# Electronic Circuits and Projects Forum

1. ## Led vr

Hi

In my circuit I want to have a red HE LED light when my 9v pp3 battery is applied the WRONG way round as a warning light. problem is the reverse voltage of the led max is 5v and this is what it needs to be when the battery is the right polarity.

I have tested the led the wrong way round and it did not damage it but it worries me having it like this as I fear it may damage it if applied long enough, so I thought of using a resistor.

Its a 12v high effciency red and reverse amps is 10 micro amps. Using ohms law 9v-5v=4v, ohms law is volts = amp x ohms. So my 4v i want to illimante in the formula is 4v / 0.00001 = 400000 ohms or 400k or 0.4M, Is this correct.

I also wondered whether it would be better to put another diode that can take 9v reverse voltage in series with it rather than use a resistor

What would you do, use a resistor (tell me if I am wrong with this rating of ohms), the other diode in series or nothing at all and let the led take the 9v in reverse. I will attach the data sheet for the led.

Thank you for any help provided and for reading.

2. If you put a huge resistor in series then when it's forward biased it wont lite up.
Everybody goes with using another diode in series.

Also it's not guaranteed that at 10uA voltage drop on reverse biased diode is 5V;
it is not a zener diode (maybe it's avalanche breakdown ).

3. The resistor is not a good idea. The current or 10 μA is the maximum current that the LED will pass at 5 V reverse. It will probably pass far less than that, and the resistor will no drop nearly that much voltage. I realise that the data sheet says that the reverse current is typically is 10 μA, but I simply don't believe it. It will be far less.

Any resistor that significantly reduces the voltage with the LED in reverse will make it just about invisible forward.

The LED will probably be fine at 9V. If you search this forum for transformerless power supplies, and running LEDs from mains, you'll see that many LEDs can stand really big reverse voltages.

What you want is a standard diode. A 1N4148 or 1N4001 can survive 50 V in reverse and will definitely prevent damage to the LED, and will only reduce the voltage by 0.7 V in the forward direction, so a small reduction in the series resistor will restore the current to what you want. You should maybe put a large value resistor, maybe 1 MΩ, in parallel with the LED to make sure that the reverse voltage is near zero even if the standard diode has a tiny bit of leakage current.

4. Thank you for your reply, thats what I was thinking but if 9v was supplied in reverse directly to the led without diode or resistor is it likely to burn out

5. Driver300
I have a few spare 1N007 diodes, will these do?

6. Originally Posted by e44-72
I have a few spare 1N007 diodes, will these do?
Yes. They 1000V ,1A diodes, usable up to few hundred Hz.

7. if 9v was supplied in reverse directly to the led without diode or resistor is it likely to burn out
Some diodes will burn out, others won't. It's a risk you probably don't want to take. Any series diode will do.

8. Thanks for the help

Driver300, if I also use a large value resistor in parallel with the led should it be in parrallel with just the led or in parrallel with both the diode and the led.

9. Originally Posted by e44-72
Thanks for the help

Driver300, if I also use a large value resistor in parallel with the led should it be in parrallel with just the led or in parrallel with both the diode and the led.
The large value resistor must be in parallel with the LED, and not the diode. It is to provide a path for any slight leakage through the diode.

You need a series resistor of a few hundred ohms to limit the current that lights the LED. That resistor has no effect when the diode is not lit.

10. Ok, thank you for the comments
If you did not use the resistor as well would it still work ok or is it essential

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