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I am measuring the voltage drop on a 1MΩ resistor with an oscilloscope, and as I understand to get accurate readings I should compensate the internal impedance of my measurement device for such a high value resistor.

It's a Tektronix TDS2024C oscilloscope, from the manual I know that it has a 1MΩ internal impedance and I'm using a 10x attenuation probe, which should have a 9MΩ resistance. According to everything I read that should add to a 10MΩ internal impedance, even though I'm not quite sure why and would like to know it a little more accurately.

So, two questions:

  1. Are there any simple methods to measure the input impedance of an oscilloscope accurately instead of taking the word of the manufacturer?
  2. How does the input impedance add to 10MΩ anyways, if the internal resistance isn't connected in series? Shouldn't it form a voltage divider for 0.91MΩ? Schematic below:

Where does the 10MΩ come from? It must be magic!

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Ignore the capacitors, cable etc. and just consider the resistors. From probe tip to ground you have 9MΩ in series with 1MΩ - that's 10MΩ! You can verify this by measuring the resistance between probe tip and ground with a multimeter (on 20MΩ scale).

Of course this only applies to DC. At higher frequencies the impedances of the capacitors and cable become significant. At 50MHz the reactance of a 10pF capacitor is about 300Ω.

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  • \$\begingroup\$ But it's not really connected in series, it kind of branches out beforehand and the resistor is in between. I'm not sure I get this... \$\endgroup\$ – I have no idea what I'm doing Oct 15 '14 at 10:12
  • \$\begingroup\$ @IhavenoideawhatI'mdoing It is connected in series. Maybe you are confused by the right hand side of the picture also shows arrows where it connects to the amplifier (not shown). The amplifier can be presumed to have Gohms of resistance i.e. it will be insignificant. \$\endgroup\$ – Andy aka Oct 15 '14 at 10:52

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