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Question
Figure shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true?
At Vi = 0.4 V, transistor is in active state.
At Vi = 1 V, it can be used as an amplifier.
At Vi = 0.5 V, it can be used as a switch turned off.
At Vi = 2.5 V, it can be used as a switch turned on.
Options
a and c
a, c and d
b and c
b, c and d
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Solution
b, c and d
Explanation:
According to the above graph transfer characteristics of a base biased common emitter transistor, we note that.
- When Vi= 0.4 V, the output voltage remain same and there is no collection current. So, the transistor circuit is not in an active state.
- When Vi = 1 V (This is in between 0.6 V to 2 V), the transistor circuit is in an active state and when input is increasing output is decreasing because when CE is used as an amplifier input and output voltages are 180° out of phase. Then it is used as an amplifier.
- When Vi = 0.5 V, there is no collector current. The transistor is in a cut-off state. The transistor circuit can be used as a switch to be turned off.
- When Vi = 2.5 V, the collector current becomes maximum and the transistor is in a saturation state and can be used as a switch turned on state.
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