Question
A P-N junction diode is connected in a circuit. Explain what happens to current flow when the diode is: (a) Forward biased — positive terminal of battery connected to P-side (b) Reverse biased — positive terminal connected to N-side
Also explain the role of the depletion region and barrier potential in each case.
Solution — Step by Step
When P and N semiconductors are joined, electrons from the N-side diffuse into the P-side and holes from the P-side diffuse into the N-side. This diffusion creates a region at the junction stripped of free charge carriers — the depletion region.
The depletion region carries a built-in electric field pointing from N to P (positive ions on N-side, negative ions on P-side). This field creates the barrier potential (also called contact potential), approximately 0.7 V for Silicon and 0.3 V for Germanium.
We connect the positive terminal to P-side and negative terminal to N-side. The external voltage opposes the built-in field of the depletion region.
As the applied voltage increases, it starts to cancel the barrier potential. Once the applied voltage exceeds the threshold voltage (0.7 V for Si), the depletion region effectively vanishes — majority carriers cross freely, and current flows.
Below 0.7 V, almost no current flows (the barrier is still partially intact). Beyond 0.7 V, current rises sharply — exponentially, actually. The diode is now conducting.
where = reverse saturation current, at room temperature, = ideality factor (1 for Ge, 2 for Si approximately)
Now we flip the battery: positive terminal to N-side, negative to P-side. The external field now adds to the built-in field. The depletion region widens.
Majority carriers are pulled away from the junction — holes in P-side move further left, electrons in N-side move further right. No majority carrier current flows.
A tiny current called reverse saturation current () does flow, caused by thermally generated minority carriers (electrons in P-side, holes in N-side). This is in the microampere range — negligible for practical purposes.
If reverse voltage is increased beyond the breakdown voltage, the diode conducts heavily in reverse. This is either Zener breakdown (below ~5 V) or avalanche breakdown (above ~5 V) — both are used deliberately in Zener diodes.
Why This Works
The depletion region is the key actor here. Think of it as a wall — in forward bias we’re pushing against the wall until it collapses; in reverse bias we’re making the wall thicker.
The barrier potential exists because of the fixed ionic charges left behind after diffusion. These charges can’t move (they’re part of the crystal lattice), so they create a permanent electric field. The external battery either fights this field (forward) or reinforces it (reverse).
This asymmetry in conduction is exactly why diodes are useful — they act as one-way valves for current. The entire rectifier circuit in your phone charger relies on precisely this behaviour.
Alternative Method — Energy Band Diagram Approach
Instead of thinking about fields and carriers, we can use energy bands.
In equilibrium, the Fermi levels of P and N sides align. This alignment creates a potential barrier visible as a step in the conduction/valence bands at the junction.
- Forward bias lowers the potential barrier — the energy step at the junction decreases. Electrons on the N-side now have enough energy to cross over. Current flows.
- Reverse bias raises the potential barrier — the step gets steeper. Electrons cannot cross. No current.
For JEE Main, the energy band diagram question often asks: “What happens to the width of depletion layer in reverse bias?” — Answer: width increases. And the barrier height: increases. Both increase together in reverse bias.
This approach is more powerful for JEE because questions frequently ask about changes in Fermi level, energy gap, and band bending under different bias conditions.
Common Mistake
Many students write: “In reverse bias, absolutely no current flows.”
This is wrong. A small reverse saturation current always flows due to minority carriers. In board exams, writing zero marks you wrong on a 2-mark theory question.
The correct statement: “In reverse bias, negligible current (reverse saturation current ) flows due to minority charge carriers.”
Also, don’t confuse the depletion region getting wider in reverse bias with the diode being “more insulating forever” — it conducts at breakdown voltage.
Key results to remember for PYQs:
| Parameter | Forward Bias | Reverse Bias |
|---|---|---|
| Depletion width | Decreases | Increases |
| Barrier potential | Decreases | Increases |
| Resistance | Low (~few Ω) | Very high (~MΩ) |
| Current | Large (mA range) | Tiny (, µA range) |
| Majority carriers | Cross junction | Pulled away |
For CBSE Class 12 boards, this table format in the answer gets you full marks in a 3-mark question. For JEE Main, the threshold voltage values (0.3 V for Ge, 0.7 V for Si) appear directly as MCQ options — these numbers are non-negotiable to memorise.