Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Ze

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Q. Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below : (Zener breakdown voltage is 15 V)

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★est

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Concept understanding — Zener Diode Voltage Regulation

Intuition First: Why a Zener Diode Can "Hold" a Voltage

Imagine a pressure relief valve in a water pipe. If the water pressure gets too high, the valve opens just enough to let some water escape, keeping the pressure downstream constant — no matter how much the upstream pressure fluctuates. A Zener diode does the same thing, but with voltage.

In a normal diode, reverse bias blocks current completely — until you push too hard and destroy it. A Zener diode is specially designed so that when you reverse-bias it past a certain voltage (the Zener voltage, $V_Z$ ), it doesn't break — it conducts steadily without damage. And here's the magic: once it starts conducting in breakdown, the voltage across it stays almost exactly at $V_Z$ , even if the current through it changes dramatically.

That flat voltage is what makes it a regulator.

The Precise Statement

Important

A Zener diode, when reverse-biased into its breakdown region, maintains a nearly constant voltage $V_Z$ across its terminals over a wide range of reverse currents. In a circuit, this allows it to hold a fixed output voltage despite variations in the input supply or the load current — provided the current through the Zener stays within its safe operating limits.

How It Works in a Circuit

The simplest Zener regulator looks like this:

A series resistor $R_s$ connects the unregulated input $V_{in}$ (which is higher than $V_Z$ ) to the Zener diode.
The Zener is reverse-biased, with its cathode toward the positive supply.
The load $R_L$ is connected across the Zener diode.

The resistor is the workhorse. It drops the excess voltage $(V_{in} - V_Z)$ and limits the current. The Zener does the regulation.

Case 1: Input voltage rises

If $V_{in}$ increases, the current through $R_s$ tries to increase. But the Zener is already in breakdown — it simply absorbs the extra current, and the voltage across it stays at $V_Z$ . The load sees no change.

Case 2: Load draws more current

If the load resistance decreases (more current needed), that current comes from the Zener branch. The Zener current drops by exactly the same amount. The total current through $R_s$ stays the same, so the voltage drop across $R_s$ doesn't change — and $V_Z$ remains constant.

Note

The Zener acts like a variable resistor that automatically adjusts its resistance to keep the voltage fixed. When the load demands more current, the Zener "gives up" some of its own current. When the load demands less, the Zener takes the extra.

The Key Formula

I_Z = \frac{V_{in} - V_Z}{R_s} - I_L

Where:

$I_Z$ is the current through the Zener diode
$I_L = V_Z / R_L$ is the load current
The first term is the total current through $R_s$

For regulation to work, $I_Z$ must always stay between $I_{Z(min)}$ (the minimum current needed to keep the Zener in breakdown) and $I_{Z(max)}$ (the maximum current the Zener can handle without overheating).

The Two Critical Limits

A Zener regulator fails if: …

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The Zener diode's heavy doping creates a thin depletion layer, enabling a sharp breakdown at a low voltage. In the given circuit, the Zener current is found by subtracting the load current from the total series current, giving 5 mA.

Why heavy doping in Zener diodes?

The n and p regions of a Zener diode are deliberately made much more heavily doped than ordinary p-n junction diodes. This is the key design choice that makes Zener breakdown possible at a low, precisely controlled voltage.

When doping levels are high, the number of charge carriers (electrons and holes) available for recombination at the junction is large. As a result, the depletion region — the zone depleted of free carriers — becomes very thin, typically just a few nanometres wide. A thin depletion layer means that even a modest reverse-bias voltage creates an extremely high electric field across it (since field strength = voltage / distance). This intense field is strong enough to directly pull electrons out of their covalent bonds in the depletion region, a process called Zener breakdown. The breakdown occurs at a sharply defined voltage (here, 15 V) and the diode can safely conduct in reverse without being destroyed, as long as the current is limited by an external resistor.

Watch out

A common mistake is to think Zener diodes break down because of avalanche multiplication (like ordinary diodes at high reverse voltage). That's wrong — Zener breakdown is a field-emission effect, not an impact-ionisation effect. The heavy doping is what makes the field strong enough for this to happen at low voltages.

Finding the Zener current in the circuit

The circuit is a simple Zener shunt regulator: a 20 V DC supply, a 250 Ω series resistor, a Zener diode (breakdown 15 V), and a 1 kΩ load resistor all in parallel with the Zener.

Step 1: Recognise the operating condition.

Since the supply voltage (20 V) exceeds the Zener breakdown voltage (15 V), the Zener diode will break down and regulate the voltage across itself — and therefore across the load — at a constant 15 V. This is the fundamental behaviour of a Zener regulator: the diode "clamps" the voltage.

Step 2: Find the total current from the supply.

The series resistor $R_s = 250\ \Omega$ drops the excess voltage. The voltage across it is:

V_{R_s} = V_{\text{supply}} - V_Z = 20 - 15 = 5\ \text{V}

Using Ohm's law, the total current flowing through $R_s$ (which then splits between the Zener and the load) is: …

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A Zener diode is a specially designed p-n junction diode intended to be operated in the reverse breakdown region without being damaged (heavily doped junction). In this region, the voltage across the diode stays almost constant over a wide range of reverse currents. This property is exploited to hol …

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Q.Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below : (Zener breakdown voltage is 15 V)

Why heavy doping in Zener diodes?

Finding the Zener current in the circuit

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