The Diode: Forward Bias, Reverse Bias & V-I Characteristics

The Semiconductor Diode

A semiconductor diode is a p-n junction with metallic contacts at its ends for applying external voltage — a two-terminal device. In its symbol, the arrow points in the direction of conventional current under forward bias (from p to n).

The external voltage V alters the equilibrium barrier. Two cases:

Forward Bias: Opening the Gate

Connect p to positive, n to negative. The applied voltage drops mostly across the depletion region (its resistance dwarfs the bulk regions'). V opposes the built-in $V_0$:

• Effective barrier: $V_0 - V$ — lowered.
• Depletion width: decreases.

Small V → only the most energetic carriers cross → small current. Larger V → barrier falls further → current grows rapidly.

Minority carrier injection

Under forward bias, electrons from n cross to p (becoming minority there) and holes from p cross to n — minority carrier injection. At the junction edges the injected minority concentrations soar, and the resulting gradients diffuse carriers toward the far contacts, sustaining the current. Total forward current = hole diffusion current + electron diffusion current — typically milliamperes (mA).

Reverse Bias: Closing the Gate

Connect n to positive, p to negative. Now V adds to the barrier:

• Effective barrier: $V_0 + V$ — raised.
• Depletion width: increases.

Majority-carrier diffusion is crushed. What remains: minority carriers (electrons on p, holes on n) that wander near the junction get swept across by the field — the drift current, a few microamperes ($\mu$A).

Why reverse current is voltage-independent

Even a small reverse voltage sweeps every available minority carrier across; the current is limited by the minority-carrier concentration, not the voltage — hence the flat reverse saturation current. (Drift also exists under forward bias but is negligible beside the mA injection current.)

Breakdown

At a critical breakdown voltage $V_{br}$, the reverse current rises sharply — a tiny voltage increase causes a huge current change. Unless an external circuit limits the current below the rated value, the junction is destroyed by overheating (equally true for excessive forward current). General-purpose diodes are never operated beyond the reverse saturation region.

The V-I Characteristic

Measured with a variable supply (potentiometer), a milliammeter for forward bias and a microammeter for reverse (NCERT Fig. 14.16):

• Forward: current negligible until the threshold / cut-in voltage — ~0.2 V (Ge), ~0.7 V (Si) — then rises exponentially.
• Reverse: flat $\mu$A saturation until $V_{br}$.

The diode conducts one way only: low forward resistance, enormous reverse resistance — the property behind rectification.

Dynamic resistance

$$\boxed{r_d = \frac{\Delta V}{\Delta I}}$$

[JEE Tip] NCERT Example 14.4: from the Si characteristic, between 10 and 20 mA, $\Delta V$ = 0.1 V → $r_{fb} = \dfrac{0.1}{10 \text{ mA}} = 10$ ohm; at V = -10 V, I = -1 $\mu$A → $r_{rb} = 10^7$ ohm. Forward tens of ohms, reverse tens of megohms — a $10^6$ asymmetry.

Solved Examples

Example 1: Diode resistances from the graph (NCERT Example 14.4)

From the Si diode characteristic: at I = 20 mA, V = 0.8 V; at I = 10 mA, V = 0.7 V; at V = -10 V, I = -1 $\mu$A. Find the forward resistance at $I_D$ = 15 mA and the reverse resistance at $V_D$ = -10 V.

Solution:

1. Forward (dynamic): $r_{fb} = \dfrac{\Delta V}{\Delta I} = \dfrac{0.8 - 0.7}{(20 - 10) \times 10^{-3}} = \dfrac{0.1}{10^{-2}} = 10$ ohm.
2. Reverse: $r_{rb} = \dfrac{10}{10^{-6}} = 1.0 \times 10^{7}$ ohm.
3. Takeaway: ohms forward, tens of megohms reverse — six orders of one-way-ness, read straight off the curve.

Example 2: Which way is forward? [Board Rapid]

A battery's positive terminal connects to the n-side of a diode. Identify the bias and the expected current scale.

Solution:

1. Positive to n = reverse bias (forward needs positive to p).
2. Barrier rises to $V_0 + V$; only minority-carrier drift flows.
3. Current: microamperes (reverse saturation), essentially independent of V until breakdown.
4. Takeaway: read the wiring first; the current scale (mA vs $\mu$A) follows automatically.

Example 3: Barrier arithmetic [NEET Numerical]

A Si junction has $V_0$ = 0.7 V. Find the effective barrier under (a) 0.5 V forward bias, (b) 5 V reverse bias.

Solution:

1. (a) Forward: barrier = $V_0 - V$ = 0.7 - 0.5 = 0.2 V (depletion narrows).
2. (b) Reverse: barrier = $V_0 + V$ = 0.7 + 5 = 5.7 V (depletion widens).
3. Takeaway: forward subtracts, reverse adds — two signs carrying the entire bias story.

Example 4: Why applied voltage falls on the depletion region [Conceptual]

Justify NCERT's statement that the external voltage drops almost entirely across the depletion region.

Solution:

1. The depletion region has no free charges — its resistance is very high.
2. The p- and n-bulk regions teem with carriers — low resistance.
3. In a series chain, voltage divides in proportion to resistance: nearly all of V appears across the high-resistance depletion layer.
4. Takeaway: that's why bias acts directly on the barrier — the external volts go exactly where the physics happens.

Example 5: Minority carrier injection [Board Conceptual]

Describe what 'minority carrier injection' means under forward bias.

Solution:

1. Forward bias lets electrons from n cross into p — where they are minority carriers; likewise holes from p into n.
2. At the junction boundary the minority concentration on each side rises far above its remote value.
3. The resulting concentration gradients drive diffusion toward the contacts — this diffusive march IS the forward current.
4. Takeaway: forward current is injected-minority diffusion on both sides; total = electron + hole diffusion currents (mA scale).

Example 6: Reading the meters [Board Practical]

Why does the V-I measuring circuit use a milliammeter in forward bias but a microammeter in reverse?

Solution:

1. Forward current: injection-driven, typically mA.
2. Reverse current: minority-drift saturation, typically $\mu$A — a milliammeter would read zero.
3. Matching meter to scale is essential to see the curve at all.
4. Takeaway: the meter choice itself encodes the diode's $10^3$-plus current asymmetry — asked directly in practicals-based questions.

Example 7: Cut-in voltages [NEET Rapid]

State the threshold (cut-in) voltages of Ge and Si diodes and what happens beyond them.

Solution:

1. Ge: ~0.2 V; Si: ~0.7 V.
2. Below cut-in: current rises 'very slowly, almost negligibly'.
3. Beyond cut-in: current increases significantly (exponentially) for tiny voltage increments.
4. Takeaway: the knee of the forward curve — the pair (0.2, 0.7) is among the most-quoted numbers in the chapter.

Example 8: Why reverse current saturates [JEE Conceptual]

Explain why the reverse current is nearly independent of the applied reverse voltage.

Solution:

1. Reverse current is fed by minority carriers near the junction.
2. Even a small reverse voltage is 'sufficient to sweep the minority carriers' across — the field is never the bottleneck.
3. The supply of minority carriers (set by thermal generation, i.e. temperature) limits the current — not the voltage.
4. Takeaway: saturation current depends on temperature, not V — so it doubles roughly every 10 °C, a favourite twist.

Example 9: Breakdown and ratings [Board Conceptual]

What happens at $V_{br}$, and how can the diode survive it?

Solution:

1. At the breakdown voltage the reverse current increases sharply — a slight voltage rise causes a large current change.
2. If the external circuit limits the current below the rated value, the junction survives (Zener diodes exploit exactly this regime).
3. Unlimited current → overheating → destruction; the same fate awaits excessive FORWARD current.
4. Takeaway: ratings are thermal limits; breakdown is not inherently fatal — un-limited breakdown is.

Example 10: Ideal-diode circuit logic [JEE Circuit]

An ideal diode (0 V drop forward, open reverse) is in series with a 200 ohm resistor and a 2 V battery, positive to p. Find the current; then reverse the battery.

Solution:

1. Forward: ideal diode = closed switch: $I = \dfrac{2}{200} = 10$ mA.
2. Reversed: diode = open switch: I = 0.
3. (With a real Si diode: $I = \dfrac{2 - 0.7}{200} = 6.5$ mA forward.)
4. Takeaway: the on/off switch model solves most JEE diode circuits; refine with the 0.7 V drop when the question supplies it.