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I want to drive 3 LEDs in Series at constant current of approx 900mA - 1000mA.

Will this circuit work? The LEDs are Vf 3.55V @ 1000mA.

LED Diagram

R1 = 0.25W 100k Ohm Resistor
R2 = 2W 0.47 Ohm Resistor
Z1 = 4.7V 1.3W 5% Zener Diode
Q1 = Fairchild KSP13 Darlington NPN 30V 500mA 625mW transistor
- Datasheet: https://www.fairchildsemi.com/ds/KS/KSP13.pdf
Q2 = Vishay IRF830PBF Power Mosfet 500V 4.5A 75W
- Datasheet: http://www.vishay.com/docs/91063/91063.pdf

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2 Answers 2

up vote 2 down vote accepted

You don't need the zener diode because Q1 will limit the gate drive if too much current is about to be drawn. I think this circuit is intended for a 12V dc supply

This looks like it should run from a 12V dc supply just fine. With 1A flowing through the 0.47 ohm source resistor this should just about start to turn on the BJT transistor which will hold the gate voltage at the right amount to maintain the 1A thru the LEDs.

Overall volt drop across the MOSFET will be about 0.9V (from a 12V dc supply) so it'll dissipate about 1W in heat - maybe consider a small heatsink.

If the dc supply is more like 15V dc then the MOSFET will get considerably warmer and maybe you should consider using a buck type constant current regulator.

EDIT - check also the maximum current that the LEDs can take - you might be running them too close if the max current is only a bit above 1 A. If this is the case, increase the 0.47 ohm resistor to maybe 0.68 ohm. A little bit of careful testing might be needed.

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Thanks for your help. Yes, a heatsink is designed in the application, however, not mentioned on the circuit and yes, it is from a 12VDC suppy. –  fizzy drink Feb 10 '14 at 12:33
As for the Zenner, it is used for future proofing for a PWM module. –  fizzy drink Feb 10 '14 at 12:33
@fizzydrink - check my edit section in case you are running too close to the LED maximum current. –  Andy aka Feb 10 '14 at 12:43

Maybe, sort of. Looks to me like it won't work that well. In particular, it might give more like 2A, (well if the particular MOSFET happens to have a low \$V_T\$, otherwise it won't be regulating via Q1 at all).

I'd increase the gate protection zener to 12V 500mW, use a regular BJT for Q1 rather than a Darlington, and start off with a higher value for R2 (around 0.65 ohms 1W or 2W).

Edit: I'm not kidding about increasing the gate protection zener. A 4.7V 1300mW zener might only allow 3.5 or 4V at tens of uA. Subtract 1.x volts for the Darlington and you're left with less than 2.5-3V for a MOSFET that requires 2-4V (that's what my reference above to \$V_T\$, is alluding to) in order to pass a mere 0.25mA. It really needs to be able to see about 6V between gate and source to assure 1A output current, meaning more than 7V referred to ground.

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Yeah, I've got some hither Ohm resistors ordered as well for this reason. My calculations required a 0.5 Ohm, so the 0.47 was closest, but was obviously less. So I might go for something around the .66 (as in .33 in series) or maybe the .75 –  fizzy drink Feb 10 '14 at 12:41
@fizzydrink You'll need a much higher value resistor if you want 0.9~1A with the Darlington as Q1. It's always easier to go up in current from a value that's a bit too low rather than replace all those LEDs, y'know. –  Spehro Pefhany Feb 10 '14 at 12:43
So play it safe and start with around 1 Ohm and work my way down? –  fizzy drink Feb 10 '14 at 12:44
Or maybe 2 Ohms, however, will that still start the circuit? –  fizzy drink Feb 10 '14 at 12:44
Are you staying with the Darlington? You can go as high as you want, within reason, 2 ohms would be fine even with the Darlington (should give about 0.5-0.6A, guesstimated). –  Spehro Pefhany Feb 10 '14 at 12:46

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