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Electronics and Communication Engineering - Electronic Devices and Circuits

Exercise : Electronic Devices and Circuits - Section 2
31.
The static characteristic of an adequately forward biased P-n junction is a straight line, if the plot is of __________ Vs → versus
log I Vs log V
log I Vs V
I Vs log V
I Vs V
Answer: Option
Explanation:

Diode current equation

I = Is(eV/Vr - 1) = eV/Vr - 1

= = eVs/VR

=

Vs = VR .

32.
In an n channel JFET
ID, IS and IG are considered positive when flowing into the transistor
ID and IS are considered positive when flowing into transistor and IG is considered positive when flowing out of it
ID, IS, IG are considered positive when flowing out of transistor
IS and IG are considered positive when flowing into transistor and ID is considered positive when flowing out of it
Answer: Option
Explanation:

All currents are assumed positive when flowing into JFET.

33.
The intrinsic carrier concentration of silicon sample at 300 K is 1.5 x 1016/m3. If after doping, the number of majority carriers is 5 x 1020/m3. The minority carrier density is
4.5 x 1011/m3
3.33 x 104/m3
5 x 1020/m3
3 x 10-5/m3
Answer: Option
Explanation:

34.
A diode is operating in forward region and the forward voltage and current are v = 3 + 0.3 sin ωt (volts) and i = 5 + 0.2 sin ωt (mA). The average power dissipated is
15 mW
about 15 mW
1.5 mW
about 1.5 mW
Answer: Option
Explanation:

The contribution of sine terms to power dissipation is zero.

35.
Two identical silicon diodes D1 and D2 are connected back to back shown in figure. The reverse saturation current Is of each diode is 10-8 amps and the breakdown voltage VBr is 50 v. Evaluate the voltages VD1 and VD2 dropped across the diodes D1 and D2 assuming KT/q to be 25 m V.
4.983 V, 0.017 V
- 4.98 V, - 0.017 V
0.17 V, 4.983 V
- 0.017 V, - 4.98 V
Answer: Option
Explanation:

According to figure, D2 is forward bias, D1 is reverse biased. Reverse saturation current I0 = 10-8A. in clockwise direction.

For Diode D2 Is = I0

Here Is = I0

eqv2/kT = 2 or ev2/0.026 = 2

V2 = 0.018 V

Drop across D1 = V1 = 5 - 0.018

4.98 V

By KVL in mesh, VD1 = -4.98 V, VD2 = 0.018 V.

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