Power & Energy
Why does my zener regulator's voltage drop?
A zener regulator's output drops when the series resistor can't pass enough current. The resistor feeds the load and the zener together, and the zener only holds its voltage by taking whatever the load leaves over. Ask for more than the resistor can supply, by loading it harder or letting the input sag, and nothing is left for the zener. It goes quiet, and the output falls with the battery.
The current budget
The resistor sets the current coming in:
Take a 12 V battery, a 5.1 V zener, a 100 Ω series resistor and a 100 Ω load. The resistor passes (12 − 5.1) ÷ 100 = 69 mA. The load at 5.1 V takes 51 mA. The zener takes the other 18 mA and holds the voltage steady. As long as Iin stays above the load's current, the load never notices the battery changing.
When the battery sags
At 11 V the resistor still passes 59 mA: the zener takes 8 and the load stays at 5.1 V. At 9 V it passes only 39 mA, less than the load wants. The zener stops conducting, the resistor and load become a plain divider, and the output falls to about 4.5 V. The edge is where Iin equals the load current:
Live experiment
Let the battery sag
A battery feeds the series resistor R1, with a 5.1 V zener and a 100 Ω load side by side after it. Pick R1, then let the battery sag from 12 V. With 100 Ω the load holds 5.1 V down to about 10.2 V; 1 kΩ and 220 Ω starve it from the start; 47 Ω holds lower but runs the zener hot.
“Holding a voltage” is lesson 6 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 5 before it first.
When the resistor is too big
Swap the 100 Ω for 1 kΩ and only 6.9 mA gets in, far below the load's 51. The regulator never starts: the divider gives the load about 1 V. The zener's 5.1 V is a ceiling, not a promise.
The other end: heat
A smaller resistor holds lower, but every milliamp the load doesn't use burns in the zener. With 47 Ω, 147 mA comes in and the zener takes 96 mA, about 0.5 W, at its limit. Unplug the load and it takes everything. Check the resistor too: 100 Ω with 6.9 V across it makes nearly half a watt. For bigger loads, a linear regulator wastes less, because it draws only what the load needs.
Free lesson
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.
“Holding a voltage” is lesson 6 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 5 before it first.
Or start from the very first lesson, no account neededWork it out
Practice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: Only 7 mA can get through, so the voltage collapses. 6.9 V ÷ 1000 Ω is 7 mA, less than the load wants, so the zener gets nothing and stops regulating. The load pulls the voltage down to about 1 V.
- Why not “The zener holds 5.1 V as before”? A zener can only hold its voltage while current is left over for it. 7 mA can't even feed the load.
- Why not “The zener overheats”? With so little current coming in, the zener gets none at all, so it runs cold.
- Why not “The load gets 12 V”? The 1 kΩ resistor drops most of the battery's voltage, so the load gets about 1 V, not 12.
Show the answer
Answer: Stays at 5.1 V, as long as enough current still comes through. Less current comes through the resistor, and the zener simply takes less of it. The load keeps its 5.1 V until the zener runs out of margin.
- Why not “Rises”? Less battery voltage can't raise the output. The zener takes a little less current and the load stays put.
- Why not “Drops to 0”? The load still gets plenty of current. Even a starved regulator only sags; it doesn't drop to 0.
- Why not “Drops to 4 V”? At 11 V the resistor still passes 59 mA, more than the load's 51, so the zener keeps holding.
Show the answer
Answer: Drops to about 4.5 V: too little current gets in, and the zener goes quiet. (9 − 5.1) ÷ 100 is only 39 mA, less than the load's 51, so there's nothing left over for the zener and it stops holding. A zener regulator only holds while R1 can pass more than the load wants: with 100 Ω, down to about 10.2 V.
- Why not “Gets nothing at all”? Current still flows through the resistor into the load. It's just too little to hold 5.1 V.
- Why not “Stays at 5.1 V”? At 9 V only 39 mA gets through, less than the load's 51 mA, so nothing is left for the zener.
- Why not “Rises to 9 V”? The series resistor is still in the way, sharing the 9 V with the load.
“Holding a voltage” is lesson 6 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 5 before it first.
Questions people ask
What is the minimum input voltage for a zener regulator?
The zener voltage plus the drop across the series resistor at full load: Vin(min) = Vz + I(load) × R. For a 5.1 V zener, a 100 Ω resistor and a 51 mA load that's 5.1 + 5.1 = 10.2 V. Allow a little more so the zener keeps a few milliamps of its own.
Why is my zener regulator's output below the zener voltage?
The series resistor isn't letting enough current through for the load, so none is left for the zener, and it stops conducting. The resistor and the load are then just a voltage divider. Use a smaller resistor, a higher input voltage, or a lighter load.
Why does my zener get hot when nothing is connected?
The resistor passes a fixed current, and the zener takes whatever the load doesn't. With no load it takes all of it: 69 mA at 5.1 V is 0.35 W. Size the zener's rating for the no-load case.
Related topics
- How does a Zener diode work?A diode built to conduct backwards at a set voltage, then hold it: the simplest regulator.
- How does a linear voltage regulator work?A zener sets the voltage and a pass transistor carries the current: the circuit inside every 7805 and LDO.
- Voltage divider explainedTwo resistors that split a voltage: the formula, potentiometers, and why a load drags the output down.
- Electrical power explained: P = V × IP = V × I: how fast energy turns into heat, and why a resistor has a power rating.