1
\$\begingroup\$

I was reading about current to voltage converters and simple opamp current-meters. I found myself in confusion comparing them to passive current meters.

Active current-meter:

enter image description here

As you see above the Iin passes through R and the voltage drops to -Iin*R as Vout. So Vout is linearly changing and with Iin.

And below is measuring current without an opamp:

Passive current-meter:

enter image description here

Above the same logic, the Iin passes through R and Vout is linearly changing with Iin.

Can someone explain what is the point/benefit to use opamp here? What is the advantage of using opamp/active version over the passive one?

\$\endgroup\$

3 Answers 3

5
\$\begingroup\$

In your first circuit, so long as the current is small enough and it changes slowly enough, the negative feedback circuit keeps the inverting input of the op-amp at 0 V, so the source sees effectively 0 load impedance.

In the second circuit, the source sees a load impedance equal to the resistor value, which could change the current it produces. Furthermore, the higher the resistance value (increasing the resolution of whatever is measuring the voltage across the resistor), the greater the resistance seen by the source, so the greater the effect on its output.

the "current source seeing some load impedance" [issue]... is a bit confusing since the load resistance will be in series with the current source and why would it have affect on the current. A simple illustration would help a lot if you have time,

schematic

simulate this circuit – Schematic created using CircuitLab

If Rsense is 0, then all of the source current goes through Rsense and the measurement does not disturb the thing being measured.

If Rsense is non-zero, then some of the source current is diverted internally through the source's internal conductance, and the measured current is lower than it would have been if measured with a 0-resistance sensor.

\$\endgroup\$
4
  • \$\begingroup\$ You wrote "In the second circuit, the source sees a load impedance equal to the resistor value, which could change the current it produces." Why would the current change? \$\endgroup\$
    – user1245
    Commented Apr 15, 2017 at 16:57
  • 1
    \$\begingroup\$ Because real current sources have nonzero output conductance. \$\endgroup\$
    – The Photon
    Commented Apr 15, 2017 at 17:00
  • \$\begingroup\$ As far as I understand there are two issues here. One of them is if one measures the output voltage, the measuring device's input resistance will have affect on the measurement. This depends on how big the Rmeter(in second figure) as well. The second issue is the "current source seeing some load impedance" before the input to the opamp. I think I understand the first one, but the second one is a bit confusing since the load resistance will be in series with the current source and why would it have affect on the current. A simple illustration would help a lot if you have time, thanks \$\endgroup\$
    – user1245
    Commented Apr 15, 2017 at 17:06
  • \$\begingroup\$ With the circuit diagram and a few notes below it makes a lot difference in understanding. Thanks \$\endgroup\$
    – user1245
    Commented Apr 15, 2017 at 17:26
3
\$\begingroup\$

A DC coil current meter has a certain loss of winding resistance often rated at 20kΩ/V which yields 50uA full scale. Shunt R's are then added across for precise ratios to scale the full scale current.

An OpAmp due to the negative feedback will have an input impedance of Rf/Aol for the feedback R and open Loop Gain of 1e6, thus provides lower loss and higher gain and of course also can convert I to V at a wide frequency range unlike a DC coil meter and also very low currents with high gain and can be bipolar AC to DC with latter stages of precision rectifiers.

Whatever the source impedance is, the current sensor must be much lower impedance to sense all of the current.

schematic

simulate this circuit – Schematic created using CircuitLab

Obviously it depends if you want a DMM output or an analog meter output for choice and AC uses a rectifier in coil meters, scaled to measure rms/peak or rms/avg so unless a pure sine may be in error and DC for rms/pk measure twice the actual in AC V or A mode in analog meters.

\$\endgroup\$
6
  • \$\begingroup\$ Let's say Iin is coming from a current source in both cases. What is the difference? I dont get why opamp makes it better. is it possible to illustrate this in your answer? \$\endgroup\$
    – user1245
    Commented Apr 15, 2017 at 16:22
  • \$\begingroup\$ Current sources do not exist. Always some compliance limits. The VirtualGround of the OpAmp imposes a constant value (V=0) to the current, providing more accuracy. At least at low frequencies. \$\endgroup\$ Commented Apr 15, 2017 at 16:56
  • 1
    \$\begingroup\$ Current source Zout must always be >>> Zin for accuracy as an impedance divider loss (loading) affects results. \$\endgroup\$
    – D.A.S.
    Commented Apr 15, 2017 at 17:21
  • \$\begingroup\$ @TonyStewart.EEsince'75 When there is no resistor before the inverting input, I thought the input impedance of the inverting amplifier circuit Zin would be zero. But you find it by dividing Rf/Aol. In this case 33Ohm. In many texts they show this as Zin=Rin which would be zero in this case since there is no Rin. What do you say about it? \$\endgroup\$
    – user1245
    Commented Apr 15, 2017 at 17:24
  • 1
    \$\begingroup\$ Zero input Z assumes infinite DC gain and thus Vin =0 ideal OA. In reality it depends on Aol which for OA's is large like 1e5 to 1e7 \$\endgroup\$
    – D.A.S.
    Commented Apr 15, 2017 at 17:29
3
\$\begingroup\$

Tony has things correctly stated. I thought I'd just add another way of writing similar things without getting mired. Let's redraw the circuits:

schematic

simulate this circuit – Schematic created using CircuitLab

I've drawn the regular analog meter movement on the top, and its equivalent. Here, you can see that the meter does actually inject a potentially significant resistance in series (which may upset the measurement, depending on circumstances.) The analog meter movement here presents \$500\:\Omega\$ of DC resistance.

I've drawn the opamp circuit on the bottom, and its equivalent. Here, you can see that the effective series resistance may be very significantly diminished, now. (\$A_O\$ is the open loop gain of the opamp.) The open loop gain of the opamp would have to be as bad as 100 to present the same series resistance. But that's not the usual case. Open loop gains are much, much higher in even the cheapest opamps. (I've neglected offset voltage and current and bias currents, to focus on the central point.)

By the way, you can often use much cheaper analog panel meters as a result, as well, and still get significant reductions in the impact on a circuit under measurement using an opamp. And opamps, even expensive ones, are often a lot cheaper than a very high quality analog D'Arsonval movement will be. I suspect this has largely killed wide-scale production of sensitive D'Arsonval movements, despite the somewhat more recent advent of very powerful Neodymium permanent magnets.

\$\endgroup\$
2
  • \$\begingroup\$ Sometimes I deal with 4...20mA current loop output by various transducers. I was doing the data acquisition as voltage just by a 250Ω ±0.1% shunt resistor. As shown here: dataq.com/products/accessories/r250/4-20ma-current-loop/… But do you think better to use opamp/active current to voltage converter for better precision? \$\endgroup\$
    – user1245
    Commented Apr 15, 2017 at 19:52
  • \$\begingroup\$ It depends on current noise , EMI , cable loss and cable type (UTP or TP, etc) and length and where your ground is for CM noise. If you flow all the current thru the loop you can use single ended TIA. If remote shunt R then use differential voltage INA. (TIA=transimpedance Amp, INA = Instrument Amp) or a simple coil-meter \$\endgroup\$
    – D.A.S.
    Commented Apr 15, 2017 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.