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Design an audio power amplifier

  1. Nov 6, 2015 #1

    PhysicoRaj

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    Hello,
    I was dissecting an old TV when I came across a set of speakers. They're 8 ohms, though I'm not sure if they're 8w or 6w..

    I decided to make my own portable speakers (mono), which could be connected by means of an aux cable to my smartphone.

    I don't know what's the output voltage of my smartphone headphone jack.. So please shed some light on that..

    Basically I need to design a 6W power amplifier.. I have 9v batteries (Vcc).

    I did some basic calculations and found out the quiescent current can fry my bias network if I used a Class A.
    So I thought of a class B, and since I don't want Trafos, I am determined on a complementary pair push pull.

    What power transistor would suit my application the best? How do I design the circuit.. Can the push pull provide both voltage and current gain? Do I need any other stage before this push pull?
    Is cross-over distortion a major problem?

    Thanks for your time and help..
    Regards
     
  2. jcsd
  3. Nov 6, 2015 #2

    Svein

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    Last edited by a moderator: May 7, 2017
  4. Nov 6, 2015 #3

    PhysicoRaj

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    Thanks a lot.. They can make my work a lot compact. I may use this..
    But if the design using push pull isn't too complicated, I would like to try my luck on making my own, or even, out of curiosity to know how to design such circuits.
    So any info on the design is also welcome..
    Thanks again.
     
  5. Nov 6, 2015 #4

    Svein

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    Applause!

    I can direct you to some pages, but please start with something simple and low power. When you have done that, I can give you some links to higher power amps.
     
  6. Nov 6, 2015 #5

    PhysicoRaj

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    Thanks :) I noted down some circuits. As I observed it seems that almost all the circuits had a preamplifier before the complementary pair.. Is that necessary? Because the output of a jack is low, I may be needing some gain before the power amp.. With this cleared I can get started.
     
  7. Nov 7, 2015 #6

    PhysicoRaj

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    What's the use of Q2 in the attachment?
     

    Attached Files:

  8. Nov 7, 2015 #7

    Svein

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    Q2 is the driver for the output stage.
     
  9. Nov 7, 2015 #8

    PhysicoRaj

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    Ok Thank you.. I decided to drive a push pull using OpAmp as preamplifier (I think it does the role of Q1 and Q2)..and raise the Q point of the push pull using a preset.

    Thanks for the help!
     
  10. Nov 8, 2015 #9

    Svein

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  11. Nov 8, 2015 #10

    PhysicoRaj

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    Hmm I'll look into it.. Ty
     
  12. Nov 9, 2015 #11

    PhysicoRaj

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    Opamp supplies have always confused me. The LM383 needs a +Vcc and ground in place of the -Vee as in the case of LM741.
    When I'm dealing with batteries, how do I connect the LM383? Can I give the battery negative to ground?
    http://www.reconnsworld.com/audio_8wattamp.html
     
  13. Nov 9, 2015 #12

    davenn

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    yes

    many op-amps work using a split rail +V, 0V, -V

    in the case of a single rail supply ( + and -) the - equals GND /negative of battery or other PSU
     
  14. Nov 9, 2015 #13

    PhysicoRaj

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    OK thanks.. Because in my lab we worked with the op amps in +V 0 - V and the instructor used to warn us not to interchange the 0 and - V.. So I was doubtful about the single rail system.
    Is a battery different from a dual power supply? Ex, what would be the -ve of a 12v battery at, 0 or - 12v? Because testing out dual supplies, my multimeter reads 24v between the terminals. That means a battery should be of +12,0 form.. Am I right?
     
  15. Nov 9, 2015 #14

    davenn

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    Many of the dual rail op-amps will happily work using a single rail supply
    others are specifically designed to use with a dual rail ( split supply) only ... .the datasheet for the particular op-amp gives this info

    yes ... it is a single rail supply

    the negative terminal is 0V relative to the positive terminal

    that's correct and you have an associated 0V terminal. so with your multimeter negative ( black) lead on the 0V term. and the red lead on the +12V term. you will see +12V on the meter
    and with your multimeter negative ( black) lead on the 0V term. and the red lead on the -12V term. you will see -12V on the meter

    you can make a dual rail supply from a single or 2 batteries like this .....

    single battery or other single rail supply

    Simple split frm single rail.gif

    2 batteries .....

    dualsypply1a.GIF

    Note that in the first one, two resistors are being used to produce a voltage divider
    this is for small currents only as the resistors produce current limiting
    so if you had, say, a 12V battery, you could produce a +6V, 0V, -6V supply

    the second dual rail supply uses 2 batteries with the addition of voltage regulator IC's to produce a stable output


    cheers
    Dave
     
  16. Nov 9, 2015 #15

    PhysicoRaj

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    Ah.. That's enough explanatory :)
    And one last thing.. How do I know the output of my smartphone headphone connector? DC level, peak AC, etc? Because most LM383 circuits set a gain of 100, and I want to be able to vary that gain to drive the 8 ohms speaker at 6-8 watts.

    8 Ohm speaker running at 6W means approx 10V peak output voltage. So I need to know the input so that I can play with the gain.
     
  17. Nov 10, 2015 #16

    Svein

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    Put a potentiometer in front of the input.
     
  18. Nov 10, 2015 #17

    davenn

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    yes :smile:

    just like this, using the circuit you linked to....

    LM383 Amp.JPG



    Dave
     
  19. Nov 10, 2015 #18

    PhysicoRaj

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    But wouldn't that be just like attenuating the input and having same gain.. What if I needed more than the full input voltage, I need more gain right? So I thought knowing the output of the jack would be better.

    I searched everywhere but in vain.. I can't find the 3.5mm jack specs.

    Thanks..
     
  20. Nov 10, 2015 #19

    davenn

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    Yes.
    That is the way it is done on pretty much every amplifier ever built
    so if the amp has a gain of 100, you can work out the output power relative to the input power

    you ONLY get max output from the LM383 ( or whatever other amp) WHEN its input is driven fully

    you cannot overdrive the input to the amp it will just produce an unlistenable distorted output and possibly damage the amp chip
    the full output of the headphone jack of your phone/media player could easily have enough output to overdrive the amp chip
    so that is why the volume control ( attenuator) goes on the input to the amp
     
  21. Nov 10, 2015 #20

    davenn

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    what jack specs ?

    the max output from your smartphone ?
    that will be in the phone spec sheet
     
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