Audio Amplifier. Which class should I use?

Sorry sir. I don't get your this point please explain. You said differential amplifier with op amp will have four resistors. But in my attachment there are only two input resistance and these resistance are used for offset, I think so. Is my given attachment is wrong?

And yes there is a differentiator amp too I can see that.

And I am reviewing your given circuit. And want to say that output from U1A is at the input of U2A (at pin 2(inverting) and at pin 3(non-inverting)). Does this not make it differential amplifier?

And you said a sentence "capacitor is in series with input". Does I understand by this that in differential amplifier there is no capacitor at input?

Thanks.

goldsmith said:

If we select the gain as large as 1000 for these simple op amps ( for an stage ) the result will be awful ! think about it , please :
Av*BW = a constant number ( BW = band width ) . so if you increase the gain , the BW will decrease . it means that when you need large gains you should use some stages as a chain !

Click to expand...

Yes, this is important for him to understand. Maybe I can explain a little more with a picture.

This picture shows how the gain of a TL074 changes with frequency. The red line is the gain of the opamp itself, with no feedback resistor. We see that the gain is very very high at low frequencies, but gets smaller as the the frequency gets higher.

Now when we make an inverting amplifier, we set the gain with two resistors. The green line shows what happens if we choose a gain of 1000. We get a gain of 1000 at low frequencies, but the gain is less above 3KHz and the audio will sound dull, with too little treble. This is because we can never get a gain higher than that shown by the red line (the gain without feedback).

If we change the resistors to set the gain to 30, then we get the result shown by the blue line. This is very good, the response is flat to nearly 100KHz. If we connect two such amplifiers together we will get gain = 30 * 30 = 900 with a very good frequency response.



---------- Post added at 00:08 ---------- Previous post was at 00:01 ----------

Shayaan_Mustafa said:

And want to say that output from U1A is at the input of U2A (at pin 2(inverting) and at pin 3(non-inverting)).

Click to expand...

Pin 3 is connected to ground, so there is no signal there.

@godfreyl

Thanks for this great explanation. I have understood what goldsmith want to make me understand.
This is very nice explanation. Now I have understood. We should use chain stages to gain gain for long range of frequencies.
As red line gets lower at 10Hz, green line gets lower at 3KHz and blue line gets lower at 100KHz. So if we add more stages then we can get gain for more range of frequencies. Right?

Also one more thing. As you said for 30*30=900. This implies two stages of each has gain of 30. So why don't we do this for green line.
I mean as green line is giving gain of 60 then if we add chain stages then we can get gain for long range of frequencies too as we did for blue line i.e. 60*60=1200. Can't we do that?
And similarly for red line?

And yes pin 3 is grounded.

Thanks for nice explanation.

Hi Shayaan

Sorry for the confusion. Gain in the graphs is in decibels. Did your teacher not explain that either? Decibels are a useful because it allows us to write big values of gain with not very big numbers. Often we just say dB instead of decibels. here are some examples:

Gain = 1 is 0 dB
Gain = 10 is 20 dB
Gain = 100 is 40 dB
Gain = 1000 is 60 dB

If gain is 10 times higher we add 20 decibels. Also if gain is 10 times lower we subtract 20 decibels. So if Vout/Vin = 0.01, we can say gain = -40dB.

Mathematically, decibels = 20 * log(gain).
Some other examples that are easy to remember:
a) Gain of 2 is almost exactly 6dB.
b) 10dB means the gain is slightly more than 3.

Now let's look at the graphs again.
The Red line shows a gain of 106dB, which means gain = 200000.
200000 = 2*10*10*10*10*10
106 = 6+20+20+20+20+20

The Green line shows gain = 1000 = 60dB
The Blue line shows gain = 30 = 29.5dB (almost 30)
I think the blue one confused you because the two numbers are the same.

With two "blue" amplifiers connected together, we can say the total voltage gain = 30 * 30 = 900 or we can say the total gain is 29.5dB + 29.5dB = 59dB.

With two "Green" amplifiers connected together, we can say the total voltage gain = 1000 * 1000 = 1000000 or we can say the total gain is 60dB + 60dB = 120dB.

Now you see why it can be easier to use decibels - with a big gain like 1000000, it is easy to make a mistake and write the wrong number of zeros at the end, or to make a mistake counting the zeros when we read it. When we write 120dB, it is harder to make a mistake.

If we look at the slope of the lines at high frequencies we see that if the frequency is 10 times higher, then the gain is 10 times lower (20dB less). This is the slope of a simple first order filter. It is often called a slope of 6dB / octave or 20dB / decade.

Regards - Godfrey

Sorry for the confusion. Gain in the graphs is in decibels. Did your teacher not explain that either? Decibels are a useful because it allows us to write big values of gain with not very big numbers. Often we just say dB instead of decibels.

Click to expand...

First of all don't sorry because I know this unit. My teacher explained that.

I have clear ideas from dB. So I understand.

As you said above the blue line is very good (in 2nd last post, post#62). I think you said so because in this condition amplifier maintains it gain over wide range of frequencies i.e. upto 100KHz than green(upto 3KHz) and red(upto 10Hz). Right?

But if we see gain shown by blue line is less i.e. 30 although it covers wide range of frequencies but it is less comparatively than red and green. Right?

And you also said that if we use chain stages then we can increase gain and this will maintains more wide range of frequencies. Right?

But can we do same thing with green or red line? So this way if we use amplifiers with gain shown by green line as chains then we can get gain more than shown by blue line and over wide range of frequencies.

Try to understand what I am trying to say. I don't have words in English, as it is not my native language.

So kindly try to understand.

Thanks for help.

Hi shayaan
Well ! i'm happy , because you got my and godfreyl's meaning .
So , lets continue :
in the other terms , tell me , please that have you access to an oscilloscope ? ( the oscilloscope is eye of each engineer ! )
If yes , the way will be very simple but if not , there won't be any problem , but it will be a bit harder . ( just a bit )

Hello. I was waiting for you. And at last you came.

Yes I know how to use oscilloscope. I can use oscilloscope, function generator, power supplied and multi-metres too.

Yes i came back !
I know that you know how to use them , but i didn't ask it . i asked that do you have access to an oscilloscope ? perhaps in your home ?

:-(

No. not at home :-(

But I know how to use electronics simulator. Which one will you prefer me to use?

Don't worry , no problem , 9 years ago , i hadn't oscilloscope too .
Well , can you predict that how much is the maximum out put of your PC ?

no. not even a guess. is it necessary to determine the max output of my PC?

I can't guess. then how to know my output?

Yes it is important , because i don't think that you will love a large value of THD , right ? BTW : do you know the meaning behind THD ?

I think total harmonic distortion.

And what is the distortion ?

unwanted signal is know as distortion. :???:

No , Distortion can create on phase or frequency or amplitude . if you give your audio to an amplifier , and if this amplifier shift it's phase or attenuate some specific frequencies or cut or change it's amplitude it has distortion , ( PD and FD and AD )

Ohh... got it.
its mean unwanted change in signal.

And what about. pc output?

Yes , hit to the point !
And about maximum out put of PC , i think teach PC is different , but i can measure mine , and tell the result .

How can I measure mine? Is there any way to measure it?

An oscilloscope ! or ask the designer company