why is the waveform output showing like this when passed through capacitor?

Thread Starter

nisha p

Joined Dec 26, 2017
20
In the figure ,square wave is passed through 1uf capacitor .in the beginning of the wave is intact but later it becomes biphasic, ie it starts to dip into negative value.can someone explain this behavior?
 

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casteels

Joined Mar 22, 2012
9
In the figure ,square wave is passed through 1uf capacitor .in the beginning of the wave is intact but later it becomes biphasic, ie it starts to dip into negative value.can someone explain this behavior?
A capacitor blocks DC so the average of that wave should be zero
 

OBW0549

Joined Mar 2, 2015
3,394
In the figure ,square wave is passed through 1uf capacitor .in the beginning of the wave is intact but later it becomes biphasic, ie it starts to dip into negative value.can someone explain this behavior?
That is perfectly normal behavior, exactly what one would expect. What you are observing is the same thing you would see if you simply set your oscilloscope for AC Coupling.

The series capacitor combines with the scope's input resistance (usually a 1 MΩ resistor connected between its input and ground) to form a first-order high-pass filter, which blocks the DC component of a signal while letting you observe its AC behavior.
 

danadak

Joined Mar 10, 2018
4,057
The cap has an initial charge on it that decreases with time thru
leakage in the cap and energy loss of that charge in external cir-
cuit, eg. the 1K in attached example. Generally its translating
from a fixed offset to a ground reference input.

See attached.


Regards, Dana.
 

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Last edited:

ebp

Joined Feb 8, 2018
2,332
Something very weird is going on. The output amplitude is more than twice the input amplitude and there is a 180 degree phase difference. This is not a simple RC circuit!
 

DECELL

Joined Apr 23, 2018
96
Thats right Dana. The capacitor has to increase or decrease its net charge to a voltage equal to the DC average of the input signal.
This may take a number cycles to achieve depending on: the time constant of the RC circuit, the duty cycle of the input and the source impedance. See graph, areas above and below 0V are equal.upload_2018-5-16_15-10-16.png
 
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