Calculating low frequency cutoff for this R-C microphone circuit

[kuruman]

Going back to the idea of potential divider suggested by @Steve4Physics in post #23, consider the following:
You already know that the circuit impedance is $Z=\sqrt{R^2+\dfrac{1}{(\omega C)^2}}.$
Find the AC current in the circuit assuming that AC emf is $V_{\text{in}}.$
Knowing the AC current, find the voltage $V_{\text{out}}$ across the resistor.
Find the ratio $V_{\text{out}}/V_{\text{in}}.$ It should be a function of $\omega.$
Plot the ratio against the frequency $f$ IN Hz for the numbers that are given.

Note that the charge on the capacitor will oscillate about its rest-value. We can not think in terms of charging and discharging the capacitor in simple DC terms. Treating the capacitor as fully charged or fully discharged after 5 time-constants is irrelevant - we are not dealing with the DC case.

I agree. High fidelity means that one can view the fluctuations of the capacitor voltage above and below the fully-charged value as a faithful reproduction of the air fluctuations above and below the ambient pressure caused by the traveling sound waves.

 

[songoku]

3)The crux of the Physics is in your third question. The net signal power from the microphone circuit depends upon volts and amps. If the capacitor is not given sufficient time to discharge at least halfway, the signal will be much diminished.

Why the capacitor need to discharge at least half of the charge for the signal to be significant?

2)The current flows both directions as the microphone oscillates about equilibrium

During each sound-cycle, there are 4 stages, corresponding to the 4 parts ( +-+-) of a sine curve:

i) While the pressure increases, plate-separation decreases, capacitance increases, stored charge increases, current flows (in, let’s say, the +ve direction).

ii) While the pressure decreases, returning to P_0, plate-separation increases, capacitance decreases, stored charge decreases, current flows (in the -ve direction).

iii) While the pressure continues to decreases (below $P_0$, plate-separation increases, capacitance decreases, stored charge decreases, current flows (in the -ve direction).

iv) While the pressure increases back to $P_0$, plate-separation decreases, capacitance increases, stored charge increases, current flows (in the +ve direction).

We take conventional current to flow from positive to negative terminal of battery but in this case there is an instant where the current flow from negative to terminal positive of battery. What does this mean?

Going back to the idea of potential divider suggested by @Steve4Physics in post #23, consider the following:
You already know that the circuit impedance is $Z=\sqrt{R^2+\dfrac{1}{(\omega C)^2}}.$
Find the AC current in the circuit assuming that AC emf is $V_{\text{in}}.$
Knowing the AC current, find the voltage $V_{\text{out}}$ across the resistor.
Find the ratio $V_{\text{out}}/V_{\text{in}}.$ It should be a function of $\omega.$
Plot the ratio against the frequency $f$ IN Hz for the numbers that are given.

Do you mean this?
$$I=\frac{V_{\text{in}}}{\sqrt{R^2+\frac{1}{(\omega C)^2}}}$$
$$V_{\text{out}}=\frac{V_{\text{in}}}{\sqrt{R^2+\frac{1}{(\omega C)^2}}} \times R$$
$$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{R}{\sqrt{R^2+\frac{1}{(\omega C)^2}}}$$

I got the graph to be like this:

Do we need to consider the emf of battery when finding the current?

Thanks

 

[kuruman]

That's the general idea behind the graph but you need to make a better plot. There is a lot of space that is not used vertically and there are no units or lables on the horizontal axis. Remember that the audible frequency is from 20 Hz to 20,000 Hz and that should be on your plot.

You don't have to worry about the battery because it doesn't provide AC. When there is no sound, there is no current in the circuit.

 

[Steve4Physics]

I deleted my previous reply as it was too confusing.

We take conventional current to flow from positive to negative terminal of battery but in this case there is an instant where the current flow from negative to terminal positive of battery. What does this mean?

With no sound, the charge on the capacitor is steady ($Q_0$).

When there is a sound wave, the following happens repeatedly:
- the capacitor's charge increases to $Q_0+\delta Q$; the current flows one way during this process;
- the capacitor's charge decreases to $Q_0-\delta Q$; the current flows the opposite way during this process.

This results in an alternating current with the same frequency as the sound wave.

 

[songoku]

That's the general idea behind the graph but you need to make a better plot. There is a lot of space that is not used vertically and there are no units or lables on the horizontal axis. Remember that the audible frequency is from 20 Hz to 20,000 Hz and that should be on your plot.

You don't have to worry about the battery because it doesn't provide AC. When there is no sound, there is no current in the circuit.

Just neglect the part where f is negative.

At f = 20 Hz, the value of Vout / Vin would be around 0.53. How to interpret this value?

I deleted my previous reply as it was too confusing.

With no sound, the charge on the capacitor is steady ($Q_0$).

When there is a sound wave, the following happens repeatedly:
- the capacitor's charge increases to $Q_0+\delta Q$; the current flows one way during this process;
- the capacitor's charge decreases to $Q_0-\delta Q$; the current flows the opposite way during this process.

This results in an alternating current with the same frequency as the sound wave.

Is it correct to say the capacitor undergoes the charging and discharging process? Something like when the capacitor's charge increases, it means the capacitor is being charged and when the capacitor's charge decreases, it means the capacitor is discharging?

What is the function of the battery in this circuit?

Thanks

 

[Steve4Physics]

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.

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.

 

[songoku]

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

 

[hutchphd]

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.

 

[Averagesupernova]

...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.

 

[Steve4Physics]

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.

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.

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.

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.

 

[songoku]

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

 

[Tom.G]

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

 

[Steve4Physics]

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.

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).

 

[songoku]

Thank you very much for all the help and explanation kuruman, Steve4Physics, hutchphd, Averagesupernova, Tom.G