The Zener diode as a regulator

A voltage that does not move

The voltage of a battery falls as it runs down. The voltage from a rectifier changes with the mains. A radio or a sensor wants a steady voltage. A Zener diode gives it. Its voltage stays nearly the same over a wide range of current.

Breakdown

When a normal diode is reverse biased with a larger and larger voltage, the reverse current stays tiny. At some voltage it suddenly rises. This is breakdown. A normal diode can be destroyed by it, because the current is large. A Zener diode is made for it, with heavy doping and a rating for power. In the breakdown region the curve is almost vertical. The voltage stays at the Zener voltage $V_z$.

There are two causes of breakdown.

• The Zener effect. In a heavily doped diode, the depletion layer is thin, so even a small voltage makes a very strong field, of the order of $10^6$ V/m or more. The field pulls electrons straight out of the bonds. This happens for $V_z$ below about $5$ V.
• Avalanche breakdown. In a lightly doped diode, the layer is wider. Carriers are accelerated and knock more electrons out of the bonds, which knock out others, like an avalanche. This happens above about $7$ V.

Between these values, both effects act. Diodes sold as Zener diodes use either effect, and they are all drawn with the same symbol. Their voltages range from about $2$ V to over $200$ V. The Zener diode is always joined in reverse bias. In forward bias it acts as an ordinary diode.

The Zener regulator

The circuit has an input voltage $V_{in}$, a series resistor $R_s$, and the Zener diode in parallel with the load $R_L$. The Zener diode is reverse biased.

• The Zener diode conducts if the voltage across it would otherwise be above $V_z$. Then the output voltage is $V_{out} = V_z$.
• The series resistor carries the current $I_s = \dfrac{V_{in} - V_z}{R_s}$.
• The load takes $I_L = \dfrac{V_z}{R_L}$.
• The Zener diode takes the rest, $I_z = I_s - I_L$.

If the input rises, the extra current goes through the Zener diode, and the extra voltage drops across $R_s$. If the load current falls, the Zener diode takes the extra current. In both cases the output stays at $V_z$. If $I_z$ falls to zero, the regulation stops, and the output falls below $V_z$.

Two conditions must hold.

1. $I_z$ must stay above a small minimum $I_{z,min}$ (so that the diode stays in breakdown). A few mA is normal.
2. $I_z$ must stay below $I_{z,max} = \dfrac{P_{rated}}{V_z}$, so that the diode does not overheat.

Real Zener diodes have a small dynamic resistance $r_z$ (a few ohm), so the output changes a little when the current changes: $\Delta V_{out} \approx r_z\, \Delta I_z$.

A worked example. A regulator has $V_{in} = 12$ V, $R_s = 220\ \Omega$, $V_z = 5.6$ V and $R_L = 470\ \Omega$. Find the currents and the power in the Zener diode. Then remove the load. Then let the input fall to $8$ V.

1. The series current. $I_s = \dfrac{12\ \text{V} - 5.6\ \text{V}}{220\ \Omega} = 29.1$ mA.
2. The load current. $I_L = \dfrac{5.6\ \text{V}}{470\ \Omega} = 11.9$ mA.
3. The Zener current. $I_z = 29.1\ \text{mA} - 11.9\ \text{mA} = 17.2$ mA. The power is $5.6\ \text{V} \times 17.2\ \text{mA} = 96\ \text{mW}$.
4. No load. All $29.1$ mA flows in the Zener diode. The power is $5.6\ \text{V} \times 29.1\ \text{mA} = 163$ mW.
5. Input of $8$ V. The Zener diode would need $I_s = \dfrac{8\ \text{V} - 5.6\ \text{V}}{220\ \Omega} = 10.9$ mA, which is less than the $11.9$ mA that the load wants. So the Zener diode is off. The output is $8\ \text{V} \times \dfrac{470}{220 + 470} = 5.45$ V, below $V_z$.

Set the input to $14$ V and the load to $100\ \Omega$. Is the Zener diode on? Then raise the load to $1000\ \Omega$ and look at the Zener current.

Choosing the series resistor

The input varies from $V_{in,min}$ to $V_{in,max}$ and the load current from $I_{L,min}$ to $I_{L,max}$. The worst cases give two limits.

• Smallest input and largest load current. The Zener diode must still have $I_{z,min}$:

• Largest input and smallest load current. The Zener diode must not pass more than $I_{z,max}$:

A value between the two limits works. If the limits cross, no single resistor works, and a better Zener diode or a smaller range is needed.

Line and load regulation

• Line regulation is how little the output changes when the input changes, with a fixed load.
• Load regulation is how little the output changes when the load current changes, with a fixed input.

A Zener regulator is simple and cheap. It wastes power in the series resistor and in the Zener diode, so it is used for small loads or as a reference for a better regulator.

Practise with new numbers

Try it in Python

The program calculates the output voltage and the Zener current of a regulator with $V_z = 6.8$ V and $R_s = 150\ \Omega$, for four inputs and three loads.

Common mistakes

• Joining the Zener diode in forward bias. In forward bias it is an ordinary diode, and it does not regulate.
• Saying the Zener current equals the series current. The load takes part of it. $I_z = I_s - I_L$.
• Forgetting the series resistor. Without it, nothing limits the current, and the diode burns.
• Saying the output is $V_z$ for any input. It holds only if the Zener current stays between its limits.
• Thinking the load current changes the series current. In a good regulator $I_s$ stays fixed, and $I_z$ changes to make up the difference.

Remember these

• A Zener diode works in reverse breakdown at a fixed voltage $V_z$. Below about $5$ V it is the Zener effect, above about $7$ V it is avalanche.
• Regulator: $V_{out} = V_z$, $I_s = \dfrac{V_{in} - V_z}{R_s}$, $I_L = \dfrac{V_z}{R_L}$, $I_z = I_s - I_L$.
• $I_z$ must stay between $I_{z,min}$ and $I_{z,max} = P/V_z$.
• $R_s$ lies between $\dfrac{V_{in,max} - V_z}{I_{z,max} + I_{L,min}}$ and $\dfrac{V_{in,min} - V_z}{I_{z,min} + I_{L,max}}$.

References

• NCERT Physics Class 12, Semiconductor Electronics, the sections on special purpose diodes and the Zener diode as a voltage regulator.
• Horowitz and Hill, The Art of Electronics, third edition, Chapter 2, the section on Zener diodes.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $9.1$ V Zener diode is in a regulator with an input of $15$ V and $R_s = 200\ \Omega$. There is no load. What is the Zener current?

2. A Zener diode of $6.2$ V is rated at $0.5$ W. What is the largest current it can carry?

3. The input of a regulator rises. What happens to the Zener current, the series current and the output?

4. Why is a Zener diode joined in reverse bias?