In an earlier project, I used an LED to indicate power (this 12V model). The PSU delivered +15V. I adjusted brightness with a 5k series resistor. According to the datasheet, this particular white LED has a forward voltage of 3.8V. According to the same datasheet, I slightly exceeded the max operating voltage, but it seems to have caused no problems so far.

I now want to use the same model LED (I have several left) on a different project which has a +20V PSU, and I want to match the brightness of the LEDs on the two projects. While exceeding the 12V LED's rating by using a 15V PSU might have worked, 20V seems like it's too much. I thought I'd try a voltage divider here, as follows (iCircuit screenshot, old circuit on the right, new circuit on the left):

iCircuit screenshot, old circuit on the right, new circuit on the left

According to the simulation, the LED receives the same current and voltage in both cases, and the resistors dissipate pretty small amounts of power, so should not overheat. Is this a reasonable approach?

  • \$\begingroup\$ It looks like those LED modules have a builtin current limiting resistor. As long as the current through the led doesn't exceed 20mA, you should be fine at any voltage. Have you tried measuring the current with your 15V supply? If the current is too high, you can simply add a series resistor to tune it down. 5kΩ seems high though; if the current at 12V is 20mA, you'd need 150Ω to get 20mA at 15V. \$\endgroup\$
    – marcelm
    Commented Oct 23, 2016 at 19:01
  • \$\begingroup\$ VOLTAGE DIVIDERS ARE NOT POWER SUPPLIES !!! \$\endgroup\$
    – brhans
    Commented Oct 23, 2016 at 21:43

1 Answer 1


It will work, but it is unnecessarily complex.

First, though, you are operating under a misconception: that you have somehow applied too great a voltage to the LED in your original circuit. For a given LED, voltage (at the LED, not at the wires of the unit) and current are not independent, and the panel mount LED you are using contains an internal resistor to limit the current, and is intended to be connected directly to the voltage being monitored. From the data sheet, the normal operating current is 20 mA, plus or minus a bit. Assuming a nominal voltage drop of 3.8 volts, at 12 volts the internal resistor must be $$R = \frac{V}{i} = \frac{12-3.8}{.02}= 410\text{ ohms}$$ By adding a 5k in series with this, your nominal current should be $$i=\frac{V}{R} =\frac{15-3.8}{5410} = 2.07\text{ mA} $$ which is reasonably close to your 2.38 mA. In other words, your LED has been running at 10% of nominal power and is nowhere near exceeding the nominal voltage. In fact, since the current through the 5k resistor is about 2 mA, the voltage at the LED leads (not the LED itself) will be about 15 - (5k x 2 mA) or about 5 volts, so the LED voltage is not remotely near being too great.

As a result, you can connect your LED/5k combination directly to 20 volts and you will be golden. LED current will be just about 3 mA. In fact, if you reduce your 5k to 500 ohms you'll get full brightness. Actually, 400 ohms is what's indicated, but there's no use in pushing things since you are apparently happy with the brightness which 5k produces.

  • 1
    \$\begingroup\$ Or change the 5k resistor to 6k8 or 7k5 to match the brightness as the OP enquired about. \$\endgroup\$ Commented Oct 23, 2016 at 17:55

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