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Current Meter for 12V Solar System

Hello,

I'd like to create a current meter for a small solar system that I've built. The system has a couple large deep cycle 6V batteries hooked up in series. It's mostly used in the evenings and the load is one or two amps. However, sometimes I like to use some short duration, high current loads on this system (hand power tools). These loads are on the order of 30-40A. I've looked into setting up a shunt resistor in series with load, but I don't like the idea that I could (a) cook that resistor, or (b) that it's wasting my precious solar electrons with resistive heating. So, knowing just enough to be dangerous, I've looked into using some flavor of Hall Effect sensor to measure current without introducing a resistive load. The H-E sensors that I've looked at have different sensitivities at different currents. For example, a

0-20A H-E sensor gives better resolution at, say, 3A than a 0-100A H-E sensor. Given that, I was wondering whether it would be okay to hook both 0-20A and 0-100A H-E sensors in series with my load. When the draw is below 20A, read the 0-20A sensor, when it's above, read the 0-100A. If I'm going about this all wrong, then I'd love to hear your solution.

Thanks in advance,

-mt

Reply to
marshall
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search for "shunt" about 1/2 way down the page there is a 600, 50 and 100A current shunts that might meed your needs.

Tom Woodrow

snipped-for-privacy@perilith.com wrote:

Reply to
Tom Woodrow

You can also use a current sense transformer for things like this.

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--
Regards,
   Robert Monsen

"Your Highness, I have no need of this hypothesis."
     - Pierre Laplace (1749-1827), to Napoleon,
        on why his works on celestial mechanics make no mention of God.
Reply to
Robert Monsen

Oops, you want DC. These won't work for that, they are only for AC.

--
Regards,
   Robert Monsen

"Your Highness, I have no need of this hypothesis."
     - Pierre Laplace (1749-1827), to Napoleon,
        on why his works on celestial mechanics make no mention of God.
Reply to
Robert Monsen

I don't think you need to worry about your losses in the shunt IF you start with a 50ua meter movement. Off the top of my head you might dissipate 1 watt in the shunt with a 40 amp (480watt) load. You just need to know or measure the internal resistance of the meter and calculate the shunt resistance. Or, you could buy two .01 ohm resistors put them in parallel and use a cheap digital meter on the 200mv range to measure the voltage across the resistor. This would be 40 amps full scale.

Reply to
amdx

One other method if you can tolorate just fair accuracy is to measure the drop across a battery cable, this can be measured with a very low cost 0 to

200 mv digital meter powered by a 9 volt battery. Just put a NO PB switch in the 9 volt battery lead, then you have push to read.

Jack

Reply to
Jack Hayes

Thanks very much for all the great responses!

Another twist in my scheme that I should have included in my original post is that I'd like to read the voltage on the [shunt, hall-effect sensor, garden gnome...] with a 10 bit resolution ADC. If the ADC measures from 0-5V and the shunts range is 0-50 mV, then I'm thinking that my 10 bit ADC is not going to cut it. Back to the drawing board.

Either way, ya'll have given me a lot of great ideas to work with.

-mt

Reply to
marshall

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