Practice problems

Zener diode practice problems

A Zener regulator is a series resistor and a Zener diode: the Zener holds its voltage, and the resistor takes the difference and sets the current. These ten problems find that current, split it between the load and the Zener, size the resistor and check the Zener won't overheat when the load is unplugged. Type your answer to check it, and open the working whenever you want it.

The method

R = V_in − V_ZI_load + I_Z,min

Give the Zener at least I_Z,min = 5 mA, or it goes soft and stops holding
Round the resistor down to a standard value, so the Zener isn't starved
The resistor's current is I = (V_in − V_Z) ÷ R. The load takes its share, and the Zener takes the rest
The Zener's power is P = V_Z × I_Z, at its worst with no load

The problems

Type your answer and tap Check. Every one has the working, step by step.

Problem 1 of 10
A 5.1 V Zener regulator runs from a 9 V battery. What voltage is across the series resistor?
Show the working
1. The Zener holds the bottom of the resistor at 5.1 V. The resistor takes the rest.
2. 9 V − 5.1 V = 3.9 V
Answer: 3.9 V
Problem 2 of 10
The same regulator (9 V in, 5.1 V Zener) has a 150 Ω series resistor. What current flows through the resistor?
Show the working
1. The resistor has 9 V − 5.1 V = 3.9 V across it.
2. I = 3.9 V over 150 Ω = 0.026 A, which is 26 mA.
Answer: 26 mA
Problem 3 of 10
26 mA flows through a Zener regulator's series resistor, and the load draws 18 mA of it. What current flows through the Zener?
Show the working
1. The resistor's current splits between the load and the Zener (Kirchhoff's current law).
2. I_Z = 26 mA − 18 mA = 8 mA, comfortably above a 5 mA minimum.
Answer: 8 mA
Problem 4 of 10
A 6.2 V Zener carries 30 mA. How much power does it turn into heat, in milliwatts?
Show the working
1. P = V_Z × I_Z
2. P = 6.2 V × 30 mA = 186 mW. Volts times milliamps gives milliwatts.
Answer: 186 mW
Problem 5 of 10
A regulator has 12 V in, a 3.3 V Zener and a 10 mA load. Giving the Zener at least 5 mA, what is the largest series resistor?
Show the working
1. The resistor gets 12 V − 3.3 V = 8.7 V, and must carry 10 mA + 5 mA = 15 mA.
2. R = 8.7 V over 0.015 A = 580 Ω. Round down to the E12 value 560 Ω, so the Zener gets a little more than 5 mA, not less.
Answer: 580 Ω
Problem 6 of 10
A regulator has 9 V in, a 5.6 V Zener and a 120 Ω series resistor. The load draws 15 mA. What current flows through the Zener?
Show the working
1. The resistor's current: I = 9 V − 5.6 V over 120 Ω = 3.4 V over 120 Ω ≈ 28.3 mA.
2. The load takes 15 mA, so the Zener gets 28.3 mA − 15 mA ≈ 13.3 mA.
Answer: 13.3 mA
Problem 7 of 10
The load is unplugged from that regulator (9 V in, 5.6 V Zener, 120 Ω). How much power does the Zener take now, in milliwatts?
Show the working
1. With no load, the Zener takes all of the resistor's 28.3 mA.
2. P = 5.6 V × 28.33 mA ≈ 159 mW, well under a 0.5 W (500 mW) rating.
Answer: 159 mW
Problem 8 of 10
A sensor needs 5.1 V and draws 25 mA, from a 9 V battery. Giving the Zener at least 5 mA, which E12 series resistor do you fit?
Show the working
1. The resistor gets 9 V − 5.1 V = 3.9 V and must carry 25 mA + 5 mA = 30 mA.
2. R = 3.9 V over 0.03 A = 130 Ω at most.
3. 130 Ω isn't an E12 value. Round down, so the Zener isn't starved: 120 Ω, giving 32.5 mA.
Answer: 120 Ω
Problem 9 of 10
Stretch: a regulator has 12 V in, a 5.1 V Zener and a 220 Ω series resistor. The Zener needs at least 5 mA. What is the most current the load can draw while the output stays at 5.1 V?
Show the working
1. The resistor's current is fixed by the Zener: I = 12 V − 5.1 V over 220 Ω = 6.9 V over 220 Ω ≈ 31.4 mA.
2. The Zener must keep 5 mA of that, so the load can have 31.4 mA − 5 mA ≈ 26.4 mA. Draw more and the Zener stops regulating.
Answer: 26.4 mA
Problem 10 of 10
Stretch: an 8.2 V Zener rated at 0.5 W runs from 12 V. If the load is ever unplugged, the Zener takes the whole current. What is the smallest series resistor that keeps it within its rating?
Show the working
1. The most current the Zener can take: I = P over V_Z = 0.5 W over 8.2 V ≈ 61 mA.
2. The resistor has 12 V − 8.2 V = 3.8 V across it.
3. R = 3.8 V over 0.061 A ≈ 62.3 Ω. This is the smallest safe value, so the E12 pick goes up, to 68 Ω: about 0.46 W with no load.
Answer: 62.3 Ω

Build it in a free lesson

“Zener regulators” is lesson 3 of 6 in Silicon Ridge, Electronics World 5. Free with an account. Lessons unlock in order, so you’ll do the 30 before it first.

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Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Zener regulators” is lesson 3 of 6 in Silicon Ridge, Electronics World 5. Free with an account. Lessons unlock in order, so you’ll do the 30 before it first.

Or start from the very first lesson, no account needed

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The ideas behind it

Questions people ask

Why do I round a Zener's resistor down, not up?

The resistor must pass enough current for the load and the Zener's minimum, a few milliamps. A bigger resistor passes less, which starves the Zener and the output sags. So round down to the next standard value, then check the Zener's power with no load.

How much current does the Zener take?

Whatever the load leaves. The resistor's current is fixed by (Vin − Vz) ÷ R. Subtract the load's current and the rest flows through the Zener. With no load, the Zener takes all of it.

How do I work out the power in a Zener?

P = Vz × Iz, the Zener voltage times its current. The worst case is no load, when the Zener carries the resistor's whole current. Keep that under the Zener's power rating, often 0.5 W, with some margin.