Zener diode in series?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.