Page created on 22/03/2025; updated on 23/03/2025.
How to connect LEDs in parallel properly?
Periodically, we get questions about the best way to connect LEDs in parallel, and why we sometimes get significant differences in brightness between two LEDs connected in this way. I will try to answer this question simply, but in a detailed way.
Obvious (?) solution: two LEDs connected directly in parallel
Consider the following circuit, in which two LEDs are connected directly in parallel, the whole connected to a single resistor. Let’s assume that this resistor has been calculated to obtain an IF current of exactly 20 mA (rated forward current) for each LED. So 40 mA will flow through the resistor, but it doesn’t matter much.
For the LEDs used, the manufacturer gives the minimum and maximum values for the forward voltage for the 20 mA current: VFmin = 2.7 V and VFmax = 3.1 V. The typical value is not given, but, according to the following data sheet extract, it is 2.83 V (this exact value is not important here).
Finally, let’s suppose that, through bad luck, one of the two LEDs has the Min characteristic and the other the Max characteristic.
On the following diagram, whose VF scale has been extended from 0 to 4 V, I have first extrapolated the manufacturer’s typical blue curve outside the yellow zone that corresponds to the graph given above. All the curves must obviously pass through the origin, because if 0 V is applied to an LED, the current will be 0 A.
From this typical extended curve, I have drawn the red Min and Max curves. These curves are certainly not exact, but they are plausible. In any case, they pass through the points (VFmin, 20 mA) and (VFmax, 20 mA).
As the LEDs are in parallel, they are subjected to the same voltage of 2.83 V. To obtain the current flowing through the “Min” LED, I look for the intersection of the vertical line with abscissa 2.83 V with the Min curve. I obtain IF = 33 mA. Similarly, for the “Max” LED, I obtain IF = 9 mA approximately! 33 mA on one side, 9 mA on the other - quite a difference, isn’t it?
So, OK, I’ve taken the worst case, but it’s clear that the currents can be very unbalanced between the LEDs; so they won’t light up with the same intensity.
How does adding one resistor per LED solve the problem?
Let’s take a very simple numerical example: supply voltage VCC 12 V, LED current IF 20 mA or 0.02 A, typical voltage VF 2.8 V.
- Calculation of the voltage across the resistor:
VR = VCC − VF = 12 − 2.8 = 9.2 V. - Value of resistor:
R = VR ÷ IF = 9.2 ÷ 0.02 = 460 Ω.
We know that, for 20 mA, the first LED has a voltage drop of only 2.7 V. So the voltage across its resistor will be 12 − 2.7 = 9.3 V, and the corresponding current IF1 = 9.3 ÷ 460 = 0.0202 A, or 20.2 mA.
The second LED has a voltage drop of 3.1 V. The voltage across its resistor will be 12 − 3.1 = 8.9 V, and the corresponding current IF2 = 8.9 ÷ 460 = 0.0193 A, or 19.3 mA.
As the two currents are almost equal, the light intensity will be identical for both LEDs.
Why is there such a difference between the two set-ups?
The difference is essentially due to the fact that, in the useful zone, the slope of the IF characteristics as a function of VF is very steep. In other words, a small variation in voltage leads to a large variation in current.
This voltage difference, which is very significant for LEDs, is not at all significant for resistors, and in the 2nd case, the resistors set the current for each LED individually.