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Analysing Power Amplifier circuits-General discussion

Discussion in 'General Electronics Chat' started by fantabulous68, Feb 27, 2010.

  1. fantabulous68

    fantabulous68 Member

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    I removed R15 (suggested by unclejed613).....
    I see the spiral:rolleyes:, Fixed that now
     
  2. fantabulous68

    fantabulous68 Member

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    Thanks, did that now
     
  3. fantabulous68

    fantabulous68 Member

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    -50V Single negative supply- 8 ohm load- single transistor input

    Circuit with changes.
     

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  4. dave

    Dave New Member

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  5. mneary

    mneary New Member

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    I would have thought that there would be a resistor from node 17 to VCC (base to emitter of Q5). C6 is only half a pole without it ;-)
     
  6. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I mentioned it a few pages ago. It helps Q5 to turn off and is the required load for Q7.

    EDIT:
    Now it is like the amplifiers in my 1964 Scott FM stereo receiver (that still works perfectly).
    Geez. That was 46 years ago.
     
    Last edited: Mar 19, 2010
  7. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    That probnably had a negative supply rail as well! :D
     
  8. fantabulous68

    fantabulous68 Member

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    -50V Single negative supply- 8 ohm load- single transistor input

    Added it:D

    lol, now does it look more modernized:p?
     

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  9. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    You've got a connection from the emitter of Q7 to ground, shorting it out.
     
  10. unclejed613

    unclejed613 Well-Known Member

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    i was going to mention that.... with that short there, there is now no feedback. a 4.7k or 10k would suffice. you onle need one potentiometer in the bias circuit, as the resistor from base to emitter. the resistor from base to collector should be a fixed resistor,

    the next thing you might want to do is go through and "normalize" all of the resistor and cap values to match the EIA standard values. use the E24 chart for the resistors and ceramic caps, and E6 for electrolytic caps. the table shows the base value, so when you see a 680 in the chart, that represents 6.8, 68, 680 6800, etc. depending on the multiplier. the chart lists the values from 100 to 1k. so for instance, your 400k resistor in the input stage, would not be available. the closest value from the table is 390k, and that IS available. the chart i'm describing is found here:
    http://www.logwell.com/tech/components/resistor_values.html
    or here:
    http://www.pc-control.co.uk/resistor-eia.htm
     
    Last edited: Mar 19, 2010
  11. audioguru

    audioguru Well-Known Member Most Helpful Member

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    No.
    My 46 years old stereo has a single positive supply and uses coupling capacitors to feed the speakers.
     
  12. audioguru

    audioguru Well-Known Member Most Helpful Member

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    It looks like amplifiers were 46 years ago. It even uses the very old TIP3055 and TIP2955 output transistors.
     
  13. unclejed613

    unclejed613 Well-Known Member

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    germanium transistors used negative supplies, because the output devices were PNP. the switch to positive supplies came with silicon transistors and the fact that NPN output transistors were common and PNP power transistors were difficult and expensive to make in silicon. quasi-complementary output stages using pairs of NPN output devices were common until PNP complements became more readily available. at that point, amplifiers began appearing with bipolar supplies.
     
  14. fantabulous68

    fantabulous68 Member

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    -50V Single negative supply- 8 ohm load- single transistor input

    :) Done "normalizing";

    Fixed the connection from the emitter of Q7 to ground that was shorting it out.

    Hope its fine now:D

    ignore details on V1 (AC voltage source)
     

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    Last edited: Mar 20, 2010
  15. fantabulous68

    fantabulous68 Member

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    I don't have a choice. The output transistors are fixed onto the kit. :) I'm going to build it on Tuesday.
     
    Last edited: Mar 20, 2010
  16. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    There's no problem with those transistors, they might be old, but were good back then, and are still good now.

    The circuits coming on nicely, but you could do with a zobel network on the output (8 ohm and 0.1uF in series).
     
  17. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Most of the polarized capacitors are shown connected backwards.
     
  18. unclejed613

    unclejed613 Well-Known Member

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    you still need a 10k resistor from node 14 to ground. also a 10nf resistor in series with a 10 ohm resistor from node13 to ground would help reduce the tendency to oscillate, as well as a 25uh coil in parallel with a 10 ohm resistor between node 13 and the top of C1.

    2N3055/2955 pairs were used up until the mid to late 80's in a lot of guitar amps, no shame in using them. before that it was quasi-complementary output stages with pairs of 2N3055 devices. there's nothing wrong with using them if you're not too fussy about a few tenths of a percent distortion.
     
    Last edited: Mar 20, 2010
  19. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    Very true :D

    Funny how you don't notice when it's a negative supply.
     
  20. audioguru

    audioguru Well-Known Member Most Helpful Member

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    No.
    You want node 14 to be close to the DC output voltage for it to be biased at half the supply voltage. The emitter current for Q7 is in the feedback resistor R10.
    The value of R15 (the collector resistor for Q7) should be much higher than only 820 ohms to make the emitter current of Q7 low and to make the open-loop gain high.
     
  21. mneary

    mneary New Member

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    If R10 carries the emitter current for Q7 there will always be a DC input offset. 10k from node 14 to ground would set the emitter current of Q7 to 2.5mA, eliminating offset voltage. R15 should be set to develop about 650mV from that 2.5mA, making R15 equal to 260 ohms. 270 should be close enough.

    Or, if you want R15 to stay 820 ohms, then Ic of Q7 would be .650/820=793uA, and the resistor from node 14 to ground would be 30k.
     

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