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thangapandi
Joined: 27 Aug 2012 Posts: 6 Location: Madurai
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circuit for handling 12v 150A dc current |
Posted: Mon Nov 19, 2012 2:04 am |
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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? |
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Ttelmah
Joined: 11 Mar 2010 Posts: 19368
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Posted: Mon Nov 19, 2012 3:05 am |
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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 |
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temtronic
Joined: 01 Jul 2010 Posts: 9174 Location: Greensville,Ontario
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Posted: Mon Nov 19, 2012 6:17 am |
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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. |
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ezflyr
Joined: 25 Oct 2010 Posts: 1019 Location: Tewksbury, MA
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Posted: Mon Nov 19, 2012 7:52 am |
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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 |
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Ttelmah
Joined: 11 Mar 2010 Posts: 19368
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Posted: Mon Nov 19, 2012 8:57 am |
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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 |
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John P
Joined: 17 Sep 2003 Posts: 331
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Posted: Mon Nov 19, 2012 8:44 pm |
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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! |
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ckielstra
Joined: 18 Mar 2004 Posts: 3680 Location: The Netherlands
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Re: circuit for handling 12v 150A dc current |
Posted: Tue Nov 20, 2012 1:33 am |
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thangapandi wrote: | ... to be processed by pic16f877a ? | Whenever I see this PIC model specified I think: School assignment!
Which is confirmed by: Quote: | How to step down 150A current which is able to be processed | You don't have 'to step down the current' because the PIC can't read currents! The PIC has an A/D converter which converts voltages to a binary value. So what you need is a current to voltage conversion circuit, for that see the suggestions given above.
What I'm missing in the discussion is the desired current range and resolution. |
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Mike Walne
Joined: 19 Feb 2004 Posts: 1785 Location: Boston Spa UK
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Posted: Tue Nov 20, 2012 5:35 am |
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The circuit below will convert current to a suitable voltage for your PIC.
Code: | Rfb
----WWWWW-----
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| -------- |
Rin | | | |
---WWWWW-----|-IN | |
| | OUT|------------- PIC ADC INPUT
| --|+IN |
| | | | OPA
| | --------
| Rsh |
-------WWWWW---------------------------- PIC GND = Common system 0V
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If Rsh is of the order of 100u the voltage across it will be ~15mV at 150A.
Rsh will then dissipate ~2.25W at 150A.
Assuming a school project you can make Rsh from bulk Eureka wire, soldered to copper buss bars.
12SWG (i.e. 2.5mm dia) is 90mOhm / m.
You'll need several ~1cm pieces in parallel.
The down side to using copper cable is its T/C
You'll need a low offset device for the OPA.
The ratio Rfb/Rin will define the gain.
To convert 15mV to 4.5V the gain will need to be 4500/15 = 300.
Mike |
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John P
Joined: 17 Sep 2003 Posts: 331
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Posted: Tue Nov 20, 2012 10:56 am |
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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. |
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Mike Walne
Joined: 19 Feb 2004 Posts: 1785 Location: Boston Spa UK
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Posted: Tue Nov 20, 2012 4:36 pm |
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Quote: | You must be assuming that the current flows from Gnd to some negative voltage, | Yes I am.
I'm assuming that PIC GND is connected to load -ve, and the left-hand side of sense resistor is connected to the -ve end of the supply source. In that way the OPA output will be positive relative to PIC GND. I'm also assuming that the supply rail will be measured relative to the PIC GND.
If supply -ve HAS to connected to PIC GND then, as you say, you'll have to change the OPA configuration. You may then have to correct for the volt drop across the sense resistor in measuring the supply voltage.
I was, after all, suggesting a pointer to a solution, and leaving a lot for thangapandi to work out for him/herself. I don't like to provide complete answers, and it's better for everyone concerned.
Mike
EDIT A bipolar OPA then it may need a resistor inserted in the OPA +ve input lead to equalise the impedance to both inputs. |
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levent81
Joined: 27 Nov 2012 Posts: 1
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Step down |
Posted: Wed Nov 28, 2012 10:51 am |
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Hi mike,
Have you tried the circuit in the picture?? _________________ pcb fabrication |
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Mike Walne
Joined: 19 Feb 2004 Posts: 1785 Location: Boston Spa UK
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Posted: Wed Nov 28, 2012 3:39 pm |
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Quote: | Hi mike,
Have you tried the circuit in the picture?? |
Yes. Loads of times, in different formats.
Why do you ask?
Mike |
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asmallri
Joined: 12 Aug 2004 Posts: 1634 Location: Perth, Australia
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Posted: Wed Nov 28, 2012 9:59 pm |
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Mike Walne wrote: | Quote: | Hi mike,
Have you tried the circuit in the picture?? |
Yes. Loads of times, in different formats.
Why do you ask?
Mike |
Looks to me like the PIC and OPamp ground should be on the other side of Rsh. _________________ Regards, Andrew
http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!! |
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Ttelmah
Joined: 11 Mar 2010 Posts: 19368
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Posted: Thu Nov 29, 2012 8:18 am |
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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 |
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Mike Walne
Joined: 19 Feb 2004 Posts: 1785 Location: Boston Spa UK
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Posted: Thu Nov 29, 2012 11:22 am |
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Very remiss of me.
I'm assuming a more complete circuit like this one:-
Code: | Positive
Supply
Input
----------------------------------------------------------------------
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-------
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| |
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Rfb | Load |
----WWWWW----- | |
| | | |
| -------- | | |
Rin | | | | | |
---WWWWW-----|-IN | | | |
| | OUT|------------- PIC ADC INPUT | |
| --|+IN | | |
| | | | OPA -------
| | -------- |
| Rsh | |
-------WWWWW-------------------------------------------------------------- PIC GND = Common system 0V
Negative
Supply
Input
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The left hand side of Rsh goes negative relative to +IN
The OPA output now goes positive relative to the PIC GND
If you want to reduce the effect of noise, you add a capacitor in parallel with Rfb.
Mike
EDIT having seen the locked contribution from the original poster, I'd taken this thread to be closed. |
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