I have the following circuit and I need to get the mean value of output voltage.


simulate this circuit – Schematic created using CircuitLab

This is a parallel RC circuit, powered by a current source which as you can see has the following waveform. enter image description here

I was thinking that to get the output voltage first, the input current should be represented by a series of Fourier, since it is a periodic signal.

Once this was done, it could obtain the relation between the output voltage and the input current, by means of a current divider and multiplying by the output resistance.

I'm not very related to the use of Fourier series for circuit analysis and I think that would even imply the use of phasors to get temporary expression of output voltage. So I think in this way it would be very difficult. Am I wrong or is there any other way to do it?

  • \$\begingroup\$ Over time there will be an average voltage over the capacitor, right? \$\endgroup\$ – neonzeon Mar 12 '18 at 4:46

The mean value of the voltage is the same thing as its DC value. The flip side of this is that on average no current goes through a capacitor. This means all of the DC current must pass through the resistor. Calculate the DC current from that waveform, and you should find it simple to calculate the mean voltage from there.

This circuit is simple enough that you can actually solve the differential equation in the time domain to double check this logic. Just split the period into two segments: one where the current source is off, and one where it is linearly decreasing from its max value. If you know the duration of each of these segments, you can calculate the ending voltage as a function of the starting voltage. Then simply set the start value of each to the end value of the other, and solve. Then go back and calculate the average voltage. If it's been awhile since you've solved differential equations you may need to review the basics of solving first order ODEs using integrating factors, but it's not too difficult.

  • \$\begingroup\$ So for the average voltage is the capacitor part ignored? Okay I understand that, but what differential equation are you talking about? \$\endgroup\$ – David Gonzalez Mar 13 '18 at 1:11
  • \$\begingroup\$ Yes, the key point is, mean value = DC value. At DC, a capacitor is an open circuit, and that current source outputs its mean current (that's the definition of DC). Any circuit made of basic components like capacitors, resistors, inductors, current and voltage sources, etc. can be represented using a system of differential equations, and then solved. If you don't have experience with this, it's probably not necessary to worry about it at the moment. However, an introductory circuit analysis textbook will give a thorough explanation. \$\endgroup\$ – David Moore Mar 14 '18 at 16:04
  • \$\begingroup\$ I think you don't need to solve any differential equation. The average current through a capacitor is zero, therefore the average current out of the current source is the same as the average current through the resistor. Multiply by \$ R_1 \$ to get the average voltage. \$\endgroup\$ – joe electro Nov 18 '18 at 15:32

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