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I am a young engineer and I need a help from you.

I need to design a driver circuit which must power-up a LED (OC) if the measured volatge is greater that 50V (Voltage to be measured can be upto 450V DC). The driver must take power from the measurement point.

enter image description here

As show in the circuit I put a voltage divider and voltage reference(TL) will be activated when the measured voltage is greater than 50V and it will drive the transistor which is gonna power up the LED.

The problem lies on the power supply to the transistor I need to have a collector Emitter volatge which can be used for LED power Up. So I put a zener diode regulation circuit (Break down volatge is according to the driver volatge requirment).

I want to make sure this is the perfect solution, is there any other feasible way to do this?. ( Because R13 is huge in power , is there any other way to limit the power requirment for this?- I need 2mA to drive the LED.

Thanks in advance

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  • \$\begingroup\$ Does it run from AC or DC? \$\endgroup\$ – Andy aka Oct 26 '15 at 14:26
  • \$\begingroup\$ I think the op should answer this. \$\endgroup\$ – Andy aka Oct 26 '15 at 14:37
  • \$\begingroup\$ Its on DC... I just used here the AC Source just for the testing purpose. So that I can Simulate my threshold (50V) voltage \$\endgroup\$ – Balusamy Oct 26 '15 at 14:39
  • \$\begingroup\$ So you are measuring (up to 450V) DC and all your circuit is DC powered too? \$\endgroup\$ – Fizz Oct 26 '15 at 14:43
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    \$\begingroup\$ yes it is 100 percent DC \$\endgroup\$ – Balusamy Oct 26 '15 at 15:48
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Here is my suggestion:

Put in a 2mA current source similar to the schematic below (you would want to adjust the resistor values and pick the appropriate transistors).

Use the TL431 to directly switch on the LED.

Looks like you can modify the R2+R3 path and use that to turn on the TL431.

Add a zener diode from anode of LED to ground to protect the TL431.

Then the total power dissipation is around 450V * 2mA = 0.9W plus the power through the R2+R3 path.

enter image description here

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  • \$\begingroup\$ I will try this... I will post soon about the results \$\endgroup\$ – Balusamy Oct 28 '15 at 9:11

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