Zener Diodes as Voltage References and Stabilizers Revision Notes | A Level AQA Physics

1. Why Voltage Stability Matters

Picture a sensitive sensor circuit that needs exactly 5 V to work correctly. Your battery supplies 9 V and drifts as it discharges. Without something to hold the voltage steady, your sensor misbehaves or fails entirely. That is the problem a Zener diode solves.

A Zener diode, paired with a single series resistor, can lock the output voltage to a fixed value — the Zener breakdown voltage $V_Z$ — even as the supply voltage or the current drawn by the load changes within limits. Understanding exactly how and why this works is the core of this lesson.

2. The Zener Diode: Behaviour Under Reverse Bias

You already know that an ordinary diode blocks current when reverse-biased. A Zener diode is engineered differently: it is designed to break down at a precise, controlled reverse voltage $V_Z$. Once that voltage is reached, the diode conducts in reverse and — crucially — the voltage across it stays pinned at $V_Z$ regardless of how much current flows through it (within its rated limits).

Think of $V_Z$ as a voltage ceiling that the diode enforces. Any attempt by the circuit to push more than $V_Z$ across the diode is met by the diode conducting harder — drawing more current — until the voltage is pulled back to $V_Z$. This self-regulating behaviour is what makes the Zener diode so useful as a voltage reference.

The chart above shows the reverse-bias region of the Zener I–V characteristic only. In the reverse direction, almost no current flows until the voltage reaches $V_Z$ (here approximately 5.1 V), at which point the current rises steeply while the voltage across the diode barely changes. That steep, near-vertical section of the curve is the operating region for a voltage reference circuit. (In the forward direction, the Zener diode behaves like an ordinary diode — conducting with a small forward voltage drop of approximately 0.6–0.7 V — but this region is not shown in the chart above.)

3. The Circuit: Zener Diode with a Series Resistor

The Zener diode cannot be connected directly across a supply — without a current-limiting component it would draw excessive current and destroy itself. The series resistor $R_S$ is the essential partner: it drops the difference between the supply voltage $V_S$ and the Zener voltage $V_Z$, and it limits the current to a safe value.

The standard circuit has the supply $V_S$ connected through $R_S$ to the Zener diode. The Zener is connected in reverse bias — cathode towards the positive supply — and the output voltage is taken across the Zener (from cathode to anode). The output is therefore fixed at $V_Z$.

When no external load is connected, all of $I_{R_S}$ flows through the Zener diode. When a load is connected in parallel with the Zener, the load takes some current and the Zener takes the remainder. In both cases, the voltage across the output stays at $V_Z$ — provided the Zener remains in its breakdown region.

4. Worked Calculations: Analysing a Zener Reference Circuit

Exam questions on this topic almost always ask you to calculate the current through the resistor, the current through the Zener, or to verify that the Zener is actually operating in breakdown. The method is always the same: apply Kirchhoff's voltage law around the loop, then use Ohm's law on the resistor.

Worked example: Calculating Currents in a Zener Reference Circuit (No Load)

A Zener diode with $V_Z = 5.1\text{ V}$ is connected in reverse bias with a series resistor $R_S = 390\text{ Ω}$ to a supply of $V_S = 9.0\text{ V}$. No external load is connected. Calculate: (a) the voltage across the resistor, (b) the current through the resistor, and (c) the current through the Zener diode. [3 marks]

1. $V_{R_S} = V_S - V_Z = 9.0 - 5.1 = 3.9\text{ V}$ — [M1] Kirchhoff's voltage law: the supply voltage is shared between the resistor and the Zener. The Zener holds $V_Z = 5.1\text{ V}$, so the resistor takes the remainder.
2. $I_{R_S} = \frac{V_{R_S}}{R_S} = \frac{3.9}{390} = 0.010\text{ A}$ — [M1] $3.9 \div 390 = 0.010\text{ A}$ (10 mA). Keep the unit as A for consistency, or state 10 mA explicitly.
3. $I_Z = I_{R_S} - I_{\text{load}} = 0.010 - 0 = 0.010\text{ A}$ — [A1] With no load connected, all current through the resistor flows through the Zener. $I_Z = 10\text{ mA}$.
4. $V_{\text{across Zener}} = V_Z = 5.1\text{ V}$, which equals the rated breakdown voltage. — The Zener is operating correctly in its breakdown region. If the supply were less than 5.1 V, the Zener would not reach breakdown and the output would not be stabilised at $V_Z$.

Answer: (a) $V_{R_S} = 3.9\text{ V}$, (b) $I_{R_S} = 0.010\text{ A}$ (10 mA), (c) $I_Z = 0.010\text{ A}$ (10 mA)

Now consider what happens when a load resistor $R_L$ is connected in parallel with the Zener. The load draws its own current, so the Zener current decreases — but the output voltage stays at $V_Z$ as long as the Zener remains in breakdown. This current-sharing is the key to understanding why the circuit works as a voltage reference.

Worked example: Zener Reference Circuit with a Load Resistor

Using the same circuit as above ($V_S = 9.0\text{ V}$, $V_Z = 5.1\text{ V}$, $R_S = 390\text{ Ω}$), a load resistor $R_L = 1000\text{ Ω}$ is now connected in parallel with the Zener. Calculate: (a) the load current, (b) the Zener current. [4 marks]

1. $V_{\text{out}} = V_Z = 5.1\text{ V}$ — The Zener holds the output at $V_Z$ provided it remains in breakdown. We assume this is the case and check at the end.
2. $I_L = \frac{V_Z}{R_L} = \frac{5.1}{1000} = 0.0051\text{ A}$ — [M1] The full Zener voltage appears across the load. $5.1 \div 1000 = 0.0051\text{ A}$ (5.1 mA).
3. $I_{R_S} = \frac{V_S - V_Z}{R_S} = \frac{9.0 - 5.1}{390} = \frac{3.9}{390} = 0.010\text{ A}$ — [M1] The resistor current is unchanged because $V_S$ and $V_Z$ are unchanged. The resistor still drops 3.9 V.
4. $I_Z = I_{R_S} - I_L = 0.010 - 0.0051 = 0.0049\text{ A}$ — [A1] The current from the resistor splits: some goes through the load, the rest through the Zener. $I_Z \approx 4.9\text{ mA}$. The Zener current has decreased but the Zener is still conducting, so $V_{\text{out}}$ remains at 5.1 V. [A1]

Answer: (a) $I_L = 0.0051\text{ A}$ (5.1 mA), (b) $I_Z = 0.0049\text{ A}$ (4.9 mA). Output voltage remains $5.1\text{ V}$.

5. The Zener Diode as a Voltage Reference

A voltage reference is a fixed, reliable voltage that other parts of a circuit can compare against or be powered from. The Zener circuit provides this because $V_Z$ is a property of the diode itself — set during manufacture — rather than depending on the supply or the load.

In practice, a voltage reference is used wherever a circuit needs a known, stable voltage: for example, as the reference input to a comparator, to set the threshold of a switching circuit, or to calibrate a measuring instrument. The Zener circuit is one of the simplest ways to generate such a reference from an unregulated supply.

The key condition for the circuit to work as a reference is that the Zener must remain in its breakdown region at all times. This means the current through the Zener must stay above a minimum value (typically a few milliamps) and below the maximum rated current. Choosing $R_S$ correctly ensures both conditions are met across the expected range of supply voltages and load currents.

6. Choosing the Series Resistor: Design Reasoning

Selecting the correct value of $R_S$ is a genuine design decision, and examiners sometimes ask you to justify a choice or evaluate whether a given resistor is suitable. The reasoning follows directly from the two constraints on the Zener: it must not carry too little current (or it leaves breakdown) and must not carry too much (or it overheats).

The worst case for maximum Zener current is when the supply voltage is at its highest and the load current is at its lowest (or zero). The worst case for minimum Zener current is when the supply is at its lowest and the load current is at its highest. A well-chosen $R_S$ keeps the Zener current within its safe operating range under both extremes.

• If $R_S$ is too small: the Zener current becomes very large when the load is light, potentially exceeding the maximum rated power of the diode.
• If $R_S$ is too large: the resistor drops too much voltage, leaving insufficient voltage to keep the Zener in breakdown when the load is heavy.
• A suitable $R_S$ ensures the Zener current stays between its minimum operating current and its maximum rated current across the full range of operating conditions.

7. Exam Technique and Common Mistakes

Zener diode questions reward students who are methodical. The most reliable approach is always: identify $V_Z$, find the voltage across $R_S$ using $V_S - V_Z$, apply Ohm's law to get $I_{R_S}$, then use Kirchhoff's current law to split that current between the Zener and any load. Every mark in a calculation question maps to one of these steps.

To consolidate everything, here is a concise summary of the method for any Zener reference circuit question: (1) write $V_{\text{out}} = V_Z$; (2) find $V_{R_S} = V_S - V_Z$; (3) find $I_{R_S} = V_{R_S} / R_S$; (4) find $I_L = V_Z / R_L$ if a load is present; (5) find $I_Z = I_{R_S} - I_L$; (6) confirm $I_Z > 0$ to verify the Zener is in breakdown. Every Zener calculation on an AQA paper reduces to these six steps.