Okay, this is even better than the op-amp design -- I just
looked at the Maxim datasheets real quickly and it seems
like a winner so far.  They even show the possibility of
using a PCB trace for the sense resistor.  Sweet.

The new question then is how to figure out the resistance
of a PCB trace.  From AWG/resistance tables for copper wire,
I can calculate the resistance of the copper layer with the
known thickness/width, but the solder coating adds an unknown
factor, as I haven't seen any PCB house specs on that.

Thanks,
-Neil.



-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Tal Dayan
Sent: Friday, May 10, 2002 1:24 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [EE]: Measuring automotive current....


How about measuring the voltage drop over the existing wire ? Try to connect
at
two points on the wire as far as possible from each other and if possibly,
use
some kind of a differential amplifier.

You can also find special chips for current sensing. Maxim has for an
example a bunch
of them with internal or external shunt and of various gains (e.g. 25, 50,
100).

You may also want to filter the output voltage to get rid of all kind of
noise (which
is probably very common when you get under the hood).

There are other way to measure current based on the magnetic field created
by
the current. Take a look for example at Tektronix site, they may have one
though I am sure it will not be cheap.

Tal


> -----Original Message-----
> From: pic microcontroller discussion list
> [mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Pic Dude
> Sent: Thursday, May 09, 2002 4:34 PM
> To: PICLIST@MITVMA.MIT.EDU
> Subject: [EE]: Measuring automotive current....
>
>
> Hi all,
>
> Back from a "break" which has me running on 9 fingers and 9 toes for
> a while. :-(  But fortunately, I'll be back to 10 and 10 in some weeks.
>
> Trying to figure out the best way to measure current in a car (12V neg gnd
> system).  The current path to be measured is that going into/out of the
> battery.  For now, I'll use a "safe" figure of 100A, although the current
> drain/alternator charge current should not be anywhere near that.
>
> First thought is to use a very small resistance with high current
> capacity.
> I believe that is what a shunt is, so a quick web search found me some
> shunts rated at a certain millivolt value and max current.  For example,
> 100mV and 100A.  My guess is that this means that the voltage drop at
> 100A will be 100mV, so the resistance is 0.001 ohms.  That's small !
> So I'll use these figures for now.
>
> However, with this small of a resistance, max power = 100A x 0.1V = 10W.
> Not bad.  And 100mV of voltage drop will not be a problem.  Now all I have
> to do is measure the voltage drop across the 2 resistor terminals.
>
> The questions are:
> - I have an ammeter that is "shuntless".  Does this mean that it has an
> internal resistor for voltage drop?  Or some other method of current
> measurement?  If internal resistance, how does it deal with resistance
> of the (long) connecting wires?
> - How do I get a PIC to measure -100mV to 100mV accurately?  A small
> voltage reference would be necessary, but would the PIC be accurate
> given noise, etc?  Or does the 100mV now need to be amplified?
> - How about getting an accurate resistor -- if I use an actual resistor
> instead of a shunt, can I get a resistor that small and that is accurate
> to within a couple percent?
>
> Or am I going about this all wrong?
>
> Cheers,
> -Neil.
>
> --
> http://www.piclist.com hint: The PICList is archived three different
> ways.  See http://www.piclist.com/#archives for details.
>
>

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