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Voltage divider circuit to run an LED?

Eva713

New member
Hi, all
Recently, I have a system running at 62V max, 54V nominal based on voltage divider formula. I wanted to have a little alert LED, so I was thinking about using 10, 1kohm resistors, and adding the LED bewteen the 9th and 10th (or 1st and 2nd, if it matters), to get at most 6V and just a few mA.

Is a better/easier way? Is this a really crappy way? Or is this a perfectly fine way?
 

dl324

New member
Voltage dividers are almost always the wrong way to go. The current in the divider needs to be at least 10 times the load current.

You'd be better off just using an appropriate current limiting resistor. Assuming a forward voltage of 2V for the LED and an operating current of 5mA, R = V/I = 60V/5mA = 12k ohms. Brightness will vary with voltage, but that isn't critical for your application.

If you want constant brightness, use a current source. You can make a simple one with a transistor and a couple diodes and resistors.
 
Last edited:

kg5q

New member
your best choice is to use a proper LED driver IC either a linear IC https://www.taiwansemi.com/en/product_search/linear-led-driver
or switching IC https://www.taiwansemi.com/en/product_search/dc-dc-led-lighting-ic
these are only suggestions on ones i work with on a regular basis there are hundreds to choose from. the care and feeding of the LED is critical end users expect same brightness over time and temp and voltage fluctuations and depending on the efficiency you need either setting the constant current with one resistor or a switcher is 99% of the time the proper course of action in my experience and the prices are so low why not unless you are making a disposable consumer item that if it fails nobody cares and the budget will only allow a resistor priced out to 7 decimal places.
 

Eva713

New member
Thank you for sharing ideas..
If I just use big enough resistors, it'll still work? If I use 10k in resistors, I'll get about 6mA, at 60V would be 360mW across the LED, wouldn't it? Or would it be 360mW split across all the resistors and the LED?
 

dl324

New member
With the resistor(s) in series with the LED, their power dissipations add.

The resistor(s) will dissipate P = I*V = 6mA*60V = 360mW. Assuming the forward voltage of the LED is 2V, it would dissipate 6mA*2V = 12mW. Total dissipation would be 372mW.
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