How does a series capacitor affect a square-wave on-off signal?
How does IC6's pin 4 get negative voltage or grounded?




If pin 3 is high (+5V) and C4 is discharged, both sides of C4 are high (+5V), The right hand side of C4 is held high by R10. The left hand side is held high by pin 3. There is 0V across C4.

If pin 3 goes low (-5V), C4 cannot instantaneously charge from 0V to 10V so taking it's left hand side down to -5V takes the right hand side to -5V. This -5V is seen at pin4 on IC6.

R10 then slowly charges C4 bringing the voltage at it's right hand side towards +5V.



The circuit would function the same if I arbitrarily changed the labels +5V and -5V to, for example, +10V and 0V (or GND). In which case you might or might not want to replace the word "negative" with something else.


  Do you mean if one pin of C4 is grounded, then the other pin of C4 would suck current from this pin as well?


No, See the "Differentiator" example in this tutorial

 

No current needs to flow through or across the capacitor because the voltage across it is initially unchanged. If the voltage with respect to a fixed reference point (ground) at the left hand is changed, the same change will be seen at the right hand side. No current need flow. Subsequently there are changes but these involve external flows of current.