Making a 50a+ buck converter.. maybe

[gentlywiringit*]

Tcmtech, in again with the MikeMi haha.
I'd asked the same question previously and Mike added an rc to the pwm circuit he'd already drawn for me. It didnt work though, the readout I have has to measure load resistance after the rc. Only did a quick test but the rc was obviously adding resistance. Perhaps I got it wrong? I'll test that again to be sure.
The only readout I've found that displays the info I need has two inputs, Vin and Vout. Vin displays battery voltage, on the Vout side it measures voltage under load and the loads resistance. It then calculates actual amps and wattage being used based on those. All are displayed on screen. It won't work with N channel mosfets either, not sure how but it makes the circuit so its permanently on.
As this is a fairly powerful lipo powered handheld device I'm making, I need to build in a modicum of safety. Knowing resistance, current and the amp limit of the battery is paramount. Which goes back to my last post asking if v/ir math still applies to the circuit, the math reads like the available current drops a higher percentage than voltage does pro rata. given my display is a safety measure it needs to read true.

ChrisP58. You know when the penny drops and you get that moment of clarity.. almost had that while reading your post I'm sure there's something I can take from that, time for a Google. Thanks buddy.

 

[MrAl]

Thanks Mr Al.
I've always found applying math to real world causality a problem..

OK, so, just so I understand this.. applying that math to my application, does it mean if I use 8.4v in to a .16ohm load I'll get close to a 50a current, but if I turn the voltage down via the pot to ~5v I won't get the expected ~30a current over the same load, or will the load resistance lower the Vout also? And in either case, I won't be able to use a v/ir calculator to initially set up the load resistance without accounting for the losses?

You've officially made my brain hurt

Hey tcmtech.
MikeMi kindly drew me up a pwm circuit for another project along the same lines, it'll awork great for that one. This project is specifically an exercise in actual volt control though. I'd like to compare the two to see which I prefer, and for this one I'd like to use a digital readout that doesn't work well with pwm.

Hello,

The load current simply follows Ohm's Law: I=V/R
If you turn the voltage down, the current follows in proportion.

 

[tcmtech]

Tcmtech, in again with the MikeMi haha.
I'd asked the same question previously and Mike added an rc to the pwm circuit he'd already drawn for me. It didnt work though, the readout I have has to measure load resistance after the rc. Only did a quick test but the rc was obviously adding resistance. Perhaps I got it wrong? I'll test that again to be sure.
The only readout I've found that displays the info I need has two inputs, Vin and Vout. Vin displays battery voltage, on the Vout side it measures voltage under load and the loads resistance. It then calculates actual amps and wattage being used based on those. All are displayed on screen. It won't work with N channel mosfets either, not sure how but it makes the circuit so its permanently on.
As this is a fairly powerful lipo powered handheld device I'm making, I need to build in a modicum of safety. Knowing resistance, current and the amp limit of the battery is paramount. Which goes back to my last post asking if v/ir math still applies to the circuit, the math reads like the available current drops a higher percentage than voltage does pro rata. given my display is a safety measure it needs to read true.

ChrisP58. You know when the penny drops and you get that moment of clarity.. almost had that while reading your post I'm sure there's something I can take from that, time for a Google. Thanks buddy.

What?

Drinking and keyboarding?

 

[gentlywiringit*]

What?

Drinking and keyboarding?

Not much .. made sence at the time

Shorthand version; I tested an rc circuit, it wasn't compatible.