A transistor is used as an amplifier in CB mode with a load resistance of 5 kΩ. The current gain of the amplifier is 0.98 and the input resistance is 70 Ω, the voltage gain and power gain respectively are :
(a) 70, 68.6                          (b) 80, 75.6
(c) 60, 66.6                          (d) 90, 96.6

Concept Questions :-

Applications of Transistor
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In the following circuits PN-junction diodes D1D2 and D3 are ideal for the following potential of A and B, the correct increasing order of resistance between A and B will be

(i) – 10 V, – 5V           (ii) – 5V, – 10 V
(iii) – 4V, – 12V

(a) (i) < (ii) < (iii)                       (b) (iii) < (ii) < (i)
(c) (ii) = (iii) < (i)                       (d) (i) = (iii) < (ii)

Concept Questions :-

Applications of PN junction
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The circuit shown in the following figure contains two diode D1 and D2 each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100 ohm resistance (in amperes) is

(a) Zero                     (b) 0.02
(c) 0.03                     (d) 0.036

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The contribution in the total current flowing through a semiconductor due to electrons and holes are 34 and 14 respectively. If the drift velocity of electrons is 52 times that of holes at this temperature, then the ratio of concentration of electrons and holes is
(a) 6 : 5                 (b) 5 : 6
(c) 3 : 2                 (d) 2 : 3

Concept Questions :-

Types of semiconductors
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Ge and Si diodes conduct at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the valve of V0 changes by 

(a) 0.2 V                    (b) 0.4 V
(c) 0.6 V                    (d) 0.8 V

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In the circuit shown in the figure, the maximum output voltage V0 is :

1. 0 V                        2. 5 V
3. 10 V                      4. 52 

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In the following circuit, find I1 and I2.

(a) 0, 0
(b) 5 mA, 5 mA
(c) 5 mA, 0
(d) 0, 5 mA

Concept Questions :-

Applications of PN junction
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For the transistor circuit shown below, if β = 100, voltage drop between emitter and base is 0.7 V, then value of VCE will be :

(a) 10 V
(b) 5 V
(c) 13 V
(d) 0 V

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The following configuration of gate is equivalent to

(a) NAND
(b) XOR
(c) OR
(d) None of these

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Figure gives a system of logic gates. From the study of truth table it can be found that to produce a high output (1) at R, we must have

(a) X = 0, Y = 1
(b) X = 1, Y = 1
(c) X = 1, Y = 0
(d) X = 0, Y = 0

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