Need Help in Making a Class AB Amplifier

Discussion in 'General Electronics Chat' started by setsunaseiei, Mar 26, 2009.

  1. setsunaseiei

    Thread Starter Member

    Mar 25, 2009
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    I often find the same configuration for a class AB amplifier.
    I need to know how to design a class AB amplifier.
    What the use of the two identical resistances? How do I set the bias current? How to tweak to decrease distortions?
    And also, any useful configuration with a driver using a single supply?
    I need to find as new ideas as possible aside from those i often found in wiki or some basic tutorial. I need a more complex example, derivations and some tips.
    Advanced thankies.
     
  2. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Do you have a particular Class AB amplifier configuration in mind? Are you using BJTs or FETs? To bias, just put two resistors of the same value (say, R1=R2+2.2K) in series across the +Vcc and -Vcc supplies. If the transistors are BJT you need to bias their bases .6 volts higher than their emmiters. To do this put 2 general purpose diodes in series between the the resistors above. Connect the base of the NPN to the +.7 volt junction of R1 and D1 and do the same for the lower R2-D2 junction. Finally, put two 10 ohm resistors at each emitter, join their other ends together and put a load resistor of, say, 1k. That's the complete amplifier. You can vary the resistances, but that is the basic design.
     
  3. setsunaseiei

    Thread Starter Member

    Mar 25, 2009
    12
    0
    I am using 2n2222 and a 2n2907 for the amplifier. What effect does the two resistors do to my input or output impedances, on the gain, and lastly how the bias current is set and how it affects the power output of my amp? I am using a voltage driver at the lower half connected to the base of the 2n2907. Still need help biasing it for maximum swing and no distortions.
     
  4. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    setsun, I see I made a mistake in the above post: The correct way is R1=R2=the same value, 2.2k, for example. I wish I could draw, but I can't so I'll describe a class AB amp I made not long ago.

    Rather than using bias diodes I used 4 resisters in series. Consider the top of the drawing at +Vcc=9V and the bottom at -Vcc=-9V and the 4 resistors in series from top to bottom are R1=10k, R2=820ohm, R3=820, R4=10k. Notice the symmetry. The center node between the 820 ohm resistors is therefore ground. The voltage drop across R1 is Vr1=Vcc*R1/(R1+R2)=8.3 volts. Similarly Vr2=Vcc*R2/(R1+R2) = .7 volts. This way .7 volts is at the base of Q1, your 2n2222, and -.7 volts is at the base of Q2.

    Put a 10 ohm resistor at each emitter and put their other ends together. Here put 230 ohms to ground. This represents the input impedence of the power amp it is driving. Since we have .7 volts at the base of Q1 and .6 volts is dropped between there and the emitter, the emitter is at .1 V and there is an idling current of 10mA (from .1/10 where 10 is the emitter resistor.)

    The gain, Av is approximately Av. Vpeaktopeak at the input is 2Vcc=18Vp-p. But the max p-p output voltage is less due to voltage divider action between R1 and the resistance looking into Q1. This is given by (R1//BRL)/(R2+R1//BRL)=16Vp-p

    Vpk=16/2=8V and Ipk=8v/230ohms=35mA and IDC=35mA/pi=11mA
    Each transistor uses VCE*(IDC+Iidling)=9*(.011+.01)=189mW
    The supply delivers twice this amount=.38W
    The load burns Vp-p^2/8RL=.139W
    The efficiency is (Psupply/Pload) x 100 = 37%
     
  5. setsunaseiei

    Thread Starter Member

    Mar 25, 2009
    12
    0
    Thanks. How if I'm using a 1n4148 diode instead of an 820 ohm resistors across the two bases?
     
    Last edited: Mar 26, 2009
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Have you guys heard the expression "a picture is worth a thousand words"? It definitely applies here. One of you should post a schematic. I don't have the patience to read a text description.
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    You can make a simple amplifier with no negative feedback which has high distortion.
    You can improve it with negative feedback and bootstrapping.
     
  8. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Hello, Audio Guru and Ron. I wish I had the ability to post circuits like you. Like Ron said, a picture is worth a million words.

    Audioguru, your circuits include a driver with voltage gain. It's the gain of that driver that introduces the distortion in the first place. This is especially due to the high input voltage specified for the driver. I believe this is only possible because of the emitter resistor with no bypass capacitor.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    A class AB follower will have high distortion at crossover when loaded. A high gain stage with lots of negative feedback gets rid of most of the distortion.
     
  10. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    The distortion shown by Audiogurus's first circuit, without feedback, shows nonlinear distortion, not crossover distortion. They are different. Nonlinear distortion occurs due to the nonlinear relationship between Vbe and Ic. Typically one side of the sinewave -- top or bottom -- gets fatter than the other. If you increase the amplitude of the input, Vbe, the fat part gets fatter and fatter until it's unrecognizable as a sinewave.
     
  11. setsunaseiei

    Thread Starter Member

    Mar 25, 2009
    12
    0
    wow great, audioguru. that driver is what i am looking for. How do you adjust its gain to avoid clipping? how is it DC biased?
     
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