Increase the distance between plates of a capacitor in direct current

[valeriy2222]

Hi.

I've been studying in details how a microphone works. I've found a type of microphones where capacitor is used to record sound. The idea is simple. Increasing and decreasing the distance between the plates of a capacitor (sound waves) results in the changing capacitance. Being connected to DC (direct current) power supply, it creates oscillating current which is then recorded. There is a detail which is not well explained.


The charging current.
If a capacitor is connected to DC power supply, the current flows (from + to -) building a charge on the plates and after some time the capacitor voltage become equal to power supply voltage. After that, if we decrease the distance between the plates of the capacitor, the current will start flowing again (from + to -) thus continue building the charge. No problem with understanding.

The discharging current?
When we increase the distance between the plates, the charge on the plates wants to go back to the power supply. This time it wants to go from + to +. Here, I do not fully understand whether it's really going back (but where it can go otherwise?) or not. Does it override the voltage of the power supply? A battery instead of spending energy begins to be recharged?
Electrons on one of the plates have to leave it but the only way to go is the power supply - "+ to +", "- to -".

[PLAIN]http://www.ustudy.in/sites/default/files/images/condenser%20microphone.jpg

 

[valeriy2222]

Well,

I've found some websites which tell that if we apply higher voltage to a battery, then it's going to be recharged. That means that the charge on the plates of the conductor will be reduced by going through the battery, though, I think they use a more complex scheme to make the discharge current go through a different way.

 

[uart]

Hi Valeriy. This is called a "condenser microphone" and the principle of operation is very straight forward.

The key relationship in a capacitor is that which connects charge and voltage,

$$Q = C V$$

If the resistance in the circuit is reasonably large then the charge "Q" is not able to change quickly. As the separation of the plates varies however, then C varies, hence V must also vary so as to keep Q (approximately) constant.

 

[valeriy2222]

I understand that.

My question was about where the charge (electrons) go when C decreases. It should go from the plates to ? where ? - back to the battery (or through it). That what was weird to me at first time. Now I know that batteries allow this.

 

[uart]

I understand that.

My question was about where the charge (electrons) go when C decreases. It should go from the plates to ? where ? - back to the battery (or through it). That what was weird to me at first time. Now I know that batteries allow this.

The current is relatively small and is AC so it won't significantly alter the charge on the battery. If you're using something other than a battery for the bias voltage, say something that doesn't allow the current to reverse, then you'd simply place a larger valued capacitor (eg an electrolytic) in parallel with that supply.