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Class A-B amplifier simulation help

Discussion in 'Circuit Simulation & PCB Design' started by neptune, Oct 16, 2011.

  1. neptune

    neptune Member

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    Hello,
    Here i am posting my simulation of class A-B amplifier, it is not amplifying
    please see
     

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  2. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You have a ground connection on both sides of V3 which shorts it out and likely causes the simulation to crash since the calculated current becomes V3/0.
     
  3. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    The other problems are the settings, look at these images, tell me what you think is wrong.?
     

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  4. dave

    Dave New Member

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  5. neptune

    neptune Member

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    Crutschow - i removed the ground but it still is the same.

    Eric - trans setting doesnt matter in this case ,when circuit is not working
     
  6. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    I reduced your asc file to only a new Vsrc and single resistor and it still crashes, the only conclusion is that the 'base' asc file you posted is either corrupted or its been edited in some way.??


    With LTSpice the Settings ALWAYS matter.
     
    Last edited: Oct 16, 2011
  7. SgtWook

    SgtWook New Member

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    My LTSpice crashed trying to simulate it as well.

    I took a quick look in the .asc file, but didn't see anything unusual. (sometimes there are artifacts if someone has been running LTSpice in wine on a Mac, or Ubuntu or other Unix OS)

    I reconstructed the schematic from scratch, and now it simulates.

    Give it a try.
     

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  8. neptune

    neptune Member

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    Hello,
    My simulation also crashes :? The only way of simulating is open LTspice and from inside select > open file and then select file.
    why is it crashing i'm using windows 7 , should i remake it new ?

    Hey SgT Wook your simulation is working but not amplifying.
     
  9. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    Why would you expect the transistor emitter followers to have a voltage gain, ie: to voltage amplify the input signal.??

    Have you covered transistor theory in your class work.??


    Remake with a NEW schematic, the original circuit and post, so that we can re-run it for you.


    EDIT:
    This slightly modified version of your circuit should give you a clue as to the purpose of this transistor configuration.
     

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    Last edited: Oct 17, 2011
  10. crutschow

    crutschow Well-Known Member Most Helpful Member

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    There likely is something corrupted in your original file. Best way to solve that is to completely redo the schematic from scratch as a new file as SgtWook did.

    The circuit does amplify, it just doesn't amplify voltage (see Eric's modified circuit).
     
  11. neptune

    neptune Member

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    I made this again,
    this circuit it has current gain of 1696
    I never said it will give voltage gain.
    How can we reduce cross over distortion ?
     

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    Last edited: Oct 18, 2011
  12. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    This file is CRASHING my LTS again!

    EDIT:
    Using my earlier asc file, modified the biasing
     

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    Last edited: Oct 18, 2011
  13. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Your extremely simple amplifier has no voltage gain so it has no negative feedback. All half-decent audio amplifiers and opamps have a lot of voltage gain then use a lot of negative feedback to reduce the gain which reduces the distortion to nearly nothing.
     
  14. neptune

    neptune Member

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    that distortion is not noise its cross over distortion, completely different
     
  15. neptune

    neptune Member

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    thankyou eric it is working flawlessly.
    one question - I removed Diodes here and increased the value of resistor from .22 to 1k ohm and it is working fine.
    i saw this circuit on a site and there .22 ohm resistor were not present instead only diodes were given there, i built it and as you can see failed.
    theoritically it was at correct biasing with diodes
     
  16. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You previously said it had no gain. I assumed you meant voltage gain.
     
  17. unclejed613

    unclejed613 Well-Known Member

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    i tested your latest post of the circuit... try it with three or four diodes, that will reduce the crossover distortion because the output transistors are partially biased on. the crossover distortion was because the transistors were biased off during part of the voltage swing near zero. i added a third diode and eliminated the crossover distortion. in commercial amplifiers, the diode stack is replaced by a transistor that acts as a variable "zener". the other problem with the circuit was that you had the diodes shorted by the input wire. see my modified circuit. i added R3 and R7, and changed the values of R4 and R5 to 4.7k

    please take note that when you posted your circuit you had a wire from the voltage source to the bases of the transistors, effectively taking the diodes out of circuit.
     

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    Last edited: Nov 12, 2011
  18. audioguru

    audioguru Well-Known Member Most Helpful Member

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    But the 10 ohm emitter resistors use up 67% of the amplifier's power. They should be 0.1 or 0.33 ohms, then you won't require a 3rd or 4th diode for biasing the transistors.
     
  19. neptune

    neptune Member

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    how did you calculated power of emitter resistance.
     
  20. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Assume the RMS output swing at the emitters is 100V.
    When the output goes up (+50V) then one emitter resistor uses half the voltage (25V).
    When the output goes down (-50V) then the other emitter resistor uses half the voltage (-25V).
    Then the 10 ohms load gets a continuous swing of half the voltage (50V) which is (50V squared/10=) 250W.

    Each emitter resistor has 25V but for only half the total time so each one dissipates half of (25V squared/10=) 62.5W which is 31.25W.

    The total power from the amplifier is 250W + 31.25W + 31.25W= 312.5W.
    Then the emitter resistors use 62.5/312.5= 20% of the total power from the amplifier and I was wrong saying it was higher.
     
  21. unclejed613

    unclejed613 Well-Known Member

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    to illustrate what AudioGuru is talking about, look at the figures below. in a "real-world" application of the circuit you are testing, we will use a similar emitter follower in an amplifier. you will see two traces, one with 10 ohm emitter resistors (R15, R16) you will notice that the output clips at +/-4V. this is because the amp itself is swinging almost rail to rail, but your 10 ohm emitter resistors are acting as a voltage divider with the load, reducing the output voltage. in the second trace, the emitter resistors are 0.22 ohms. the full output swing (almost all that is...) is now across the load, and not being wasted in the output circuit. the input signal level and the overall gain of the amp remained the same. also notice that there is a driver stage, and a bias transistor. the bias transistor Q4 performs the same function as the diodes in your circuit, with R9 and R10 being used to set the conduction voltage of Q4. the outputs are actually wired as darlingtons, so Q5+Q7 act as one composite transistor, and Q6+Q8 act as a composite transistor.
     

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