circuit for handling 12v 150A dc current

[thangapandi]

I need circuit or kinds for handling 12v 150A. I need to measure this voltage and current like voltmeter & ammeter using pic16f877a. How to step down 150A current which is able to be processed by pic16f877a ? Can anyone help for this?

[Ttelmah]

You don't say whether AC or DC?.
If AC, use a current transformer.
If DC, use a hall effect sensor like the ACS750SCB-200.

Critical thing is _layout_. Single connection point between the PIC circuitry and the power circuitry only. Single point ground. If you are controlling the power with a FET, remember just how high the gate capacitance of a FET this size (or multiple smaller ones paralleled together will be, and therefore the capabilities of the circuit you use to drive it. Remember also that you may want to actually slow the switching edges fractionally to control RF.
150A, is not really that large. Many tiny BLDC controllers only the size of a Iphone, will routinely handle peak currents at this level.

Best Wishes

[temtronic]

and...
be sure to protect the PIC from 'noise' or 'transients' and 'spikes' from the device you're monitoring ! 150Amps can easily kill a nearby PIC if not suppressed.
the measuring part of the project is fairly straight forward..protection is critical though.

[ezflyr]

Hi,

It sounds like you only need to measure/display 12V @ 200A, correct? To
measure the 12V, you can use a simple voltage divider to 'scale' the
voltage to a level that is compatible with the PIC A/D input. To measure the
200A DC current, a Hall Effect sensor is the way to go! I have had excellent
luck with the offerings from Honeywell. The Hall sensors I use (CSLA2CD)
require an 8VDC excitation, and output 4VDC for a full scale current input
of 72A. This sensor series also has higher current ranges!

John

[Ttelmah]

As a comment, you also don't have to use a pre-built module. I have one unit that controls over 3000A, and it has two heavy metal bus bars that carry the current. I glue a off the shelf linear output hall sensor IC directly to one of the bars and use this to measure current. Given the size of the magnetic field (about 600gauss at the sensor), you don't have to worry about the Earth's magnetic field, and sensing is really easy.....

Best Wishes

[John P]

Given that the current is enormous, chances are the conductor it flows in has some significant voltage drop. It may be possible to use this as a shunt, with taps at some distance apart which give a predictable current to voltage figure. Of course you have to plan for whatever voltage difference from Gnd exists, maybe using an instrumentation amplifier or the equivalent. And also, unlike a Hall sensor, measuring voltage drop directly will link your processor to the high-current circuit--but if you're measuring voltage, it's hard to avoid that. In theory, with voltage measurements at two points, you can solve the whole problem!

[ckielstra]

[Mike Walne]

The circuit below will convert current to a suitable voltage for your PIC.

[John P]

You must be assuming that the current flows from Gnd to some negative voltage, as otherwise the output of that op amp will be negative. I'd call it a job for an instrumentation amplifier. Or if there's plenty of voltage drop available and not much resolution is needed, just run the taps to two A/D inputs of the chip and subtract the two readings in software.

Good point that copper has a poor temperature-resistance characteristic. That's a factor in whether the simple scheme is usable or not.

[Mike Walne]

[levent81]

Hi mike,

Have you tried the circuit in the picture??
_________________
pcb fabrication

[Mike Walne]

[asmallri]

[Ttelmah]

Agree.
The node at the left of Rsh, will rise relative to the node at the right of Rsh with the layout shown. This will give a -ve output from the op-amp.

Best Wishes

[Mike Walne]

Very remiss of me.

I'm assuming a more complete circuit like this one:-