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I want to build this circuit I found at eetimes.com:

enter image description here

The idea is to use this circuit as a current sensor which outputs a voltage Vo which I can read and process using a tool like NI myDAQ.

I've come up with a few questions as I began simulating the circuit in software:

  1. How do I model the system load as a resistance? For example, if \$V_{bus}\$ is a battery for a car and everything that the battery powers is the load, how do I determine the equivalent resistance of all the car's electrical systems? Also, if the car only exists on paper, how would I calculate an equivalent resistance that way? very large system that you couldn't reasonably calculate its equivalent resistance. How would I go about determining the equivalent resistance? What if the system does not actually exist--what analytic methods could I use to determine equivalent resistance?
  2. How do I decide what resistance values to choose for Rf and Rg? Do I want to choose them such that the value of Vo is equal to the value of \$I_{load}\$ or can I choose any arbitrary values for \$R_f\$ and \$R_g\$?
  3. How small can I make \$R_{shunt}\$? Preferably, I'd like to make it as small as possible so that I lose as little power as possible from it as possible.
  4. Can I use the same source Vbus to power both the system load and the op-amp (\$V_{OA}\$)?
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  • \$\begingroup\$ The source impedance of the signal is dominated by the shunt. The impedance of the System Load does not matter. For your purposes, the System Load is just a current source. I am too lazy to answer all your questions. I do suggest you google "current sense amplifier" and see if you can make do with an off-the-shelf part instead of building your own with op-amps. Also, you may want to use high-side current sensing instead of low-side, to avoid creating two different grounds. Finally think about starting current. The starter may draw 200 amps or so. \$\endgroup\$
    – mkeith
    Oct 28, 2017 at 17:59

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This doesn't look like a great idea.

  • You have very high currents to contend with - particularly on engine starting.
  • Car grounds are dodgy at best. With corrosion, etc., subtle changes of resistance will occur.

I would recommend that you use a clamp-on Hall effect current sensor.

enter image description here

Figure 1. A DIY Hall-effect sensor. The Hall chip measures the flux induced in the core by the current in the wire threaded through the core.

enter image description here

Figure 2. A ready-made Hall-effect current sensor module.

This approach offers the possibility of complete isolation from the actual current being monitored and eliminates earthing problems and addition of a shunt and it's terminations.

You'll need to find one with an aperture large enough to handle the earth cable on your vehicle. Split clamps might be available in the size you required.

Now your questions:

  1. How do I model the system load as a resistance? For example, if \$ V_{bus} \$ is a battery for a car and everything that the battery powers is the load, how do I determine the equivalent resistance of all the car's electrical systems? Also, if the car only exists on paper, how would I calculate an equivalent resistance that way?

Just use \$ R = \frac {V}{I} \$.

  1. How do I decide what resistance values to choose for Rf and Rg? Do I want to choose them such that the value of Vo is equal to the value of \$I_{load}\$ or can I choose any arbitrary values for \$R_f\$ and \$R_g\$?

Figure out what the max output of your current sense is. Read what the desired full-scale input for your ADC is. The gain required is given by \$ A = \frac {V_{FS}}{V_{InMax}} \$.

  1. How small can I make \$R_{shunt}\$? Preferably, I'd like to make it as small as possible so that I lose as little power as possible from it as possible.

Skip it! Use the Hall technique.

  1. Can I use the same source Vbus to power both the system load and the op-amp (\$V_{OA}\$)?

Yes and it becomes simpler with the Hall sensor because there is only one ground with no shunt. Regulate and decouple well. Vehicle electrics are noisy.

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  • \$\begingroup\$ I second that, starter motor current will be YUUGE (like >100 amps) so shunts are a bad idea, hall effect is the way to go. You can even use a readymade current clamp with a voltage output like this one uk.rs-online.com/web/p/multimeter-current-clamp-adapters/… if the range and accuracy suits you. \$\endgroup\$
    – bobflux
    Oct 28, 2017 at 18:33
  • \$\begingroup\$ My mistake, I wasn't trying to imply that I was using this for a car. I just used the car example as a hypothetical to exemplify a large system with a lot of components. Please assume that the current sensor would see well under 100A. \$\endgroup\$
    – user104243
    Oct 28, 2017 at 18:59
  • \$\begingroup\$ Just for info, a car up to 2.5 litres petrol can see in the region of 500 to 600 amps on starting and a diesel of 2 litres 800A, while trucks can exceed 1000A. Just in case you decide to play! \$\endgroup\$
    – Solar Mike
    Oct 28, 2017 at 19:18
  • \$\begingroup\$ Are you sure, @SolarMike? I put a 300 A fuse on a pickup truck battery. It has not blown yet, but I have only started it a few times. The petrol motor is 7.2 liters.\ \$\endgroup\$
    – mkeith
    Oct 28, 2017 at 19:40
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    \$\begingroup\$ @mkeith reasonably - I had a snap-on current meter years ago that predated the digital clamp ones, not as accurate and no storage to look at the exact values... But I did see a truck battery terminal explode into molten lead when it was turned over with a looseterminal had to re-make the post (melt and pour the lead - had to borrow the mould). If the engine is a v8 (I have worked on a straight 8 of a similar capacity, but I don't think your pickup is made by Rolls-Royce... Anyway, a v8 is surprisingly easier to turn over that a 2.2 or 2.5 inline 4 so the current necessary comes down. \$\endgroup\$
    – Solar Mike
    Oct 28, 2017 at 20:10

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