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Zener diode (A-level only)

Zener diode reference voltage circuit

Here is how the circuit works:

  • The supply voltage VSV_SVS​ is shared between the series resistor RRR and the Zener diode.
  • The Zener diode "claims" its breakdown voltage, VZV_ZVZ​.
  • The rest of the voltage is dropped across the resistor. Therefore, the voltage across the resistor is exactly VR=VS−VZV_R = V_S - V_ZVR​=VS​−VZ​.
  • The resistor limits the total current flowing through the circuit, protecting the Zener diode.
Key Idea

Voltage distribution

In a Zener reference circuit, the sum of the voltages must equal the supply voltage.

VS=VR+VZ V_S = V_R + V_Z VS​=VR​+VZ​

Where VSV_SVS​ is the supply voltage, VRV_RVR​ is the voltage across the series resistor, and VZV_ZVZ​ is the Zener voltage.

Let's look at how AQA tests this with some calculations.

Example

Calculating the required series resistor

A circuit is built to provide a stable 5.0 V5.0 \text{ V}5.0 V reference voltage from a 9.0 V9.0 \text{ V}9.0 V DC battery. The designer uses a Zener diode with a breakdown voltage of 5.0 V5.0 \text{ V}5.0 V. The manufacturer specifies that the typical minimum operating current to keep the diode in breakdown is 2.0 mA2.0 \text{ mA}2.0 mA.

Calculate the maximum value of the series resistor RRR that guarantees the Zener diode receives its minimum operating current.

  1. Find the voltage across the series resistor. The Zener diode locks the voltage across itself at 5.0 V5.0 \text{ V}5.0 V. The remaining voltage from the 9.0 V9.0 \text{ V}9.0 V supply is dropped across the resistor:
VR=VS−VZ V_R = V_S - V_Z VR​=VS​−VZ​ VR=9.0−5.0=4.0 V V_R = 9.0 - 5.0 = 4.0 \text{ V} VR​=9.0−5.0=4.0 V
  1. State the condition for maximum resistance. From Ohm's law (R=VIR = \frac{V}{I}R=IV​), for a fixed voltage, the resistance is maximum when the current is at its minimum allowed value. The minimum current is given as 2.0 mA2.0 \text{ mA}2.0 mA.
  2. Calculate the resistance. Substitute the values into Ohm's law (remembering to convert milliamps to amps):
R=VRI R = \frac{V_R}{I} R=IVR​​ R=4.02.0×10−3 R = \frac{4.0}{2.0 \times 10^{-3}} R=2.0×10−34.0​ R=2000 Ω R = 2000 \ \Omega R=2000 Ω

So, a maximum resistance of 2000 Ω2000 \ \Omega2000 Ω (or 2 kΩ2 \text{ k}\Omega2 kΩ) can be used.

Common Mistake

Spec limits

The AQA specification states: "Use as a stabiliser is not required." This means you will only be asked to evaluate the circuit as a simple provider of a reference voltage (no external load, or very simple loads), not complex calculations involving a varying load resistor drawing different amounts of current. Focus purely on the relationship between VSV_SVS​, VZV_ZVZ​, VRV_RVR​, and the current in that single loop!

Example

Calculating power dissipation

A 12 V12 \text{ V}12 V power supply is connected in series with a 150 Ω150 \ \Omega150 Ω resistor and a Zener diode. The Zener diode has a breakdown voltage of 7.5 V7.5 \text{ V}7.5 V. Calculate the power dissipated by the Zener diode.

  1. Calculate the voltage across the series resistor.
VR=12−7.5=4.5 V V_R = 12 - 7.5 = 4.5 \text{ V} VR​=12−7.5=4.5 V
  1. Calculate the current flowing through the circuit. Use Ohm's law on the resistor:
I=VRR I = \frac{V_R}{R} I=RVR​​ I=4.5150=0.030 A I = \frac{4.5}{150} = 0.030 \text{ A} I=1504.5​=0.030 A
  1. Calculate the power dissipated by the Zener diode. We know the current through the Zener (0.030 A0.030 \text{ A}0.030 A) and the voltage across it (7.5 V7.5 \text{ V}7.5 V). Use the electrical power equation:
P=I×V P = I \times V P=I×V P=0.030×7.5=0.225 W P = 0.030 \times 7.5 = 0.225 \text{ W} P=0.030×7.5=0.225 W

(which is 225 mW225 \text{ mW}225 mW).

Exam technique

In the exam

  1. Check the bias: Always check the polarity of the diode in the diagram. To act as a reference voltage, a Zener diode must be reverse-biased (cathode connected to the positive terminal).
  2. Find the resistor voltage first: The single most common first step in any Zener diode math problem is finding the voltage dropped across the series resistor using VR=VS−VZV_R = V_S - V_ZVR​=VS​−VZ​.
  3. Watch your prefixes: Currents in Zener diodes are often given in mA\text{mA}mA (milliamps). Remember to multiply by 10−310^{-3}10−3 before using them in Ohm's law or power equations.
  4. Know the curve: If asked to sketch the I-V curve, don't forget the forward bias section! It behaves just like a normal diode for positive voltages, turning on at around 0.7 V0.7 \text{ V}0.7 V.
Self review

Check yourself

  • Can you describe the difference between how a standard diode and a Zener diode behave in reverse bias?
  • Why is it necessary to connect a resistor in series with a Zener diode when creating a reference voltage circuit?
  • If a 10 V10 \text{ V}10 V supply is connected to a 3.3 V3.3 \text{ V}3.3 V Zener diode and a 100 Ω100 \ \Omega100 Ω series resistor, what is the voltage across the resistor?
  • What does the "typical minimum operating current" represent on the I-V characteristic curve?
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Circuit schematic of a Zener reference circuit with supply Vs, series resistor R, reverse-biased Zener diode, resistor voltage Vr, Zener voltage Vz, and current I A Zener diode is designed to operate in reverse bias at a chosen breakdown voltage called VZV_ZVZ​. When the reverse voltage reaches this value, the potential difference across the diode stays almost constant even though current can change.

In the simple reference circuit, the supply voltage is shared between the series resistor and the Zener diode. This setup gives the key relation between the components in the circuit:

VS=VR+VZ  ⟹  VR=VS−VZ V_S = V_R + V_Z \implies V_R = V_S - V_Z VS​=VR​+VZ​⟹VR​=VS​−VZ​

The resistor is essential because it limits the current in the loop. Without it, the Zener current could become too large and damage the diode through excessive power dissipation.

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How is the supply voltage VSV_SVS​ distributed between the components in a Zener reference circuit?

Zener diode (A-level only) Revision Guide

  1. A Level
  2. /Physics
  3. /Zener diode (A-level only)