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What is the difference between "forward" and "reverse" voltages when working with diodes and LEDs?

I realize this question is answered elsewhere on the interwebs such as wikipedia, but I am looking for a short summary that is less of a technical discussion and more a useful tip to someone using diodes in a hobby circuit.

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    \$\begingroup\$ LED means Light Emitting Diode. \$\endgroup\$
    – Dean
    Commented Mar 2, 2011 at 23:16
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    \$\begingroup\$ I added "and LED's" as a way of capturing search queries that don't spell out the "D" :) \$\endgroup\$
    – JYelton
    Commented Mar 3, 2011 at 0:26
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    \$\begingroup\$ I removed the apostrophe from "and LED's" because I'm pedantic. :) \$\endgroup\$
    – Adam Lawrence
    Commented Feb 12, 2014 at 14:56
  • \$\begingroup\$ @Madmanguruman Most of my edits definitely qualify for that description, too. Thanks. :) \$\endgroup\$
    – JYelton
    Commented Feb 12, 2014 at 16:10

7 Answers 7

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The forward voltage is the voltage drop across the diode if the voltage at the anode is more positive than the voltage at the cathode (if you connect + to the anode).

You will be using this value to calculate the power dissipation of the diode and the voltage after the diode.

The reverse voltage is the voltage drop across the diode if the voltage at the cathode is more positive than the voltage at the anode (if you connect + to the cathode).

This is usually much higher than the forward voltage. As with forward voltage, a current will flow if the connected voltage exceeds this value. This is called a "breakdown". Common diodes are usually destroyed but with Z and Zener diodes this effect is used deliberately.

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    \$\begingroup\$ @JYelton, if you exclude zener diodes, most every diode you use will conduct when it has a "forward" voltage, and will cutoff when the reverse voltage is met. \$\endgroup\$
    – Kortuk
    Commented Mar 2, 2011 at 23:36
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    \$\begingroup\$ @Kortuk, I wish your comment was an answer and also accepted. The other answers are correct, but the point of a diode is to be a one-way check valve for current. Breaking a diode is how interesting things like Zener diodes were found. \$\endgroup\$
    – Krista K
    Commented Jun 14, 2012 at 18:16
  • \$\begingroup\$ What do you mean by 'more positive'? Can a current be 'less positive'? And what is a Zener diode used for? \$\endgroup\$
    – Agent Zebra
    Commented Aug 27, 2015 at 3:54
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    \$\begingroup\$ @AgentZebra Compared to a common reference, yes, see this picture. A Zener diode can be used as a voltage reference or a voltage limiter. \$\endgroup\$
    – AndreKR
    Commented Aug 27, 2015 at 5:43
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    \$\begingroup\$ @fukanchik When you connect the diode as described in my answer, you can expect the voltage difference between the diode terminals to be approximately the forward voltage given in the datasheet. \$\endgroup\$
    – AndreKR
    Commented Jul 15, 2016 at 3:24
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Typically the forward voltage is the voltage at which current starts to flow in the normal conducting direction (as mentioned above it's somewhere in the range 0.3-0.6v)

Reverse voltage is sort of the same thing - it's the voltage where current starts to flow when the diode is in the normally non-conducting region - this is also the point where the diode is likely to turn into a charred mess as all the internal semiconductor stuff turns to mush (choose a value somewhat larger than the largest PEAK [not RMS] AC voltage the diode will see)

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    \$\begingroup\$ "Above" is relative; answers are first sorted by vote count and then randomized. I think you meant "As mentioned by Leon". \$\endgroup\$
    – Kevin Vermeer
    Commented Mar 3, 2011 at 1:12
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Forward-bias is when the anode (the pointy part of the symbol) is positive and the cathode (the bar) is negative. Reverse-bias is when the anode is negative and the cathode is positive. A lot of current flows when the diode is forward-biased, provided that the voltage is higher than 0.6V or so for a silicon diode or 0.3V or so for a germanium device. A very small amount of current flows if a diode is reverse-biased.

If you have a DVM and some diodes, you can check it for yourself. Diode cathode leads are usually identified with a band, so if you switch the DVM to a low resistance setting, and connect the leads across the diode in both directions, you should see a low resistance in one direction and a high resistance in the other direction, provided that the DVM is supplying a high enough voltage. Some DVMs have a special diode test setting that is easier to use.

LEDs usually have a flat against the cathode lead.

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    \$\begingroup\$ Why's it called Forward-bias? why the word bias? And is that the same as forward voltage? \$\endgroup\$
    – Agent Zebra
    Commented Aug 27, 2015 at 3:55
  • \$\begingroup\$ Forward is just the opposite of reverse. It is the same as forward voltage. \$\endgroup\$
    – Leon Heller
    Commented Aug 27, 2015 at 8:09
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Excavating in 3... 2... 1...

Just so that the information is condensed here and I like to know where to find my offspring, I'd add typical forward voltages for common LED as a quick reference for everyone. (And also because I like to dig an old thread on the 18th December.)

As per Wikipedia:

Typically, the forward voltage of an LED is about 1.8–3.3 volts; it varies by the color of the LED. A red LED typically drops 1.8 volts, but voltage drop normally rises as the light frequency increases, so a blue LED may drop around 3.3 volts.

LED forward voltage quick reference

  • IR LED drops about 1.5V
  • Red: ~2V
  • Amber: ~2V
  • Yellow: ~2V
  • Green: ~2.5V
  • Blue: ~3.5V
  • White: ~3.5V
  • Laser diodes: ~1.5V but may vary a lot with wavelength (like 375nm to 3300nm)
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    \$\begingroup\$ While I agree these forward voltages are helpful, this doesn't really answer the original question which is about contrasting forward and reverse voltages with respect to diodes. You might have good response by posting a separate question, "What are some common diode forward voltages?" and posting this answer. See also this meta answer about self-answering questions \$\endgroup\$
    – JYelton
    Commented Dec 18, 2015 at 16:29
  • \$\begingroup\$ I clearly stated I just wanted to "add typical forward voltages" information. Take it as supplemental info about your very question, for instance. \$\endgroup\$
    – user59864
    Commented Dec 18, 2015 at 16:31
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    \$\begingroup\$ I'm not disputing that you are adding good supplemental info; just saying that people looking for this might find it more quickly if it's attached to a question reflecting it. (Plus you'd get some upvotes on both the question and answer, at least from me!) \$\endgroup\$
    – JYelton
    Commented Dec 18, 2015 at 16:36
  • \$\begingroup\$ Not that I'm running after reputation, well, I might consider adding a dedicated question about that. \$\endgroup\$
    – user59864
    Commented Dec 18, 2015 at 16:37
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Although you mention "voltage," I believe you mean bias. If this is correct, then "forward bias" is the application of a voltage in such a way that the diode "shows" a low resistance. "Reverse bias" causes the diode to show a high resistance.

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  • \$\begingroup\$ My question is phrased to address common (at least in my experience) expressions, but yes, bias is what is meant by the terms. Leon's answer covers this, though I think your answer explains it differently and is helpful as well. \$\endgroup\$
    – JYelton
    Commented Dec 26, 2014 at 19:25
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Forward voltage is the one that makes the diode to conduct while the reverse voltage is the one that makes a diode very bad conductor or almost an open circuit unless diode "breaks down".

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For perfect junction diodes there is a relation between the current and the voltage given by Shockley's diode equation I = Is(exp(Vd q/nkT). You can solve this to get Vd, the diode voltage, as a function of I. But when you have a resistor in series with the diode, you can NOT solve the circuit; you must use successive approximations. (Unless you have a WP-34 calculator with the built in Lambert W function, of course.) Ordinarily, you can replace nkT/q with about 0.6 volts, and saturation current Is with about 1 mA, and your voltage calculation will be in the ball park.

Google "shockley diode equation" to learn more than I can type here.

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