Question

(a) Draw the V-I characteristics of silicon diode.

(b) Explain the following terms:

(i) minority carrier injection in forward bias
(ii) breakdown voltage in reverse bias

Hint:

Describe the V-I curve with its knee at about 0.7 V and the reverse breakdown. Then define injection of minority carriers in forward bias and reverse breakdown voltage.

Answer 1

Step 1: Part (a): Sketch and Description:
Take the V axis horizontally and the I axis vertically, crossing at the origin. Put forward voltage on the right and reverse voltage on the left. Use mA for the upper half of the I axis and $\mu$A for the lower half.
For $V > 0$: the current is nearly zero until about 0.7 V (the knee voltage of silicon). After the knee it increases rapidly, almost vertically.
For $V < 0$: the current is a tiny reverse saturation current (order of $\mu$A), nearly flat, until the breakdown voltage $V_{br}$. At $V_{br}$, the curve turns sharply downward.

Step 2: Part (b)(i): Injection of Minority Carriers:
A forward bias reduces the potential barrier across the junction. Majority carriers can now cross it easily.
When electrons enter the p-region, they become minority carriers there. Holes entering the n-region likewise become minority carriers.
This process is called minority carrier injection. The injected carriers recombine near the junction, and this gives the large forward current.

Step 3: Part (b)(ii): Breakdown Voltage:
Under reverse bias, the junction field increases with the voltage. At a certain critical reverse voltage, the diode current increases very sharply with a small further rise in voltage. This voltage is the breakdown voltage.
Two mechanisms cause it: Zener effect, where the very strong field pulls electrons out of bonds, and the avalanche effect, where accelerated carriers knock out more carriers.

Final Answer:
The curve shows a knee near 0.7 V in the forward direction and a sharp current rise at the breakdown voltage in the reverse direction.