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So I'm using a Fluke i200s current clamp in combination with an oscilloscope to measure the AC current. Normal current clamps have a current output, but this one has a voltage as output. Now my question is does this clamp has the same working principle as the others(like a transformer with one primary winding), and just measures the voltage instead of the current. Or does this clamp works differently, and how?

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  • \$\begingroup\$ Have you tried reading the documentation? \$\endgroup\$
    – PlasmaHH
    Commented Sep 27, 2017 at 12:55
  • \$\begingroup\$ It probably just has a precision shunt built in. \$\endgroup\$
    – R Drast
    Commented Sep 27, 2017 at 13:00
  • \$\begingroup\$ Yes I did @PlasmaHH this is from the instruction sheet: Output level(s) 100 mV/A 10mV/A \$\endgroup\$ Commented Sep 27, 2017 at 13:04

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enter image description here

Figure 1. The Fluke I200S current clamp.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. The internal wiring.

Now my question is does this clamp has the same working principle as the others(like a transformer with one primary winding), and just measures the voltage instead of the current.

It is a true current transformer but with a shunt resistor built in. The current will create a voltage drop across the resistor and this voltage is monitored by the oscilloscope.

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  • \$\begingroup\$ Is there a reason that they use the shunt resistor? \$\endgroup\$ Commented Sep 27, 2017 at 13:35
  • \$\begingroup\$ Yes. Your oscilloscope has a 1M input impedance. If the current transformer tried to push 10 mA through that, for example, the voltage would have to rise to \$ IR = 1M \cdot 10m = 10 \ \mathrm kV \$. You don't want that. The 'scope inputs are voltage inputs, not current. A resistor is a perfect current to voltage converter! \$\endgroup\$
    – Transistor
    Commented Sep 27, 2017 at 14:00

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