Firstly, I'm not sure if sensitivity is the correct term, if anyone has a better suggestion please feel free to edit the title

So I have a motor and I want to measure the current output which I'm doing via a low ohm resistor in series. The voltage across this resistor then goes through an op-amp and finally through an A/D. I then take this digital value and plot it on a graph in real time.

The motor moves back and forwards along a rail, the problem I have found is that some rails I've bought have had burrs/defects and I have to resort to grinding every rail I buy. So I want to measure the motor current so as to look for which rails are ok and which ones need grinding as the motor should draw more current when passing over the burrs/defects.

The current circuit I have works great, however the extra current drawn when passing over the defect parts of the rail doesn't deviate too much from the standard current draw. This means that when I plot it on a graph in real time it's hard to discern where the defect parts of the rail actually are.

What I want to do is to increase the sensitivity(?) of the op-amp. So that there is a greater deviation on the output voltage. I would just increase the gain but the problem I have then is hitting the rails of my op-amp.


simulate this circuit – Schematic created using CircuitLab

AD623 Pinout

What are some other ways to increase my input to output ratio? The only idea I've had is to decrease the sense resistor value and use a higher gain.

Some values incase you need them:

Motor Voltage:  5V
Motor Current:  600mA
Sense Resistor: 0.27R
  • \$\begingroup\$ decrease your sense resistor ? That will allow you to increase your gain. \$\endgroup\$
    – efox29
    Commented Aug 23, 2016 at 13:56
  • \$\begingroup\$ As a signal visualization problem, this reminds me a lot of measuring power supply output ripple on a scope. The DC output is high and the ripple amplitude is small, maybe a 1000/1 difference. The solution there is to AC-couple the signal and crank up the vertical sensitivity, so you're only seeing changes in voltage. I'm not sure how you would do that here, just an idea in the spirit of brainstorming, for whatever it's worth :) \$\endgroup\$
    – scanny
    Commented Aug 24, 2016 at 5:04

2 Answers 2


Reducing your sense resistor and adding gain later is not going to help. What I think you need is another difference stage. (you could use an opamp or another instrument amp.) Take the output (of the first Instrument amp.) and subtract a DC voltage from it... set by a pot. Then gain that signal up to a useful level.

  • \$\begingroup\$ How would I go about subtracting a voltage? By a differential amplifier? \$\endgroup\$
    – user103993
    Commented Aug 23, 2016 at 14:41
  • \$\begingroup\$ @Hayman, sure a differential amp, an instrument amp (your AD623) or you could use an opamp in the summing configuration.. in which case you would use a negative voltage... not subtracting, but adding the negative... which turns out the same in the end. \$\endgroup\$ Commented Aug 23, 2016 at 16:10

Instead of connecting the negative input of your instrumentation amplifier to GND, connect it to a voltage just below the typical value of the positive input. You could just use a resistor/potentiometer divider and see if it's repeatable enough.

If it's too hard to reliably set the potentiometer, you might need your circuit to have a calibration step where it reads the normal voltage and supplies it to the negative input. Easiest would probably be with a ADC and DAC on a microcontroller, but you could possibly design some kind of minimum-hold analog circuit.


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