Thread: Zener Diodes...

Tweet

Ok, all you electrical engineers, here's a simple one for you. I have 4 12v 5a SLA batteries connected to provide +24, +12, 0 (common/ground), -12v, & -24v DC outputs. Using just zener diodes and biasing resistors, I need to add 18v, 9v, 3v, & -9vDC outputs. The first three I have, but I am having a bit of a time with the negative 9v. I asked one other person, but I don't agree with his answer.

My question is: How do you place the zener & resistor to yield -9vDC? Does the zener attach to the ground/common lead of the battery or the -12v lead? Does it need to be a 9v zener or a 3v?

Here is what I have so far, but I'm just not sure about R4 & D4.

Thanks, but if that drawing was accurate, it would still only answer half of my question. Is D2 a 3v zener or a 9v zener?

However, the drawing is inaccurate. The lower battery is reversed which would make the lower output positive, not negative. Also, D2 is forward biased which would make the lower output .6v (600mv).

Originally Posted by NowThisIsScary

Maybe this will help:

Yes, that fixed it. I looked through a copy of the Encyclopedia of Electronic Circuits and found the answer. If I switch R4 & D4 in my original drawing, it will be correct. Thanks for your help.

Humm,
Have you considered putting your Zener in series with your load and avoid the power loss of a current limiting resistor?

Instead of an 18V zener use a 6 volt one in series. Your load will see 24V - 6V=18V and the current flowing will only be that which your load requires.

I've never seen that in practice. Have you? I like the idea, though. I'll have to add that to my list of trial & error circuits.

It works, the only problem I can envision might be when NO load is drawn and voltage can float high to the source voltage.

I put diodes in my power "in" all the time to protect from reverse hook up. You always get the undesirable diode voltage drop. I've never done it with a zener but I think it would work better.

Just mind your load ratings on the Zener.