Condenser capacity inversely proportional to source voltage - why?

[carpetman]

Hello everyone.
I indeed like to understand the things I am learning, and for the love of me, I cannot understand why, according to the formula, capacity of a condenser in a circuit consisted of said condenser and a power source is inversely proportional to the power source's voltage. Could you please help me get to the bottom of this?

 

[the_emi_guy]

carpetman,
Condenser is a very old terminology. Is it a specific type of capacitor that you are studying that has capacitance inversely proportional to voltage? A varactor diode has this propery.

 

[carpetman]

Ah, yes, my native language uses a word similar to "condenser" for "capacitor", that's what I meant, thanks! Its capacitance is inversely proportional to voltage indeed, and I'm having trouble figuring out what the cause of this would be. I apologize for my bad English!

 

[the_emi_guy]

No need to apologize,
Look up "varactor diode" or "varicap" on the Internet. This is actual a semiconductor diode operated in reverse bias. The depth of the depletion region increases with increased reverse bias, this decreases the capacitance.

 

[nasu]

Hello everyone.
I indeed like to understand the things I am learning, and for the love of me, I cannot understand why, according to the formula, capacity of a condenser in a circuit consisted of said condenser and a power source is inversely proportional to the power source's voltage. Could you please help me get to the bottom of this?

Well, you don't say which formula but I suspect that you mean
C=Q/V, the definition of capacitance.
If this is the case, the capacitance of a given system (capacitor) is not inverse proportional to V. This would be the case if Q were a constant (which it is not). When V changes, Q changes as well.The meaning of the formula is that the ratio between Q and V is a constant and this constant is what is called capacitance.

If you mean something else, please show the formula.

 

[carpetman]

No need to apologize,
Look up "varactor diode" or "varicap" on the Internet. This is actual a semiconductor diode operated in reverse bias. The depth of the depletion region increases with increased reverse bias, this decreases the capacitance.

Thanks mate, it's a bit clearer, but still a little over my head since I don't even have high school behind me. It seems to be what I'm looking for though, so I'll definitely educate myself further on the matter. Thanks again!

Well, you don't say which formula but I suspect that you mean
C=Q/V, the definition of capacitance.
If this is the case, the capacitance of a given system (capacitor) is not inverse proportional to V. This would be the case if Q were a constant (which it is not). When V changes, Q changes as well.The meaning of the formula is that the ratio between Q and V is a constant and this constant is what is called capacitance.

If you mean something else, please show the formula.

Yes, terribly sorry, I really should have put the formula to avoid confusion over my original question, C=Q/V is the formula in question.
Though, I would like to know exactly how the change of voltage with a constant charge affects the capacitance. I can understand that, the larger the charge is, the more electrons there are to fit into the capacitor, and therefore the capacitance is larger.
However, thinking it through, I would also assume (leaving aside that the calculations wouldn't work out, more as a thought experiment) that the greater the voltage, the greater will the capacitor's capacitance would be. Let me elaborate why my mind is inclined to think that:
The greater the voltage, the greater would essentially be "electron surplus-shortage" the difference between the power source's poles. I would say that, since one side of the power source is positively charged, therefore the same panel of the capacitor (assuming it's consisted of two panels with an insulator in between, easier for me to explain the situation if we assume this type of a capacitor) would also be positively charged, and the other side being negatively charged, the greater difference would amount to a greater electric force, effectively "pulling in" more electrons to the panel, increasing the capacity, and actually increasing the charge.
I know my logic is terribly flawed, but this is the most reasonable solution I can work out when imagining how this plays out on the microscale in a real circuit. I would really love to find out what exactly is happening inside the capacitor though. Thank you very much for your input!

 

[Dickfore]

Hey OP, so if write Ohm's Law as $R = \frac{V}{I}$, or Newton's Law as $m = \frac{F}{a}$, would you say that resistance is inversely proportional to the current, and mass is inversely proportional to the acceleration?

 

[nasu]

Yes, terribly sorry, I really should have put the formula to avoid confusion over my original question, C=Q/V is the formula in question.
Though, I would like to know exactly how the change of voltage with a constant charge affects the capacitance. I can understand that, the larger the charge is, the more electrons there are to fit into the capacitor, and therefore the capacitance is larger.
However, thinking it through, I would also assume (leaving aside that the calculations wouldn't work out, more as a thought experiment) that the greater the voltage, the greater will the capacitor's capacitance would be. Let me elaborate why my mind is inclined to think that:

The greater the voltage, the greater would essentially be "electron surplus-shortage" the difference between the power source's poles. I would say that, since one side of the power source is positively charged, therefore the same panel of the capacitor (assuming it's consisted of two panels with an insulator in between, easier for me to explain the situation if we assume this type of a capacitor) would also be positively charged, and the other side being negatively charged, the greater difference would amount to a greater electric force, effectively "pulling in" more electrons to the panel, increasing the capacity, and actually increasing the charge.
I know my logic is terribly flawed, but this is the most reasonable solution I can work out when imagining how this plays out on the microscale in a real circuit. I would really love to find out what exactly is happening inside the capacitor though. Thank you very much for your input!

No, for a "normal" capacitor the capacitance does not depend on voltage. Is this simple.
And no, you cannot change the voltage when keeping the charge constant unless you change the actual configuration of your system. But then you have a different capacitor, with a different capacitance. And still independent of voltage.

 

[carpetman]

Hey OP, so if write Ohm's Law as $R = \frac{V}{I}$, or Newton's Law as $m = \frac{F}{a}$, would you say that resistance is inversely proportional to the current, and mass is inversely proportional to the acceleration?

No, for a "normal" capacitor the capacitance does not depend on voltage. Is this simple.
And no, you cannot change the voltage when keeping the charge constant unless you change the actual configuration of your system. But then you have a different capacitor, with a different capacitance. And still independent of voltage.

Aha, I see what you guys mean. Thanks!