Capacitor behavior for square wave

[Murugesh_89]

Hi,
For the following figure


When the input becomes low, the capacitor discharges and hence the output of the capacitor should be 5V. But in the CRO the output is showing as -5V.

Why?


Thanks,
Murugesh

 

[LvW]

A capacitor can be charged/discharged in a closed loop only.

 

[Murugesh_89]

can you please elaborate little more

 

[chuckey]

For your circuit to be analysed, it needs a source impedance and a load impedance. In general with any pulse wave the area above the zero volts line will equal the area below the zero volts line. As you are using a square wave (on/off = 1 :1) the output square wave will sit symmetrically about the zero volts line.
Frank

 

[MarkChristensen]

Looks to me as if the capacitor has charged up so there is 5 volts across it. Without anything hooked up the the Vout side of the capacitor, the difference between Vin and Vout will stay at 5V. There is no way for the capacitor to discharge. (Vout is just kind of following whatever Vin does, only 5 volts lower.) Try hooking a resistor up between Vout and ground. When Vin goes from 0 to 5 volts, the output will jump to 5 volts (because Vin and Vout were both at 0V, there is 0V across the capactor) and Vout will decay back to 0V. At this point, there is 5 volts across the capacitor. When the input goes from 5V back to 0V, the output will be "pushed" down to -5V (voltage across the capacitor does not change right away) and will then "decay" (in an upwards kind of way) back to 0V. You should get a positive spike, then a negative spike, then a positive spike, etc. Careful - if Vout is feeding some integrated circuit, driving some inputs below ground can destroy the chip (and you might need to use a diode to protect the chip.)