# Help understanding the output from a circuit I have built

Discussion in 'General Electronics Discussion' started by Rt2-o, Mar 22, 2015.

1. ### Rt2-o

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Mar 22, 2015
I have built a circuit that can be simplified to the layout in the link below: -

https://www.dropbox.com/s/e0jl8ds0dw0ucx0/CircuitQ.png?dl=0

When I put in a triangular wave voltage input, the output looks similar to the one I have sketched on my drawing. I am trying to understand why it behaves like this, please can someone help?

I know that for a single capacitor the output should be a square wave because I = C* d(V)/dt and d(V)/dt is constant - with an amplitude which depends on the value of C. For a single resistor I think I should see a triangular wave with zero phase shift between the voltage input and current output. I can understand that the output is some form of combination of these two effects, but I don't understand much more than that.

Please can someone help me to get a better idea of what is happening, is there any way to calculate what a and b would be if I knew the values of the capacitors and resistors without building the circuit again and measuring? I assume this would also be dependent on the frequency of the input signal because with a very low frequency I can see the current would eventually reach zero before the voltage switches sign again....

Also, should the slopes in the output be linear? They looked very slightly curved on the oscilloscope, I may have mis-read.

Thank you

2. ### Harald KappModeratorModerator

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Nov 17, 2011
Welcome to electronicspoint.

Please show us the input waveform, too. This will make it much easier to answer your question.

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Mar 22, 2015

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Sep 5, 2009

5. ### Laplace

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Apr 4, 2010
Your circuit consists of two resistors in parallel which are in parallel with two capacitors in parallel driven by a triangular voltage source with zero source resistance. So the equivalent circuit is one resistor in parallel with one capacitor. The time constant is zero because the source resistance is zero, so there are no exponential charge or discharge curves. The current through the resistor is a triangular wave, V(t)/R. The current through the capacitor is a square wave, Cd/dtV(t). The resultant waveform is the sum of the square wave and the triangular wave.

If there is any curvature in the output it would be due to non-zero source resistance.

6. ### Rt2-o

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Mar 22, 2015
Thank you Laplace! That's the answer I was looking for, I was worried that there would be something else going on here. Since the output current is just the sum of the two waveforms I can also work out the values of a and b myself from the resistance and capacitance of the components. Thank you!

7. ### Harald KappModeratorModerator

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Nov 17, 2011
Sorry, I can't.
Yesterday I could see the schematic (without logging in to Dropbox), today it's gone.

8. ### davennModerator

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Sep 5, 2009
ok no probs  