I have a 127VAC -> 24VAC transformer to power a project and I want a LED to light when my electronics is ON. I rectify it and pass through a smooth 2200uF capacitor and get a good 32VDC.
I don't have the ratings for my transformer, so I don't know the max current it can output. I used some online math I found to estimate it as 1A.

How is the best and safest way to get it down so I can power a single LED?
I know voltage divider, but I don't know how and if I have to worry about current and if I may fry my resistors. I'm also not sure about my transformer max current, so if it is less than 1A, I'd have to do something different?


  • \$\begingroup\$ Normally you would just use a resistor in series with the LED. \$\endgroup\$
    – user253751
    Feb 7 '18 at 1:26
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    \$\begingroup\$ It looks like there is a mistake in your schematic, the two wires leading to "do something" are connected to the same positive rail. Also, as Immibis says, a resistor in series is the standard solution. The size will depend on your LED current rating and forward voltage, but 3K is probably a good bet. \$\endgroup\$
    – BeB00
    Feb 7 '18 at 1:29
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    \$\begingroup\$ @BeB00: 10mA would dissipate 300mW in the resistor -- a bit much. I would drop the current in half: 6k, 5mA, 150mW. \$\endgroup\$
    – Dave Tweed
    Feb 7 '18 at 1:32
  • 1
    \$\begingroup\$ With the super brightness of 20mA LED's these days unlike 30 yrs ago. You can do this from a couple mA with a 16 Cd LED. R[kohm]=(Vdc-Vf)/I[mA] but never put in backwards as it will silently fail. \$\endgroup\$ Feb 7 '18 at 1:44
  • \$\begingroup\$ @BeB00 Yes, sorry, my schematic is wrong. Didn't see that. Sorry guys and thanks for the comments. \$\endgroup\$ Feb 7 '18 at 2:26

Simple - assume no voltage at the LED (false, but safe) choose a resistor that will run less than the rated current of the LED at 32 volts.

For an indicator, "Much less" is often fine. i.e. 20mA is almost always safe and visible. So 1600 ohms, more or less would probably be fine. (32 V / 0.020A) Check your actual part for real numbers and try to stay away from running at the limits if you want long life.

Now, at 32 Volts you'd (again, making the safe but wrong assumption of 0V across the LED) be burning 0.64W (32V X 0.020 A) in that resistor, so you'd need a 1W resistor to be safe. You could either reduce the current further, which is fine so long as it's "bright enough" to your taste, or you could use 4 800 Ohm resistors in series, where 1/4W resistors would be happy.

Per Dave Tweed's comments you might want to change that to 5 mA, 6400 Ohms (in my safe assumption model), and less than 1/4 W (but more than 1/8W)

Either way it should have minimal impact on the supply output current, which you estimate as 1000 mA From which we are stealing 20, 5 or "less than 5" if you find that 10KΩ or 20KΩ gives you enough light. Also as commented, this needs to be across the supply, not as drawn.

When running from a lower voltage supply (32 is a lot of headroom) it becomes more of an error to assume 0V across the LED, and you start to need to look at its data sheet and subtract its typical (or minimum) operating voltage from the supply voltage to get the voltage to compute the resistors for this purpose. Or, you may change to a constant-current drive scheme - but neither is really needed in this example, as the error will be fairly minor for any typical LED.

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    \$\begingroup\$ 20mA is way too much for a modern indicator LED, and your resistor will be dissipating 640mW. \$\endgroup\$
    – Dave Tweed
    Feb 7 '18 at 1:34
  • \$\begingroup\$ All the indicator LEDs I commonly stock (may not be all that modern) will happily eat 30 mA. I was editing in the power considerations as you commented. I did suggest consulting the data sheet for the actual part. First. \$\endgroup\$
    – Ecnerwal
    Feb 7 '18 at 1:42
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    \$\begingroup\$ @Ecnerwal. Many Leds in the 'high-efficiency' catagory are at full brightness at 2 mA for the 3mm package and 15 mA for the 5 mm package. You did not mention that it is the LED manufactures specs that set the limits, not the package type but the LED material as well. Your answer is too simple. \$\endgroup\$
    – user105652
    Feb 7 '18 at 2:25
  • \$\begingroup\$ Really, I didn't? "choose a resistor that will run less than the rated current of the LED at 32 volts." \$\endgroup\$
    – Ecnerwal
    Feb 7 '18 at 2:37
  • \$\begingroup\$ @Ecnerwal. 'rated current' by itself is an ambiguous term at best. The manufactures often give several current ratings, based on their MTBF results, including PWM values. A 20 mA LED is seldom run beyond 15 mA. Lifetime is 10 to 20 years. At 10 mA lifetime quadruples. At 5 mA life expectancy is 200 years of continuous operation. Your answer is too simple in that you did not elaborate on what 'rated' means. Sorry. \$\endgroup\$
    – user105652
    Feb 7 '18 at 4:30

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