What is the function of these capacitor?

[madires]

The input capacitor could be replaced with 100n film, and be 3dB down at 16Hz. 220n gives 7Hz. But replacing it for a "posh" low-ESR or film type will make no obvious difference - suppose the original has an ESR of 10 ohms; as it works into 100k, the loss will be 0.00087dB. Doubt you'd measure that, let alone hear it.

Just a brief supplement on how to calculate the bottom frequency at -3dB:

  f = 1 / (2 pi R C)

[Mark Hennessy]

The op amp inputs have nothing to do with it.  If you "charge up" the capacitor to a couple of millivolts, and you slam it with -5V, you're going to damage the capacitor.

But how, under normal circumstances, are you going to get -5V across the capacitor?

And if you did put -5V DC at the input, would that input capacitor actually be damaged?

(for the purposes of understanding, let's restrict ourselves to the input capacitor only - that is totally deterministic because we have the whole circuit in front of us. We can think about the output later if you like)

(I am trying to help. And I know from my day job that people can struggle with this topic )

All the best,

Mark

[John Coloccia]

The input cap has the - terminal looking at the input, so if you put +5 on it....or +2, or whatever, you reverse bias the cap.  At a minimum, it will behave terribly as a cap when the voltage swings positive.  The + terminal is sitting, more or less, at ground potential.  I don't know what you're asking how I get the voltage "across" the cap.  One end is looking at ground, and the other end is looking at something other than ground. 

[Mark Hennessy]

The input cap has the - terminal looking at the input, so if you put +5 on it....or +2, or whatever, you reverse bias the cap.  At a minimum, it will behave terribly as a cap when the voltage swings positive.  The + terminal is sitting, more or less, at ground potential.  I don't know what you're asking how I get the voltage "across" the cap.  One end is looking at ground, and the other end is looking at something other than ground.

No.

Whatever voltage you put on the input, the other end of the cap will have the same. Hence, the voltage across the cap will be 0V.

OK, after a while, the charge on the capacitor will change. But how long is a while? Well, to get an idea, let's calculate the time constant: 10uF times 100k is 1 second.

What is the period of the lowest audio frequency? Let's say 50ms (20Hz). So, with a healthy 20Hz sine wave playing into this, the capacitor doesn't have time to significantly alter its charge during the positive-going part of the waveform.

To prove this, get a dual input scope and connect probes either side of the cap. You will see exactly the same waveform on each side - perhaps with a small DC offset. Case closed

Above, when working out the time constant, I said 100k. Was that fair? Yes, it is. The positive end of the capacitor sees 100k to ground, plus whatever the op-amp presents.

Earlier I quoted the first "golden rule" of op-amps: the inputs are always at the same potential. The second "golden rule" is that the inputs draw no current. In other words, the input impedance is extremely high (and is further raised by bootstrapping because of NFB).

To answer the second question I posed: if you place 5V DC at the input such that the capacitor "saw" the wrong polarity (which - having just re-checked the schematic - would be +5V, not -5V - sorry!), then that would be in series with 100k. The worst-case current would be tiny: just 50uA. Would that damage the cap? Based on my experience, I very much doubt it.

I've seen dozens of 40 year old capacitors with 3 or 4 volts of incorrect polarity (caused by a silkscreen error), and they've always checked out fine - I don't even bother to change or reverse them... (I can provide more details if you like)

[Mark Hennessy]

You guys are scaring the OP off with those advance discussions , looking at the original question and also OP's previous posts , I suspect he is just a novice in electronics.

Yes, I agree that the OP is probably new to this, but speaking as someone who teaches electronics, I do think that this is something that is pretty basic; the idea of separating out the DC conditions and AC signal is fundamental to understand any circuit that deals with signals. And, as we're seeing, it is a source of confusion, even among experienced people. In my experience, the earlier you deal with this topic, the easier everything else becomes.

Perhaps the mods should move these posts to a new thread? I'm new to this forum, so don't know the "form"

Mark

[John Coloccia]

If you had a pure audio signal, such as when coupling one stage into another, you wouldn't need the input cap.

[sync]

The world is not perfect. There are small DC offsets. But 1V reverse bias doesn't damage an electrolytic cap. This circuit is used billion times in audio gear and the caps survive.

[ejeffrey]

Electrolytics can work perfectly well if properly applied. They're actually very, very common..

Sure, a 10µF electrolytic is less expensive than a 2.2µF MKS.  Old measurements of different cap types for usability in audio applications showed following rank list:
1. KP/MKP/Styroflex/KS/MKS
2. MKT/KC/MC (about 3 times worse than 1.)
3. electrolytics (about 100 times worse than 1.)

That depends strongly on what you are using it for.  You don't want electrolytic capacitors in feedback networks or filters because their value can change dramatically with time and temperature, changing the frequency response.  For input coupling caps the limitations of electrolytic caps are not important and they will work fine.

[madires]

That depends strongly on what you are using it for.  You don't want electrolytic capacitors in feedback networks or filters because their value can change dramatically with time and temperature, changing the frequency response.  For input coupling caps the limitations of electrolytic caps are not important and they will work fine.

I found an old Elektor article about that topic (http://www.die-wuestens.de/iz/audiokondensatoren.pdf). Unfortunately it's in German but you'll see the measurement setup they have used. I'm not an audio expert but I've done a internship in a high-end audio EE company (no audiophoolery!) while studying and learned a lot. They had two amazing EEs, one is an absolute expert in analog audio stuff and electrolytic caps for input coupling was a no-no. If you took some audio circuit diagram from an electronics magazine to him, he made some changes whithin 5 minutes and told you to take measurements of the original design and then of his modified one. Guess which one always won!

[sync]

Thanks for the PDF. The distortion of electrolytics was too low to measure (<0.005%) . But there was a difference using the square wave. Unfortunately it lacks information. Rise and fall times. What does 1% difference mean. What are the frequencies of the difference. And I'm missing a inductance compensation.

[huu_tri0101]

Thank all of you for helping me out.
First, this is really a working circuit, I draw from a Pioneer amplifier A-N702.It's use +/-12V for opamp 4558s. This circuit is just apart of big circuit. the output of this 4558 will be the input of another opamp (JRC2068 for volume adjust) then another 2 JRC4558 (for bass and treble adjust) then final opam JRC2068 to drive some small BJTs, and BJTs drive 4 mosfets.

After reading all your post. I learn a lot of things. Thank again guys.
So, I should change the electrolytic capacitor to film capacitor for lower the THD, reliability and better high pass filter ?
Can I change the value of capacitor from 10uf to 2.2uf ? what will change when I do that(cut off frecquency, input current,...) ?

And why when we calculate the cut off frequency, we calculate at 3 dB?
how can we calculate/measure the impedence of input and output ?
Sure I will change the opamp, any suggestion ?
Thank you

[amyk]

Unless the electrolytics have dried out significantly and lost most or all of their capacitance you will not hear a difference. Otherwise the bass will be gone.

[Monkeh]

Sure I will change the opamp, any suggestion ?

Yes, leave them alone. There's nothing wrong with the opamps you have.

[madires]

Can I change the value of capacitor from 10uf to 2.2uf ? what will change when I do that(cut off frecquency, input current,...) ?

Actually you could use any value down to 220nF for that circuit.

And why when we calculate the cut off frequency, we calculate at 3 dB?

For a lot of stuff the frequency limits are specified for -3dB.

how can we calculate/measure the impedence of input and output ?

It's based on the in/output circuitry and the OPamp. The input impedance of your buffer amp is about 50k (set by the input resistors).

[Orpheus]

+/- 3dB is an arbitrary cutoff but there is *some* rationale behind it. +3dB voltage or current means 2x (-3dB means a factor of 1/2). To be more precise: +/- 3.0103 dB = (10 log 2) dB = 2x the voltage.

In general situations, engineers may not want to worry much about factors of one-point-something, so they often set the limit at "this difference is not just 1.x". Hence 3dB = factor of 2 = the cutoff

[Hideki]

There is more rationale for it. Here's an image of a (first order) lowpass filter response. Highpass looks the same, except you have to flip the curve left to right.



If you look at the blue curve, the response of the first order filter falls off at 20 dB per decade (or ~6 dB per octave) once you get past the cutoff frequency. Following the slope of the curve up from the bottom right corner, the "ideal" curve would hit the 0 dB point at exactly the given cutoff frequency.

However, a real filter doesn't behave like two straight line segments. There is a smooth curved transition between no attenuation and attenuating 20 dB per decade. At the cutoff frequency the response is at the half-power pont, ~3 dB down.

[Mark Hennessy]

+3dB voltage or current means 2x (-3dB means a factor of 1/2). To be more precise: +/- 3.0103 dB = (10 log 2) dB = 2x the voltage.

Not quite. -3dB is half the power, not the voltage.

A halving of the voltage would be -6dB. You'd get that with two equal value resistors.

The -3dB point occurs when the capacitor has the same impedance as the resistor in the filter. To calculate the impedance, use Zc = 1 / (2pi times FC)

So on the face of it, it might seem like the voltage would half when the capacitor has the same impedance as the resistor. But, when you factor in the phase angle caused by the reactance of the capacitor, the change is actually 0.707 (1/sqr-root(2)), which is -3dB.

Oh, and the input impedance of this circuit is 50k, as there are two 100k resistors in parallel.

Take it from me - I've been doing this stuff for all my life, and professionally all my working life: there is no point whatsoever changing the op-amp in this circuit. It won't sound any different (at best), and you won't learn anything from doing it. Build your own gear - and then debug it. Correctly selected and correctly implemented op-amps do not have a "sonic signature".