I am trying to connect 6 led on to a 3v battery.

It is 1mm led’s with colors 2xred, 2xyellow and 2xwhite. Each color connected in series.

Problem is once red and yellow light up, the white pair shuts off.

What is the best way to connect? What resistor or value do i need? A disgram would help me much. Thank you in advance.

  • \$\begingroup\$ Can you provide the part number and/or datasheets for the battery and LEDs? \$\endgroup\$ – SolveEtCoagula07 Aug 20 '18 at 17:30
  • 1
    \$\begingroup\$ @Dene: My article Battery and LED without resistor may be of interest in helping your understanding. \$\endgroup\$ – Transistor Aug 21 '18 at 15:37

A non-perfect voltage source like a battery with a high Equivilant Series Resistance, and an load like an LED (diode) work like a see-saw. As the current goes up, the voltage drop across ESR and diode goes up, so the current goes down. As the current across the diode goes down the voltage needed goes down until there is a nice balance.

The 2 white leds in series with two 3V coin cells work because the balance is found at a higher voltage and lower current, enough to be visible. The 6V drops as the current goes up, so the leds likely see 5.1V at 5mA (random numbers). But add the two red leds in parallel, and their lower forward voltage drop cause more current to go through, which means the battery has a higher voltage drop. You may be seeing 4V at 15mA. Since the red leds are conducting 4V at the same node/connection point of the white leds, both parallel led pairs will receive 4 volts. This is not enough to significantly light the leds.

Basically, you are drawing so much current from the battery that the voltage drops below the white leds visible output point. Solution is to find a battery without a high ESR (and low capacity), or reduce the ammount of current you need. You could use resistors to limit your current draw for each led to 1mA. This would still be visible, but not particularly bright.



I am using CR 2032

Two CR2032 cells do not have the capacity to power 6 LEDs very brightly for any reasonable amount of time.

They will work, just a matter of how brightly for how long.

You need to carefully select the LEDs you use. You want the most mcd intensity. At 20 mA you can get over 20,000 mcd for red and over 15,000 mcd for yellow. You can run these LEDs a 1 mA and they will still be bright.


You must not be using any resistors.

Each LED needs its own current limiting resistor. And they cannot be connected in series.

You problem with the white, the required forward voltage is about 1 V higher for white than red and yellow.

The problem is the forward voltage of the white LED is likely greater than 3V.

First deal with the red and yellow. I am taking the safe road and assuming the LEDs are 20 mA and the red and yellow forward voltages are 2V.

enter image description here
Source: Hobby Hour LED Calculator

Now for the white you can try a 10Ω resistor and check the brightness. You may be able to get away with no resistor so if the 10Ω is too dim you can try no resistor.

The red and yellow need resistors for sure.

You can measure the voltage across the leads of the red and yellow LEDs to get the actual forward voltages, then use those voltages in the resistor calculator. Not likely to make that big of a difference.

If you want the brightness of each LED to be at the same intensity you will need to adjust the current based on the lumen output of each color.

  • \$\begingroup\$ if i intend to use two white led, should i put it in series with one resistor or in parallel with individual resistors? would it still be a 10 ohm resistor? \$\endgroup\$ – Dene Zee Aug 21 '18 at 13:36
  • \$\begingroup\$ THis answer failed to recognize your title that says button, because as I stated there is no solution with resistors when the CR2032 battery is 110Ω and the IR LED is <20Ω while Vf white is >> Red Yellow just like a lower voltage zener. This incremental resistance is often given different names for different parts. For dielectrics like batteries and caps it is most often called ESR, but for semiconductors; sometimes called Rin,Rs,Zzt,Rce,Rπ, but I prefer to use the same term, ESR even if the physics reasons are different. ESR=ΔV/ΔI. No solution is possible with resistors here. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 21 '18 at 15:27
  • \$\begingroup\$ @TonyEErocketscientist batteries have Internal Resistance not ESR. Why you prefer to use ESR is a mystery and it's incorrect terminology. The datasheet shows the IR is between 10 and 20Ω not 110Ω. See the green line labeled IR on the discharge curve graphic. A button cell will work. A CR2032 will likely not have the energy capacity to light 6 LEDs. \$\endgroup\$ – Misunderstood Aug 21 '18 at 15:44
  • \$\begingroup\$ @DeneZee you can try a 10-30Ω resistor on the white and 50Ω resistor on the others. You still have the problem that the white LEDs very likely require more than 3V for proper operation. \$\endgroup\$ – Misunderstood Aug 21 '18 at 16:16

Even though you didn't mention the battery size, as soon as you mention button, you can think of a 3V cell with high ESR value.

CR2032 Maximum discharge current: 3 mA.
Maximum pulse discharge current: 15 mA.
ESR = 110 Ohms (Energizer)

LED's are like Zeners with an incremental resistance at knee voltage and forward voltage at rated current (in this case the LED is ) less than 20 Ohms. Thus the battery voltage will drop when the 2.1V Red or Yellow LEDs are applied.

This prevents the 3V White LED threshold from operation at the same time when the battery voltage is expected to be < 2V.

Also keep in mind this size cell can barely drive 1 LED let alone 2 or 3.


simulate this circuit – Schematic created using CircuitLab

There is no solution to drive both R/Y with W LEDs using resistors since the battery ESR is too high.

  • \$\begingroup\$ Why would you assume CR2032? The OP said "battery" not "button". I think you mean internal resistance (IR) not ESR. The datasheet shows the IR curve between 10 and 20Ω not 110Ω. BTW LEDs are nothing like Zeners. \$\endgroup\$ – Misunderstood Aug 21 '18 at 13:28
  • \$\begingroup\$ Sorry, just to clarify I am using CR 2032 \$\endgroup\$ – Dene Zee Aug 21 '18 at 13:54
  • \$\begingroup\$ @Misunderstood At least you agree with my answer that IR<20Ω, but everything else you said was wrong. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 21 '18 at 15:15
  • \$\begingroup\$ You did not say IR<20Ω. What is the R1 100Ω resistor? You said ESR[sic] = 110Ω where the datasheet shows 10-20Ω. Your 20Ω referred to the "incremental resistance at knee voltage" which is totally wrong on so many levels. Where do you come up with this stuff. A zener is NOT like an LED!!! \$\endgroup\$ – Misunderstood Aug 21 '18 at 15:30
  • \$\begingroup\$ misunderstood, sorry you made so many wrong assumptions, I simplified it for you with more details. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 21 '18 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.