The Miracle Cure For My Test Load Measurements!

[scumbagatheist]

It's no secret that I'm not an electrical engineer. At best, I am a professional tinkerer. Nobody wants to hear the autobiography.

So, I was having trouble getting good readings on a boost driver and I consulted every thread I could find. I've read hundreds of posts regarding how many diodes to select in your dummy load, Vf changes at high current, charts and graphs, you name it.

What I never came across (and was unable to articulate my dilema) was a comment post, or what have you, that dealt with MEASURING YOUR VOLTAGE across your load. Well, the light bulb FINALLY lit up!

I have Lazeerer or Angelos to thank for this epiphany, nay, satori!
I caught him on a bad day and he was in no mood to be questioned about the quality of his individually-tested drivers. And, who can blame him, considering noobs??

He saw I did my homework and was so very kind as to take his valuable time to make a 10 minute long video, answering my question about how to measure your voltage use on your test load! He is quite thoughtful!!

So, once I had that final piece of knowledge- How to test the voltage draws on the diode in my dummy load-I am now confident in my ability to accurately test and adjust any driver that comes my way! I am terribly excited about it.

So, as a big thanks to him, and in the hope that this helps so many like myself, I present you with the video he produced for me. (I hope he doesn't mind!)

Enjoy and share your knowledge!

 

[Cyparagon]

There's nothing wrong with using math to predict the voltage. You can't assume diodes are going to always be 0.7V though; it's higher than that at higher current and the datasheet will reflect that as well. If you look up the current in the instantaneous forward voltage graph, you'll see it'll be around 0.85V. 1.9A on 1 ohm is 1.9V + 0.85*4 = 5.3. Take one out: 5.3 - 0.85 = 4.45 which is rather close to what he was getting.

At 1.4A, that's 1.4V on the resistor + 4*0.81(from graph) = 4.6V which is still a good approximation.

Two of the (many) limitations of your typical $2 test load is that the resistor voltage increases with heat, and the diode voltage decreases with heat. One way to mitigate these problems is to use a lower value shunt and use diodes rated for many times the current you put through them (or heat sink them).

 

[scumbagatheist]

I agree with your point. But, let's call it the best of both worlds. Being able to TEST your math, and knowing how to make those measurements was what was eluding me. Once I had that piece of information (and realizing where I misplaced .9A), I now feel comfortable being able to make accurate measurements and adjustments in the future.

For that, I am thankful.

 

[dethlore]

Yep, that's exactly what I was getting at in the other post that both you guys commented in. It's nice to see a visual of it and confirm I understood it right.

Thanks for the vid! I hope others find their way to this before they fry any diodes because of a lack of understanding about their test loads and high current.

EDIT: Just did some playing around with testing methods tonight and realized something that hasn't really been mentioned as much. The types of connections you are using makes a big difference. At this point in electrical DIY projects, you should know that any conductor has some amount of resistance. You will always get a much more accurate reading when there is less "wire" in the circuit. There wasn't as much a difference in voltage mode, but if you use the Ammeter mode to triple check your driver, get as solid and close of a connection as you can.

Test Data:
Driver set to 2.11A at 4.56V out
Test Load set at 3 1N5401 and a 15W 1-Ohm Resistor
DMM standard 30" cable with 18" alligator clip to reference point and 24" connector from Driver to Load:
Voltage across load: 4.52V
Voltage across Resistor: 2.10A
Ammeter Current: 1.47A

DMM standard 30" cable with 18" alligator clip to reference point and 18" alligator from Driver to Load:
Voltage across load: 4.54V
Voltage across Resistor: 2.10A
Ammeter Current: 1.75A

DMM with 18" alligator clip to reference point and 18" alligator from Driver to Load:
Voltage across load: 4.56V
Voltage across Resistor: 2.11A
Ammeter Current: 2.10A

It seems like there's a threshold around 18" where there isn't enough loss to matter, but beyond that there is a huge loss in current measured by the Ammeter.

Just my 2 cents that might clear any other confusions about readings that aren't where they should be.