In a part of my circuit, there will be three LEDs (of the same type) in parallel. Is it true that they will not automatically get the same current and have the same brightness?

Somewhere I read about that.

The LEDs are: UV

  • Voltage Typ. 3,4V - Max. 4,0V
  • Current: 20mA Typ., 30mA Max

Up to now they do not have series resistors individually, but all three have one.

The whole thing will be an ultraviolet light show, regulated by a Raspberry Pi via 16 GPIOs, each of which will activate a mini panel of three LEDS.

  • 2
    \$\begingroup\$ Yes! Try to simulate two parallel ones with different Vf within the range given. \$\endgroup\$
    – winny
    Commented Jul 28, 2020 at 10:14
  • \$\begingroup\$ Related: electronics.stackexchange.com/a/424035/73158. \$\endgroup\$
    – Transistor
    Commented Jul 28, 2020 at 10:17
  • \$\begingroup\$ So should I put for each an individual resistor? I guess the current will be similarly if the resistors are as high as possible? \$\endgroup\$ Commented Jul 28, 2020 at 10:27
  • 1
    \$\begingroup\$ Yes, you should. \$\endgroup\$
    – winny
    Commented Jul 28, 2020 at 10:34

3 Answers 3


Yes, as a load such as your UV LEDs, when connected in parallel will consume the same amount of driving voltage by the virtue of parallel circuit combination but will consume varying levels of current depending upon the internal & external resistances associated with the UV LEDs & the circuit itself.

Hope this answers your question. If you have any other queries hit me up :)


The V vs I curve will be slightly different for each LED even if they are the same type. This is simply due to manufacturing tolerances. If the LEDs are from the same batch, you should be able to get away with a single current limiting resistor by virtue of the fact that as that each LED has an internal resistance (albiet small). This means if one LED starts to consume more current, its voltage will rise a bit, in turn reducing the current.

Current sharing based on the internal resistance of LEDs is not the most accurate though. If you want accurate current sharing, use a separate resistor in series with each LED.


Given the exponential nature of the relationship between current and voltage, even the smallest variation in forward voltage results in significant current change.

Even among "matched" LEDs, there is some variability due to the intrinsic non-ideal characteristics.

A simple way to look at this is by examining the well known exponential curve for a diode:

$$I_D = I_S e^{V_D/V_T}$$

Where \$I_S\$ is the scale current, \$V_D\$ is the forward voltage across the diode, and \$V_T\$ is the thermal voltage at room temperature (\$V_T=26\text{mV}\$).

Say now that you have two diodes, "matched" (\$I_S\$ the same for both) diodes with currents \$I_{D1}\$ and \$I_{D2}\$, then the ratio between those two comes back to be:

$$ \dfrac{I_{D1}}{I_{D2}}=e^{\frac{V_{D1}-V_{D2}}{26\text{mV}}}$$

Which means that if \$V_{D1}-V_{D2}=26\text{mV}\$, that the ratio of \$I_{D1}\$ to \$I_{D2}\$ is 2.7. A difference of 60\$\text{mV}\$, represents a ratio of 10, you get the idea. It shows that one of the diodes will be taking up a lot more current than the other pretty quickly. Now imagine that \$I_S\$ isn't exactly the same for the diodes, which is the real case, then you could see that the ratio of the \$I_S\$ will scale the current ratio further.

You may ask why should there be a difference in the voltage if the diodes are place in parallel and that goes back to the non-ideal characteristics (like \$I_S\$ being different). Think of each of the diodes having some internal resistance in series, in that case the perfect parallel connection situation falls apart.

All this analysis could have been done without using the exponential model, and instead using a linear approximation of the exponential model around the nominal \$V_F\$ of the diode--that would more clearly show the "internal" resistance since the model would have the form \$V_D = V_{F0} + I_D R_{Di}\$, where \$R_{Di}\$ is the slope of the linear approximation of the exponential I-V plot around the nominal voltage \$V_{F0}\$.

Hope it helps.


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