Increase the distance between plates of a capacitor in direct current

AI Thread Summary
The discussion revolves around the operation of a condenser microphone, which utilizes a capacitor to convert sound waves into electrical signals by varying the distance between its plates. When connected to a DC power supply, the capacitor charges until the voltage equals that of the supply, and changes in plate distance create oscillating currents. The main point of confusion is regarding the discharging current when the plate distance increases, specifically where the charge goes and whether it can flow back to the power supply. It is clarified that batteries can allow this flow, although the current is small and primarily AC, which does not significantly impact the battery's charge. Understanding these principles is essential for grasping how capacitors function in audio applications.
valeriy2222
Messages
14
Reaction score
0
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
 
Last edited by a moderator:
Engineering news on Phys.org
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.
 
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.
 
Last edited:
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.
 
valeriy2222 said:
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.
 
Very basic question. Consider a 3-terminal device with terminals say A,B,C. Kirchhoff Current Law (KCL) and Kirchhoff Voltage Law (KVL) establish two relationships between the 3 currents entering the terminals and the 3 terminal's voltage pairs respectively. So we have 2 equations in 6 unknowns. To proceed further we need two more (independent) equations in order to solve the circuit the 3-terminal device is connected to (basically one treats such a device as an unbalanced two-port...
suppose you have two capacitors with a 0.1 Farad value and 12 VDC rating. label these as A and B. label the terminals of each as 1 and 2. you also have a voltmeter with a 40 volt linear range for DC. you also have a 9 volt DC power supply fed by mains. you charge each capacitor to 9 volts with terminal 1 being - (negative) and terminal 2 being + (positive). you connect the voltmeter to terminal A2 and to terminal B1. does it read any voltage? can - of one capacitor discharge + of the...
Thread 'Weird near-field phenomenon I get in my EM simulation'
I recently made a basic simulation of wire antennas and I am not sure if the near field in my simulation is modeled correctly. One of the things that worry me is the fact that sometimes I see in my simulation "movements" in the near field that seems to be faster than the speed of wave propagation I defined (the speed of light in the simulation). Specifically I see "nodes" of low amplitude in the E field that are quickly "emitted" from the antenna and then slow down as they approach the far...
Back
Top