I am currently measuring the current draw of a PCB and have managed to get down into the uA range. I am able to read 8uA when I use my TENMA 72-7780 as a current sensor.

However when I use a 7R5 current sense resistor and two scope probes either side of the resistor, both referenced to the ground connection of the N93CX power supply I get a ~600uA calculated current draw from V = IR, 0V005 = I 7R5 .

I had a look at the input resistance of my probes ( P6060 ) and they are 10MR, which should only give ~0.3uA draw. The power supply has a ~0.5m cable from it to the 7R5 resistor. The scope probes are ground referenced by their clips to the end of the ground cable.

Does anyone have any idea where this 600uA calculation error is coming in?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Is your current sense resistor in the high side? That's the only time it makes sense to do it the way you are. That's probably the source of the error - the amplifiers and attenuators in the two channels aren't perfect. Probably easier to measure low side. That's one probe and a ground connection. \$\endgroup\$ – JRE Oct 31 '17 at 10:03
  • \$\begingroup\$ Try swapping the two scope channels over to see if the readings are the same. \$\endgroup\$ – Finbarr Oct 31 '17 at 10:03
  • \$\begingroup\$ Yeah, the resistor is on the high side. I measured a voltage drop of 8mV, or 1mA, one way and 6mV, or 800uA, the other way. \$\endgroup\$ – ConfusedCheese Oct 31 '17 at 10:16
  • 1
    \$\begingroup\$ Does 3V751 means 3.751V? And 0V005 means 0.005V or 5mV. \$\endgroup\$ – StainlessSteelRat Oct 31 '17 at 11:47
  • \$\begingroup\$ Yeah, apologies it's a convention I learned. \$\endgroup\$ – ConfusedCheese Oct 31 '17 at 12:51

Calibrating the gain on each channel is critical for differential mode (DM) with this very low DM/CM ratio. 7mV/3.7V * 100%= 0.19%. You cannot do this by eye on the scope so a DSO calibration is key and may not be easy to adjust.

For resolution this and power dissipation reasons, a 1% max Vdrop is computed for current sense R or 75mV is often standard. This raises your load regulation error by 1% which is tolerable.

Thus I would suggest a 75 Ohm R and expect a 53mV drop with 700uA drain.

| improve this answer | |
  • \$\begingroup\$ Sorry, I am a little confused, there are alot of 75 related numbers about and i'm not sure if there is confusion. You say to use a 1% max Vdrop resistor, which for 3V7 would be 0V37. I then take my max expected current load, say 5mA, and calculate the current sense resistor from that, i.e. 74R. Which is nearly equal to your 75R quoted resistor? \$\endgroup\$ – ConfusedCheese Oct 31 '17 at 11:29
  • 1
    \$\begingroup\$ Also, most oscilloscopes have 8 or 10 bit ADCs. Using 4 V range and 10 bit resolution, the LSB is 4mV. It is not suitable for this kind of measurement. \$\endgroup\$ – TemeV Dec 26 '18 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.