Computer technician struggling with electronics: building an amplifier.

[hitech95]

I think you need a very basic and practical course like this one:

http://www.learnerstv.com/Free-engineering-Video-lectures-ltv052-Page1.htm

Ehm... Ok... 
Problem: the stupid Flash Player...
EDIT:
It 'a shame, the pronunciation is very bad.

[Richard Crowley]

At that particular point in the circuit (C_i) the "load impedance" consists of R_i.
That is because the right end of R_i is "virtual ground". Hopefully you learned that from studying op-amp circuits.
Since we have a different DC reference in the source AC signal (the audio from your TDA7439) than the LM4880 wants, we must use a DC-blocking AC-coupling capacitor (C_i)
The series capacitor C_i and the parallel resistance R_i form a high-pass filter.
So we must chose a value for C_i which is large enough to pass-through the lowest frequency of interest.

"Roll-off" refers to the decrease in the signal because of a filter function. 
Most of these terms are adequately defined and illustrated on Wikipedia:  http://en.wikipedia.org/wiki/Roll-off
It is good practice to attempt to look up unfamiliar terms on Google or Wikipedia and study the explanations for yourself.
Then if you don't understand the explanation, tell us what you don't understand and where you found it.
We will be happy to help you understand the terms. 
But simply asking for an explanation without first trying to discover it yourself makes people think you are lazy and wanting us to do the work for you.

Standard "audio frequency band" is 20Hz to 20KHz.
So we would like to have the circuit pass everything from 20Hz to 20KHz without reducing the audio ("roll-off")

The series capacitor C_i and the parallel resistance R_i also form a "voltage divider".
But the "voltage division" depends on what frequency you are talking about.
A very simple voltage divider with two equal resistors will divide the voltage in half.
So, at a single frequency, if we select a capacitor whose reactance is equal to the load impedance,
then we have a 50% voltage divider at that frequency.  50% is 3dB in terms of voltage.

If I go to that RFC calculator page and plug in 20000 ohms, and 20Hz, it says that a 0.397885 uF capacitor is 20K ohms at 20Hz.
So If I go back and select 20000 ohms, and use a "standard" capacitor value like 0.5uF, it says that at 15.91Hz, a 0.5uF capacitor = 20K ohms.
That means that your frequency response will be down 3dB at 16Hz which sounds very good to me for a headphone amplifier.
If you used a 1.0uF capacitor, you would have a 3dB-down frequency of 8Hz.  And it is unlikely that your headphones will reproduce 8Hz.

And you can use the same technique for selecting the output capacitor (C_o) assuming you know the impedance of your headphones.

[hitech95]

At that particular point in the circuit (C_i) the "load impedance" consists of R_i.
That is because the right end of R_i is "virtual ground". Hopefully you learned that from studying op-amp circuits.

Yea, you're right!

Since we have a different DC reference in the source AC signal (the audio from your TDA7439) than the LM4880 wants, we must use a DC-blocking AC-coupling capacitor (C_i)
The series capacitor C_i and the parallel resistance R_i form a high-pass filter.
So we must chose a value for C_i which is large enough to pass-through the lowest frequency of interest.

So far, I'm there!

"Roll-off" refers to the decrease in the signal because of a filter function. 
Most of these terms are adequately defined and illustrated on Wikipedia:  http://en.wikipedia.org/wiki/Roll-off
It is good practice to attempt to look up unfamiliar terms on Google or Wikipedia and study the explanations for yourself.
Then if you don't understand the explanation, tell us what you don't understand and where you found it.
We will be happy to help you understand the terms. 
But simply asking for an explanation without first trying to discover it yourself makes people think you are lazy and wanting us to do the work for you.

My fault, I was studying for an exam and I did not have time to check.
I apologize if I gave the wrong impression. I'm here to learn, not to make you lose time.

Standard "audio frequency band" is 20Hz to 20KHz.
So we would like to have the circuit pass everything from 20Hz to 20KHz without reducing the audio ("roll-off")

The series capacitor C_i and the parallel resistance R_i also form a "voltage divider".
But the "voltage division" depends on what frequency you are talking about.
A very simple voltage divider with two equal resistors will divide the voltage in half.
So, at a single frequency, if we select a capacitor whose reactance is equal to the load impedance,
then we have a 50% voltage divider at that frequency.  50% is 3dB in terms of voltage.

If I go to that RFC calculator page and plug in 20000 ohms, and 20Hz, it says that a 0.397885 uF capacitor is 20K ohms at 20Hz.
So If I go back and select 20000 ohms, and use a "standard" capacitor value like 0.5uF, it says that at 15.91Hz, a 0.5uF capacitor = 20K ohms.
That means that your frequency response will be down 3dB at 16Hz which sounds very good to me for a headphone amplifier.
If you used a 1.0uF capacitor, you would have a 3dB-down frequency of 8Hz.  And it is unlikely that your headphones will reproduce 8Hz.

And you can use the same technique for selecting the output capacitor (C_o) assuming you know the impedance of your headphones.

Oh, now I understand the reasoning!

Thanks, and sorry if I ask stupid questions.
Unfortunately, I am realizing that what I learned is useless on the practical side.

[Richard Crowley]

My fault, I was studying for an exam and I did not have time to check.

Studying for your exam is probably more important than this!

I apologize if I gave the wrong impression. I'm here to learn, not to make you lose time.

It is better to show that you are willing to learn for yourself and ask us questions you can't find for yourself.

Thanks, and sorry if I ask stupid questions.

The questions are not "stupid". There is a saying in English "The only stupid question is the one that you didn't ask."
But it is sometimes difficult to understand what you know so we know what words to use.

Unfortunately, I am realizing that what I learned is useless on the practical side.

No, that is probably not true. It simply does not apply to THIS particular problem. 
But it may be very useful information for OTHER problems.

[hitech95]

My fault, I was studying for an exam and I did not have time to check.

Studying for your exam is probably more important than this!

I apologize if I gave the wrong impression. I'm here to learn, not to make you lose time.

It is better to show that you are willing to learn for yourself and ask us questions you can't find for yourself.

Thanks, and sorry if I ask stupid questions.

The questions are not "stupid". There is a saying in English "The only stupid question is the one that you didn't ask."
But it is sometimes difficult to understand what you know so we know what words to use.

Unfortunately, I am realizing that what I learned is useless on the practical side.

No, that is probably not true. It simply does not apply to THIS particular problem. 
But it may be very useful information for OTHER problems.

In a moment of madness I had the idea that the gain varies depending on the input signal. This means that the output of the TDA must always have a signal at line level.
This it also needed to calculate the gain of the headphones amplifier.

My question is: I have to make a circuit that automatically adjust the gain? If so I have to make digital  (The TDA has a GAIN register).

Now, however, a problem arises: What solution?
I decided to sample the signal after the MUX of the TDA, calculate and adjust the gain in the chip as a result of the calculations.
Could it be enough? There are simpler solutions?

This project is becoming more complicated than expected. But I'm learning so many new things.
Thanks, hitech95.

[Richard Crowley]

Yes I think you are still suffering from madness.
If you want to make a practical audio amplifier circuit for speech/music listening, there is no reason for automatic gain.
Adjusting the gain to maintain a constant output signal will effectively RUIN most speech or audio signals.
"Natural" sound that we hear normally has a wide variation in levels. If it was all the same level, it might drive us into madness.

If you want to make an AGC (automatic gain control) there are many more factors that you have not even thought about yet.
Like attack and release time, etc.  Again suggest at least reading about it first: http://en.wikipedia.org/wiki/Automatic_gain_control.

We often use automatic gain when we are recording and producing recordings.
But it is rarely or never used at the end of the "signal chain" when replaying the recordings.

[hitech95]

Yes I think you are still suffering from madness.
If you want to make a practical audio amplifier circuit for speech/music listening, there is no reason for automatic gain.
Adjusting the gain to maintain a constant output signal will effectively RUIN most speech or audio signals.
"Natural" sound that we hear normally has a wide variation in levels. If it was all the same level, it might drive us into madness.

If you want to make an AGC (automatic gain control) there are many more factors that you have not even thought about yet.
Like attack and release time, etc.  Again suggest at least reading about it first: http://en.wikipedia.org/wiki/Automatic_gain_control.

We often use automatic gain when we are recording and producing recordings.
But it is rarely or never used at the end of the "signal chain" when replaying the recordings.

Thanks for the explanation.

I have a question about this scheme:

From what I understand C_S is a bypass capacitor, and if I understand it takes to combat distortion and should be made of ceramic materials.

Instead C_B takes to achieve ground  V_DD / 2. I do not understand what kind it should be: electrolytic or ceramic. 
The two input capacitors I have chosen are 470nF polypropylene.
Those output are 470uF electrolytic.

I'm going crazy.
Some components are available on farnell, some on RS and others DigiKey... 
I want to order some chips to test them on the breadboard and then make the PCB ...

Thanks, hitech95.

Bye, hitech95.

[Richard Crowley]

From what I understand C_S is a bypass capacitor, and if I understand it takes to combat distortion and should be made of ceramic materials.

C_S is there to suppress any high-frequency oscillation that might happen. The materials are not critical.  Its only purpose is to "short" (conduct) any high frequency AC directly to ground.  Note that it should be located physically very close to the chip pin 8 and have a good, low-impedance path to ground (pin 4).

Instead C_B takes to achieve ground  V_DD / 2. I do not understand what kind it should be: electrolytic or ceramic. 

The type of capacitor is not critical. However, note that a 1uF non-polar capacitor will be quite large compared to a 1uF electrolytic, especially since it is operating at such a low voltage.

[hitech95]

From what I understand C_S is a bypass capacitor, and if I understand it takes to combat distortion and should be made of ceramic materials.

C_S is there to suppress any high-frequency oscillation that might happen. The materials are not critical.  Its only purpose is to "short" (conduct) any high frequency AC directly to ground.  Note that it should be located physically very close to the chip pin 8 and have a good, low-impedance path to ground (pin 4).

Instead C_B takes to achieve ground  V_DD / 2. I do not understand what kind it should be: electrolytic or ceramic. 

The type of capacitor is not critical. However, note that a 1uF non-polar capacitor will be quite large compared to a 1uF electrolytic, especially since it is operating at such a low voltage.

Thanks again.
This is a draft of the PCB.

I hope that is correct, (Or at least in part). I've never done anything with analog signals ...

[Richard Crowley]

Until you confirm that the schematic diagram is correct, laying out a PC board is premature.
Did you share the schematic diagram?  I don't remember seeing it.

Your PC board image shows that it is not complete. There are many parts missing and/or not routed.
That is what all those fine yellow lines show (they are "error messages")

[hitech95]

Until you confirm that the schematic diagram is correct, laying out a PC board is premature.
Did you share the schematic diagram?  I don't remember seeing it.

Your PC board image shows that it is not complete. There are many parts missing and/or not routed.
That is what all those fine yellow lines show (they are "error messages")

The missing part is the digital section.
A connector for the front panel, and connectors to relays. There is also a PCF8574T. (Needless) In the previous version the processor was on the front.

The Schematic: Link
A connector for the front panel, and connectors to control Rale. There is also a PCF8574T. (Needless) In the previous version the processor was on the front.]Link[/url]

EDIT: borken URL