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Sometimes I need to calibrate devices which have current outputs. Typically the devices output 4...20mA currents and I have been using a calibrated 250Ohm precision resistor(goes to a single-ended data acquisition channel) to measure current output( measure of course as a voltage and then divide by the resistor value).

This is very common way of measuring such current loops. But my concern is that none of the current sourcing devices I encounter have specified output impedances in their manuals/datasheets. Here is an example device.

As you see there is no output impedance mentioned in the manual. So I was thinking to be on the safe side maybe it is better to use an active circuit to minimize the possible error due to the current source's output resistance and the 250Ohm shunt combination: enter image description here

I also see some noise time to time. So my aim is to measure the current as precise as possible and the error should not be more than 1%. In this case I'm not even sure if it would be waste of time to make an active circuit. The problem is I don't know these device's output impedances so that I could estimate the error when shunted with 250Ohm.

What is output impedance of such current output devices roughly? Is it worth to worry about precision here and to use an active circuit? I would appreciate any ideas.

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    \$\begingroup\$ Why are you worried about the output impedance? Provided the device delivers the correct current under the right conditions it's the device's job to worry about it, not yours. \$\endgroup\$ – Finbarr Apr 18 '17 at 9:10
  • \$\begingroup\$ It all depends on the sensor, how accurate is that sensor ? The example device has a +/-3% accuracy (1st table, 3rd line) so trying to get down to 1% is pointless for this device. Also, for this device the load resistor must be 300 ohms or less, so use that. You cannot get better accuracy by making an "ideal" current measurement setup. \$\endgroup\$ – Bimpelrekkie Apr 18 '17 at 9:10
  • \$\begingroup\$ If that helps you and is practical, you could do the measurement with 2 different Rsense values and solve for Isource and Gsource. \$\endgroup\$ – vrleboss Apr 18 '17 at 9:12
  • \$\begingroup\$ Ok but when are active current to voltage converters are used then? If the current sources always have extremely high output impedance why to use an active device such as this: myclassbook.org/wp-content/uploads/2017/03/… ? \$\endgroup\$ – atmnt Apr 18 '17 at 9:19
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    \$\begingroup\$ No, it's because the feedback mechanism would compensate for the opamp's output impedance. \$\endgroup\$ – Finbarr Apr 18 '17 at 9:52
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The device you're especially considering lists:

Load ≤ 300Ω

as requirement. Others give a maximum output voltage (which, in this case, would be 20 mA · 300 Ω = 6 V), yet others actually do list a source conductance in the sensor datasheets.

These numbers typically are the result of both limited supply voltage and limited current drive capabilities within the sensor, and are just as good as giving you a source conductance – you wouldn't use a very small-valued measurement resistor, anyway, because that would make your measurement hard to make precise.

So, your 250 Ω is a good choice – close enough to the maximum load resistor, and small enough so that it makes sense for your measurement setup.

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If you worry about any possible error due to source-resistance vs load resistance, don't. Because, as you might guess, current sources have theoretically infinite impedance (A huge value in practice). So, for current-to-voltage conversion, load resistance becomes unimportant in this case --but, of course, it's important for accuracy.

I've no idea about expected resolution; but 250Ohms seems quite acceptable for 1mA-20mA range with 0.01mA resolution/accuracy (2.5mV/step).

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  • \$\begingroup\$ I see so you mean the output impednace by default is crazy high. But when are active current to voltage converters are used then? If the current sources always have extremely high output impedance why to use an active device such as this: myclassbook.org/wp-content/uploads/2017/03/… ? \$\endgroup\$ – atmnt Apr 18 '17 at 9:20
  • \$\begingroup\$ @user134429 You're thinking too much in "it is always like this". There is no general truth, what you have to do depends on the circumstances. Some sensors/circuits do need the active opamp based load, others don't. All current sources have some output impedance. Question is, how does it compare to the resistor used for measuring that current. If the output impedance is 100x higher than the measurement resistor, you have 1% error. "Crazy high" or "very high" has no meaning to engineers. 100x higher than... does mean something. \$\endgroup\$ – Bimpelrekkie Apr 18 '17 at 9:52
  • \$\begingroup\$ @FakeMoustache So I need to be sure these devices have at lease 250x100=25k output impedance for 1% accuracy measurement? Since they dont write anything I assume its higher than 25k. If I can be sure about it I wouldnt use an active circuit. \$\endgroup\$ – atmnt Apr 18 '17 at 9:59
  • \$\begingroup\$ @user134429 Personally, I'm using active circuitry for measuring currents when I need a wide range. For example, I designed an instrumentation amplifier (INA) based circuit for measuring currents from 1mA to 7A with 1mA accuracy using single 20mR shunt resistor. About output impedance, maybe it's not given in the datasheets since they could be extremely high (1MOhm for example). \$\endgroup\$ – Rohat Kılıç Apr 18 '17 at 10:00
  • \$\begingroup\$ I assume its higher than 25k Don't assume, better to "not know" or measure it like suggested above, use 250 ohms and 125 ohms and see if there is a difference. \$\endgroup\$ – Bimpelrekkie Apr 18 '17 at 10:09
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You can ignore \$R_{source}\$ because the output current (going out of your dashed box) is controlled to be the value you expect, and that's all that matters (just as you can ignore series resistance of a controlled voltage source); of course this holds only within certain boundaries, i.e. if \$R_{sense}\$ is not too large.

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