to investigate the variation of current(i) with pd(v) for a semiconductor diode

Question 1

how did the student vary and measure the potential difference?

Answer 1

a rheostat or variable resistor. by adjusting the rheostat, the resistance of the circuit is adjusted. the voltage is being divided between the thin metallic conductor’s resistance and the rheostat’s resistance, and when adjusted or varied, results to lowering/increasing the voltage

Question 2

what is the function of the 330 ohm resistor in this circuit

Answer 2

the resistor protects the diode from damage that would result if too high a current does not flow through the diode and destroy it

Question 3

what does the graph tell you about the variation of current with potential difference for a semiconductor diode?

Answer 3

in forward bias, when the applied voltage is less than the junction voltage for the diode, the depletion layer still exists and the current will be small. when the applied voltage exceeds the junction voltage, the current increases rapidly

Question 4

estimate from your graph the junction voltage of the diode.

Answer 4

the point on the graph where the current begins to increase rapidly

Question 5

the student then put the diode in reverse bias and repeated the experiment.
1. how would a student connect the diode in reverse bias?

Answer 5

change the milliammeter for a micrometer (because current flowing will be very small), and place the microammeter in series with the diode, the voltmeter across both pieces of equipment (as the resistance of the microammeter is tiny compared to the resistance of the resistance of the reverse biased diode, so virtually all potential diff is across the diode)

Question 6

explain the graph obtained for the diode in reverse bias.

Answer 6

in reverse bias, only a small amount of current(leakage current) is able to flow. if the voltage becomes too high (the breakdown voltage for the diode), a very large current will flow which destroys the diode