I want to turn an LED indicator on when current is flowing through a path. If there's no current the LED will be off, and when there's current, the LED would turn on.

I've been experimenting with placing a shunt resistor in series with the measured load, and the voltage reading indeed relates to the current load, on the mV scale around 6 to 60 mV's. I'd like to turn a LED on when there's any voltage on either side of the shunt resistor. For this I've unsuccessfully tried to amplify the voltage with a 741 op-amp by "supposedly" measuring the difference between the voltage and setting a gain of 1000's so that the output saturates the maximum voltage available for the op-amp. Here's an overview of what's going into the op-amp according to pinout:

  • V+ : +5v
  • V- : -5v
  • +op : One side of the shunt resistor
  • -op : The other side of the shunt resistor

Output: Always 4.5v no mater what load the line with the shunt resistor has.

So what would I need to do for the 741 to return 4.5v when pins across the shunt resistor measures "any" voltage and 0v when there's no voltage across the shunt resistor. In programming style:

if (shuntdifferential > 0)
    output = 4.5v (anything that would emit light from LED)
   output = 0v (anything less than what would emit light from a LED)

The question is: How can I set up the amplifier to do this? I've got +12, +5, 3.3, 0v -5v, -12v from my power supply available. Note that I'd like to know if it's possible to read current from any rail from the psu with the same circuit. Also, maybe I'm looking into some weird solution when there could be a simpler solution for my purpose, recommendations?

Please don't be harsh on me if I'm doing something wrong, the reason I'm asking for help is because i want to learn. Thanks for your time and help :)

  • 6
    \$\begingroup\$ Could you please post a schematic of what you currently have? It would help us a lot. \$\endgroup\$
    – AndrejaKo
    Commented Apr 21, 2011 at 20:32

4 Answers 4


If you want a simple solution, you could look for "current sense amplifier" ICs, which are designed specifically for this purpose.

The approach you describe is using the op-amp as a comparator, which is a viable strategy, but it will behave unpredictably when the input voltage difference is around 0V, so somehow you need to introduce an offset in the comparison.

There is a sample current sense circuit in figure 27 of this application note. The quick way to analyze that circuit is to use the rule of thumb that an op-amp in a negative feedback configuration (meaning with a connection from the output back to the \$-\$ terminal) will make the voltages at its inputs equal. I can explain in more detail if it would be useful.

To create a current threshold detector from that circuit, you could add a comparator on \$V_{OUT}\$, or you could remove the \$R_3\$ / \$R_5\$ divider and connect the \$-\$ terminal directly to the low side of the sense resistor \$R_1\$. The \$R_2\$ / \$R_4\$ divider would then set the trip point of the detector.


"... when current is flowing through a path"

That's pretty vague. You don't say what the current range is or what the voltage supply is. My guess is that the input voltages to your opamp are either too high, or too low, depending on the position of your shunt. Too close to the rails anyway.

In the following I'll assume that your current range is from a few \$\mu\$A to hundreds of mA.

That poses a dilemma for the shunt resistor. You'll need a high resistance to detect the lowest current, but that would cause a too large voltage drop at the highest currents. A current mirror is the solution. It will allow you to drop any voltage you like without troubling the voltage supply for your load. A 10k\$\Omega\$ shunt for instance will give you 100mV at 10\$\mu\$A. At currents greater than 1mA the lower end of the shunt will go to -5V, but that's not a problem; we want to detect current, not measure it.

Let's place the current mirror on the low side, and the shunt goes in the mirror path on the high side. Connect both sides of the shunt to a differential amplifier. The resistive dividers will shift the shunt's high voltage (well, near V+) to ground level. Feed the output of the amplifier to a comparator, which can compare with a low threshold. Comparators often have an open collector output, so they can drive the LED directly. Don't forget the series resistor.


Here's another idea which may be simpler if you have needed tool-chain or would be good to consider when you do get it.

Get a simple microcontroler with analog comparator. Connect one side of the comparator to the ground and the other to the shunt resistor. After that you could keep comparing the voltages and have the microcontroller turn on the LED when there's voltage on the pin.

I can't recommend any specific microcontroller right now, but I'm sure that your favorite manufacturer would have one. Also small microcontrollers are very cheap and you might be able to find one which could replace the op-amp and not increase the price of the project considerably. Also don' forget that there are tons of other stuff a microcontroller could do, so with some planning, you could replace two or more devices and make up the cost.

On the other hand, if you never worked with microcontrollers and don't have needed tools, the introduction into that area could get expensive or troublesome.


This is simple, why don't you use an INDI-LINK LED current indicator? check it out on U-Tube working range +/- 3mA to 10 Amperes AC or DC, line powered, works regardless of the line voltage ( as long as you have above 350mV available within the circuit) Inni-link will also display the polarity of a current and also give 2 volt free opto-isolated outputs if you need them to go to a data system maybe. Very cheap solution and the only way to make electrical systems transparent for their working condition.

  • \$\begingroup\$ A schematic of how this circuit is connected would be helpful and links to associated pages would also be useful. Telling people to look at youtube does not seem that helpful. \$\endgroup\$
    – Kortuk
    Commented Jan 12, 2013 at 16:02

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