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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?

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2 Answers

up vote 10 down vote accepted

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.

enter image description here

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.

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It's to power a single quad-op-amp chip, so not a lot of power is needed. –  Majenko Jun 23 '11 at 13:56
    
That makes alot more sense, and gives me exactly the -5V I need - thanks. –  Majenko Jun 23 '11 at 14:00
    
@Matt - what's the opamp? Doesn't it work with 12V? –  stevenvh Jun 23 '11 at 14:18
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Could you flip the right side diagram upside down? I like positive voltages higher on the page than negative, in general. –  endolith Jun 23 '11 at 16:16
4  
@endolith - sorry, no can do. Just an image I found on images.google.com. Can't you turn your monitor upside down? ;-) –  stevenvh Jun 23 '11 at 16:25
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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.

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