BJT amp to drive an 8 ohm load

Discussion in 'General Electronics Chat' started by automagp68, Dec 6, 2015.

  1. automagp68

    Thread Starter Member

    Nov 13, 2011
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    Hey guys


    So i have build many a BJT amps but I've never tried to make one that supports an 8 ohm load with a 6 volt swing

    Any ideas on this?

    Right now i just have a few CE ganged up to get my gain correct.

    If i use a CC as an output stage with the coupling cap i loose all my voltage swing because I'm obviously blocking DC

    Any one got any ideas on how to accomplish this in a simply fashion with out a lot of fancy stuff

    This is just a for fun project. I have seen many designs but they all include very small swings, say 10mv or so. I need 6 volts across

    [​IMG]


    Here is what i got so far. Please excuse the diff amp on the left its from another project that i had built so I'm just using it.
    Im getting the gain from my CE stage and trying to get current from the CC stage but as i said before when i put the cap in across the 8 ohm load i loose all my swing and if i take the cap out the DC current in the CC shoots through the roof of course.

    And ideas? a basic direction perhaps? As i said above I'm not really familiar with power amplifier design at all
     
    Last edited: Dec 6, 2015
  2. automagp68

    Thread Starter Member

    Nov 13, 2011
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    Also i should add, Fets are an options if that helps. Those are also available to me
     
  3. dl324

    Distinguished Member

    Mar 30, 2015
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    BJTs will be easier to bias. Show your existing schematic/ideas. What is the frequency range? Will the input have any DC bias?
     
  4. Veracohr

    Well-Known Member

    Jan 3, 2011
    549
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    You can use a CC output stage but it needs higher quiescent current than what your schematic shows. Reduce R7 a lot. Also increase the coupling cap to at least 1000uF.

    Or you can look up class AB output stages for another approach.
     
  5. automagp68

    Thread Starter Member

    Nov 13, 2011
    81
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    Thanks for the replies.

    I updated the picture above.

    So i i reduced it to say 300 ohms giving me an increased IE.

    I also upgraded the cap value

    Any ideas on how to get 6 volts across the speaker?
    I am not understanding how I'm gonna see 6 volt swing with the coupling cap still there
     
    Last edited: Dec 6, 2015
  6. dl324

    Distinguished Member

    Mar 30, 2015
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    Q4 needs to provide your 6V signal swing. Q5 is an emitter follower (common collector to some) that provides a low impedance to drive the speaker. C2 decouples any DC bias from Q5 and passes your 6V AC signal.
     
  7. automagp68

    Thread Starter Member

    Nov 13, 2011
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    ok i think i understand what your saying.

    So the Q4 stage is where my swing comes from. Now my AC is no longer small signal its large signal but still acc thus allowing it to pass through c2

    ok thanks that was helpful to understand. now how to get it to work lol
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    6V peak ? And do you know that V_max_negative_output_voltage = Ieq * Re||RL
     
  9. crutschow

    Expert

    Mar 14, 2008
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    It's difficult to get that kind of output into 8Ω using just an emitter follower unless the follower has a very high bias current.
    You should go to a push-pull output stage, which is a common-way to drive a speaker.
     
  10. automagp68

    Thread Starter Member

    Nov 13, 2011
    81
    0

    Thanks

    I do understand that is typically the way its done but i don't have that experience and don't have the time to learn it right this second.
    This was just a small experiment simulation only.

    6v peak swing yes
    Amplitude at this time is unimportant just as long as it swings 6 volts
     
  11. kyka

    New Member

    Jun 7, 2015
    22
    1
    Ok, here's the analytical approach.

    If by "voltage swing" you mean peak-to-peak value, then at some point the output transistor will have to supply 3/8=375 mA. Let's assume that the output transistor has enough gain and proper biasing to do that (it's not difficult at all). The problem is the negative cycle. At that point the coupling capacitor must get discharged in a controllable manner in order to follow the input signal. This can't be done, because its peak current will be 375 mA which will force the output transistor to get cutoff and the discharging process will be abrupt.

    I ran a simulation on LTSpice and here's what came out. I biased the transistor to the same DC emitter current.

    [​IMG]

    The only way to fix this problem, without changing the topology, is by reducing the emitter resistance. See the next picture,
    [​IMG]
    The only problem now is that the transistor conducts almost an amp of DC current and you must make sure that your transistor can handle it.
     
  12. crutschow

    Expert

    Mar 14, 2008
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    Also it will dissipate about 8W at idle so the transistor will need to be a power type (such as a 2N3055) mounted on a good heat sink.
     
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