I have two LED's in parallel with different forward voltages, and want to know how much current flows through each of them. They have one series resistor connected before being split. Like so:

enter image description here

As LED's don't follow Ohm's law, I'm not sure how to calculate the current through each LED. I thought I should treat the LED's like voltage sources and apply KVL loops, but I'm still stuck.

enter image description here

  • \$\begingroup\$ If you are asking this as a practical question (i.e. you want to build this circuit), you should always give each LED its own current-limiting resistor. As @Andy aka mentioned, they won't be perfectly matched, and you will likely end up burning one of the LEDs. \$\endgroup\$ – Hari Ganti Aug 18 '14 at 21:06
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    \$\begingroup\$ related thread: Is paralleling diodes a bad idea? \$\endgroup\$ – Nick Alexeev Aug 19 '14 at 3:04
  • \$\begingroup\$ What they all said - AND/BUT. LEDs have an (modified) exponential voltage / current curve which is shown in data sheet by all good manufacturers (and some bad ones). The LEDs do NOT have a fixed current draw - it varies with voltage. When you place two LEDs with different Vf/If curves in parallel they stabilise at a point where the resistor drop produces a voltage at which the LED current sums result in the voltage supplied via the LED. While that is trivially obvious it is also (almost) profound :-). .... \$\endgroup\$ – Russell McMahon Aug 19 '14 at 23:46
  • \$\begingroup\$ .... The system is dynamic. You CAN model it but probably easier and almost always good enough is a little iteration using the VI curves for each LED. You can easily calculate provided V_LED with a given current from: [ V_LEDS = Vsupply - I_LEDS x Rseries]. Plug various V_LEDS into the two VI curves until the currents match what the above formula gives. Takes a minute or few. \$\endgroup\$ – Russell McMahon Aug 19 '14 at 23:48

If two LEDs with different forward voltages are connected as shown, then for idealized electronic parts, the LED with the higher Vf will allow no current through it, and will not light up at all. The LED with the lower Vf will be the only one lit up.


simulate this circuit – Schematic created using CircuitLab

To understand this better, note that the voltmeter as shown above will read 2.4 Volts, the forward voltage of LED1, and that is insufficient to light up LED2.

To calculate current drawn from the battery (first diagram in the question), the voltage drop across the 100 Ohm resistor passing said current, must equal the difference between supply (5 Volts) and Vf (2.4 Volts):

$$I = \frac{V}{R} = \frac{5.0 - 2.4}{100} = 0.026 A = 26 mA$$

LED1 will also thus have 26 mA flowing through it, and LED2 will have 0 mA.

When using real world components, the behavior is marginally different. Note the V-I graph for this 2.7 Volt blue LED:

V-I curve for LED

Even though the datasheet indicates a forward voltage of 2.7 (typical) to 3.6 Volts, the actual current it will allow at 2.4 Volts, shown by the red line, is just under 1 mA going by the graph. Of course, the graph is an approximation. Even two LEDs from the same production batch will have slightly different actual V-I curves, with temperature variation adding yet another set of variables.

Be that as it may, this ~ 1 mA current through LED2 will reduce the current drawn by LED1 by approximately the same amount, if one were to simplify things somewhat. The exact currents through the two LEDs can only be determined experimentally, due to the environmental and manufacturing variables affecting the various parts.

  • \$\begingroup\$ I'm not sure i'd call the difference between 0 and 1ma "marginally", 1mA is enough to get noticable light out of most LEDs. \$\endgroup\$ – Peter Green May 10 '19 at 14:29

If the two LEDs were absolutely perfectly matched to each other then they could share the same resistor. Because they are not perfectly matched in VI characteristic one might appear a little brighter than the other because it will tend to hog more of the current.

To avoid this, it's usually deemed preferable to use a resistor for each LED but, despite this, some LEDs will just appear brighter but (statistically) less so than if the LEDs all shared one resistor.

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    \$\begingroup\$ NOTE: LED's (and diodes, BJT's) have a negative temperature coef so the additional issue with a single burn resistor is thermal runaway (more of a concern for higher power LED's) \$\endgroup\$ – JonRB Aug 18 '14 at 11:18
  • \$\begingroup\$ How do some driver ICs get away with using a single resistor of relatively high (compared to a typical ~200Ohm or so) resistance value, greater than 10k, to limit the current for all LEDs? \$\endgroup\$ – sherrellbc Aug 18 '14 at 13:24
  • \$\begingroup\$ @sherrellbc: The driver ICs you refer to use the resistor to control internal current sources, and will have one current source for each LED output. \$\endgroup\$ – Peter Bennett Aug 18 '14 at 15:53
  • \$\begingroup\$ I tend to work on the assumption that for "identical" indicators you can get away with a shared (between 2 LEDs) resistor, but for illumination (where you want near-max current) you can't. For the indicators, assume LEDs that can run at 20mA max. Pick a resistor to provide 10mA per LED. If 1 LED disappears in a puff of smoke, the survivor is still within spec, also the initial current is low enough to reduce significant self-heating. However if you're designing a board, there's no point skimping on resistors. \$\endgroup\$ – Chris H Aug 19 '14 at 8:45

Often the spec sheets will contain IV plots, you may get an idea by matching the voltage and adding the currents together and doing a few successive approximations. If the LEDs are of different colors, for example, they will have significantly different forward voltages, and the higher-votage (usually shorter-wavelength) LED will pretty much be off.

In general, you will need to solve a system of equations:

$$V_{supply} = R_1(I_{led1}+I_{led2}) + V_{led}$$ $$I_{led1} = f_{1}(V_{led})$$ $$I_{led2} = f_{2}(V_{led})$$

where f1 and f2 are functions expressing the respective IV characteristics of the LEDs. You can find an approximate solution using the plots, or, if you're interested, you could use a mathematical model (e.g. see Diode modelling article in Wikipedia) and find a symbolic solution or an approximate numerical one, using essentially the same successive approximation method as you would with the plots.

On a more practical note, you need to use separate ballast resistors if you want both LEDs to work. You may also be able to feed LED1 (2.4V) off of LED2 through a small ballast resistor, especially if LED2 is a high-britghtness, high-current diode.


The "cheater" answer is to put a low ohm resistor (eg 1 ohm) in series with each led, measure the voltage across each and calculate the relative currents with good ol' Ohms law.

  • \$\begingroup\$ +1 not because it answered the question well (it didn't) BUT doing this will greatly enhance LED current draw. \$\endgroup\$ – Russell McMahon Aug 19 '14 at 23:39

In the early 80's, I made an invention based on this "bad" idea - a 3-LED zero voltage indicator. It is interesting to see how it operates. The LED 1 was green (VF = 2.5 V) while LED 5 and LED 7 - red (VF = 1.5 V). The base resistor 8 can be omitted; the ratio between the resistances 2 and 3 can be changed but their sum should be kept constant.

Zero voltage LED indicator


Using the circuit above, you will need to know three values in order to determine the current value.

  • \$R\$ = Resistor
  • \$V_f\$ = LED forward voltage drop in Volts (found in the LED datasheet)
  • \$V_s\$ = supply voltage

Once you have obtained these three values, plug them into this equation to determine the current :

$$I = \frac{V_s-V_f}{R}$$


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