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I have a device and the output ac voltage is 1-2Vp-p with very low current. I made a voltage quadrubler with a storage capacitor(output) in order to rectify the voltage and observe how much time is needed to charge the capacitor.

So, while the device's current is very low, if I connect my oscilloscope to the storage capacitor and let the device operates for two hours I will never see the charge waveform because the capacitor discharges through the probe.

Could you suggest me something to stop the flow of the current through the probes. I thought to connect a resistor parallel to the capacitor and then connect the probes but it didn't work.

I upload a rough layout circuit

Thanks

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    \$\begingroup\$ You basically need to design a measurement front end to increase the input resistance. Have you considered instrumentation amplifiers? \$\endgroup\$ – loudnoises Jan 23 '18 at 15:38
  • \$\begingroup\$ A resistor in parallel with the capacitor would make things worse, did you mean series? \$\endgroup\$ – evildemonic Jan 23 '18 at 16:08
  • \$\begingroup\$ What are your capacitor values? What is the frequency of your generator? What diodes are you using? \$\endgroup\$ – Andy aka Jan 23 '18 at 17:27
  • \$\begingroup\$ yes on series i meant. \$\endgroup\$ – Anti Jan 23 '18 at 17:50
  • \$\begingroup\$ capacitor 4x10uF and 1x0.47uF, diodes 4X 1N4007 \$\endgroup\$ – Anti Jan 23 '18 at 17:51
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One option is to try and increase the input impedance to the measurement stage. An oscilloscope with a 10x probe has an input impedance of around 10 MΩ, but you could boost that another factor of 100 or more by using an instrumentation amplifier. For example, the AD8295 has an input impedance of 100 GΩ. This may not be enough depending on how slow the voltage charges, but will give you a time constant 10000x longer than that of an oscilloscope probe.

For illustrative purposes consider the following:

schematic

simulate this circuit – Schematic created using CircuitLab

You connect the two inputs across the output of your device, and end up with a single ended output you can connect to your oscilloscope.

If you don't need it to be differential then you can simply use a high-impedance op-amp configured as a non-inverting amplifier, which will give you the same benefits but at less cost.

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The simplest solution is: do not use a scope.

You talk of a two hour timeframe for your measurement: a multimeter polled at 1 Hz or less is probably more than enough for your needs, and multimeters have a much higher impedance than your typical (or high end) scope.

If this is not enough just throw a relay in series and close it when you want to measure.

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  • \$\begingroup\$ This is the method we are currently using. But I would like something else. Thanks for the reply. \$\endgroup\$ – Anti Jan 24 '18 at 1:53
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You can choose RC values so that a 10:1 scope probe will have negligible effect compared to the leakage of the caps. This is more valuable, IMHO.

You must learn the theory to understand the practical failures and simple solutions of your experiment. Please define your variables and goals.

  • The Signal source impedance (Rs), frequency, f and peak voltage Vp.

  • the diodes must have part numbers, so the forward voltage, Vf can be subtracted from Vp.

  • the desired load resistance or current.
  • for practical reasons Vp must be much greater than Vf so 1~2V is unpractical.
  • when you don't have a good lab, use a good simulator.
    • my example and learn to use it. enter image description here When Diodes only conduct during peaks their forward resistance is divided approximately by the duty cycle. This means instead of a normal T=RC rise time, it increase by 1/d, the % duty cycle.

There exists a quasi-linear gain, g between the rise time of this Cockcroft-Waltron multiplier and the signal frequency and the f=g/RC time constant. You can find this by theory and verify by experimentation with a simulator. enter image description here ref wiki

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