In a common emitter (CE) amplifier having a voltage gain G, the transistor used has trans conductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance of 0.02 mho and current gain of 20, the voltage gain will be:
1. 1.5G
2. G
3. G
4. G
The output \((X)\) of the logic circuit shown in the figure will be:
1. \(X= \overline{A\cdot B}\)
2. \(X = A\cdot B\)
3. \(X= \overline{A+ B}\)
4. \(X=\overline{\overline{A}} \cdot \overline{\overline{B}}\)
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is:
1. \(0.75~\text{A}\)
2. zero
3. \(0.25~\text{A}\)
4. \(0.5~\text{A}\)
Transfer characteristics [output voltage () vs input voltage ()] for a base biased transistor in CE configurations are as shown in the figure. For using the transistor as a switch, it is used:
1. In region III
2. Both in the region (I) and (III)
3. In region II
4. In region I
The symbolic representation of four gates is shown as:
Pick out which ones are for AND, NAND, and NOT gates, respectively.
1. (i), (iv), and (iii)
2. (ii), (iii), and (iv)
3. (ii), (iv), and (iii)
4. (ii), (iv), and (i)
A transistor is operated in a common emitter configuration at Vc =2 V such that a change in the base current from 100 to 300 produces a change in the collector current from 10 mA to 20mA. The current gain is:
1. 75
2. 100
3. 25
4. 50
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is:
1. 500
2. 1000
3. 1250
4. 50
To get an output Y = 1 from the circuit shown below, the input must be:
1. A=0 B=1 C=0
2. A=0 B=0 C=1
3. A=1 B=0 C=1
4. A=1 B=0 C=0
1. \(6000~\mathring{A}\)
2. \(4000\) nm
3. \(6000\) nm
4. \(4000~\mathring{A}\)
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