# Putting Zeners in Parallel

• xcvxcvvc

#### xcvxcvvc

What happens when you put two 5.1 V Zeners in parallel? Someone told me that if the output on my Zener regulated power supply was off, doing so would help make my output closer to 5.1 V. Before I take the advice of a student, I'd like to hear it from a more experienced person from here. I am cautious, because my book neglects the subject and google finds nothing on it. This absence of information could signify unpredictability or predictably useless outputs when putting two zeners in parallel. Besides, two zeners in parallel ideally behave like two voltage sources in parallel, which is a bad thing.

Diodes, like any components, have a random spread in their operating values. The extent to which they share current depends on how closely they are matched compared to the width of the knee in their I-V curves. In practice, the diode with a lower Zener voltage will carry most or all of the current and the other will just sit there with little to do.

What happens when you put two 5.1 V Zeners in parallel? Someone told me that if the output on my Zener regulated power supply was off, doing so would help make my output closer to 5.1 V. Before I take the advice of a student, I'd like to hear it from a more experienced person from here. I am cautious, because my book neglects the subject and google finds nothing on it. This absence of information could signify unpredictability or predictably useless outputs when putting two zeners in parallel. Besides, two zeners in parallel ideally behave like two voltage sources in parallel, which is a bad thing.

In addition to marcusl's comments, keep in mind that real Zener diodes are rated at their voltage for some particular test current. If you have less than that current passing through them, you get a [STRIKE]higher[/STRIKE] lower voltage.

So putting two Zeners in parrallel will [STRIKE]raise[/STRIKE] lower the output voltage slightly. It's more important to just make sure that the Zener current matches the datasheet number, in order to get as close as possible to the rated Zener voltage.

BTW, do you know what the tempco of a 5.1V Zener diode is? Why is it special? (Quiz Question)

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In addition to marcusl's comments, keep in mind that real Zener diodes are rated at their voltage for some particular test current. If you have less than that current passing through them, you get a [STRIKE]higher[/STRIKE] lower voltage.

So putting two Zeners in parrallel will [STRIKE]raise[/STRIKE] lower the output voltage slightly. It's more important to just make sure that the Zener current matches the datasheet number, in order to get as close as possible to the rated Zener voltage.

BTW, do you know what the tempco of a 5.1V Zener diode is? Why is it special? (Quiz Question)

the temperature coefficient changes the zener's on voltage basically linearly. for diodes with zener breakdown (V < 5.6), it is negative. for avalanche, it is positive. By the way, i fixed my nominal 5v output ripple by increasing the capacitance from 57 to 57 + 22. The ripple went from ~56mV to ~45mV under max load conditions. With no load, my power supply never had difficulty(5mV ripple)

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In addition to marcusl's comments, keep in mind that real Zener diodes are rated at their voltage for some particular test current. If you have less than that current passing through them, you get a higher voltage.

So putting two Zeners in parrallel will raise the output voltage slightly. It's more important to just make sure that the Zener current matches the datasheet number, in order to get as close as possible to the rated Zener voltage.

BTW, do you know what the tempco of a 5.1V Zener diode is? Why is it special? (Quiz Question)

I must disagree Berkeman. Zeners are not negative resistance devices. Lower current will result in lower voltage across the device, not greater.

The answer to the quiz question can be found here. http://en.wikipedia.org/wiki/Zener_diode

I must disagree Berkeman. Zeners are not negative resistance devices. Lower current will result in lower voltage across the device, not greater.

Yes, you are correct. Thanks for catching that. I was picturing the Zener plot (like the one in your link), but did not interpret it right for this question. Duh. I've gone back and corrected my post with strikeout so it hopefully doesn't confuse anybody.