# Zener diode in series?

Is it OK to use a zener diode in series? Normally I see them across the output of a power source to limit the voltage to the value of the zener.

Putting it in series instead should reduce the voltage by the value of the zener, yes? (so a 5V zener on a 12V supply would give a 7V output).

The thing is, I have a -12V supply and I need a -5V (or near, -7 will do - I have added a couple of Si diodes to drop the voltage a little more) supply from it - BUT - I don't want to lose the -12V supply (that is needed for other parts of the circuit), and I don't have any negative regulators, only positive.

I have tried it and it appears ok, but will it cause problems if I leave it like that?

• So, what did you end up with? I have similar problem - I need to drop about 0.5V out of 5V for the boost DC-DC to work properly. The current is 500mA. I don't see why I need power dissipating resistor if zener itself can do this job. The only question - what is better, simple diode with suitable forward drop or zener with same reverse drop? The answer by @Mike seems to be that zener works just fine. Commented May 14, 2018 at 18:42
• I dunno if you'd find a half-volt zener, though. But if it doesn't need to be exactly 0.5 V and can drift a bit with temperature, you could just use a regular silicon diode, forward biased. Those have forward voltage drops of about 0.6 V. Check the diode's datasheet for a V_F vs. I_F graph and see what V_F is at 500 mA. Also remember that 0.6 V at 500 mA is 300 mW, so the diode needs to not burn up while continuously dissipating that power. Commented May 30, 2018 at 5:50
• If you're willing to tolerate more parts and a larger size, you can use a source-follower circuit. Take the left-hand circuit in steven's answer, then connect an N-channel enhancement-mode FET's gate to Vout+, its drain to Vin+, and then its source becomes your new Vout+. This can handle a lot more current than a zener by itself and also works across a wide range of load currents. The trick will be picking the resistor to provide enough current to the zener at minimum Vin+, and not burn too much power at maximum Vin+. There's negligible current through the gate. Commented May 30, 2018 at 5:55

A regulator would be the best solution, but a zener (one 'n') is ok, at least if you don't want to draw too much power from the regulated voltage. You don't place the zener directly on the -12V, but use a series resistor to limit the current. It's this series resistor which dictates how much current you can draw. The -12V will still be available.

To calculate the resistor value, you have to know how much current your load will draw. Suppose this is 1 mA. Also suppose the zener diode needs 10 mA. That's 11 mA through the resistor. Voltage drop is 12V - 5V = 7V. Then R = 7V / 11 mA = 640$\Omega$.

edit
You may think that in my example I'm exaggerating a bit to have 10mA for a zener if the circuit would require only 1/10th of that. But you'll find that zeners are often specified at much higher currents, like 50mA. Certain newer zeners are specified at much lower currents, these ones only need 50$\mu$A.

YAE (Yet Another Edit)
The reason why you don't want to use the zener in series to get the voltage drop is that especially at low currents the reverse voltage may be much lower than the rated value. This diode for instance is specified at 5.1V @ 50mA, but only drops 1V at 10$\mu$A.

• It's to power a single quad-op-amp chip, so not a lot of power is needed. Commented Jun 23, 2011 at 13:56
• That makes alot more sense, and gives me exactly the -5V I need - thanks. Commented Jun 23, 2011 at 14:00
• @Matt - what's the opamp? Doesn't it work with 12V? Commented Jun 23, 2011 at 14:18
• Could you flip the right side diagram upside down? I like positive voltages higher on the page than negative, in general. Commented Jun 23, 2011 at 16:16
• @endolith - sorry, no can do. Just an image I found on images.google.com. Can't you turn your monitor upside down? ;-) Commented Jun 23, 2011 at 16:25

In a couple rare instances, I've used a Zener diode to get a supply voltage down to the input range of a regulator. For example, I had an off-the-shelf transformer that was working fine in a switching power supply design except for the bias power winding, which had too many turns and generated 34 volts after rectification.

Since bias current requirements were rather low (a few milliamps), and bias voltage required for the chip was 12-32 volts (it has a built-in linear regulator), the previous engineer used a resistor and 30 volt Zener combination like the one in stevenvh's answer.

While the circuit worked, its quiescent current was twice what was expected. We looked at the circuit with the thermal camera (Fluke Ti25--awesome tool if you can afford it) and the Zener was glowing hot.

So we changed the circuit to use a series 10 volt Zener, reverse biased, to get the voltage down to about 24 volts, below the chip's maximum. The built-in regulator does the rest. We just had to make sure there was a minimum amount of current through the diode, but that wasn't difficult.

Safest you handle the situation by setting the Zener parallel with the load. You calculate the resistor and Zener combination so that when your load takes zero Amps, your zener is able to handle the current without burning.

Example your load takes from 0mA to 100mA current. Supply Voltage is 12V. Needed Voltage is 6V.

You need a Zener Pd=5W to handle the current when load takes no current.

Resistor value is (needed Voltage drop)/(max load current) so in this case 6V/0,1A = 60 ohm. Check the power dissipation of the resistor (PdR). PdR = Iload * Iload * R = 0,1A * 0,1A * 60 ohm = 0.6W.

So 6V 5W Zener and 60 ohm 1W resistor are needed exact values without the the fear of overheating.

You can use a resistor value of 56 ohm(easier to find from shop). The Zener can handle the extra current.

• Welcome to EE.SE. Did you notice that the OP (original poster) accepted an answer as being correct seven years ago? Commented Nov 4, 2018 at 15:58

In my opinion the best thing to do in Mike example is to add another 5 volts zener in series before the resistor so that the power lost in the zener and resistor is minimized. The resistor needs to drop only 1 volt at maximum load. R value is 1/0.1 =10 ohms. The added zeners maximum current at full load is the maximum load current=0.1A The original zener maximum current at no load 1v/10 =0.1A As you see we dont have any current lost in the the zener or the resistor that exceeds the circuit requirements. The circuit will run cooler and less power will be lost. These are only thoughts but not trsted so i might be wrong.