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I have the following schematics that consists two 40106 Schmitt trigger circuits. The left hand side one has bigger capacitor then the right hand one.

I know that the right hand one will oscillate 10 time faster then the left hand one. My question is regarding the charge on the capacitor? the 10uF will accumulate more electrons on his plates? meaning more voltage? I know this not true because both circuits fed with the same 9V battery. So how to explain the charge on the plates in relation to the size of the capacitors ?

enter image description here

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2 Answers 2

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Q = I×t = C×U.

So it takes longer for the larger capacitance to charge up to same threshold voltage.

Larger capacitance at same voltage holds more charge.

Which is apparent from the RC time constant of a RC circuit being R×C.

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A simple way to look at it is capacitor plate area.

A capacitor with 10 times the capacitance has ten times the plate area (other things being equal) so it can store 10 times the charge (electrons) at the same voltage.

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  • \$\begingroup\$ Thanks. I also wondered why is the need of the negative lug of the capacitor to be connected to ground? electrons cannot pass between the two plates. Why the charge can be just accumulate on the one plate? \$\endgroup\$
    – frequency_abovetime
    Commented Nov 13 at 4:39
  • \$\begingroup\$ @frequency_abovetime That's not how capacitors work. The current flows through the capacitor, resulting in an almost perfect balance between positive charge on one plate and negative charge on the other. Electrons carry current in metals (well, aluminum is tricky...), but current is not generally carried by electrons. The current between the capacitor plates is "displacement current", not carried by electrons. \$\endgroup\$
    – John Doty
    Commented Nov 13 at 13:03
  • \$\begingroup\$ @JohnDoty - from what I read the current can pass through the capacitor rather a charge can accumulate on his plates(?) so when the circuit is off the voltage at the upper plate is 0V so no charge. when powering on the circuit the output goes HIGH and current can start flow from the output to the capacitor plate building charge on it? (negative? positive?) so what is the relation with the lower plate of the capacitor in that matter? \$\endgroup\$
    – frequency_abovetime
    Commented Nov 14 at 4:15

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