Test load/Blitzlinear Problem

[Jstr]

Hi y'all, so I've been trying to set my driver (Blitzlinear) to 790 mA for a BDR-209 build, but something doesn't add up with my test load. After I attached leads and connected it to a pair of 16340 batteries, I get my 790 mA using the usual 6 diodes in series (dropout should be ~6.8 V at that current according to ARG's graph). But then I had a genius inspiration to use the ohmmeter to measure the actual resistance of the "1 Ohm" resistor, and turns out it is actually 1.4 Ohms. I confirmed this with two different multimeters. So using the voltage drop across the resistor (.79 V), the current should actually be closer to 564 mA, a huge difference. Here's the weird part: when I connect the ammeter in series with the test load, I get 810 mA, an increase rather than the expected decrease in current, which makes absolutely no sense. Which should I trust? 564, 790, or 810 mA?



Here's some additional information about my experience with this test load because I have used this load for a while without checking the resistance.

This will be my second time using this diode (same driver, same current). That diode got :gun: (because I underestimated the heat output of this driver and ran it too long). One thing to support the current actually being ~564 mA is the power output of the last diode was just under one Watt (with G-2), which fits the power curve much better, albeit on the efficient side. I would expect around 1.2 Watts at 790 mA.

I have measured the current of several constant current x-drives, which I have used. Two were spot on 1.8A with the test load but I think I only got about 1.8W with one m140 and for the other, I got 2.3W from an m462. Then I measured a CC-boost at 1.79A and that gave me a pretty incredible 2.4W from an m140, but here this was a boost and the reading was with 4 diodes on my load, a 5.4V load compared to a 4.5V actual load, so current may actually have been higher once I attached the LD. But then there were some 2.4A x-drives that measured below spec, three at 2.15-2.16A and those gave me 2.65W, 2.68W, and 2.7W.

One more thing, the driver gets extremely hot quickly. Without a heatsink, thermal protection kicks in after ten seconds. If it is only 564 mA, that heat is somewhat unbelievable.

I would normally want to trust the math and Ohm's Law, but the ammeter in series reading higher really bothers me. Also, that would mean my x-drives were very under spec (1.54A from 2.4A drivers, and then my diodes were way underdriven and impossibly efficient).


Thanks for reading and any input would be great.



I used a Radiant x4 LPM and the same G-2 lens for all power tests. Also just want to add, I have had no problems with the driver.

 

[Cyparagon]

Re: Test load/Blitzlinear Conundrum

turns out it is actually 1.4 Ohms. I confirmed this with two different multimeters.

Let me stop you right there. What is the resistance of your multimeter leads? When you short them together, what is the resistance readout?

 

[DTR]

Re: Test load/Blitzlinear Conundrum

I would stop you again. 790mA? I usually suggest 600mA. 700mA if you want to live on the wild side. 790mA I really would not suggest that.:beer:

 

[Hiemal]

Re: Test load/Blitzlinear Conundrum

That, and you're using a linear driver. The greater the voltage difference is between the output and the input the more heat is generated.

Two fully charged li-ion cells should be about 8.4 volts. 8.4 volts - 6.8 volts = 1.6 volts.

Wattage is 1.6 * 0.79, which gives about 1.2 watts of dissipation. Without heatsinking, on a tiny board that could be enough to trip thermal protection since the heat has nowhere to go.

But, definitely check your multimeter leads for parasitic resistance. It can mess up low resistance readings pretty badly if you don't compensate for it.

 

[Jstr]

Re: Test load/Blitzlinear Conundrum

That, and you're using a linear driver. The greater the voltage difference is between the output and the input the more heat is generated.

Two fully charged li-ion cells should be about 8.4 volts. 8.4 volts - 6.8 volts = 1.6 volts.

Wattage is 1.6 * 0.79, which gives about 1.2 watts of dissipation. Without heatsinking, on a tiny board that could be enough to trip thermal protection since the heat has nowhere to go.

But, definitely check your multimeter leads for parasitic resistance. It can mess up low resistance readings pretty badly if you don't compensate for it.

Sorry, I forgot to mention, the input voltage I measured at 7.7V because the batteries were not fully charged. And the driver itself uses about half a volt, so the difference is actually closer to .4V and dissipation=.3W

So how exactly do I check for parasitic resistance? Compensate? I used two different meters to be sure.


DTR, don't people run them around 750mA? I thought that the max suggested current was limited by the microboost's "<1.2 rule"?

Thanks for the replies! Also, what do you think about the different reading from the ammeter in series?

 

[Cyparagon]

Re: Test load/Blitzlinear Conundrum

So how exactly do I check for parasitic resistance?

Did you read my post?

 

[Jstr]

Re: Test load/Blitzlinear Conundrum

Oh no, sorry about that, I must have skipped over it cuz it looks like it would be in your sig. Only saw the gif. Thank you for that. Yeah it was .4 ohms and yeah I feel stupid now. I think I am going to lower the current anyway to be safe and just hopefully I can make this diode last with a nice heatsink. Thanks again!