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FrozenGate by Avery

Test Load Fail...Grrrrrr

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brianhud1991

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I've read posts upon posts and none directly answer to the issue I'm having. I built a test load for the 2W m140 diode I purchased. It's 4 N5404 (3A max) with a 1 Ohm resistor(10A max) in a series hooked to the 3A Blitzbuck driver. My battery source is two C123A batteries. ( checked via my DMM, they read 2.8 volts each)

I figured it would be enough volts to the driver to get the required mA's but I guess I was wrong. When I measure across the resistor i get between 08.5 to 09.6 mAs.... I checked with the mV and I got 0.9. now that doesn't seem right at all. I can adjust the pot but it doesnt make a huge difference w so ever. I'm trying to get it set at 1500mAs to break the 2W range with the 402-G-lenses I have. Any suggestions? I'm a bit lost, spent a damn near 2 hours trying to figure this out

P.S. I'm not the best with this stuff, used to working on cars myself. I can use all the help I can get.
 
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are you in the high range solder jumper on the board?

did you look at the voltage of the batteries WHILE under load?

Why arent you using lithium ion?

Dont put the ammeter across the load resistor either, only voltmeter.
 
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Yeah, for that output voltage, you'll need more than 5.6V, most likely. Especially if the batteries are under load.
 
You should try powering it with Lithium Ions fully charged, and Voltage across the resistor is always measured in mV's. The main reason of using a 1.0Ohm 1% resistor in test load is to be able to read mV=mA readings, If you are getting 0.9mV something doesnt feel right.. Make sure connections are right.
 
By C123A you mean CR123A right? 2.8V? They sound burned out or the protection circuit kicked in. LiIon batteries should be at least 3.6 volts. If they're < 3V they've been overly discharged and are dead (or the protection circuit kicked in).
 
Yes Fiddy is correct they are 3V max. I didn't take into account how much the driver actually consumes. I'll pick up some ultrafire rechargeable, the ones I have are weak I realized. It's only putting out 3.5 volts across the driver.

Okay so basically one mV across a 1 ohm = 1 mA, that makes sense. So I use mV across the resistor and mA if I was to see desolder one of the legs and have the DMM in circuit? Sounds dangerous, ill stick with mV lol.

I found some 3.6Vs but with a max of 800mah, not a lot of wiggle room for increasing power output.
 
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BTW I'm not using the high range solder jumper. I realize if I do that or make contact my lowest would be 1.8A. Thanks guys! I'll get those batteries and hit ya back with an update
 
brianhud1991 said:
Yes Fiddy is correct they are 3V max. I didn't take into account how much the driver actually consumes. I'll pick up some ultrafire rechargeable, the ones I have are weak I realized. It's only putting out 3.5 volts across the driver.

Okay so basically one mV across a 1 ohm = 1 mA, that makes sense. So I use mV across the resistor and mA if I was to see desolder one of the legs and have the DMM in circuit? Sounds dangerous, ill stick with mV lol.

I found some 3.6Vs but with a max of 800mah, not a lot of wiggle room for increasing power output.
Click to expand...

The whole point of the test load is to be able to easily measure the current without interfering with the circuit. You definitely want to use the voltage setting and measure across the one ohm resistor. That's why it's there.

As for the batteries, you would be hard pressed to find a better option than lighthound and the AW IMR for that setup. Unless you want to run 18650's and then I think there are some sellers here that can help.
 
It's my understanding that wire wound resistors can have inductance that could kill a buck/boost driver.

That's why they make non inductive resistors.

This is also why Dr. Lava states that lead wires should be no longer that 3" between driver and laser diode, because of inductance.


brianhud1991 said:
Excuse me, its a 10W Resistor, wire wound from radioshack
Click to expand...
 
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EVERYTHING has a small parasitic inductive value - even a mm of wire, resistor lead, or PCB trace. The inductance of a wirewound resistor is still very low, especially for low values like this. Even if it wasn't, it only serves to smooth out the current and cannot harm the driver.
 
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