# measure the current of a short pulse

#### Emam

Jul 7, 2014
63
Hello,

I have a pulse generator, generating a voltage pulse of duration 2-3 milliseconds.
The voltage is around 6.5 V when I connect a special PCB to this generator.
I would like to measure the "current" which goes through this PCB.
I think with multimeter I can not because its too fast.
I dont have current probes for seeing the current directly on the scope.

Can I do something else?
Thank you

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,822
You can put a small resistor (e.g. 1Ω) in series with the pulse generator and the PCB. Measure the voltage drop across the resistor with your oscilloscope. from I=V/R you get the current.

#### Emam

Jul 7, 2014
63
Thank you. But can I see it on a multimeter? since its too fast?

#### GreenGiant

Feb 9, 2012
842
You can set the multimeter current reading to min/max (obviously you want max) but depending on how fast your multimeter is it may or may not be able to get the true peak of it

The other option is to try calculating out the current based on the resistance across the +6.5V and Return that the pulse generator is connected to, this will only work if its a simple circuit and doesn't have anything that changes resistance when the circuit comes on.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,822
Thank you. But can I see it on a multimeter? since its too fast?
With the series resistor you don't need currrent probes for the scope, you use the ordinary voltage probes.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,822
You can set the multimeter current reading to min/max (obviously you want max) but depending on how fast your multimeter is it may or may not be able to get the true peak of it
In theory: yes. In practice: no. No multimeter will be fast enough to record such a short pulse.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,891
You may have problems trying to measure the voltage across a current-sensing resistor with an oscilloscope if the resistor is placed in the pulse output line to the PCB. There will be a large common-mode 6.5 V pulse at the resistor and you want to measure the voltage drop across the resistor. This could be done using A and B channels on the 'scope with the input selected to subtract B from A, provided the sensitivity is high enough to observe the difference and the common-mode signal doesn't overload the A and B channel inputs. For example, if the current expected is 1 mA and you are trying to measure that across a 1 Ω resistor, the drop across the resistor will be 1 mV, but this will be sitting on a common-mode pulse of 6.5 V. Cranking up the sensitivity of the A and B inputs sufficiently to "see" a 1 mV differential signal may overload those inputs in the presence of the 6.5 V common-mode signal.

OTOH, some 'scopes have differential input capability with a large common-mode voltage range. With this type of 'scope it wouldn't be a problem.

In my experience it is better to avoid trying to measure differential signals in the presence of a large common-mode signal. There are two ways to do this. The easiest way is to place the current-sensing resistor in the return or common lead of the signal source. Then you can use a single-ended 'scope channel and connect the "ground" lead of the probe to the common lead of the signal source and measure the voltage across the resistor. This only works if the commons of the signal source and the PCB load are not electrically connected.

The other method, which I cannot recommend because of other issues, is to "float" the oscilloscope with a good isolation transformer. A "good" isolarion transformer will have a Faraday shield between the primary and secondary windings, low parasitic capacitance between the primary and secondary windings, and low leakage (high resistance) between the primary and secondary windings. Such transformers are not cheap. A simple power transformer with a 1:1 turns ratio is not an acceptable substitute.

If you proceed with the isolation transformer approach to "floating" your oscilloscope, the current-sensing resistor is placed in series with the output of the signal source, but now you need to use only a single input channel to the 'scope. Connect the 'scope probe to one end of the resistor and the probe "ground" to the other end. The reason I don't recommend this approach is it depends on the characteristics of the isolation transformer, particularly leakage resistance and parasitic capacitance, to obtain an accurate signal. And it is all too "easy" to accidentally "short" the oscilloscope "ground" to a real ground, like power-line or chassis common, and thereby destroy the isolation, rendering whatever measurement is made invalid.

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