Vf of SST-90 at 8A?

greencardigan

Newly Enlightened
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Jan 16, 2007
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Hi All,

I have question about the forward voltage of the SST-90 LED in my recently finished Mag Mod.

It's powered by 24 AMC7135 linear regulators from 3 parallel IMR26500 cells.

Measuring the current at the tail gives me 8.2A with fresh batteries.

However, I can only measure the LED forward voltage to be 3.2V. Isn't that too low for 8A current?? A Vf of 3.2V suggests only a 2.5A drive current???

I'll try measuring it again tonight. Maybe I did something wrong??
 
I think you can consider yourself lucky to have such a low vf:thumbsup:. If it were 3.7v or higher, I don't think your light would even run in regulation. If I'm not mistaken, v in needs to be .12v above emitter vf for 7135 based drivers to run in regulation. If your battery setup gives a voltage under load of 3.7v and you had a vf of 3.7v, you would be in trouble.
 
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I think the 3 parallel 26500 will hold above 3.7 volts at 3A for a fair time.

More testing is required I think.
 
According to the datasheet, the typical vf at 3.15A is 3.2v and at 9A it's 3.87v.
 
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I took my light apart to measure the Vf again.

Batteries under load = about 3.8V
Voltage drop across AMC7135 stack = 0.6V = about 5 Watts being burnt by the regulator.
Vf = 3.2V

I only have 1 working multimeter so I cant measure Vf and current draw at the same time. But am definitely getting 8A draw from the batteries.
 
Is there any downside to having a low Vf? The LED will be burning less Watts but am I still getting the same amount of light?

Is a low Vf LED more efficient than a high Vf LED?
 
A couple of things could be going on, but one factor is that the Vf of a diode drops when they are hot vs cold, and an LED is a type of diode.

The Vf is dependent on so many things
- current flow
- junction temperature
- burn in conditions
- minor variations in the mfg process

This is why constant current drivers are recommended when ever possible. This is easier said than done at 8 amps.

As far as efficiency (lumens per watt), higher temperatures mean a lower Vf, but also less efficient light emission.

In a strange way, LEDs will do some self balancing with the battery pack as it runs down. The Vbat will drop, which lowers the current, which lowers the LED Vf, which will increase the current again. This can be modeled if one is inclined to do so, but it takes a pretty serious algorithm to do it.

Your measurements are probably not too far off. If you want another way to determine the current, you can put a very small resistance (0.1 - 0.001 ohm) sense resistor in series with the current flow, and measure the voltage drop across the resistor. It is still not a trivial measurement for most hobby people, because even a small resistance changes the number, and the voltage drop is small, making the measurement right on the edge of viable. (at least for my simple DMM)

24 AMC chips in parallel - now that is seriously pushing the concept. :)
 
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