Good day, 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?
you only need a series resistor.
Hi, I have a bunch of 1k, 1/4W resistors, so 10 of those will work fine. I guess that 6mA at 60V is only 360mW, which might be a bit high. I could always use a lot more resistors and experiment by reducing it until I get to a suitable brightness.
Lernend B. wrote: > have a system running at 62V max, 54V nominal > I wanted to have a little alert LED, You want, thst the LED turns off if the voltage drops below 54V ? You need a much sharper transition, so you need some amplifing circuit. People usually take TL431, but it can withstand only 36V. So a voltage divider is also part of the circuit.
MaWin wrote: > but it can withstand only 36V That is not a problem in connection with cascode schema.
As Helge said: you only need a series resistance. A resistor parallel to the LED is not required. As you calculated correctly, for 6mA at 60V you need a 10k resistor which will dissipate 360mW. To spread that you could use multiple resistors, either in parallel (i.e. 10 x 100k) or in series (i.e. 10 x 1k) or any combination which gives 10k. If the resisistors are equal, the power will be spread equally, i.e. with 10 resistors, each one dissipates 36mW. > Is this a really crappy way? Well, you waste 360mW to light a LED with 10-20mW ...  actually 60V minus the Vfwd of the LED (1.8-3V)
The series resistor drops voltage to control the current through the LED - the current through the LED is the important parameter. The forward voltage of the LED must be used to calculate the series resistor. Red LEDs have a forward voltage of about 2.2 volts. White LEDs have a forward voltage of around 3.7v - it depends on the color and chemistry of the LED. Is that right?