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Firstly, I am a ME seeking advice from EE's. I have a Quadvolt voltage data acquisition box from Omega (P/N: OM-CP-QUADVOLT-30V) http://www.omega.com/pptst/OM-CP-QUADVOLT.html, that I intend on using to monitor current through a 100A 50mV current shunt in a 12 volt solar system that powers a light.

I want to monitor current in the positive conductor between the battery and the solar/light controller. My problem is that the Omega box cannot handle reverse polarity. I am concerned that when the system switches from charging to drawing power from the battery, the current direction will switch and therefore (I would assume) the polarity at the current shunt would switch, correct?

If my logic is correct I want to be able to protect the Omega box. I had landed on using an N-Channel MOSFET in series with the negative side conductor from the current shunt to the Omega box. Then I got lost in all the terminology. Since the voltages I am monitoring are so low, on the order of 50mV, will a MOSFET protect the box without causing noticeable voltage drop? Will negative 9-13 volts be enough to actuate the switching device and protect the Omega box in a reverse polarity situation?

Found this recommended on another page https://www.digikey.com/product-detail/en/SI4838DY-T1-E3TR-ND/SI4838DY-T1-E3TR-ND/1656477. Would this work for my application? Thanks so much for the help! I'm lost when it comes to this stuff!

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You said you are going to use a 50 mV shunt with a 34 V, 16 bit ADC. My first comment is that because the shunt output range is such a small portion of the ADC range, you will not get much resolution in your readings.

But specific to your question, the specs for your Omega ADC says that it can measure down to -2 volts. So when the polarity reverses on your shunt, the -50 mV is still well within the range of the ADC converter. You should not have a problem.

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  • \$\begingroup\$ Thanks for pointing that out. With the 4 input channels I am measuring current from the solar panel to charge controller. Current from the charge controller to the battery. And voltage of solar input and charge output. So solar will never change polarity. The battery leg will, but it will only be a 6A draw, so 6A/50mV=-120mV. So I should be good. Thanks again. \$\endgroup\$ – Jason P Jun 26 '17 at 16:08

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