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I need to measure the ripple current in a buck converter. In general, I add leads to the inductor and use a current probe. My issue now is that the inductor I am using has such a low inductance that the leads are almost doubling the total inductance. If I can't use a current probe, I would prefer to make the measurement with only one voltage probe. The only way I can think to do this is add a high precision resistor between my output capacitor and ground. I'm concerned though that in order to get an acceptable resolution from my oscilloscope, the resistor will be large enough that it adds substantial voltage ripple at the output. Are there any other methods anyone has had success with?

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  • \$\begingroup\$ What are the numbers we are talking about here? nanohenries? mhz? Is your load resistive? Then you could just use that as the shunt. \$\endgroup\$ – PlasmaHH Feb 5 '16 at 15:09
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    \$\begingroup\$ Use kelvin configuration: take look at this microsemi.com/document-portal/doc_view/… \$\endgroup\$ – R Djorane Feb 5 '16 at 15:28
  • \$\begingroup\$ Brett, ultimately what are you trying to achieve? I note than on a previous question you have tried to inject current into an inductor so, what is it that ultimately you are trying to measure or determine? \$\endgroup\$ – Andy aka Feb 5 '16 at 15:32
  • \$\begingroup\$ If your ripple current is in single amps, and your scope measures in half mV accuracy with decent noise floor, only tens of miliOhms in series with your inductor may be needed to get somewhat of an impression. Don't forget you need two probes if your current isn't at ground potential, though. \$\endgroup\$ – Asmyldof Feb 5 '16 at 15:33
  • \$\begingroup\$ @PlasmaHH I am talking ~100nH. Currently, I am running the circuit no load and only looking at the ripple (synchronous buck converter) \$\endgroup\$ – Brett Prudhom Feb 5 '16 at 17:15
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You could use a hall-effect current clamp sensor.

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    \$\begingroup\$ Hall effect devices use mag fields to measure current. How will the hall effect device, in close proximity to the localized, stray magnetic field emanating from the inductor, perform do you think? \$\endgroup\$ – Andy aka Feb 5 '16 at 15:25
  • \$\begingroup\$ Hall effect probes are not that fast .They respond in the microsecond range .At 20 KHz maybe but not at the unspecified frequencies that allow 100nH to be used. \$\endgroup\$ – Autistic Feb 5 '16 at 21:25
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Having in mind a standard Buck with a capacitive output filter, one straight forward option would be measuring the capacitor and load current at the same time and add it via the oscilloscope. If you perform a De-skew to compensate time delays on the current probes you may have acceptable results.

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If you dont want to do the current probe then you are talking a low value resistor like the others have said .Im assuming that your output and input voltages are low because you are using 100nH .Hence the current sense resistor has to be a low value so converter operation is not affected .Ballparking between 10 and 100milliohm will allow the converter to operate normally.Clearly your expected probe voltages will be low .The sensitivity of the scope will be adequate but probe noise pickup will be bad obscuring the expected triangle signal .What I do here is use a BNC connecter and RG58U coax instead of the scope probe .Most scopes dont have a low impedence termination switch ,Sure termination is a good thing but I generaly dont bother because my cable is only half a meter long and I get inflicted with otherpeoples strange scopes .If you do this you will get a proper clear waveform .Now off the subject have you considered using an air cored coil which off course wont saturate .!00nH should be easy to do in air .

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Sense the current to transform it into a voltage hence a transimpedance op-amp. See the LTC6102, you can simulate circuits using LTSpice.

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