A sealed box has three terminals A, B and C. Inside, two identical germanium diodes (each with a forward thres

NCERT Exemplar · Q20

0.7\,\text{V}

(i) with A positive and B negative — essentially no conduction (current stays near zero);

(ii) with A negative and B positive — no current until about

0.7\,\text{V}

, then a rising straight line whose inverse slope

\Delta V/\Delta I

1000\,\Omega

;

(iii) with B negative and C positive — no current until about

0.7\,\text{V}

, then a steeply rising conducting curve;

(iv) with B positive and C negative — no conduction;

(v) with A positive and C negative — no conduction;

(vi) with A negative and C positive — no current until about

1.4\,\text{V}

, then a steeply rising conducting curve. From these characteristics, determine the arrangement of the two diodes and the resistor between A, B and C.

Yanam CbseLong· 5mImportance★★★★★

92% · 34/37 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — P N Junction Biasing

P-N Junction Biasing: The First Meeting

Imagine you have two rooms separated by a door. One room is full of people who want to leave (electrons in the N-side), and the other room is full of empty chairs (holes in the P-side). The door is initially locked — that's the depletion region, a zone with no free charges. Now, what happens if you push on the door from one side, or pull it from the other? That's exactly what biasing does to a P-N junction.

The Unbiased Junction (The Starting Point)

When you bring P-type and N-type semiconductors together, something interesting happens at the boundary. Electrons from the N-side (which has extra electrons) rush toward the P-side (which has extra holes). Holes from the P-side rush toward the N-side. They meet and recombine — an electron falls into a hole, and both disappear as free charges.

This leaves behind a region near the junction with no free charges — just fixed positive ions on the N-side (where electrons left) and fixed negative ions on the P-side (where holes left). This is the depletion region. It acts like a tiny battery: the N-side is positive relative to the P-side, creating a built-in electric field that stops further diffusion. The voltage across this region is about 0.7 V for silicon, 0.3 V for germanium.

Note

The depletion region is not a physical gap — it's a region depleted of mobile charge carriers. The crystal is still continuous.

Forward Bias: Pushing the Door Open

Connect the positive terminal of a battery to the P-side and the negative terminal to the N-side. This is forward bias.

What happens? The external battery's positive terminal repels holes in the P-side toward the junction. The negative terminal repels electrons in the N-side toward the junction. Both carriers are pushed toward each other — they overcome the built-in electric field. The depletion region shrinks.

When the applied voltage exceeds about 0.7 V (for silicon), the depletion region becomes so thin that carriers can cross freely. A large current flows. The junction is now "on" — like a switch closed.

Watch out

Never apply forward bias without a current-limiting resistor. The junction has very low resistance once forward-biased, and the current can destroy it instantly.

Reverse Bias: Pulling the Door Shut

Now reverse the battery: positive to N-side, negative to P-side. This is reverse bias.

The positive terminal attracts electrons from the N-side away from the junction. The negative terminal attracts holes from the P-side away from the junction. Both carriers are pulled apart. The depletion region widens.

The built-in electric field is now reinforced by the external field. Only a tiny current flows — the reverse saturation current, caused by thermally generated electron-hole pairs in the depletion region. This current is typically in the nanoampere range and is almost independent of the applied voltage (until breakdown).

The junction is "off" — like a switch open.

Important

In reverse bias, the depletion region acts as an insulator. The junction blocks current flow except for a negligible leakage current.

The Precise Statement

I = I_S \left( e^{\frac{qV}{nkT}} - 1 \right)

This is the Shockley diode equation. Here:

$I$ = diode current
$I_S$ = reverse saturation current (very small, typically $10^{-12}$ to $10^{-15}$ A for silicon)
$q$ = electron charge ( $1.6 \times 10^{-19}$ C)
$V$ = applied voltage (positive for forward bias, negative for reverse bias)
$n$ = ideality factor (1 for ideal, 1–2 for real diodes)
$k$ = Boltzmann constant ( $1.38 \times 10^{-23}$ J/K) …

Why this formula?

Why a PN Junction Biases the Way It Does

A PN junction is not a resistor. Its current-voltage behaviour is fundamentally asymmetric — and that asymmetry comes directly from the physics of the depletion region and the energy barrier it creates.

The Unbiased Junction: A Built-in Barrier

When P-type and N-type semiconductors are joined, holes from the P-side diffuse into the N-side, and electrons from the N-side diffuse into the P-side. This diffusion leaves behind fixed, charged ions: negative acceptor ions on the P-side, positive donor ions on the N-side. These ions create an electric field that opposes further diffusion.

The result is a depletion region — a zone with no free carriers — and a built-in potential $V_0$ across it. This potential acts as an energy barrier that prevents net current flow at equilibrium.

Important

At equilibrium (zero bias), the net current is zero. The diffusion current (due to concentration gradient) is exactly balanced by the drift current (due to the built-in electric field).

Forward Bias: Lowering the Barrier

Apply a positive voltage $V$ to the P-side relative to the N-side. This external voltage opposes the built-in potential. The net barrier becomes:

V_{\text{net}} = V_0 - V

The depletion width shrinks. More importantly, the energy barrier for majority carriers (holes from P-side, electrons from N-side) is reduced. A larger number of carriers now have enough energy to cross the junction.

The current that flows is diffusion current — carriers injected across the junction become minority carriers on the other side, where they recombine. The relationship is exponential because the number of carriers with energy above the barrier follows a Boltzmann distribution.

I = I_0 \left( e^{qV / kT} - 1 \right)

Here $I_0$ is the reverse saturation current (very small), $q$ is the electron charge, $k$ is Boltzmann's constant, and $T$ is absolute temperature.

Why the exponential? The fraction of carriers with enough energy to surmount a barrier of height $q(V_0 - V)$ is proportional to $e^{-q(V_0 - V)/kT}$ . At equilibrium ( $V=0$ ), this gives a current that exactly cancels the drift current. When $V>0$ , the barrier drops, and the net current becomes proportional to $e^{qV/kT}$ .

Reverse Bias: Raising the Barrier

Apply a negative voltage to the P-side relative to the N-side. Now the external voltage adds to the built-in potential:

V_{\text{net}} = V_0 + V_R

The barrier becomes higher. The depletion region widens. Diffusion of majority carriers is almost completely suppressed. The only current that flows is a tiny reverse saturation current $I_0$ , carried by minority carriers (electrons from the P-side, holes from the N-side) that are swept across by the electric field.

This current is essentially constant with voltage because the number of minority carriers is fixed by thermal generation — it does not depend on the barrier height once the barrier is large enough to block majority carriers. …

Start your 14-day free trial to unlock the full solution →

The conducting characteristics fix the layout: A-B conducts (at $0.7\,\text{V}$ , with $1000\,\Omega$ in series) only when A is negative; B-C conducts (at $0.7\,\text{V}$ ) only when C is positive; and A-C needs $1.4\,\text{V}$ — two diode drops — to conduct. This is a Y (star) network: a diode from A and a diode from C meet at an internal node O, with the $1000\,\Omega$ resistor joining B to O.

Reading the characteristics

Each measurement uses only two terminals (the third floats), so current flows through the two internal components linking those terminals.

A-C needs $1.4\,\text{V}$ to conduct (vi): $1.4=0.7+0.7$ , so the A-to-C path contains two diodes in series. Hence both the A-arm and the C-arm are diodes, and the resistor must be on the B-arm.
A-B (i, ii): blocks when A is $+$ , conducts at $0.7\,\text{V}$ with inverse-slope $1000\,\Omega$ when A is $-$ . So the A-arm diode conducts when A is negative (its cathode faces A), and the $1000\,\Omega$ resistor is in series on the B-arm.
B-C (iii, iv): blocks when C is $-$ , conducts at $0.7\,\text{V}$ when C is $+$ . So the C-arm diode conducts when C is positive (its anode faces C), again with the B-arm resistor in series.
A-C (v, vi): blocks when A is $+$ /C is $-$ ; conducts when A is $-$ /C is $+$ at $1.4\,\text{V}$ — the sum of the two diode drops, with the resistor not in this path.

The arrangement

Introduce an internal node O:

A-O: a diode with its cathode towards A (conducts when A is negative w.r.t. O).
C-O: a diode with its anode towards C (conducts when C is positive w.r.t. O).
B-O: the $1000\,\Omega$ resistor.

Equivalently, the two germanium diodes are connected in series between A and C, both pointing the same way (anodes towards C), their common point being O, and the $1000\,\Omega$ resistor runs from B to O.

Consistency check …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ19 — Suppose a 'n'-type wafer is created by doping Si crystal having 5 × 10…→Next questionQ1 — (a) Write the truth table for the combination of the gates shown in th…

Reading the characteristics

The arrangement

Consistency check …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days