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Author Topic: [SOLVED] TL082 preamp: help please!  (Read 9629 times)
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Roma - Italy
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As teacher you are not only tough but unfair:

- you didn't track any score regarding the second ohm's law
- my rambling about point 3 was more to demonstrate is such a huge subject with so many implications that I hardly believe I could use a forum post to even just list them all: you should have allowed at least half point for admitting the subject is so vast
- I admit my answer at point 2 was incorrect, but I believe 9 out of 10 would have understood I meant to say "current flows more in the easiest path"... And that was quite clear in the context of the remaining part of the answer

Anyway, I'm now working on your answers: at least I got the reason behind the two diodes!
Regarding the high pass filter question, I knew you were creating a voltage divider to get a 2.5V reading point, but why isn't that an high pass filter as well? Because the AC is not swinging to negative?
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Manchester (England England)
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As teacher you are not only tough but unfair:
Maybe that is why not all my students passed their exams.  smiley

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but why isn't that an high pass filter as well?
Well it does have a frequency response like anything involving a capacitor. But it's function is not to act as a filter, therefore the components are not critical to it's performance. It is meant to pass all audio frequencies and so in terms of the circuit is not a filter.

Arguing about this docks you 5 marks.  smiley-razz
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Roma - Italy
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I haven't had the opportunity to prototype this circuit yet due to some components I'm still missing, but I've studied it over and over, so here a few considerations/questions/doubts:

  • I still don't get why C4 R6 do not configure as an high pass passive filter. I mean, I know R5 and R6 are a voltage divider and I understand C4 removes the DC component from the output, but doesn't the two configure a high pass filter as well? I'm wondering about this because the cut-off frequency of such filter would be around 160Hz which is well within the frequency range I'm interested to process: as a consequence I'm wondering if I need to change the R5 R6 values (let's say 10K) to avoid an unwanted filtering effect (new high pass filtering cut-off frequency would be 15Hz)
  • With regards to C2 and C3 I now understand (thanks Mike) their goal is to remove noise that might get introduced by the DC voltage source, but if I'm willing to accept that noise (temporary at least) can I prototype the circuit without those two components?
  • I understand C1 is there to prevent DV voltage to reach the AC source, but I don't understand how come such event can occur, unless we consider the possibility of a failure within the IC suddenly connecting the supply DC voltage to the AC input

I will post here as soon as I'll get the missing components and I'll be able to produce the suggested circuit.

As usual, thanks to everybody who will have the patience and the time to answer.
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Manchester (England England)
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I still don't get why C4 R6 do not configure as an high pass passive filter.
As I said it  has a frequency responce. If you define anything with a frequency responce as a filter then yes it is a filter. But the rest of the world will not call it a filter. This is mainly due to the fact that it's function is not to filter a signal but to do something else. Any audio amplifier has a frequency responce, it wil only amplify frequencies over a certain range, so you might call it a band pass filter but no one else will. It is important that you use the same words as the rest of the world if you want to communicate with them.

Without C2 & 3 the power supply might oscillate or your chip might oscillate so making the whole circuit not work.

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understand C1 is there to prevent DV voltage to reach the AC source, but I don't understand how come such event can occur,
One end of the microphone is connected to your vertuial ground. Any DC offset on the output of the op amp will produce the corresponding DC offset on the -ve input which will then put it through your microphone, thus destroying or damaging it. Remember a speaker and microphone are essentially the same thing so any DC signal will push the diaphragm to one side distorting any input.
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Roma - Italy
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Any audio amplifier has a frequency responce, it wil only amplify frequencies over a certain range, so you might call it a band pass filter but no one else will. It is important that you use the same words as the rest of the world if you want to communicate with them.

I get that and I understand the filtering effect is not the goal of those two components, I'm though concerned about the unwanted effect and by the fact the cut-off frequency (with my obviously erroneous calculations  smiley-lol) is going to be 159.15Hz which is right in the middle of my frequency range of interest 50-200Hz.

If my calc of the cut-off frequency is correct (I just considered C4 and R6) then I might wish to bump the two voltage divider resistors up to 10K to push that cut-off away from my range (down to 15.9Hz).

I don't know if my reasoning makes sense (it appears it does from your previous comment) and if my calculations are correct (I doubt they are, otherwise it would mean I'm starting to understand something on this topic  smiley-eek-blue).

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Without C2 & 3 the power supply might oscillate or your chip might oscillate so making the whole circuit not work.

Which means you do not suggest to even try the circuit without those? I'm having difficulties in sourcing capacitors here in this s**th**e where I live.....

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One end of the microphone is connected to your vertuial ground. Any DC offset on the output of the op amp will produce the corresponding DC offset on the -ve input which will then put it through your microphone, thus destroying or damaging it. Remember a speaker and microphone are essentially the same thing so any DC signal will push the diaphragm to one side distorting any input.

Got it, thanks!
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Which means you do not suggest to even try the circuit without those? I'm having difficulties in sourcing capacitors here in this s**th**e where I live.....

Then what values of capacitor are available there in "s**th**e"?
If you have the 1uF (C4) then get more of those for the 4.7u (C3): place 4 or 5 in parallel.

Maybe you can live without the 100uF (C2).
Again, depending on what you have available, there may be an alternative.
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the cut-off frequency
Do you know what that means? The frequency where the capacitave reactance equal the resistance is the 3dB point. It is not the point where no more signal is let through.  If this is not good enough Then you are better off using a larger capacitor than lowering the resistance this is to make the vertuial ground more robust than it would be at 10K. But you can up it to 10K. If you want.
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Roma - Italy
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Then what values of capacitor are available there in "s**th**e"?
If you have the 1uF (C4) then get more of those for the 4.7u (C3): place 4 or 5 in parallel.

Maybe you can live without the 100uF (C2).
Again, depending on what you have available, there may be an alternative.


When i went to the local store I was able to find only a 47pF capacitor, I had to unsolder other capacitors from an old radio. I would prefer to avoid unsoldering SMD components and I don't have any other old stuff around to unsolder .

I'm waiting for an assorted set of capacitors to be delivered from China, but that takes a bit  smiley-mr-green
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A capacitor of 47pF is hardly any capacitor at all so it is a waste of time trying to use it.
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Roma - Italy
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the cut-off frequency
Do you know what that means? The frequency where the capacitave reactance equal the resistance is the 3dB point. It is not the point where no more signal is let through. 

I do understand that, at least to a certain extent (I have no doubt I'm ignoring a lot of factors that play a role on that). I''ve plotted the bode diagram of that and I wouldn't be concerned if it wasn't right in the middle of my frequency range of interest. At 50Hz I will have around 10dB which, if I understand the meaning of that value, I'll have 1:10 of the original voltage... back to 250mV.

Please take in consideration my goal is to measure frequence and amplitude of two sine waves in [50,200]Hz
To me that pair of components represent an issue, unless I'm mistaken in my reasoning.

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If this is not good enough Then you are better off using a larger capacitor than lowering the resistance this is to make the vertuial ground more robust than it would be at 10K. But you can up it to 10K. If you want.

What is the difference? If I increase the voltage divider resistors value I will lower the current in the Aduino side of the circuit (don't know about other effects), I don't know the effect of increasing the capacitor value though
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Increasing the capacitor value lowers the  beak frequency as does increasing the resistor value. By using lower values of resistor you make more current flow through the resistors and so the vertuial ground is more robust. That is to say it is less suceptable to noise and disturbance.
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Roma - Italy
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Increasing the capacitor value lowers the  beak frequency as does increasing the resistor value. By using lower values of resistor you make more current flow through the resistors and so the vertuial ground is more robust. That is to say it is less suceptable to noise and disturbance.

Ok, I get it. When I'll have my final design though I'll have to balance such robustness with battery consumption though.

Is there a current value under which it wouldn't be advised to go? I mean, I belive it will depend on how much noise is going to be acceptable and the operating frequency (or frequency range) so if you can point me to some sort of information regarding this or just give me the solution  smiley-grin smiley-lol
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Increasing the capacitor value lowers the  beak frequency as does increasing the resistor value. By using lower values of resistor you make more current flow through the resistors and so the vertuial ground is more robust. That is to say it is less suceptable to noise and disturbance.

Greater capacitor doesn't mean greater current draw at a specific frequency?

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Greater capacitor doesn't mean greater current draw at a specific frequency?
Sorry I don't know what you mean.
What current are you talking about?
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Greater capacitor doesn't mean greater current draw at a specific frequency?
Sorry I don't know what you mean.
What current are you talking about?

We are going off topic, my bad.

A capacitor in an AC circuit has a capacitive reactance which depends on frequency and capacitance.
Reactance is like resistance in DC circuits: it opposes current flow.

Because reactance is inversely proportional to capacitor capacitance and voltage frequency, with a fixed frequency reactance diminishes if capacitance is increased.

If reactance diminishes then current increases: in a purely capacitive circuit a capacitor of infinite capacity or a source of infinite frequency imply an infinite current flow....
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