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Simple question on capacitor charging

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smartsarath2003

smartsarath2003

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Hai,
I got a simple doubt,when I am explaining myself. Can anyone explain me the output wave form at Point A. I thought its half-wave off-set at 2.5V. Is that correct
Thanks
 

Re: Simple question on cepacitor charging

This is a linear circuit so that superposition is valid. The potential at point A is the sum of half the supply 5 V and all of the input AC signal (assuming that it is not so low in frequency that there is some voltage drop across the capacitor.) So you will get at the output the 10 V p-p with an offset of +2.5 V. This will make 7.5 V on high and -2.5 V on the low side of the sine wave.
 

Re: Simple question on cepacitor charging

In an actual circuit, the +7.5v and -2.5v also depend on the resistor values and the 5v source. Not all 5v sources would appreciate +7.5v nor -2.5v being applied to it's output.
 

Re: Simple question on cepacitor charging

This is a linear circuit so that superposition is valid. The potential at point A is the sum of half the supply 5 V and all of the input AC signal (assuming that it is not so low in frequency that there is some voltage drop across the capacitor.) So you will get at the output the 10 V p-p with an offset of +2.5 V. This will make 7.5 V on high and -2.5 V on the low side of the sine wave.
Click to expand...

So its 2.5V off-set of the input. Thats what I couldn't explain my-self how its happening.
The cepacitor only starts charging when ever there is a current flowing through it, which happens only when the input side is greater than 2.5V. So Va should be at 2.5V, untill Vin>2.5V. After Vin>2.5V, Va fallows Vin as C is charging and discharging upto 2.5V again.Is there anything I got it wrong with this explanation?
Thanks
 

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