Help with Zener diode

I managed to zap a zener diode in a circuit and cannot read its value. It is a fairly simple circuit. It is basically a bank of 14 diodes each with a resistor of 300 ohms in series. The zener just goes across positive and negative near the input connector. Im assuming that it it there for overcurrent protection? I do not know the input voltage of the circuit. Is there a way to figure out what value to replace it with? What would happen if I did not replace it?. The diode is unreadable. I would also assume that the circuit would work without the diode.

Thanks, Tracy

You could test one of the (hopefully) good remaining diodes in the circuit. Apply voltage in small steps (0.2V at a time, for example) and start noting when the voltage applied to the input side of the resistor and the other side of the resistor starts to become non-negligible (5mA or so, so a voltage difference of 1.5V across the resistor).

If you don't replace it, you lose the protection of the zener. It'll work fine until you apply too much voltage, then whatever it is the zener is protecting may be damaged. The circuit should work without the zener as long as you don't apply overvoltage.

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Unfortunately there is only one zener on the board. After thinking about it, Im stating to wonder if its not there for voltage regulation. It is simply wired across power and ground for the whole circuit.


Oops, I thought you meant the "14 diodes" were just like the zener diode that smoked.

If it goes across positive and negative without any resistor it's probably just for protection, not voltage regulation. Do you know anything about the expect input voltage to the circuit? You could then choose a zener with a slightly higher value (though I'm wondering how effective this circuit is without a fuse or resistance in series and if you already managed to zap it...not such great protection).

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Know nothing about the input voltage. Ive gotta assume it must be about 5v based on what it is. 15 LED and each with its own resistor all protected by this one zener across positive and negative. Ill post a couple pics

If you look at the backside of the board the zener is the bottom let two solder pads. The power input is the 2nd and 3rd one up from the bottom left corner.

The zener doesn’t have a series resistor of its own, so if the input > VZ then it’ll conduct like a son-of-a-gun (it’s not there as a zener regulator.)
What’s the value of those resistors? Can’t differentiate the colours very well. Anyway, each is in series with two LEDs. If they’re orange-black-brown, 300?, then your LED current would determine a rough zener voltage. Two LEDs, 4V + (10mA * 300?) = 7V: a 9V zener. If it got to 9V, then there’d be 15mA through each LED string.

you are correct...they are 300 ohm resistors. I think that I could barely make out part of the digit 7 on what was left of the diode body. So you are saying that it is there for circuit protection not regulation? What happens if I leave it out? Other than no circuit protection?

Tell us what that is, preferably with a model/part number as well, and we can probably find the exact specs for it.

It is the LED board out of a Corvette heads up display. I can’t buy just the card or I would. I can put 5v on the input and it works fine without the diode but I want to make sure that I can leave it out. It’s a big job to put the display in the dash.

There’s no resistor, on-board, to limit I**Z** if it went into conduction (Vin > VZ. It’s sort of there sacrificially, if some over-voltage situation persisted.
The board is just a bunch of LEDs, strings of two each, in parallel. Unless it’s going to be in a situation where VIN has the potential (forgive the pun?) to stray > 9V, I see nothing to bust a gut over.

Thats what I needed to hear…Ill leave it out and install it as is.