Class AB power amp. calculations

Discussion in 'Homework Help' started by kilwan, Jun 8, 2013.

  1. kilwan

    Thread Starter New Member

    Jun 8, 2013
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    So I need to calculate quiescent points of both transistors, current gain, voltage gain, output resistance input resistance and the efficiency of a diode biased class AB power amp.
    Now ... if you ignore Ib1 and Ib2 the diode currents should be (Vcc-0,6)/R3 (2mA in my case).

    UceQ1=UceQ2=Ucc

    IcQ1 and IcQ2 I tried calculating like this:

    Ud1+Ud2=Ube1+Ueb2
    Utln(Id1/Is1)+Utln(Id2/Is2)=Utln(Ic1/Is3)+Utln(Ic2/Is4)
    'cos of Id1=Id2=Id, Is1=Is2=IsD and Ic1=Ic2=IcQ, Is3=Is4=IsT (if we ignore Ib1, Ib2)
    IcQ^2=(IsT^2/IsD^2)*Id^2
    IcQ=(IsT/IsD)*Id

    I looked up the datasheets for 1N4148 and the transistors and it doesn't add up ... I would like some input from you guys.

    Plus, any idea how to calculate the efficiency? I'm familiar with the efficiency calculation of a class B amp. and how to get those 78,5% but how do you calculate in the dissipation in a class AB amp?
     
    Last edited: Jun 8, 2013
  2. tubeguy

    Well-Known Member

    Nov 3, 2012
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    If this is a homework project, were you given this circuit to analyze ?
    It's not workable in the real world because the transistors shown can't reliably supply enough current to drive an 8 ohm load.

    That being said, if you assume a perfect 0.7 volt drop for each diode and a perfect 0.7 volt drop for each base-emitter junction then what would be the idle current ?

    What would be the impedance of the 100uf capacitor at 1khz ?
     
    Last edited: Jun 8, 2013
  3. kilwan

    Thread Starter New Member

    Jun 8, 2013
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    0
    Actually, my homework project is to first theoretically analyze any form of a class Ab amp.(quiescent points of both transistors, current gain, ...), so I chose the diode biased one for simplicity sake. After I'm done with that I need to chose a BJT and verify all of the calculations in a simulation.

    For the second question, I think I see what you mean. But, if the diodes where totally identical to the transistors (at least the junction areas), then Id1=Id2=Ic1=Ic2. I tried to simulate that using transistors in a diode connection and got sim1.

    I removed the capacitor, if I got it right it's resistance is way to high compared to the load.

    Btw, I looked up some other ways to provide current and I seems to me a pnp current mirror would be ok. The only problem is the load current should be Io=(beta/(2+beta))*Iref. Only for some reason I can't get that(Sim2). I would appreciate some hints.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In real word we don't use this type of a biasing network with discrete transistor.
    We add emitter resistor and use a Vbe multiplier Q1.
    [​IMG]

    And we set small idle bias current on the bench.

    As you already see we are unable to solve for quiescent points because we don't know Is current. Also don't forget about Early effect.

    As for your current mirror circuit try to add two 100Ω resisters in BJT's emitter.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Don't yuu have a prblem if I_VAS+ and I_VAS- are not sufficiently well matched? Or do you typically use sources that have sufficiently low ouput impedance such that the two sources will compromise readily? The only time I ever actually built a circuit using a Vbe multiplier I just used resistors for the two current sources. This was many moon ago, but IIRC it had the effect of moving the DC level of the output around due to the mismatch, but since this stage was AC coupled to the next, it didn't matter and was small enough not to cause other problems.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I only use one current source I_VAS+ or I_VAS-. And instead of the second current source I use CE amplifier.

    But almost 95% of a discrete audio power amplifiers use Vbe multiplier thermally coupled to the output transistors to avoid thermal runaway and set idle bias current.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    I don't see the relevance of the second part in response to me using resistors for the current sources. The Vbe multiplier can still be thermally coupled to the output transistors. How does using resistors in place of the two current sources change that?
     
  8. kilwan

    Thread Starter New Member

    Jun 8, 2013
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    So I did more research, found out that you can't really use standard analysis on these kind of circuits. You basically need to use graphical analysis, the DC sweep function in spice, draw load lines and stuff... I did all of those and my calculations match the multisim simulation. I showed it to my prof. and he said that it was O.K. Now the last step I needed to do was prototyping the circuit. I made the circuit on a breadboard and went to the college lab.

    So here is the circuit:
    http://postimg.org/image/jbfq1jogf/

    And here is the breadboard:
    http://postimg.org/image/t9rmvc7tz/
    http://postimg.org/image/6vfgpbg8n/

    I should have gotten something like Vce1=15V and Vce2=15V
    I got Vce1= 9V and can't really remember Vce2.
    I used two DC sources that have (+),(GND) and (-) ports. For the input I used a function generator with (+) and (-) ports. Now I didn't even get to adding the function generator since my DC measurements weren't correct, so there really was no point in trying to do AC measurements. I hoped you guys had some tips on what to do, since I at a total loss here. I checked all of the connection, all of the resistors can handle the power .... Maybe I created a short somewhere ?

    Would really appreciate some input
     
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