# Current sensing resistor selection

Discussion in 'General Electronics Chat' started by russpatterson, Oct 14, 2010.

1. ### russpatterson Thread Starter Member

Feb 1, 2010
351
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I'd like to design a current sensing solution that will measure up to 20 amps. I understand one of the approaches is to measure the voltage drop across a resistor to to deduce the current flow. The part I'm having trouble with is figuring out how to spec that resistor.

The goal is to use a very small resistor, I've seen values from 10mOhm to 100 mOhm. I'm assuming the lower case m is for "milli"-Ohm. To make things simple assume a system with:

V = 15 Volts
I = 20 Amps
Rsense = 1 Ohm

How do I calculate the voltage drop across the 1 Ohm resistor?
V=IR gets 20 which is obviously wrong

I know that the load resistance must factor in but overall I don't get how to figure out what power rating the resistor needs. All the current must flow through the resistor.

P=VI gets 300 watts, obviously wrong too.

What am I missing?

Jul 7, 2009
1,585
141
If you had 20 A through a 1 Ω resistor, there must be 20 V across it, exactly as v = iR predicts. Thus, a 15 V potential cannot get a 20 A current through a 1 Ω resistor.

For currents in the range of 10 to 20 A, you'd likely want a sense resistor in the mΩ range. Thus, a 1 mΩ resistor would have a voltage drop of 20 mV if 20 A were flowing through it. You'd then calculate the power being dissipated by the resistor by squaring the current and multiplying by the resistance; in this case, you'd get 400*0.001 or 0.4 W. You can also use iV to get (0.02)(20), which is also 0.4 W.

A handy thing to remember is that 1 foot of 10 gauge (AWG) copper wire at room temperature is 1 mΩ. 1 foot of 12 gauge copper wire is 1.59 mΩ. This can be handy to make shunts with resistances on the order of a mΩ. You can use insulated wire, bend the piece in half, then wrap the doubled wire around a cylinder to minimize the inductance of the shunt (probably not terribly important for a DC application).

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3. ### schmitt trigger Active Member

Jul 12, 2010
46
8
One can actually get current sense resistors in the micro-ohm range.
As a rule of thumb, one must not drop more than 1% of the available voltage thru the sensing resistor. In other words, 150 mV.
Doing the math, this equals 7.5 milliohm...go for a 5 milliohm resistor.

The voltage will of course, be way too low to be useful. Therefore, a good opamp with low offset voltage is required to amplify it to an useable level.

Another route is to use a hall-effect current sensor. The ACS712 or ACS713 would work for you.

4. ### russpatterson Thread Starter Member

Feb 1, 2010
351
16
Thanks for the replies. So the voltage drop across the resistor is the same no matter what the source voltage is, correct?

Given the example, the 20mV drop is 0.133 % of the source voltage so that's well below 1%. Why not always use the smallest resistor possible? Is it in the sensitivity of the op-amp? Is that the limiting factor?

Regarding the magnetic solution with the Hall Effect sensor. Do you run a trace below the part on the board? 20 amps through a trace that fits under a 8-SOIC package doesn't sound like a good idea. I've looked at the data sheets for a bunch of those Hall Effect sensors and it's unclear how you're supposed to get the part in range of your current.

What are the trade offs? More money for the Hall Effect sensor vs. the resistor and op-amp solution but there's less power loss?

5. ### Markd77 Senior Member

Sep 7, 2009
2,803
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If you look at the ACS712 datasheet current flows through the chip. You just have a big trace that runs into pins 1 and 2 and another big trace that runs out of pins 3 and 4.

6. ### timrobbins Active Member

Aug 29, 2009
318
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You can use the smallest shunt resistance that you can obtain that also allows you to amplify to the level you need with suitable noise and accuracy. Not a trivial design process if you have tight accuracy requirements, or some other significant issue, or need to document the whole design.

7. ### DonQ Active Member

May 6, 2009
320
11
One option for a low value resistor is a measured length of wire. Tables of resistance per length for different gauge wires are widely available, then the math to get the resistance you want is easy. This can work from fractional amps to more than anyone should ever be messing with! You can also do this with a trace on a circuit board. Calibrating is just a matter of moving one of the sense leads back and forth along the wire/trace.

You can also get current shunts that consist of a calibrated section of conductor with 'Kelvin' or 4-wire connections. These are generally called out in 'voltage per amp' values. The one I use commonly is a '50mV/Amp' (which just happens to work out to 1V/20A. It is a piece of conductor (looks brassy) about 1/4 inch wide, 1 inch long, and about 1/32 inch thick, on a mount with a couple of big nuts for the current, and a couple of small screw terminals for the sensing circuit. Calibration is by filing a shallow notch in the long side.

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8. ### marshallf3 Well-Known Member

Jul 26, 2010
2,358
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Calibration is by filing a shallow notch in the long side.

Just having some fun because I know there's some of the older gang reading this thread.

In the early days virtually all common resistors were carbon and if you didn't have the proper value you could "trim one up" in value by filing down through the body and a little of the carbon element with a triangular file.

One of my early "pizza and beer money" jobs in college was working as the TV tech for a place that also sold the complete line of GC/Calectro/Motorla HEP components so obviously we were the source of parts for all the students and their projects. On occasion you'd get an engineer in there, 5 pages of computer printout and all, showing that he needed 8.3405K and 28.1376K resistors. Most would accept the explanation and facts of life that resistors come in common values but some wouldn't budge, that's when we'd draw out the concept of filing a resistor.

9. ### schmitt trigger Active Member

Jul 12, 2010
46
8
Oh boy; you've stirred some fond memories.
I did purchase plenty of Calectro stuff for my early student projects.
HEP was also popular. If I remember properly General Electric did have some useful components here and there too.

I dreamed about owning a B&K 3 digit multimeter.

10. ### russpatterson Thread Starter Member

Feb 1, 2010
351
16
Thanks for the replies. Regarding the Hall Effect approach and the ACS712 part mentioned. In the data sheet I can see that there are two input leads IP+ and two output leads IP-. So 20 amps flows through that part with those skinny 8-SOIC legs? That's not going to burn up or at least have a significant voltage drop due to resistance? 16 guage wire is rated for 22 amps for chassis wiring and that's a lot thicker than two 8-SOIC leads.

Also I couldn't find where it was rated for 20 amps. Is it the:

Overcurrent Transient Tolerance : IP 1 pulse, 100 ms 100 A

or

Optimized Accuracy Range : -20 to 20 Amps

Are those the characteristics that specify maximum and optimum current through the part?

11. ### marshallf3 Well-Known Member

Jul 26, 2010
2,358
201
They really put Radio Shack to shame - even in those days when Radio Shack actually sold some parts. One of my favorite parts of the job (when I didn't have to make a service call) was to keep track of the inventory and what was coming out as new, I made darn sure to carry everything they had to offer with of course an exception for the high-end HEP RF power transistors.

Everyone else had a Simpson 260, I had a Triplett 630NA - far better analog meter. I think my first digital was a B&K bench top but in those days accuracy just wasn't all that important and a good mirrored scale analog could easily get you to x.xx