I am trying to measure the current consumed by my circuit using an MDO4104 oscilloscope. I have soldered a 0.01 ohms resistor in series with the Vcc. The 2 ends of the resistor is are probed using 2 different probes of the said scope. Then I am using the math function of the scope to do a difference. Then this difference value is divided by the resistor value (0.01) to give me the current. The selected resistor is of 10W. I am expecting something like a peaks of 10Amp or so.

Now the 1st end of scope is Vccb (blue) and the other end is Vccg (green). Ideally Vccb > Vccg.


simulate this circuit – Schematic created using CircuitLab

In the oscilloscope I am doing a math operation of Vccb - Vccg. Ideally I must divide it by 0.01R. I understand that. But just the difference should be very small is what I assume. Just for me to analyse the drop across the sense resistor.

The scope screen shot is as shown below -

Scope screenshot

Please note I have removed the dc bias of 24V to analyse the voltages carefully. Is this analysis or method correct ? Also, if you notice the peak (red). It shows almost 0.25V (scale is 50mV). So, the current at that instant is 0.25/0.01 = 25Amps. Now, I am using just 2 22AWG wires for this. 22AWG can source a max of 7Amps/wire. So we get a max of 14Amps sourcing via the wires.

If so, why doesnt the wire burn out or get damaged ? I understand that this peak is momentarily. If so, what is the max time that this wire can sustain till it burns out (if we continuously supply 25Amps).

How do i calculate the max time for sustaining the 25AMps of 22AWG(or for that matter any gauge) wire.

  • \$\begingroup\$ Measurement uncertainty. Can you float your scope and measure with just one channel? Can you increase R1? Do you have access to a hall effect current probe? The time constant for copper is several seconds. \$\endgroup\$ – winny Jul 15 '17 at 11:37
  • \$\begingroup\$ And what will that achieve ? Floating means no ground ??? \$\endgroup\$ – Board-Man Jul 15 '17 at 11:38
  • \$\begingroup\$ It means you remove most if your measurement uncertainty by not subtracting one noisy measurement close to zero by another noisy measurement close to zero. Also, why not just measure in the low side? \$\endgroup\$ – winny Jul 15 '17 at 11:40
  • \$\begingroup\$ Sorry am a bit confused. Could you kindly elaborate please ? \$\endgroup\$ – Board-Man Jul 15 '17 at 11:41
  • 1
    \$\begingroup\$ Question 11 and 16 here: allaboutcircuits.com/worksheets/basic-oscilloscope-operation \$\endgroup\$ – winny Jul 15 '17 at 11:47

Attempt to measure mV-range signals under the 24-V bias with a 8-bit scope and individual (and therefore different) probes is a failing proposition. There will be too much noise. There are basically three options:

  1. Use a dedicated active differential probe, but they are pretty expensive. To accommodate 24V offset you would need a TDP1500 probe (+-25V, $4K price), or high-voltage probes like P5200A (eBay ~$700);

  2. Built your own sensor using something like INA199 or ZXCT1107/09/10 from Zetex' enter image description here

The above circuit is literally all you need, an IC, and a scaling resistor. You might need to calibrate this circuit with known stable loads before taking actual measurements.

  1. put the shunt resistor into ground wire as suggested by "winny", if the load allows this small 20-30mV bias. Subtracting readings from two different probes still be subject to general uncertainties of low-volt measurements.

The answer to sustainable or impulse capability of a copper wire is given here, on stackexchange.


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