Zetex ZXSX310 question

TMorita

Enlightened
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Sep 5, 2006
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I just took apart an Ultrafire WF-801, and the driver circuit appears to use a Zetec ZXSC310. The stock LED is a K2, so the drive current may be too high for a SSC P4. So I'm trying to drop the current.

The inductor appears to be 100 uH, and the sense resistor is labeled "R020". The transistor is labeled "7n2", and the diode is labeled "Z51" or maybe "ZS1". There's also two 100 uF capacitors on the board.

So, is the resistor 20 mOhms? Any idea of the drive current?

Toshi
 
ARC mania said:
The converter is a step down which has a 1.11A output current to the LED. Its fine as is to drive a SSC-P4. If you want more info, go to http://www.zetex.com/3.0/application.asp?app=11&top=3&curr=21 and click on DN81 - Lighting design note handbook and scroll down to DN-73. The parts list and everything you need to know is there.

ARC mania

Well, uhh...no.

I replaced the K2 with an SSC P4, and it killed the P4.

So, I definitely need to change the sense resistor.

Toshi
 
Did you replace the entire star or just the emitter? I have had to reseat SSC emitters back to the star from DX with arctic adhesive because the thermal compound it came with was causing problems.
 
Randy Shackleford said:
Did you replace the entire star or just the emitter? I have had to reseat SSC emitters back to the star from DX with arctic adhesive because the thermal compound it came with was causing problems.

The WF-801 doesn't use a star. Just the emitter on a heatsink.

Toshi
 
Did you isolate the slug on the SSC P4? That could be a problem as Luxeon is one of the few LEDs where the slug is isolated.

I don't think the drive level should hurt the SSC.
 
dat2zip said:
Did you isolate the slug on the SSC P4? That could be a problem as Luxeon is one of the few LEDs where the slug is isolated.

I don't think the drive level should hurt the SSC.

The base was precoated with AA.

I'm confusesd why people think the drive level should be ok. If the drive level was 1 amp or less, wouldn't they have used a Lux III and saved the extra cost of a K2?

Toshi
 
To be more precise the 0.020 ohm resistor sets the peak inductor current and since the regulator is a boost converter then the input current will be higher than the output current assuming Vin < VLED.

The amount of current to the LED varies with this driver depending on the Vin and the Vf of the LED.

Assume 3V input and a 3.4VF LED. If we simplify the boost peak current equation to average inductor current then if inductor average current is 1A then the LED current will be the ratio of Vin to (Vf + Vdiode). Vin = 3V, Vf = 3.4 and Vdiode ~= 0.4 3V / (3.4 + 0.4) = .789 X Iavg = 0.789A.

As Vin drops the ratio changes and at 2V the output current is now 2V / (3.4 + 0.4) = 0.526A

If your LED Vf is higher then the LED current will be less and if your Vf of the LED is less then the current will be higher.

These are all simplified boost converter equations, but, good enough for approximation and will be relatively close to actual measured results.

Wayne
 
dat2zip said:
To be more precise the 0.020 ohm resistor sets the peak inductor current and since the regulator is a boost converter then the input current will be higher than the output current assuming Vin < VLED.
...
Wayne

This doesn't sound right. The WF-801 is a 2xCR123 light. So it should be a buck converter, right?

Toshi
 
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