How Can I Accurately Measure Voltage Drop across a Capacitor?

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SUMMARY

The discussion focuses on accurately measuring voltage drop across a capacitor during a leakage current test. A simple subtraction of initial voltage from the final voltage is insufficient due to the exponential decay characteristic of capacitors. Instead, the voltage can be modeled using the equation V_t = V_{ref} * e^{-t/(RC)}, where V_t is the voltage after time t, V_ref is the reference voltage, R is the resistance, and C is the capacitance. This approach allows for precise calculations of the parallel resistance affecting leakage.

PREREQUISITES
  • Understanding of capacitor behavior and exponential decay
  • Familiarity with the formula V_t = V_{ref} * e^{-t/(RC)}
  • Knowledge of measuring voltage and time in electrical circuits
  • Basic concepts of leakage current and parallel resistance
NEXT STEPS
  • Research methods for measuring capacitor leakage current accurately
  • Learn about the effects of parallel resistance on capacitor discharge
  • Explore advanced capacitor modeling techniques using circuit simulation tools
  • Study the implications of time constant (RC) in capacitor circuits
USEFUL FOR

Electrical engineers, circuit designers, and technicians involved in testing and measuring capacitor performance and leakage characteristics.

simoncastle
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Hi all,

Here's the problem!

I am attempting to create a leakage current test suite. Basically it centers around cutting the voltage reference supply to a capacitor and measuring the voltage drop. Similar to the one in the attachment.

I figured that to determine the drop it would be a simple case of V at start minus V after time delay.

However a colleage told me that because of the exponential decay of capacitors this simple equation wouldn't be enough to accuratly determine the voltage drop.

Any ideas?

Thanks!

- Si
 

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I'm assuming you want to measure the parallel resistance (which causes the charge to leak) of the capacitor.

You can use an exponentially decaying function for the voltage:

<br /> V_t = V_{ref} \ e^{-t/(RC)}<br />
where Vt is the voltage after t seconds. Solve the equation for R after measuring all other quantities (Vt, Vref, t, and C)
 

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