Question

Draw a circuit diagram of a p-n junction diode in reverse-biased condition and show a graph between the variation of current and voltage related with it. What is meant by avalanche breakdown?
OR
What is the difference between p-type and n-type semiconductors? A silicon diode and a load resistance of 500 Ω are joined in series with an alternating voltage source of 20 V. The forward resistance of the diode is 10 Ω and the barrier voltage is 0.7 V. Find out: (i) the peak current through the diode (ii) the peak voltage across the load.

Hint:

Reverse bias: p to battery negative, n to positive; reverse current is tiny until breakdown, then rises sharply by avalanche (impact-ionisation chain). For the diode: peak current \( =(V_0-V_b)/(R_f+R_L) \), then load peak voltage \( =I_0 R_L \).

Answer 1

Option 1

Step 1: Reverse-bias circuit.
Connect the p-side of the diode to the negative terminal and the n-side to the positive terminal of a battery, in series with a microammeter. (Circuit: battery negative → p-region, battery positive → n-region, so the diode symbol arrow points opposite to the conventional current the battery would drive.)

Step 2: What happens inside.
The applied voltage pulls holes and electrons away from the junction, so the depletion layer widens and its potential barrier increases. Majority carriers cannot cross, so only a very small reverse (leakage) current due to minority carriers flows, and it stays almost constant with voltage.

Step 3: I-V graph.
Plot current (y-axis) against voltage (x-axis). In the reverse (third) quadrant the curve runs almost flat and very close to the voltage axis (tiny current of the order of microamperes) until a particular voltage called the breakdown voltage \( V_{br} \) is reached. At \( V_{br} \) the reverse current suddenly shoots up steeply (the curve drops almost vertically).
Description of the graph: a nearly horizontal line just below the axis in the reverse region, then a sharp vertical fall at \( V=V_{br} \).

Step 4: Avalanche breakdown.
When a lightly doped junction is strongly reverse biased, the minority carriers are accelerated by the strong field and gain enough kinetic energy to knock out more electrons from the atoms by collision (impact ionisation). These freshly released carriers again collide and free still more carriers, so the number of carriers multiplies rapidly like an avalanche. The reverse current therefore rises very sharply at the breakdown voltage. This is called avalanche breakdown.
\[\boxed{\text{Reverse current stays tiny, then rises sharply at } V_{br}\ \text{(avalanche multiplication of carriers)}}\]

Option 2

Step 1: p-type vs n-type semiconductor.
p-type: formed by doping with a trivalent impurity (e.g. Al, B, In); it creates extra holes, so holes are the majority carriers and electrons the minority carriers. n-type: formed by doping with a pentavalent impurity (e.g. P, As, Sb); it supplies extra free electrons, so electrons are the majority carriers and holes the minority carriers. Both are electrically neutral as a whole.

Step 2: Given data.
Peak source voltage \( V_0 = 20\ \text{V} \), load \( R_L = 500\ \Omega \), diode forward resistance \( R_f = 10\ \Omega \), barrier voltage \( V_b = 0.7\ \text{V} \). (The stated 20 V is taken as the peak value of the a.c. source.)

Step 3: Peak current formula.
In the forward half cycle the diode conducts. The barrier voltage drops 0.7 V, and the rest is shared by \( R_f \) and \( R_L \) in series:
\[ I_0 = \frac{V_0 - V_b}{R_f + R_L} \]

Step 4: Substitute and compute (i).
\[ I_0 = \frac{20 - 0.7}{10 + 500} = \frac{19.3}{510} = 0.0378\ \text{A} \approx 37.8\ \text{mA} \]

Step 5: Peak voltage across the load (ii).
\[ V_{L,0} = I_0\,R_L = 0.0378 \times 500 = 18.9\ \text{V} \]
\[\boxed{I_0 \approx 37.8\ \text{mA},\qquad V_{L,0} \approx 18.9\ \text{V}}\]