Class AB Audio Amplifier With Class A Driver - Questions

Discussion in 'Analog & Mixed-Signal Design' started by elec_dude, Mar 26, 2018.

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  1. elec_dude

    Thread Starter New Member

    Mar 31, 2017
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    I built this amplifier circuit and it works but I have some questions.
    Component values are as follows:
    R1 = 100 K ohm; R2 = 15K ohm; R3 = 1K ohm; R4 = 100 ohm
    C1 = 1 uF; C2 = 1000 uF
    D1 = D2 = 1N4148
    Q1 = 2N3904; Q2 = TIP31C; Q3 = TIP32C
    Supply Voltage = 12 V
    Load = 8 ohm speaker

    DC Readings:
    VCQ1 = 6.90V; VBQ1 = 1.10V; VEQ1 = 0.40V; VCEQ1 = 6.57V
    VCQ2 = 11.94V; VBQ2 = 8.11V; VEQ2 = 7.50V; VCEQ2 = 4.48V
    VCQ3 = 0.00V; VBQ3 = 6.90V; VEQ3 = 7.50V; VCEQ3 = 7.51V
    IcQ1 = 0.11mA; IcQ2 = IcQ3 = 6.75mA

    Questions:
    1) Should VEQ2 = VEQ3 not be 6.00 Volts?
    2) According to my calculations, I expected VCQ1 = 5.30V, VBQ1 = 1.60V
    and VEQ1 = 0.90V. I checked the resistor values and they are pretty
    close to what they should be. I must be missing something.

    Thanks in advance for any comments.
     
    Last edited: Mar 26, 2018
  2. MrChips

    Moderator

    Oct 2, 2009
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    The proof of the pudding is in a real test, not a simulation.

    Adjust R1 and/or R2 until you get even distortion on both positive and negative excursions at maximum output levels.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well in perfect amp yes. But keep in mind that you do not know the exact value of a beta and Vbe. Also resistors tolerance.
    Tto get Ve2 = Ve3 = 6V you need around 5.4V at Q1 collector (VBQ3).
    Have you include the base Q1 base current in your calculations?

    Ib1 ≈ (Vth - Vbe)/(Rth + (β+1)*R4)

    Vth = Vcc * R2/(R1 + R2) = 1.56V

    Rth = R1||R2 ≈ 13kΩ

    Ib1 ≈ (1.56 - 0.66V)/(13kΩ+ (120+1)*100Ω) ≈ 36μA

    and

    Vb1 = 36μA*(120+1)*100Ω + 0.66V = 1.1V

    And Ie1 ≈ (1.1V - 0.66V)/100Ω ≈ 4.4mA
     
  4. crutschow

    Expert

    Mar 14, 2008
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    What are the values of R5 and R6?
     
  5. elec_dude

    Thread Starter New Member

    Mar 31, 2017
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    Actually, I removed these from the real circuit temporarily. Sorry, forgot to mention that.
     
  6. crutschow

    Expert

    Mar 14, 2008
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    So were they out when you made the measurements?
     
  7. elec_dude

    Thread Starter New Member

    Mar 31, 2017
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    Yes, they were out when I made the measurements.
     
  8. elec_dude

    Thread Starter New Member

    Mar 31, 2017
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    If I disconnect the driver circuit from the rest of the circuit, the voltage
    on Q1's base is 1.60V as expected. However, when the rest of the circuit
    is connected, the voltage drops to 1.10V. Why would this base voltage
    drop when the voltage divider (R1 & R2) produces the correct voltage?
     
  9. crutschow

    Expert

    Mar 14, 2008
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    Here's an LTspice simulation of the bias currents and voltages for no R5 and R6 (set to 1 micro ohm).
    The collector current for Q1 and Q2 is very high in the simulation (.23A) because of no R5 and R6.

    upload_2018-3-26_11-33-14.png
     
  10. Audioguru

    Expert

    Dec 20, 2007
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    1,289
    Audio amplifiers almost always bias your Q1 transistor from the center point of the output transistors like I show R1 so that the DC negative feedback adjusts the center point voltage.
    You do not need R4 because then my new placement of R1 does the AC negative feedback that R4 did before along with the source impedance of the signal source.
    Lookup "bootstrapping" then your R3 will be two series resistors with a capacitor to the center point to their junction.
     
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  11. ebp

    Well-Known Member

    Feb 8, 2018
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    '"The collector current for Q1 and Q2 is very high in the simulation (.23A) because of no R5 and R6."

    To elaborate on crutchow's comment:
    Without R5 and R6 the no-signal current though the output transistors is set by the voltage across D1 and D2 in a logarithmic relationship. Neither Vbe nor Vf of the diodes is a well-known quantity and varies from unit to unit. They are are all close to each other because they are all silicon PN junctions, but they not perfect matches. If Vbe of the transistors happens to be on the low side and Vf of the diodes on the high side, the quiescent current through the transistors can be quite high. If the relationship is the opposite way, it could be quite low. R5 and R6 provide a form of negative feedback, sometimes called emitter degeneration, that helps overcome the mismatch. R4 serves the same function for Q3.

    If you have LTspice you could take crutchow's circuit and add a voltage source in series with D1 and D2. If you play with the value of that voltage with different values of R5 and R6 you will get a feel for the effect.
     
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  12. elec_dude

    Thread Starter New Member

    Mar 31, 2017
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    I downloaded LTSpice. There is no transistor symbol. No instructions?
     
  13. sghioto

    Well-Known Member

    Dec 31, 2017
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    This is the same thread you started back in January, are we still doing homework?;)
     
  14. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    @elec_dude ,Watch your language or you won't be long here.

    Bertus
     
  15. elec_dude

    Thread Starter New Member

    Mar 31, 2017
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    Don't you dare threaten me. You tell that creep to stop harassing me all the time.
    I am not a student and this is not homework and he has been told that before. His big mouth is going to get him into a whole lot of trouble. Kick him out and you owe me an apology!
     
  16. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Then You will be banned for one day in a moment to cool down.

    Bertus
     
  17. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    User permanently baneed for repeating profany.

    Bertus
     
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