Circuit capacitance

Replies

• 

  reachrkata

  One way is to give a step input to the circuit and calculate the output based on the rise time at the output.
  
  Perhaps you can share the circuit that you built and we can help you exactly.:smile:
  
  - Karthik
• 

  e2saleh

  Thanks reachrkata,
  
  Applying step input is something that was considered, so the capacitance would be: C = -t/(R*ln(1-Vo/Vi)), which I derived from the exponential behaviour of the voltage across charging capacitor, i.e Vo = Vi*(1-e^(-t/RC))
  
  Nonetheless, my concern was about R. Imagine if we are measuring the self-capacitance of a resistor, so it is just a potential divider with two resistors in series, so would this R be the sum of those two resistors? or just the first one would be considered as R? that is where my concern coming from.
  
  I did few practical experiments and it seems the first resistor is the dominant R, but I still prefer if there is an equation to confirm that.
  
  Thanks
• 

  shreyasm89

  Your method is correct but please can you divulge some more details?Like if you are using a potentiometer then what is the resistance value?Because self-capacitance of that resistance is going to be so small that you need to be very accurate.Also, the self capacitance of a 'resistor' will come into play only when your signal frequencies enter the RF range(above 30Mhz).Have you used RF frequencies?Then you might as well check out the equivalent RF cicuit of a resistor.
• 

  e2saleh

  It is high resistance in general. Lets say we are measuring the self capacitance of 100 MOhm resistor. Imagine the circuit is 10MOhm resistor in series with 100MOhm both are supplied with DC, and we want to work out the self capacitance for the second one (100M).