Resistor for LED diode in button

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Darum_Sevid

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Hi

I bought a button with a LED diode inside. Working voltage of the diode is 9-24V DC.
I want to use it to switch on the bms module in my e-bike battery. I will power the LED diode directly from it. The voltage of a fully charged battery is 54.6V, and of a discharged - 36V. What resistor should I use to power the diode when the voltage will decrease with the discharge of the battery?
 

Solution
1) Learn Ohm's law.
2) I seriously doubt that your battery discharges to -36V.
3) Your question makes no sense. Do you want to know what the resistor should be to illuminate the diode when the battery is fully charged? Try 1.5K
4) The LED will glow at 55V and it will glow at +36V. What are you actually expecting it to do?
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1) Learn Ohm's law.
2) I seriously doubt that your battery discharges to -36V.
3) Your question makes no sense. Do you want to know what the resistor should be to illuminate the diode when the battery is fully charged? Try 1.5K
4) The LED will glow at 55V and it will glow at +36V. What are you actually expecting it to do?
 
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Solution
It is no minus but dash. Voltage is +36 volts . This is the lowest voltage that bms cuts off. I would like the diode to work on these voltage.
 

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Then what makes you think my "-" was minus and not a dash?

Assume the LED draws 20 milliamps at 24V input, and assume the LED drops 2V. That means the internal series resistor is 1100 ohms. To draw 20mA at 55 volts you'll need a total resistance of 2650 ohms. So, as I said, add a 1.5K resistor.
 
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1) You put ”-” closer to the number so I thought you wrote minus 36V, but nevermind.
2) Yes, led draws 20mA
3) I will try to put this 1,5K ohm resistor.

Thank you for reply
 

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You should check the LED current at e.g. 24 V DC to determine the internal resistor value. According to 20 mA current spec we would expect 1 to 1.1 k, but it may be higher. In this case, the internal resistor may be overloaded when applying 55V with external 1.5 k resistor. It may be a good idea to reduce the LED current by increasing the external resistor.
 
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