POWER AMPLIFIER

Discussion in 'General Electronics Chat' started by tam72f1, Oct 9, 2014.

  1. tam72f1

    Thread Starter New Member

    Aug 19, 2014
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    Hi all!
    I find this design on the Internet and i want to make it.I don't understand mission of some components.
    Do transistor Q911 and Q913 bias??? and why?
    The missison of D13 D15 D17, why they are paralled with R927???
    Can anyone help me???
    My knowledge is very very basic to know all it!
     
  2. alfacliff

    Well-Known Member

    Dec 13, 2013
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    the series connected diodes are there for bias, foreward biased diodes have a fairly constant voltage drop. and a little variance for temprature.
     
  3. tam72f1

    Thread Starter New Member

    Aug 19, 2014
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    Are two transistor (Q911 and Q913) biased???
     
  4. #12

    Expert

    Nov 30, 2010
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    Yes.......
     
  5. tam72f1

    Thread Starter New Member

    Aug 19, 2014
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    Do you can calculate the current through resistor R929, R923, R927????
    I think that the current through R929 is (VD901+VD903)/150R and current through R923 is ( VD909+VD911)/150R.
    Can anyone help me calculate it?
     
  6. donpetru

    Active Member

    Nov 14, 2008
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    The current through the resistor R929 is calculated like this: I = 0.6 [V] / R929 = 0.6 [V] / 150 [Ohm] = 0.004 [A] = 4 [mA]. This current is the same as the current through the resistor R923 and ~R927.
    And the voltage on resistor R927 is approximately equal to: Vbias = 3 x 0.6 [V] = 1.8 [V] because 0.6 [V] is the diode voltage drop (0.6 ... 0.7V theoretically, but in practice we consider the minimum value, 0.6V respectively).
    Knowing the bias voltage, then you can calculate the current through the transistors Q911 and Q913 (in stationary operation mode). For example, collector current through transistor Q911 is calculated like this: IQ_911 = (0.5 * Vbias ) / R941 = 0.5 * 1.8 [V] / 470 [Ohm] = 0.0019A = 1.9 [mA] (same current flows through Q913).
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    donpetru
    Well I think that Q901 and R923, R921 D909, D911 form a constant current source. I = 0.6V/180ohm = 3.3mA;
    Also notice that 1.8V/390 1ohm = 4.6mA but Q9101 provides only 3.3mA and Q907 only 4mA. So without any signal the current will be equal to
    3.3mA.
     
  8. donpetru

    Active Member

    Nov 14, 2008
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    Jony130,
    I don't see 180 Ohm resistors ???? Later edit: I see now: R923.
    Restores calculations using 150 Ohm resistors and then you'll see that my calculations are quite accurate.

    LATER EDIT: In the diagram posted above will be R929 = R923, which initially is wrong. In fact, there are some mistakes in the schematic. In first post above is a purely theoretical schematic.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    What about R923 ?
     
  10. donpetru

    Active Member

    Nov 14, 2008
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    I edit my post above. Read that one again. Yes, R923 is 180 Ohm, but I think it's wrong.
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Ok, but even if we assume Ic= 4mA , then I still don't see how can voltage drop across 390 resistor be equal to 1.8V. And in real life audio amps we always use R929 < R923.
     
  12. donpetru

    Active Member

    Nov 14, 2008
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    R929 and R923 values depends on the structure of the amplifier input stage. If the input stage is a fully differential input stage (which isn't designed in the schematic from the first post), then R929 = R923.
    Voltage drop across R927 will be imposed on the three diodes: D913, D915 and D917, regardless of the current through R929 or R923. If you don't believe this, then I recommend you build circuit diagram from the first post above and then you will convince yourself about that.
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You, but you have to redesign the circuit.

    Well yes, but I'm sure that if Iq = 4mA then the voltage drop across R927 will be less than 1.8v (much closer to 1.5V).
    Also I don't understated how this "IQ_911 = (0.5 * Vbias ) / R941 = 0.5 * 1.8 [V] / 470 = 1.9mA" can be true? Why 0.5*1.8V ?
    If Vbias = 1.8V then from KVL we have Vbias = VbeQ911 + Ie*R941 + Ie*R943 + VebQ913
    V_R941 + R943 = Vbias - 2Vbe
     
  14. tam72f1

    Thread Starter New Member

    Aug 19, 2014
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    I'm sorry because i don't give you some valuse of D921 and D923. They are zener and they have a voltage is 10V for each!
    I calculate: Ie923=I913 915 917 +Ie929. and i also agree voltage drop across R927 is 1.8V. But in the schematic Zener is 10V
    I don't understand why they are design it!!!!
     
  15. donpetru

    Active Member

    Nov 14, 2008
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    ohshit72f1,
    Schematic from the first post above is somewhat one theoretical problems (from school)? I tend to think that, because it has a few mistakes, so there can not be a practical applicability of it. Or could implement practically, but a few things need to be corrected.
    Those Zeners+diodes D919, D925 may be missing, but they have more keeping role in SOA limits, the operating points of the transistors Q913 and Q911 when they are working in overload or, for various reasons, there is a short circuit on the output amplifier. Those Zener diode, not choose anyway, there are several criteria after which to choose. In BJT transistor audio schematics, there are other more effective solutions, but in audio amplifiers schematics with MOSFET (ie. Vertical MOSFETs), use those zener can be a good compromise.

    Jony130,
    Why 0.5*1.8V ? Because 0.5 mean only half of the bias voltage, since Q911 and Q913 will see half of the bias voltage, with resistors: R941 and R943. I recommend you build assembly and then you'll see this things (I mean values, after measurements).
     
    Last edited: Oct 11, 2014
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you know the Ohms law ? V = I*R ?.
    In circuit show in first post Icq = 4mA. This means that the voltage drop across 390Ω resistor cannot be larger then 4mA*390Ω = 1.56V.
    In this amp this voltage will be smaller then 1.56V. Why? Because not all Icq=4mA current will flow through the 390Ω resistor, the small part of this 4mA will also flow through the diodes.
    So how can you say that the voltage drop across 390Ω is 1.8V? Since Q909 don't provide the enough current to this to be true.

    So, you're trying to say that KVL don't "work" in this case? How sweet.

    Also I build this circuit

    4.PNG

    And the measurements are as fallow.
    VR5 = 0.57V
    VR6 = 1.45V
    VR7+VR8 = 0.348V
    And DC offset = 0.33V.

    So from this we can tell that
    Icq2 ≈ 3.8mA
    And Icq4 ≈ Icq5 ≈ 17.4mA ( using your equation is should be (0.5 * 1.8V)/20Ω = 45mA)

    And the AC voltage gain is equal to Av ≈ R2/R1 ≈ 20V/V

    NewFile0.PNG
    Ch1 = Vo
    Ch2 = Vin
     
  17. donpetru

    Active Member

    Nov 14, 2008
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    I have considered a theoretical 0.6V voltage drop across the diode. In reality, this value can vary greatly from one schematic to another. In fact, things depend a lot on the diodes used.
    For example, some manufacturers of diode guarantees a 0.6V drop only if a certain minimum current passing through the diode. If this current is not reached, the voltage drop across the diode can be less than 0.6V (which you basically determined).
    And if you do not have those at least 0.6V across the diode, the diode can not do their proper function in the schematic. For this reason it is very important to polarize diodes with their current minimum, which varies from one diode to another.
    Finally, one thing is certain: the proposed schematics in this topic are not optimized schematics, they are just a matter of theoretical analysis.
     
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