Calculating low frequency cutoff for this R-C microphone circuit

In summary: Yes, you want the output voltage be in the audible range unattenuated. ...electrical energy. The higher the frequency of the vibration, the higher the electrical energy that will be converted.
  • #36
songoku said:
Is it correct to say the capacitor undergoes the charging and discharging process
It is misleading because it suggests that the capacitor alternates between 'fully' charged and 'fully' discharged. That's not what happens.

The capacitor charges slightly and discharges slightly. Call these 'charge oscillations'.

The charge on the capacitor when there is no sound is ##Q_0##. With sound, the charge might oscillate between ##0.999Q_0## and ##1.001Q_0## for example. The changes are small.

songoku said:
What is the function of the battery in this circuit?
When there is sound, there are charge-oscillations in the capacitor. As a result, there is alternating current through the resistor. Without a battery, the capacitor would be uncharged - there would be no charge oscillations. So there would be no current.

So the function of the battery is to maintain a charge on the capacitor.

Edit - minor.
 
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  • #37
Steve4Physics said:
So the function of the battery is to maintain a charge on the capacitor.
From the point of view of the battery, what is the meaning (or maybe consequence) of the alternating signal produced in the circuit?

At one time, the current will flow clockwise, and at the other time, it would be anticlockwise. Let say when the current flows clockwise the capacitor is slightly charging so it means that the capacitor is being charged by the battery. When the current flows anticlockwise, the capacitor is slightly discharging so it means the capacitor discharges through the resistor and battery? Is the battery being charged by the capacitor?

Thanks
 
  • #38
I believe this is an electromechanical problem, with most of the physics (at least low end response) governed by the coupling of the pressure wave into the microphone diaphragm. Then the unknown mechanical inertia makes it a parametric oscillator and any simple "RC" analysis is likely misguided.
 
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  • #39
hutchphd said:
...and any simple "RC" analysis is likely misguided.
Absolutely. This is what is known as a condenser mic. The capacitor is considered the source. So as a source it's output impedance is likely lowered as the frequency increases. Simple analysis shows that around 30 hertz is the knee and the frequency response is flat as the frequency increases.
-
The OP claims to have no interest in RC high pass at this point.
I don't study high-pass RC filters.
I don't understand how all of the questions can be answered without having a grasp of a basic RC filter.
 
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  • #40
songoku said:
From the point of view of the battery, what is the meaning (or maybe consequence) of the alternating signal produced in the circuit?
I don't understand your question.

songoku said:
At one time, the current will flow clockwise, and at the other time, it would be anticlockwise.
Correct. This happens repeatedly. That's why you get an alternating current through the resistor (and therefore an alternating voltage across it) when there is sound.

songoku said:
Let say when the current flows clockwise the capacitor is slightly charging so it means that the capacitor is being charged by the battery.
Ok - for half of each cycle.

songoku said:
When the current flows anticlockwise, the capacitor is slightly discharging so it means the capacitor discharges through the resistor and battery? Is the battery being charged by the capacitor?
Yes - for the other half of each cycle.

But thinking in those terms will not help you understand part c) of the original question (if that's what you are trying to do). E.g. see @Averagesupernova comments in Post #39.
 
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  • #41
Steve4Physics said:
Yes - for the other half of each cycle.

But thinking in those terms will not help you understand part c) of the original question (if that's what you are trying to do). E.g. see @Averagesupernova comments in Post #39.
I am not trying to understand question (c) since from all the replies I know I won't be able to understand without knowing some other things outside what I have learned.

The best I can guess would be from the graph suggested by @kuruman and also your reply. The ratio of Vout / Vin is around 0.5 from the graph, matching what you said that at 20 Hz the output will be half of what it would be at high frequency and it is still significant (although I don't understand why the ratio of 0.5 is still significant)

I am trying to clear some doubts in my mind about the circuit. What happens to the battery when it is being charged by the capacitor? I think the voltage of the battery is constant so when it is being charged, what does the effect?

Thanks
 
  • #42
Yes, an 'ideal' (or perfect) battery has a constant voltage. If you have ever had to replace batteries in a flashlight when the the light gets dim, you have seen that a real battery is not 'ideal'.

There are also batteries that can be recharged, they are used in cars and in cell phones.

Even a flashlight battery can be slightly recharged a few times.

Recharging a battery reverses the chemical process that supplies power used in normal use.

So the answer to your question is the power from the capacitor very slightly recharges the battery.

I hope this helps!

Cheers,
Tom
 
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  • #43
songoku said:
although I don't understand why the ratio of 0.5 is still significant
The filter doesn't have a sharp cut-off: it doesn't completely block all frequencies below some exact value and it doesn't completely transmit all frequencies above that value. For a signal of frequency, ##f##, the fraction of the signal transmitted is a smoothly varying function of ##f##.

So, for convenience, we can call the cut-off frequency the value of ##f## when some fraction of the signal's amplitude gets transmitted. We could choose, for example, 0.1 or 0.9 as the fraction - but 0.5 is a ‘natural’ choice.

songoku said:
I am trying to clear some doubts in my mind about the circuit. What happens to the battery when it is being charged by the capacitor? I think the voltage of the battery is constant so when it is being charged, what does the effect?
While a (real) battery is being charged, the voltage is greater than the battery's 'own voltage' (greater than the battery's emf to use the correct terminology).
 
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  • #44
Thank you very much for all the help and explanation kuruman, Steve4Physics, hutchphd, Averagesupernova, Tom.G
 
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