Semiconductor Diode MCQ Level – 2 - Physics MCQ

During non-conducting period, the voltage across D₁ is V_m. = -2Vm.
The correct answer is: 

Under reverse bias (applying a negative voltage to the P-side with respect to the N-side), the potential drop (i.e., voltage) across the depletion region increases. Essentially, majority carriers are pushed away from the junction, leaving behind more charged ions.

 The two zener diodes when connected in series will double the breakdown voltage.

For the first position half of the input signal, the current pass through the diode, hence no current flows through the resistor R, hence, no output voltage is observed and capacitor is charged upto voltage V. For second half (negative), diode blocks the flow of current, hence current pass through resistor R but in opposite direction, hence the potential difference observed across R comes input from signal-V and from the capacitor again in opposite direction, thus making it-2V.
The correct answer is:

The avalanche effect is observed when reverse biased voltage is more than the breakdown voltage.
The correct answer is: the reverse voltage exceeds the breakdown voltage

The circuit looks like a voltage doubler. At the peak of positive half cycle D₁ is forward biased and D₂ is reverse biased. Ideally, this charges C₁ to the peak voltage V_m , with polarity as shown below. At the peak of the negative half cycle, D₁ is reverse biased and D₂ is forward biased. Because the source and C₁ are in series, C₂ will try to change towards 2V_m
The correct answer is: 

For step-down transformer,
E₂ < E₁


(44/π)

I = V / R =  5 / 2.5 kΩ = 2 mA 
For an ideal diode when it is conducting the potential drop across its terminal is zero

In case of reverse biased, the resistance of an ideal p-n diode should be infinity. But practically infinite is not possible and so resistance becomes very high.
The correct answer is: high

Holes will diffuse to right and electrons to left junction initially. Positive holes neutralize the acceptor ions near the function in the p-type silicon and disappeared across junction. Similarly, electrons will be neutralized across n-type by combination with holes of p-type due to diffusion. Thus, across p-type and n-type, there will be lack of majority charge carriers and excess of minority charge carriers. Then, majority charge carriers decreases as we move away from junction.
The Correct answer is