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Forum: Analog Circuits High voltage divider circuit to run an LED?


von Lernend B. (Company: KAI Electro Semiconductor Inc.) (lernend05)


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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?

von Helge (Guest)


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you only need a series resistor.

von Lernend B. (Company: KAI Electro Semiconductor Inc.) (lernend05)


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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.

von MaWin (Guest)


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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.

von Dietet (Guest)


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MaWin wrote:
> but it can withstand only 36V

That is not a problem in connection with cascode schema.

von foobar (Guest)


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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[1] 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 ...



[1] actually 60V minus the Vfwd of the LED (1.8-3V)

von Lernend B. (Company: KAI Electro Semiconductor Inc.) (lernend05)


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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?

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