Does actual voltage (potential) affect capacitance?

[kex]

If I replace a 16 volt rated capacitor of 47uf with a 64 volt rated capacitor of 47uf, will
the working capacitance be the same or does this depend on the specific design and
materials of the capacitor?

I'm thinking that the capacitance is proportional to the voltage applied and that the
replacment capacitor should be 4 times 47uf to get the same circuit characteristics.
However I have not seen any discussion to that effect. I just remember or perhaps
mis-remember my elementary phyics.

 

[f95toli]

The voltage rating of a capacitor is just a parameter that tells you at which voltages it is safe to use; a 47 uF capacitor always has a capacitance of 47uF no matter what the voltage; but if you try to to apply 60V to a 16V capacitor there is a pretty good chance that it will explode (and if it is an electrolytic capacitor it can be a pretty loud bang).

Note, I am NOT joking here; capacitors DO actually explode if you apply voltage much higher than what they are rated for; it is usually not dangerous but you should nevertheless be careful (some of the chemicals inside them are quite nasty and you don't want them in your eyes).

 

[kex]

I hear what you are saying. It's just that I'm questioning it. I'm too analytical to accept
the easy answer.

According to 2 different Physics books I have (which are rather old),

"The capacitance is the electric charge that must be added per unit increase in potential."

Of course, this is for a parallel-plate capacitor with no dielectric. Even so, with a dielectric,
or without, I'm seeing this relationship expressed as an equation in both texts:

C = Q/V, where C is capacitance, Q is charge (coulombs) and V is voltage (potential)
difference between the two plates.

This equation implies that if the voltage is reduced, the capacitance increases. (That's the
opposite of what I thought.)

While I realize that I'm talking physics here, I posted this in an engineering forum because I want a real world answer. Has anyone actually measured the difference (or sameness)?
There must be some studies or even spec sheets on the internet that address this issue.

Are there dielectrics used in capacitors which give the same capacitance regardless of how
small the voltage as long as the voltage does not exceed the manufacturer's specification?

 

[Corneo]

Are there dielectrics used in capacitors which give the same capacitance regardless of how
small the voltage as long as the voltage does not exceed the manufacturer's specification?

Tantalum capacitors have this type of property. Ceramic cap's with dielectric x7r, tend to lose its capacitance as you bias it closer to the manufacturer's voltage rating. There is a science behind capacitors, the best information I have found when dealing with capacitors reliability is articles from the vendor AVX. There is a lot of information though.

 

[Proton Soup]

Of course, this is for a parallel-plate capacitor with no dielectric. Even so, with a dielectric,
or without, I'm seeing this relationship expressed as an equation in both texts:

C = Q/V, where C is capacitance, Q is charge (coulombs) and V is voltage (potential)
difference between the two plates.

This equation implies that if the voltage is reduced, the capacitance increases. (That's the
opposite of what I thought.)

i think i see what you're saying. you're thinking of C as a variable, when it's a constant. what actually happens is that when V is reduced, so is Q. it would be a bit like interpreting R=V/I as "when current is reduced, resistance increases". but resistors are more intuitively "obvious" so you don't do that.

what you may want to think of the capacitor as is a balloon that you can store Q in. as you increase the pressure, more Q goes in. but at some point, too much tension causes it to pop. that tension just below the pop is your rated voltage.

 

[kex]

I think I understand where my problem was. I was equating capacitance with charge.
Not so! Capacitance is the charge per voltage. As the voltage increases, so must the
charge and the capacitance remains the same (as long as the device does not break
down!) C remains constant because that's the way capcitance is defined.

On the other hand, the assumption is that the permittivity or dielectric constant remain
constant regardless of the potential pressure or voltage as long as there is no physical
breakdown of the device. I'm really curious to see some actual lab tests done on various capacitor types. My guess is that there is very little change in permittivity
unless voltage limit is exceeded.

OK, I feel my question has been answered. Thank you all.

 

[cabraham]

Usually, capacitance doesn't vary with voltage. But, with low grade ceramics, such as Z5U, and Y5V, C will decrease as V increases. To a lesser extent, X5R and X7R display this tendency. The premium ceramic dielectric, C0G, is free from this problem.

Film caps are also free from this issue as well as electrolytics.

Claude

 

[StgIIIOvrDvn]

Usually, capacitance doesn't vary with voltage. But, with low grade ceramics, such as Z5U, and Y5V, C will decrease as V increases. To a lesser extent, X5R and X7R display this tendency. The premium ceramic dielectric, C0G, is free from this problem.

Film caps are also free from this issue as well as electrolytics.

Claude

Although this is on the surface a little off topic, it is an example of behaviors of real devices. The low grade ceramics that Claude mentioned also exhibit a fair degree of pressure capacitance drifts resulting in "Microphonic effects" making them an undesirable selection in amplifier and sensing circuitry with dc or near dc bandwidths.