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Old 01-02-2009, 06:50 AM #1
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Default Daedal Driver Current Equation

What is the equation I can use to figure out diode current from input voltage, resistance, and diode voltage drop?

edit: more specifically, I'm an idiot, and I fried my phr-803t because I don't know how to test current (I put ammeter inline with diode, I would guess this isn't how to do it). How much current was the diode getting if the resistance was 4 ohms and running off a 9v? Also, how do I properly test current with an ammeter (instead of a 1 ohm resistor)?


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Old 01-02-2009, 08:09 AM #2
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Default Re: Daedal Driver Current Equation

Ignore the input voltage, work on the reference voltage.

To calculate the resistor needed for a given current, take 1.25 (the ref voltage) and divide it by the current. So say you want to drive the diode with 110 ma's. 1.25 divided by .110 = 11.3 ohms.

Another way you could do this is to take 1.25 and divide it by the resistance, in your case 1.25 divided by 4 = 312ma.....Ouch!

I would always measure the current with a test load before soldering on my LD, a final check can be done, once the LD is soldered in, by measuring the current draw from the battery with a 1ohm resistor.

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Old 01-08-2009, 03:32 PM #3
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Default Re: Daedal Driver Current Equation

So this would be how I would limit the current on a DDL driver? like to limit at 110ma, it would be 1.25v/.110ma = 11.3ohms (so I should use ~10 ohm resistor in series with the pot), and to limit to 400ma (long die open can) it would be 1.25v(is this the same for most LDs?) / .400 = 3.2 (so a ~3 ohm resistor or equiv. [3 10ohm in parallel give 3.33] in series with the pot)
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Old 01-08-2009, 04:24 PM #4
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Default Re: Daedal Driver Current Equation

Quote:
So this would be how I would limit the current on a DDL driver? like to limit at 110ma, it would be 1.25v/.110ma = 11.3ohms (so I should use ~10 ohm resistor in series with the pot), and to limit to 400ma (long die open can) it would be 1.25v(is this the same for most LDs?) / .400 = 3.2 (so a ~3 ohm resistor or equiv. [3 10ohm in parallel give 3.33] in series with the pot)
Correct

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