Class B push-pull amplifier.

Discussion in 'General Electronics Chat' started by Vorador, May 30, 2013.

  1. Vorador

    Thread Starter Member

    Oct 5, 2012
    87
    1
    Hi,

    I'm having some difficulty understanding the path the AC input current would follow during each half-cycle in a class B push-pull amplifier (involving one NPN and one PNP BJT).

    I'd really appreciate it if anyone would be so kind to draw current directions during each half-cycle through each path and each BJT of the simplest of class B push-pull amplifier.

    Thank you for your time and help!
     
  2. BreadCrum6

    New Member

    Aug 17, 2011
    18
    3
    Visit the link below, it shows the current path (yellow dots) for a Class B amp. I'd attempt to explain it but I think I'd make it even more confusing.

    __http://www.falstad.com/circuit/e-pushpullxover.html
     
    Last edited: May 30, 2013
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  3. Dodgydave

    Distinguished Member

    Jun 22, 2012
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  4. Vorador

    Thread Starter Member

    Oct 5, 2012
    87
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    Java SE 7 U doesn't work for me so I couldn't see your link, BreadCrum6. Thanks a lot, though!

    Dodgydave, I found that link very helpful. Thank you very much!

    Edit: I got your link working BreadCrum6 and wow, it's a fantastic visual aid! I never visualized it like this. Thank you so much!

    I also checked the improved push-pull amplifier simulation in BreadCrum6's link which has two diodes connected to each BJT's base. I don't understand how the input current is able to follow in a direction opposite to the diode and into base of the top BJT, since diodes allow current in one direction only..
     
    Last edited: May 31, 2013
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Input current don't need to flow through the top diode. The base current is provide by 220 resistor. The input signal "modulates" this current in "rhythm" of ac input voltage.

    See this example of a common emitter amplifier.

    [​IMG]

    At first we bias the transistor in linear active region.
    Next we connect the input signal 2V peak to peak AC signal.
    The DC voltage at base is equal Vb=2.6V, and emitter dc-voltage is 2V.
    So if we apply the ac signal to the base. The dc-voltage will be change from 2.6V + 1V = 3.6V and 2.6V - 1V = 1.6V. So the base voltage will change from 3.6V to 1.6V in "rhythm" of a ac input signal.
    These changes will result that the emitter voltage will also change. From 3V to 1V. This will result the change in emitter and in collector current by (3V) 0.9mA to 0.3mA (1V). And this change in collector current will cause change in VRc voltage, between 9V to 3V. We have a three times larger change in VRc voltage because Rc is three times larger then Re resistor.
     
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  6. Vorador

    Thread Starter Member

    Oct 5, 2012
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    Wow. Fantastic explanation, Jony130! I feel like I'm finally starting to understand transistors! :D

    You've made it very clear what's happening in the circuit example you gave, But in the "improved" push-pull amplifier, shouldn't the base voltage always be fixed at 0.7 V because the diode voltage usually doesn't vary much from this point?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No.
    Without input signal voltage at T1 base is equal to 0.6V and - 0.6V at T2 base.

    [​IMG]

    But now we connect +2V at the input terminals. D1 voltage drop will shift this voltage up. So T1 base voltage is now equal to 2.6V. And T2 base voltage is equal +1.4V.

    [​IMG]

    The voltage at emitter is equal to 2V. So now R1 provides 7.4mA of current. The 0.2mA is the base current, so the rest of a current (7.2mA) flows through D1 diode.
    The voltage drop across R2 resistor is equal to
    VR2 = (1.4V - (-10V)) = 11.4V. So R2 current must be equal to:
    IR2 = 11.4V/1K = 11.4mA
    But R1 resistor provides only 7.2mA. This additional current (4.2mA) need for R2 resistor will be provide by the Vin input source.

    As you can see Vin voltage source don't provide any current for T1.
    Vin is only loaded by D2 and R2.
    Also my diagram don't show "all secrets" hidden in this "improved" push-pull amplifier.
     
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  8. Vorador

    Thread Starter Member

    Oct 5, 2012
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    Jony130, you are absolutely amazing! You've no idea how greatly I appreciate you taking out your time to help me here with all these diagrams and everything.

    Thanks a heaps! :)
     
  9. screen1988

    Member

    Mar 7, 2013
    310
    3
    Hi Jony,
    How can you know T2 is OFF?
    I can't explain it. Veb = 0.7V and Vec = 12V and therefore I think it will conduct.
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I'm glad to hear it

    But in this circuit Vbe = -0.6V. So knowing that Vbe is equal to -0.6V.
    Can you tell me in which state BJT is?

    In reality the answer is not so simple. And this is why I assume that T2 is off.
    When I started the analysis, I simply treat BJT as a base current controlled collector current source.
    Ic = β * Ib with constant Vbe.

    [​IMG]

    [​IMG]

    This current control model woks good most of the time. But not this time.
    To full understand how this "improved" push-pull amplifier work we are force to use a voltage control mode. In fact the BJT is a Vbe controlled current source.
    According to Shockley's equation collector current is externally controlled by the Vbe voltage Ic = Is*(e^(Vbe/Vt) - 1).

    Without the input signal D1 and D2 provide bias voltage for T1 and T2.

    VD1 + VD2 = Vbe1 +Vbe2

    This bias voltage cause that a small idle bias current flows from the top NPN transistor through the bottom PNP transistor.
    If all device have a symmetrical and identical characteristic. The idle bias current will we equal to
    Id1 = Id2 = Ic1 = Ic2 = (10V - 0.6V)/1K = 9.4mA
    But as input signal increase the top NPN transistor current also increase.
    And Vbe1 also must increase. If so the Vbe2 voltage must decrease as much as Vbe1 increase. So the bottom PNP transistor current decrease. Because we have a kirchhoff's loop here.

    [​IMG]

    Vd = Vbe1 + Vbe2
    and if Vd is constant. Vbe1 rise reduce Vbe2 voltage.
    And we usually assume that T2 is in cut-off region (on positive half-cycles of the signal,) if Iload > 2 * idle bias current. In our case 2*9.4mA = 18.8mA.
    VLoad = 18.8mA*100Ω = 1.88V
    So for Vin > 1.9V we can treat this circuit as a ordinary CC (emitter follower) amplifier.
     
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  11. screen1988

    Member

    Mar 7, 2013
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    Brilliant!
    I enjoy reading your posts and your passion.
    Can you tell me about Ic1 = Ic2 = (10V - 0.6V)/1K = 9.4mA? I can see that Id1 = Id2 = 9.4mA not Ic1, Ic2.
    I can't figure out it. I have reread your post #7 and see that in case of Vin = 2, Vbe1 is still equal to 0.6V but Ic now is 20mA (increasing). There must be something I missed here.
     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you know how current mirrors work?

    The picture in post 7 don't include Shockley's equation. To understand how this improved" push-pull amplifier work when both transistor conduct small idle bias current. You need to think of it a as collector current is externally controlled by the Vbe voltage. So to increase collector current we need to increase Vbe voltage.
     
    Last edited: Jun 3, 2013
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  13. screen1988

    Member

    Mar 7, 2013
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    Yes.
    I know Ic1 = Ic2 but don't know why Id1 = Ic1.
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    D1 and T1 act just like a current mirror.
    So if we assume Vd1 vs Id1 characteristic is exactly the same as Vbe1 Vs Ic then:
    Id1 = Ic1 (without any input voltage).
     
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