If I had an LED that had a forward voltage of 2.2 volts with a forward current of 20mA and I somehow created a battery that was exactly 2.2 volts and I put that LED, and only that LED, on that battery, what would the current be?

I'm just confused how diodes/LED's work. I understand usually you would have say a 5V battery and the resistor that you put in series with the LED would control the current. I'm just trying to better grasp LED characteristics.


  • \$\begingroup\$ You could measure the voltage on a LED at 20mA, then supply exactl that same voltage to it and have it take 20mA. However, even slight deviations from that voltage and changes in LED temperature will have proportionately large effects on the LED current. \$\endgroup\$ Commented Dec 1, 2011 at 20:18
  • 1
    \$\begingroup\$ Get a data sheet and look at the diagrams. \$\endgroup\$
    – starblue
    Commented Dec 1, 2011 at 22:05
  • \$\begingroup\$ @OlinLathrop - I think you may have meant 'disproportionately'. 'Proportionately' sort of suggests linearity. \$\endgroup\$
    – JustJeff
    Commented Dec 2, 2011 at 0:02
  • \$\begingroup\$ @JustJeff: Oops, yes. Too late to edit now. \$\endgroup\$ Commented Dec 2, 2011 at 0:12

2 Answers 2


In theory, this would work, and you could get 20mA. However, this is a very fragile system that you describe. If something shifts slightly, you won't get your desired current. For example, you would need to control/know the following:

  • The temperature that the diode operates at, possibly accounting for self-heating
  • The exact voltage that a diode draws 20mA at your given temperature (note that the datasheet will probably give a "nominal" value or a tolerance - you would need to know the exact voltage.
  • Your power supply (battery) would need to be much more precise than is practical for just driving a LED

The problem is that diodes change their current dramatically with a very small change in voltage. This can be seen in the Shockley diode equation:

$$\Large I=I_s ( e^{\frac{V}{n V_T}}-1) $$

This shows that the current (I) varies exponentially with the applied voltage (V). So while it's possible to apply a fixed voltage to a diode and get a precise current, it's hard. Diode current is relatively easy to control in current mode, as you can make a rough current source with a resistor and enough voltage headroom. This is what is happening when you have a resistor in series with your diode at 5V. An alternative is a constant current sink, which is easy to do on an IC. These show up as LED driver chips that can sink a programmed current, and they work well too.

  • \$\begingroup\$ odd... my gut instinct would have been that the diode would create a short circuit, since there is nothing to limit the current. \$\endgroup\$
    – Tevo D
    Commented Dec 1, 2011 at 20:19
  • \$\begingroup\$ Thanks for the great info. When you say an LED driver chip can sink a programmed current, what do you mean? When I've read up on LED drivers, it seems they still need a resistor thrown in there. Thanks \$\endgroup\$ Commented Dec 1, 2011 at 20:21
  • \$\begingroup\$ The intrinsic characteristics of the diode limit the current. Again, since there is an exponential relationship between voltage and current, it would definitely be difficult to use a diode this way. Eventually (if the diode doesn't fuse) the series resistance of the diode would limit the current. \$\endgroup\$
    – W5VO
    Commented Dec 1, 2011 at 20:23
  • \$\begingroup\$ Learn something new every day... or lots of new things on a good day. Will have to ponder this idea some more. Thanks \$\endgroup\$
    – Tevo D
    Commented Dec 1, 2011 at 20:28
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    \$\begingroup\$ @TylerDeWitt I'm referring to chips such as the TI TLC5940, et. al. that basically act as switchable current sinks. There may be a resistor to set the reference current, but there is no resistor in series with a LED. \$\endgroup\$
    – W5VO
    Commented Dec 1, 2011 at 20:33

It will be obvious on consideration that

  • "An LED drawing 20 mA when you applied 2.2V would draw 20 mA when you applied 2.2 Volts" :-)

ie in you ideal situation you have stability. But your question shows that you are well aware that in the real world this ideal situation would not apply. Realising this is a good start.

As Starblue said, looking at the datasheet is a good idea.

Below is the voltage versus current curve for a reasonably typical LED. The makers say that it nominally is rated at 100 mA at 3.2V but a look at the curve show that as drawn it drops 3.3V at 100 mA. The table and the curve are both meant to be "typical" values - it's a bad start when a datasheet disagrees with itself even slightly, but it's unimportant here an demonstrates the inexact relationship between Vf and If in practice for a randomly chosen LED. The data sheet says Vf could be as low as 2.9V and as high as 3.5V at 100 mA.

Look at the curve and note what happens for a typical LED if Vf is changed from 3.3V (=100 mA) to 3.4V. At 3.7V it draws 200 mA, at about 4.05V = 300 mA and at 4.4V it draws 400 mA.

That is, for a change of Vf from 3.3V to 4.4V =~ a 33% increase in Voltage, current goes from 100 mA to 400 mA.

A few simple exercises, which will greatly help your understanding if you do them yourself, will give you a much better feel for what happens in real life. Try working out the stable state on this curve for a fixed power supply voltage and a series of resistor values. Then try a fixed resistor value and a series of power supply voltages.

Tell us if this helps your understanding.

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