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Common base current gain of a NPN transistor is 0.99. The input resistance is $ 1000 \Omega $ and load resistance is $ 10, 000 \Omega $. The voltage gain in common emitter mode is
$ \beta = { \alpha \over 1 - \alpha } and A_v = \beta { R_L \over r_i} $
In forward bias made, the P.N junction diode resistance will
In forward bias mode, the P-N junction diode allows current to flow easily, resulting in a lower resistance. Therefore, the resistance of a P-N junction diode in forward bias is 'less'.
To obtain OR gate from NOR gate, you will need
To obtain an OR gate from a NOR gate, you need two NOR gates. The first NOR gate inverts the input, and the second NOR gate combines these inverted inputs to produce the OR function.
The ratio of concentration of electrons and holes in a semi-conductor is 7 /5 , then what is the ratio of currents is 7/ 4 , then what is the ratio of their drift velocities ?
$ v_d = { I \over nAe } $ $ \therefore v_d \alpha { I \over n } $
In a P-type silicon, which of the following statement is true ?
In a P-type silicon, holes are the majority charge carriers, and the dopants are trivalent atoms. This is because trivalent atoms create 'holes' by accepting electrons.
A zener diode used as voltage regulator is connected . (i) in forward bias (ii) in reverse bias (iii) in parallel with load (iv) in series with load
A Zener diode is used as a voltage regulator by operating it in reverse bias. In this configuration, the Zener diode is connected in parallel with the load to maintain a constant voltage across the load. This is because in reverse bias, the Zener diode maintains a constant voltage (Zener voltage) across it once the breakdown voltage is reached.
PASSAGE : A n-p-n transistor is used in common emitter made in an amplifier circuit. A change of $40 \mu A$ in the base current changes the output current by 2mA and 0.04V in input voltage. The input resistance is
$ r_i = { \triangle V_{BE} \over \triangle I_B} $
PASSAGE : A n-p-n transistor is used in common emitter made in an amplifier circuit. A change of $40 \mu A$ in the base current changes the output current by 2mA and 0.04V in input voltage. The current amplification factor is
$ \beta = {\triangle I_C \over \triangle I_B }$
PASSAGE : A n-p-n transistor is used in common emitter made in an amplifier circuit. A change of $40 \mu A$ in the base current changes the output current by 2mA and 0.04V in input voltage. If a load of $ 6k \Omega $ is used, then the voltage gain of the amplifier is
$ A_V = \beta { R_L \over r_i } $
An amplifier has voltage gain $A_V$ = 1000. The voltage gain in dB is
voltage gain on dB = 20 $ log_10 A_0 $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.