Having only an ordinary oscilloscope that shows voltage, is there any way to modify the circuitry or do some intelligent guessing to get an idea of what the current wave form looks like?

In my case, I'm pulsing a coil with square voltage waves and I want to see the resulting current being caused by inductance and parasitic effects to get a sense of how well my coil is constructed and what exactly it's doing to the waveform. I'm particularly interested in its behavior during discharge during the low portion of the square wave.

  • 5
    \$\begingroup\$ Put a small resistor in series and measure voltage across it: You'd have to ensure though that the resistor itself is not going to significantly modify the circuit behavior. \$\endgroup\$ Commented May 29, 2013 at 18:47
  • \$\begingroup\$ Using a series resistor is what I theorized, but it didn't work. But apparently I messed up because I measured the voltage across the inductor instead of across the resistor. Woops. :P \$\endgroup\$
    – JamesHoux
    Commented May 29, 2013 at 19:11
  • \$\begingroup\$ If the inductor has a ground connection then you should be OK to put the small valued resistor in the ground connection. Your scope will naturally connect to ground and the junction of inductor/resistor. Resistance value depends on inductor value so more info please. \$\endgroup\$
    – Andy aka
    Commented May 30, 2013 at 7:12

1 Answer 1


There are many ways you might go about this. If you have a resistor, or any approximately ohmic thing in series with the current you wish to measure, then you can measure the voltage over the resistance and calculate the current by Ohm's law:

$$ I = E/R $$

This gives you the current in the resistor, which must be equal to the current in anything else in a series circuit with that resistor. (Kirchhoff's current law)

It's somewhat likely that you have a MOSFET in your circuit. A MOSFET that's on is almost like a resistor, and depending on how accurate you need to be, you may be able to treat it as one. I've done this many times for sensing motor current in H-bridge drivers where accuracy wasn't critical.

If you don't have a MOSFET or you need better accuracy, then you can always add a very small resistor, small enough to not significantly change the operation of your circuit.

If you have a capacitor in series, then you can measure the voltage across it and calculate the current by:

$$ I(t) = C\dfrac{dV(t)}{dt} $$

The current and voltage in an inductor are also related by:

$$ V(t) = L\dfrac{dI(t)}{dt} $$

Though in this case by measuring the voltage you only get the rate of change of current, which doesn't really tell you the magnitude of the current, but maybe you don't need to know that.

If none of that works for you, you can measure AC current with a current transformer. Put a couple turns of wire around a toroid core, then pass the conductor in which you want to measure the current through the core. There are commercially available current transformers that are already calibrated, or you can make your own and calibrate it with a known AC current source.

Or, you can look at hall effect sensors, which can measure DC currents. There are commercially available modules which contain a straight-through conductor, the sensor, and an amplifier all in one package. Usually not good for small currents, but a good way to measure large currents without the losses that would come from a series resistance.

After having established a relationship between current and voltage by any of these methods, you can now observe the voltage with your oscilloscope, and know what the current must be.

  • \$\begingroup\$ While this is an outstanding overview of current measurment techniques, the question wasn't about measuring current. It was actually about visualizing the waveform of the current on an oscilloscope. \$\endgroup\$
    – JamesHoux
    Commented May 29, 2013 at 20:17
  • 3
    \$\begingroup\$ @Jim This answer describes many ways to convert a current into a voltage, so you can put your oscilloscope probe on them. Bam, visualized. \$\endgroup\$
    – Phil Frost
    Commented May 29, 2013 at 20:45

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