Connect with us

Electronics n00b, how does this circuit function?

Reply to Thread

Discussion in 'General Electronics Discussion' started by Solidus, Aug 6, 2012.

Scroll to continue with content
  1. Aug 6, 2012 #1
    Solidus

    Solidus

    349
    4
    Jun 19, 2011
    Okay, if it wasn't given away by the title, I'm an electronics newbie. I want to learn how electronics work, and I'm trying to piece together what I can.

    Now, it should be said that those of you that have followed posts of mine before should be assured this is just me asking questions out of learning and curiosity, and that this is not going to turn into my usual borderline-insane or just not-so-feasible project (look at my thread start history and you should gain some idea of what I'm talking about :p)

    Anyway, I was looking at this schematic:

    [​IMG]

    and I'm trying to piece together what does what. Obviously I know that resistors increase resistance, the function of pots and grounds, etc., but it's the electronic components that I need help on.

    I was wondering about this about possibly building a DJ mixer, but that's aside from the point of this post.

    Does the op-amp (OPA132) in the circuit simply amplify the audio signal? It doesn't split the audio signal frequency-wise in any way?

    That being said and if so, does it hold true that if I were to split any audio signal (regardless of intensity) across three resistors of varying resistance, would I generate three differing "bands" of signal across each? (of course, being very weak without amplification?)

    The resistors depicted are of 1.8K, 3.6K, and 11K types. Does each resistor serve to "filter" out the other two bands, allowing for the modification of the one band being selected?
     
  2. Aug 6, 2012 #2
    Harald Kapp

    Harald Kapp Moderator Moderator

    11,163
    2,550
    Nov 17, 2011
    IC1 has R1=R2=100kOhm therefore it operates only as an inverting unity gain amplifier (meaning voltage at IC1, pin 6 has the same magnitude but inverse sign of the voltaeg at In).
    IC2 is also an inverting amplifier. The fedback loop of IC2 contains three different filters. So you can adjust the level of feedback for each of the three frequency bands separately. So the gain of IC2 becomes frequency dependent. The "splitting" of the audio is therefore done within the feedback loop. The filtering is not done by the resistors alone but by the combination of resistors and capacitors. A resistor is (as far as this circuit is concerned) a fairly freqeuncy independent component.
    If you look closely you'll notice the different combination of resistors and capacitors for each band (not only the values but also the way the components are interconnected).

    Here is a tutorial on RC filters to gain some understanding.
     
  3. Aug 6, 2012 #3
    KJ6EAD

    KJ6EAD

    1,114
    159
    Aug 13, 2011
  4. Aug 7, 2012 #4
    (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,481
    2,830
    Jan 21, 2010
    Perhaps you should read the article linked to by KJ6EAD and ask us for clarification about aspects you don't understand.

    If you understand nothing, then we've got more work to do, but at least we'll know what you don't know (or some of it)
     
  5. Aug 9, 2012 #5
    Solidus

    Solidus

    349
    4
    Jun 19, 2011

    I've taken a look at the article (I was actually reading it as I was posting it), but what is it about the circuit itself and its feedback loop that actually carries out the filtering?

    (I'm reading the article Harald posted in the meantime so upon the next series of posts I should have some idea)
     
  6. Aug 9, 2012 #6
    Solidus

    Solidus

    349
    4
    Jun 19, 2011
    Which one is the feedback loop? Is that the one connected to line 2 of IC2?
     
  7. Aug 9, 2012 #7
    Harald Kapp

    Harald Kapp Moderator Moderator

    11,163
    2,550
    Nov 17, 2011
    The feedback of an OpAmp is generally used to set the gain of the amplifier.

    If the feedback is frequency dependend, the gain also becomes frequency dependend.
    By putting three different filters in the feedback loop (low-pass, band-pass and high-pass) the frequency response becomes adjustable in these three bands.
     
  8. Aug 9, 2012 #8
    Harald Kapp

    Harald Kapp Moderator Moderator

    11,163
    2,550
    Nov 17, 2011
  9. Aug 9, 2012 #9
    Solidus

    Solidus

    349
    4
    Jun 19, 2011
    Okay. So the feedback loop is connected through the op amp and the variable pots?
     
  10. Aug 9, 2012 #10
    Harald Kapp

    Harald Kapp Moderator Moderator

    11,163
    2,550
    Nov 17, 2011
    It is
     
  11. Aug 9, 2012 #11
    Solidus

    Solidus

    349
    4
    Jun 19, 2011
    Okay. So let me know if I have this right.

    Applying what I read in that article to this circuit,

    The output of the amplifier is dependent upon the difference of the two inputs, and so because of the bypass (correct me if that's the wrong term) loop at IC1, depending on the voltage through, the amplification at that line is variable.

    If the line signal is stronger, the ratio is less, and so amplification is reduced. It is a self-limiting amplification circuit.

    Am I on the right track?
     
  12. Aug 9, 2012 #12
    Harald Kapp

    Harald Kapp Moderator Moderator

    11,163
    2,550
    Nov 17, 2011
    Sorry, no, only partly.

    Think of the OpAmp as an ideal amplifier with unlimited gain. The OpAmp will try to make the difference between the "+" input (V+) and the "-" input (V-) zero.
    The OpAmp will try to adjust the output voltage such that it counters the input voltage of the complete circuit such that the difference V+ - V- = 0 V. The feedback circuit determines which output voltage is needed to achieve this by using a part of the output voltage to counter the input voltage. What part is used is determined by the configuration of the feedback circuit.

    Do read the tutorial, it can't be explained in a few words.
     
  13. Aug 12, 2012 #13
    Solidus

    Solidus

    349
    4
    Jun 19, 2011
    Oh, that makes sense.

    I'm reading the article, it's a bit hard to decode for someone like me who's new to electronics, but I'm piecing it together. I will post again up here later today with questions and clarifications.

    On a side note, for a project today, how would I cut 20V down to about 12-14V while preserving (or increasing) amperage? I know it's off-topic in regards to this, but it's a necessary provision for a project I am going to be attempting today.
     
  14. Aug 12, 2012 #14
    Harald Kapp

    Harald Kapp Moderator Moderator

    11,163
    2,550
    Nov 17, 2011
    Is it AC, then use a transformer.
    Is it DC, then use a step-down-switching converter. But that is not a good beginner's project... If you go for this option, try to buy a suitable kit.
     
  15. Aug 12, 2012 #15
    Solidus

    Solidus

    349
    4
    Jun 19, 2011
    It's DC current...and I can see by your link that I don't feel like going through the calculations. I can do them, but I don't want to (haha)
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Similar Threads

  1. How does a multimeter's continuity function work?

  2. Function Generator Circuit

  3. Circuit function help

  4. Function Generator Circuit.

  5. How can I get this switch to function

  6. N00B who needs help....

  7. Function of an electronic circuit

  8. Help design circuit from transfer function

Loading...
Electronics Point Logo
Continue to site
Quote of the day

“”

-