Class B Push Pull Amplifier. Need guidance and confirmation.

Discussion in 'Homework Help' started by maam_13, Nov 29, 2009.

  1. maam_13

    Thread Starter New Member

    Mar 23, 2008
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    Hi!

    Im still newbie to this analog circuit design, so i have something to ask about the circuit in the attachment below. The circuit is Push Pull Class B Amplifier with AB Characteristic.

    My questions are:

    1)The purpose of mirror currents is to provide a current into to the collector of T5,right? For what purpose is it actually and how could i calculate the value of the current at the collector of T5? T7 and T8 are actually diode-connected Transistors, which have a role to eliminate the crossover distortion,right?

    2)If the crossover distortion can be eliminated with the T7 & T8, what is the purpose of the negative feedback connection from Vout to the OP, if the switch in opposite direction from the one in the circuit. ( I assume that the feedback can eliminate the crossover distortion too)

    3) How could i calculate the gain of the OP exactly?

    4) Both Re1 and Re2 could reduce the power dissipation at the T1 and T2 right?

    Thanks!!

    Sorry for many questions, i have read a a lot and need some confirmation and guidance about this circuit.
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Hello maam

    It is normal to post your attempts if these are homework questions so that you can learn by discussing them.

    Since this is a complicated circuit I will offer some hints to start with.

    The voltage gain comes from the op amp. The transistors form a voltage follower with a high current gain but a voltage gain of slightly less than 1. T5/T1 and T6/T2 form what are termed Sziklai pairs. This is a compound emitter follower formed from complimentary pairs of complimentary pairs. The rest is just fancy biasing.

    You cannot affect crossover distortion by feedback. This is because crossover distortion occurs at or near the zero crossing point of the waveform, where there is by definition no signal to feedback!

    Yes the emitter resistors reduce the available output, but also apply some local negative feedback.

    The diode connected transistors biasing the output pairs are not enough to offset the base emitter voltage of the output pairs plus the emitter resistor voltages completely so they reduce, but do not eliminate crossover.

    http://en.wikipedia.org/wiki/Sziklai_pair
     
    Last edited: Nov 29, 2009
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    What?? Sziklai pair? Where?
    In this schematic there is no Sziklai pair, only common emitter follower (T1 and T2)
    And yes, you can reduce crossover by applying negative feedback. Crossover region is reduces from about 1.4V to 1.4V/Open loop gain
     
  4. maam_13

    Thread Starter New Member

    Mar 23, 2008
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    Thanks for the reply both of you. =)

    Can I assume that the 1.4V is the dead band at Vin-Vout graph? Okay, so when it reduces from about 1.4V to let say 0V, thats mean, the nonlinearity of graph Vin-Vout will be disappear to, am i right? And what do you mean by open loop gain? how could i calculate it? I am not clear.

    Tq all!
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well yes, something like this.

    Open loop gain is a gain of a amplifier without negative feedback.

    And remember that T3 and T5 is a simply current mirrors which provide a current for output emitter follower and current for bias transistors diodes.
    And to determine this current note that T3 is also diode connection BJT.
    So applying KVL you can easily calculate IcQ5
    Vcc≈Ic*Re3+Vbe+Ic*Rref1
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    The crossover distortion will be low and can be eliminated by increasing the current in the diode-connected transistors.

    The current-source transistors have a high output impedance that is not needed since the opamp has such a low output impedance. Therefore many parts can be eliminated.
     
  7. maam_13

    Thread Starter New Member

    Mar 23, 2008
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    sorry again, i have another question.

    The circuit has 2 supply voltages, that are Ucc1 und Ucc2, from what i've read, normally there is only 1 supply voltage, but here, we need 2 supply voltages to increase the efficiency right? How is it can happen? Can anybody explain to me, i cant find the answer after i've searched in book and internet.
     
  8. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Firstly my apologies,I was too hasty looking at your circuit.

    Of course the mirrors form active loads for the op amp and supplies for the output pair. There are no compund pairs.

    However I disagree with guru in this respect. The OP42 requires a relatively high value load (≈ 10k from tha data sheet) so they are not unecessary, although there are other ways to achieve this.

    To answer your question about the power supplies.

    Firstly it is not unusual to have different supplies for the output stage and earlier stages, however normally the output has higher voltage supply which is dropped via resistors to feed the earlier stages at lower current.

    However in this case the differnetial is the other way round. Looking at the extract from that OP42 data sheet shows why. Then op amp max swing is about 26 volts at 30 volts total supply, thereby loosing about 4 volts. Doing things this way will allow enough voltage to fully drive the output stage.

    I do not agree that you can alter the crossover distortion of a pure class B amplifier one iota by feedback. The dead zone is just that. A zone of voltage near zero signal level where there is zero amplifier response.
    Zero response means, by definition, zero output. Zero output means by definition zero feedback.

    Guru is correct in stating that by suitable biasing (via the diode connected transistors) you can bring the two edges of the dead zone together to come very close to eliminating the crossover, but the amplifier is no longer in pure class B. It is also very difficult to achieve a good result in practice because even matched complementary pairs are not identical mirror analogs. The slopes of the transfer curves will not match at the very least. This last problem can be overcome by negative feedback. Overbiasing also brings problems as it can cause over response in the output.

    Finally I would guess that this circuit is a teaching circuit to allow you to test the effect of various changes and ideas so experiment away.
     
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