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I read statements like ‘An LED has a fixed voltage drop regardless of the current running through it’, but looking at the following examples (using arbitrary numbers), can someone confirm that the following are true/correct -

  1. If I had a 12V battery, 500 Ohm resistor and an LED with a forward voltage of 2V, the current in the circuit would be 20mA ((12 – 2)/500). The current-voltage curve of an LED shows a small increase in current would equal a tiny increase in forwarding voltage, so for example if I wanted the current to be 40mA, the LED would have a tiny increase in voltage drop?

  2. If I had a 3V battery, no resistor and an LED with a forwarding voltage of 2V, what happens? KVL states that the total voltage drop of the circuit must equal the power source. So in simplistic terms, does the battery just pump out as much current until the LED is forwarding 3V, and the current of the circuit is whatever the current-voltage curve of the LED says is at 3V?

  3. If both of the above are true, then is the statement regarding LEDs having a fixed voltage drop, is it more correct to say that an LED has roughly a fixed voltage drop when a similar amount of current is running through it?

  4. …and finally, and this may never happen or even be possible, for example a 20V battery and just an LED, apart from the LED disintegrating, would it ever be possible for the power supply to not be able to supply enough current so that the LED can reach a voltage drop of 20V, and therefore breaking KVL, what happens in that situation?

I am fairly new to this stuff, and not looking for a too technical explanation, but really just a confirmation of the above questions, and if they are wrong a simple explanation as to why that is.

Thanks.

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  • \$\begingroup\$ Hi @stefandz, that was my question I asked a few days ago. I know then you mentioned some of the stuff that's in this question, but I've added to it and just wondered if someone could go through it each point and make sure my understanding is correct. Thanks. \$\endgroup\$
    – RJSmith92
    Commented Aug 16, 2015 at 13:05
  • \$\begingroup\$ Single specific design questions are required. \$\endgroup\$
    – Leon Heller
    Commented Aug 16, 2015 at 13:08
  • \$\begingroup\$ "breaking KVL" that's rather difficult. Kirchhoff's laws are the fundamental equations of circuit theory and are an approximation of Maxwell's equations, the foundation of all electromagnetism. The only way for Kirchhoffs laws to break is when you work outside the assumptions behind them, i.e. the circuit is much smaller than the wavelength of the signal. At DC the wavelength is virtually infinite, so at macroscopic level KVL at DC is always valid (at microscopic level, or more precisely, at quantum mechanics level it's another story). \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Commented Aug 16, 2015 at 14:38
  • \$\begingroup\$ You can model a diode more accurately than with just a fixed voltage drop. However, for design work you need to allow for the fact that each individual diode is actually slightly different, so that using a more detailed model doesn't usually improve your design in a meaningful way. \$\endgroup\$
    – The Photon
    Commented Aug 16, 2015 at 16:01

1 Answer 1

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  1. Yes

  2. No, practically speaking this will not happen because the 3V battery is not an ideal voltage source. It has a non-ideal source impedance, sometimes referred to as its internal resistance, which we can model with a resistor. For example, a CR2032 has an internal resistance of around 15 ohms and a forward voltage of around 3V. If you connect a 2V LED to this, you would expect around (3-2)/15 = 66mA of forward current. Depending on the battery chemistry and physical dimensions, this internal resistance could be larger or smaller. You can read more about internal resistance here.

  3. Not really - if you're putting enough current through an LED to significantly change its forward voltage (i.e. from 2V to 3V, a 50% increase) you're almost certainly going to have released its magic smoke.

  4. If the power supply cannot supply the required current, this is due to one of two things. It could be the internal resistance situation described above, where the rest of the voltage is dumped across the non-ideal source resistance of the supply. Alternatively, it could be due to an active current limiting circuit within the supply which dynamically reduces the output voltage of its output in order to maintain a maximum output current. The second regime is actually the basis of constant-current drive circuits which are commonly used to drive current mode devices such as LEDs!

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  • \$\begingroup\$ Thanks again @stefandz, much appreciated answering all those. Regarding point 3, OK, maybe not a 50% increase but small differences with a small changes in current? Regarding point 2, don't know much about this yet and will read more later, but quickly, in your example with the 3V battery and an LED with a 2V drop at 20mA, the current would be around 66mA and the LED would have slightly higher voltage drop than 2V. If i then used battery with 5V and an internal resistance of 25 Ohms (just made it up), the current would be (5-2)/25 = 120mA and the voltage drop of the LED would be a bit higher? \$\endgroup\$
    – RJSmith92
    Commented Aug 16, 2015 at 17:23
  • \$\begingroup\$ Yes, but at 120mA you're starting to get beyond the absolute maximum rating of a normal LED (again, this is LED specific) and likely to get into thermal runaway where the voltage drop starts to decrease with temperature increase (counterintuitive, but that is because the LED is a semiconductor device and you are giving the charge carriers higher mobility), you get higher current draw and this runs away until the LED goes pop! You know, given your interest, if I were you I would buy a grab-bag of resistors, a bag of cheap LEDs, a power supply and try out some of these scenarios for yourself :) \$\endgroup\$
    – stefandz
    Commented Aug 16, 2015 at 17:32
  • \$\begingroup\$ I might just to do that :) One last thing that I forgot to ask regarding your example with the 3V battery and 2V LED. The LED will still have voltage drop of around 2V at 66mA, is the 3V battery now only pushing with a voltage of around 2V so KVL is broken? \$\endgroup\$
    – RJSmith92
    Commented Aug 16, 2015 at 22:32
  • \$\begingroup\$ I'll be honest with you - KVL never gets broken until you're way up in RF territory, as @LorenzoDorati mentions in a comment above. The 3V battery is no longer a 3V battery at this stage - it's a 3V battery with an internal resistance (inside its casing) dropping 1V so that at its output you're only getting 2V. That is to say there's 3V inside, 1V dropped internally and 2V available externally. \$\endgroup\$
    – stefandz
    Commented Aug 16, 2015 at 22:39
  • \$\begingroup\$ Brilliant, thanks for the quick response as always, much appreciated! \$\endgroup\$
    – RJSmith92
    Commented Aug 16, 2015 at 23:01

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