I’m trying to use an ESP8266 to control a fairy light string via a relay. The lights look like this, with 40 tiny LEDs that seem to be wired in parallel. I don't see a resistor along the wire anywhere.

I would like to be able to power the lights off the ESP board so I can rely on the integrated USB power port. The development board I'm using is an ESP-12F which has both 3.3v and 5v pins.

There is a battery and timer box included with the lights that runs off 4.5v worth of AA batteries. Measuring the voltage at the contacts where the light string connects to the timer PCB, I see 2.7-2.8v. I can't find a data sheet to get the exact 𝑉𝑓. I'm opting to use the ESP's 3.3v pin to hopefully waste less energy as heat since it's closer to the LEDs' required voltage.

My first thought was to build a voltage divider with two resistors, and measuring the voltage across the second resistor was exactly 2.8v as expected. But powering the light string from that is so dim it’s basically unusable. When I connect it directly to 3v3 and GND pins, it's perfectly bright, so it seems microcontroller has the current to power it. I wouldn't want to power it this way for long though to not shorten their lifespan.

I recently learned that voltage dividers don’t really work for LEDs because their intensity is current-driven. How do I properly change the 3.3v supply voltage down to 2.8v that I can safely give to the light string, without dimming them? The original power source supplies only 2.8v with full brightness so I know it’s possible somehow.

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    \$\begingroup\$ How much current does the LED string take? What value of voltage divider resistors you used? Also, you don't really know how the battery box drives the string. It might use PWM to have pulses of high peak current and thus even if you measure 2.8V, the string appears brighter than with 2.8V DC. \$\endgroup\$
    – Justme
    Commented Feb 20, 2023 at 5:10
  • \$\begingroup\$ My ammeter functionality on my multimeter isn't working so I can't tell the exact current draw. I'll invest in a new one if required. I tried voltage divider resistor combos using output from this site: ti.com/download/kbase/volt/volt_div3.htm I tested with a 10 and 56 ohm resistor, among others that I will have to check back on because I'm away from the hardware right now. Good point about the PWM thing, didn't know that was even a remotely common practice to drive multiple LEDs. The meter was showing consistent voltage but I imagine voltage is changing too quickly to be picked up \$\endgroup\$
    – drcomputer
    Commented Feb 20, 2023 at 5:23
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    \$\begingroup\$ But you can just measure the voltage over the 10 ohm resistor to find out current through it. Also adding a total of 66 ohms to 3.3V is 50mA of wasted current through resistors. And the voltage will only be 2.8V if the LEDs are not connected, because any output current will drop more voltage in the 10 ohm resistor so the voltage was not 2.8V but less when LEDs connected. Which you could have also measured how much it really is. So it might not be PWM drive after all. Oh and if the ESP has a linear regulator, all voltage drops waste equal amount of energy as heat no matter what. \$\endgroup\$
    – Justme
    Commented Feb 20, 2023 at 5:34
  • \$\begingroup\$ Understood. Thanks for y'all's patience as I'm fairly new to electronics. I will test the voltages and implied current tomorrow when I'm home. So I need to measure the voltage across both resistors while the light string is connected to the same contacts, or do I just measure across one? Then take that voltage and divide it by 66 to find the current draw of the LEDs across that resistor? \$\endgroup\$
    – drcomputer
    Commented Feb 20, 2023 at 5:51
  • \$\begingroup\$ No. Don't divide by 66 for now. Just measure the voltages over each resistor separately. \$\endgroup\$
    – Justme
    Commented Feb 20, 2023 at 6:18

1 Answer 1


You can not make a voltage divider without taking into account the load current. Since the LEDs were not as bright as expected, it is assumed the voltage was measured across the voltage divider before connecting the string of 50 white LEDs?

You can probably find a silicon diode that will drop the voltage by .7V and handle the current of 50 white LEDs. 3.3V - 0.7V = 2.6V which is close to 2.8V.

White LEDs are really blue or violet LED painted with phosphorus. Blue LEDs require more current. Perhaps in a range of 10mA to 20mA each. It is difficult to say what the total current consumed by these 50 LEDs. Take care not to exceed the current capacity of the 3.3V regulator.

  • \$\begingroup\$ Correct, the voltage given was measured prior to putting load on the resistor. Additionally, I wanted to test the current but the amp meter functionality on my multimeter is currently broken. I will have to get that fixed to test what the current is. \$\endgroup\$
    – drcomputer
    Commented Feb 20, 2023 at 5:12
  • \$\begingroup\$ Would a Schottky diode with a slightly higher voltage drop work in place of the silicon one or is there a specific reason to use the silicon type? \$\endgroup\$
    – drcomputer
    Commented Feb 20, 2023 at 5:17
  • \$\begingroup\$ Normally there is a separate fuse protecting the low current DVM measurement circuit. I try to remember to test the current flow with the high range current setting before to make sure it is in the mA range. Then switching to the lower DVM current range setting. Looks like a Schottky Diode has about a 0.3V drop. So you might try 2 of these in series. It is usually the type of material used to make the diode that dictates the forward bias voltage drop. Silicon is about .7V and Germanium about .3V. \$\endgroup\$
    – st2000
    Commented Feb 20, 2023 at 13:32

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