh and
\(\frac{dF_{1}}{dP_{1}}=0.02P_{1}+17\, and\frac{dF_{2}}{dP_{2}}=0.06P_{2}+19\)-
Solution
\(\frac{dF_{2}}{dP_{2}}\times 1.0=25\Rightarrow 0.06 P _{2}+ 19 = 25\)
⇒P2= 1000 MW⇒PL= B11 P21
⇒B11= 0.001
\(\frac{dP_{L}}{dP_{1}}=2B_{11}P_{1}(0.02 P_{1}+ 17)\frac{1}{(1-2B_{11}P_{1})}=25\)
⇒(0.02 P1+ 17) = 25(1–2B11 P1)(0.02 P1+ 17) = 25(1–2 × 0.001 P1)0.02 P1+ 17 = 25 – 0.05P1
⇒P1= 114.2 MW
Pde= P1+ P2– PL
Pde= (114.28 + 100) – (0.001)(114.20)2
Pde= 201.22 MW
An n-type GaAs semiconductor is doped with Nd= 1016 cm-3 Na= 0. The minority carrier lifetime is Tp0= 2×10-7s. If a uniform generation rate, g’= 2×1021 cm-3 s2-1, is incidenton the semiconductor then,the steady-state increase inconductivity will be __________
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Solution
For n-type GaAs,
Δσ= e(µn + µp) (δp)
In steady-state δp= g' Tp0. So, the steady-state increase inconductivity is
Δσ= (1.6 × 10-19) (8500 + 400)(2 × 1021) (2 × 10-7)
= 0.57 (Ω–cm)-1
The minimum small-signal diffusion resistance of an ideal forward-based silicon pnjunction diode at T =300 K is to be rd= 48 Ω. The reverse saturation current is Is= 2×10-11 A.The maximum applied forward-bias voltage that can beap plied to meet this specification is __________
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Solution
rd= 48 Ω
So, gd=\(\frac{1}{r_{d}}\)=0.0208
Also gd=\(\frac{I_{D}}{v_{t}}\)
or,ID= (0.0208)(0.0259)= 0.539mA
Since,ID=Is exp\(\frac{V_{a}}{V_{t}}\)
Hence,Va= VtIn\(\left (\frac{I_{D}}{I_{S}} \right )\)=(0.0259)In\(\left ( \frac{0.539\times 10^{-3}}{2\times 10^{-11}} \right )\)
= 0.443 V
In the infinite plane y = 6m, there exists a uniform surface charge density of \(\left ( \frac{1}{600\pi } \right )\mu c/m^{2}\). The associated electric field strength is _______ V/m.
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Solution
In the Infinite plane y =6 m
Surface charge density ρs=\(\frac{1}{600\pi }\mu c/m^{2}\)
E=\(\frac{\rho _{s}}{12\varepsilon _{0}}\widehat{a}_{x}\)
E=\(\frac{10^{-6}}{600\pi }\times \frac{1}{2\times 10^{-9}}\times 36\pi \times \widehat{a}_{x}\)
∴E = 30 V/m
An oscilloscope has input resistance of 1MW and input capacitance of 45 PF. It is used with a compensated 20 : 1 attenuation probe. What are the probe parameters and what is the effective resistance at probe tip?
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Solution
Input Resistance Ri= 1MW Input Capacitare Ci= 45pF;
Attenuation Probe = 20: 1
Rp= (20– 1)Ri
Rp = 19 × 1 MΩ
Rp= 19 MΩ
\(C_{p}=\frac{C_{i}}{20-1}=\frac{45}{19}=2.37pf\)
Reff= 20 Ri
Reff= 20 × 1MΩ
Reff= 20 MΩ
Consider the following statements relating to a circular discrotating in a traverse magnetic field B Wb/m2 as shown infigure. The generated emf across outer rim A and centre 0 is proportional to
1.Angular velocity
2.Flux density
3.Cube of the Radius of the disc
4.Square of the radius of the disc Of these statements which are correct.
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Solution
1. The generated emf α Angular velocity (w rad/sec)
2. The generated emf µα Flux density
3. The generated emf µα Square of radius of the disc.
The two power stations A and B are located close together.Station A has 4 identical generaters each rated as 100 MVA and having inertia constant of 9 MJ/MVA. Station B has 3 atternators of identical rated as 300 MVA6MJ/MVA. The equivalent inertia constant of the both machines on 150 MVA base is ______
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Solution
Ha1new =\(\frac{9\times 100}{150}\)= 6
Ha1new =\(\frac{9\times 300}{150}\)= 12
Hcp= Ha1 ep+ Ha2 ep
Hep = (6 × 4) + (12× 3) = 60
A 3-phase star-connected alternator with Zs= j8W, delivers 200 A at power factor 0.8 lagging to 11 kV infinite bus with stram supply unchanged, Assume rotational losses to remain constant. The percentage increase or decrease in the excitation emf, necessary to raise the PF to unity is ________.
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Solution
A-3 phase star-connected alternator with Zs = j8W, de-livers 200 A at power factor 0.8 lagging to 11 kV infinite bus
\(V_{b}=\frac{11000}{\sqrt{3}}=6351V\)
Ia= 200 (0.8 –j 0.6)
E = (6351+ j 0) + 200 (0.8– j0.6) (j 8)
E = 7422.2 Ð9.93°
E'sin δ'= E sin δ= 1291.47
E cos δ'= 6351
δ'=\(tan^{-1}\left [ \frac{1291.47}{6351} \right ]\)
E'= 6478.7 V
Percentage decrease
=\(\frac{E-E'}{E}\times 100\)
=\(\frac{7422.2-6478.7}{7422.2}\times 100\)
= 12.12%