SEMICONDUCTOR

Vin < 0.3V

1. In the circuit shown below, D1 and D2 are silicon diodes
with cut-in voltage of 0.7 V. Vin and Vout are input and output V0 = Vin + 0.7
voltages in volts. The transfer characteristic of the circuit is Case III:
D1 0.3V < Vin < 1.7V
Vin Vout D1 → OFF and D2 → OFF
V0 = 1V
D2 1 kW 2. In the circuit shown, the breakdown voltage and the
maximum current of the Zener diode are 20V and 60 mA,
respectively. The values of R1 and RL are 200 W and 1 kW,
1V respectively. What is the range of Vi that will maintain the
Zener diode in the ON state?
R1
Vout
1 Vi RL
1
(a)
Vin
1 0.3 1.7 (a) 24 V to 36 V (b) 22 V to 34 V
(c) 20 V to 28 V (d) 18 V to 24 V
Vout
Ans. (a)
Sol.
(b) –0.7 1 RL
Vin IL
0.7 IZ
1 IR
Vi RL

Vout
1 1 20 − 0

=IL = mA 20mA
(c) 1
Vin
IR = IZ + IL
1 –1.7 1.7

Vi min = IR min R1 + VZ
Vout
IRmin = 0 + IL = 20mA (Q IZ(min) = 0)

Vi,min = 20 × 10 × 200 + 20 = 4V + 20V = 24V
–3
1

Vimax = IRmaxR1 + VZ
(d) Vin
0.3 1.7
IRmax = IZmax + IL = 60mA + 20mA = 80mA
1
Vimax = 80 × 10–3 × 200 + 20 = 16V + 20V = 36V
\ Vimin = 24V; Vimax = 36V

Ans. (a) 3. The network shown below represents which of the following
Sol. Case I: gate
Vcut-in voltage = 0.7V
If Vin > 1.7 V Y
For diode D1: Vin – 1V > 0.7
D1 → ON and D2 → OFF A
B
Vin > 1.7V
V0 = Vin – 0.7
Case II:
If Vin < 0.3 V (a) AND gate (b) NAND gate
For diode D2: 1 – Vin > 0.7 (c) NOR gate (d) XOR gate
D1 → OFF and D2 → ON Ans. (c)
 Sol. Y = ( AB + AB ) + ( AB )
4− x x 3x 3x
=i + ⇒ 4 − =i ⇒ 4 =i +  (1)
1 2 2 2
= ( AB + AB ) ⋅ ( AB )
4− y y−0 3y 2i + 3 y
+ i= ⇒ 2 + i= ⇒ 2= (2)
= ( AB ⋅ AB ) + ( AB ) ⋅ AB
2 1 2 2
=0 + ( A + B )( AB ) Substracting (2) from (1)

3
= ( A ⋅ AB ) + ( B ⋅ A ⋅ B ) 4 − 2 = 2i + ( x − y )
2
= AB + AB
0.9
2= 2i +
= ( A + B)
= AB 2

 3.1   3.1 
4. The output Y of the logic gate is ⇒= 2i  ⇒ =i  A
 2   4 
= 0.775 A
A 6. In the circuit shown. Vs is a 10 V square wave of period,
T = 4 ms with R = 500 W and C = 10 mF. the capacitor is
Y
initially uncharged at t = 0, and the diode is assumed to be
ideal. The voltage across the capacitor (VC) at 3 ms is equal
B to ____ volts (rounded off to one decimal place).
R
(a) A + B (b) AB
+10
(c) A + B (d) AB
0 T/2 T VS C VC
Ans. (a)
–10
Sol. Y = ( A + AB ) ⋅ ( B + AB )
t=0
= ( A + AB ) ⋅ ( B + AB )

Ans. [3.3]
= AB
Sol.
= A+ B Vi

= A + B +10

5. In the network shown below, find the current through the 0 t

2 ms 4 ms
ideal germunium diode. The cut in voltage of germinium –10
diode is 0.3V
1W 2W
Vcap
Ge
0 t
2W 1W 2 ms 4 ms

t = RC = 500 × 10 × 10–6 = 5 ms

T
(a) 0.570 A (b) 0.470 A 0 < t < : Vcap = Vf + (Vi – Vf)e–t/t
2
(c) 0.375 A (d) 0.775 A
= 10 + (0 – 10)e–t/RC
Ans. (d) −2×10−3
Sol. At t = 2 ms, Vcap =
10 − 10e 5×10 =
−3
3.296 3.3V
1W x 2W For 2 ms < t < 4ms, diode is OFF and capacitor has no path
to discharge. Hence, at t = 3 ms, Vcap = 3.3V.
4V 0.3 V 7. In the circuit shown, V S is square wave of period T
with maximum and minimum values of 8 V and –10 V,
2W y 1W respectively. Assume that the diode is ideal and R1 = R2 =
50 W. The magnitude of average value of VL is ______ volts
(rounded off to one decimal place).
4V
 R1 0.7 V
i1 = 0
+8 V
i2 0.7 W
0 T/2 T i3
VS R2 VL
0.4 W
–10 V 5W 1W
0.3 V
Ans. [3]
Sol. When VS = 8 V diode is in reverse bias
50 W e
0.7 i2 = 0.7 ⇒ i2 = 1 A
0.4 i3 + 0.3 = 0.7 ⇒ i3 = 1 A
VS 50 W VL for minimum e, i1 = 0
5(i1 + i2 + i3) + 0.7 + 0.3 = e
10 + 1 = e or e = 11 V
8 × 50 9. The current in D1 if the emf of the battery is twice the
=VL = 4 V minimum emf found in previous question is
50 + 50
(a) 2.0 A (b) 2.1 A
If VS = –10V, diode is in forward bias and VL = –10V
(c) 2.2 A (d) 2.4 A
Ans. (c)
VS Sol. 5(i1 + 2) + 0.7 + 0.3 = 22
VL
5i1 = 21 – 10 = 11
i1 = 2.20 A

Average value of VL 10. The maximum value of e for which current does not flow
Area 4 × 0.5T + (−10) × 0.5T through any one of the diodes
= = = −3V (a) 1.81 V (b) 1.91 V
Time period T
(c) 2.11 V (d) 2.01 V
Ans. (d)
PASSAGE
In the network shown below, D1 is an ideal silicon diode, D2 and Sol. The current through each diode will be zero when p.d. across
D3 are ideal germanium diodes. D3 is less than 0.3 V
D1 ⇒ 1 × i < 0.3
⇒ 5i + 0.7i + 1i = e
⇒ e ≤ 6.7 × 0.3 = 2.01 V
0.7 W

5W D2 0.4 W
D3 1W

e
8. The cut-in voltages of silicon and germinum diodes are
0.7 V and 0.3 V respectively. The minimum emf e of the
battery for which current starts flowing in D1 is
(a) 11 V (b) 13 V
(c) 15 V (d) 17 V
Ans. (a)
Sol. Current through D1 will start flowing when p.d. across it
becomes 0.7 V