simple 2 npn amp wont work

Discussion in 'General Electronics Chat' started by chunkmartinez, Feb 13, 2013.

  1. chunkmartinez

    Thread Starter Senior Member

    Jan 6, 2007
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    For fun I've constructed this simple amp: http://www.eleccircuit.com/simple-preamplifier-circuits-by-transistors/

    The one with 2 bc547s. It dosn't work at all. Yes I have studdied transistors and biasing and design etc but i'm still a newb to it.

    Is this circuit bs? I built it exactly as it should be. The grammar on that page is bad so maybe I shouldn't expect a good result? Btw, I used an 8 ohm .1w speaker on the output and a laptop for the input. Is it possible the laptop voltage signal is much too small? My oscope said the audio voltage from my laptop was under a volt in the millivolt range but I cannot recall the actual voltage.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    That is a preamplifer to amplify low level signals to a higher level. It is designed to feed the high impedance input of another amplifier, not the low impedance of a speaker. For that you need a power audio amplifier.
     
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  3. chunkmartinez

    Thread Starter Senior Member

    Jan 6, 2007
    180
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    my mistake i'm very new to amp circuits..
     
  4. SPQR

    Member

    Nov 4, 2011
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    Do you know that it is "high impedance" because of the capacitor in the output?
     
  5. Brownout

    Well-Known Member

    Jan 10, 2012
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    No. The output impedance is mostly due to R3.
     
  6. #12

    Expert

    Nov 30, 2010
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    For me, the clue is mostly R6. No matter how hard Q2 works, the output voltage can only be raised as much as R6 will allow (compared to the load resistance).
     
  7. Brownout

    Well-Known Member

    Jan 10, 2012
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    Yeah, R6. I missed the part where the OP said the "two transistor amp" and I didn't look past the first circuit on the link.
     
  8. #12

    Expert

    Nov 30, 2010
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    Sorry. Didn't mean to shoot you in the foot.
     
  9. SPQR

    Member

    Nov 4, 2011
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    Ok, so the 10k between the + and collector.

    I understand about "impedance matching" in the sense that you should connect a low impedance stage to a high impedance stage (and vice versa), but I'm really not sure how to determine what part of a circuit make it "high" or "low" impedance.

    The work continues...
     
  10. #12

    Expert

    Nov 30, 2010
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    Well..voltage is always a voltage compared to someplace else, and impedance is low or high compared to something else. Most of the time, the 250k pot in an electric guitar is considered high impedance. I designed an amazing, low noise preamp with an input impedance of over 400 meg. Compared to that, the guitar volume control is low impedance. It's all relative.
     
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  11. MrChips

    Moderator

    Oct 2, 2009
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    Look at it from a different perspective - power.
    Suppose you connect a 10V source into an 10Ω load (just to round up from 8Ω). The current through the load is 1A. The power delivered to the load is given by P = I x V = 10 watts.

    10 watts is a lot of music power, but this is just an example.

    So let's reduce the voltage to 1V.
    Current = 100mA
    Power = 100mW, which is a reasonably small amount of listening power.

    Now, can your transistor circuit deliver 100mA into the 8Ω load?
     
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  12. SPQR

    Member

    Nov 4, 2011
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    Ooooooooooooooooooooohhhhhhhhh!
    So instead of looking at the "R" part of good ol' Ohm's law, look at the IE side!!!!
    A Eureka! moment!
    Thanks very much!
     
  13. #12

    Expert

    Nov 30, 2010
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    Love to see the light bulb go "on" in somebody's head.
     
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  14. chunkmartinez

    Thread Starter Senior Member

    Jan 6, 2007
    180
    1
    It would try to sink 125 mA, correct?

    Thanks for the help guys, can anyone tell me if they think this circuit could supply 100mA to a power amp? This power amp is what I want to feed(excluding the opamp and putting this circuit in it's place) http://forum.allaboutcircuits.com/showthread.php?t=80687

    I am sure I would have to redesign it to match the impedence right? Is the circuit in the latest link I posted(with the opamp) a pre-amp and power amp together, the op-amp beeing the pre-amp? Also, A pre-amp and driver are the same, correct?

    EDIT: After reading and thinking about it, I think I figured something out. The pre-amp stage is to amplify the voltage for the output because push/pull power outputs do not amplify voltage. The output section is needed to supply the current that the following equation demands: preamplified signal voltage/speaker impedance = Output sections current demands which the preamp section cannot supply. Is this correct?
     
    Last edited: Feb 16, 2013
  15. tubeguy

    Well-Known Member

    Nov 3, 2012
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    Yes, you're getting it.

    To calculate voltage and current requirements for a given load use ohms law for power.

    P=I^2 x R and P=E^2 / R

    For 10 watts power into an 8 ohm load:

    10/8 = I^2.
    1.25 = I^2
    1.12 = I

    10*8=E^2
    80 = E^2
    8.94 = E

    So 8.94 volts at 1.12 amps ~ 10 Watts
    Try it with higher power requirements.
    Try with 4 ohm load and notice the difference.
     
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  16. SPQR

    Member

    Nov 4, 2011
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    Again, I thank you.
    When I see "impedence matching" I'm thinking of a resistor/coil/capacitor circuit with a given impedence "X"
    that needs to match the impedence "X" in the next stage (which may be partially true).

    But really, for us newbies, it should be called "power matching".

    Further, this is another example of "working backwards" -
    look at your output device, and then build everything proximal to it according to its requirements.

    Superb discussion - thanks again to all the experts!
     
  17. MrChips

    Moderator

    Oct 2, 2009
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    It's the same difference. Impedance matching is desirable when you want maximum transfer of power. This occurs when impedances are matched.
     
  18. MrChips

    Moderator

    Oct 2, 2009
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    Here is an example in impedance matching.

    Suppose you built an audio amplifier with an output impedance of 5kΩ and wanted to connect this to an 8Ω speaker. You can use an audio matching transformer with
    turns ratio = sqrt(5000/8) = 25

    Suppose you wanted 500mW output = 1/2 W.

    2V output would give 2 x 2 / 8 = 1/2 W.

    To get 2V output at the speaker your amplifier would have to generate 2 x 25 = 50V output!

    Power generated at the 5kΩ output = 50 x 50 /5000 = 1/2 W
    same as power at the speaker, i.e. maximum power transferred.
     
    Last edited: Feb 16, 2013
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  19. tubeguy

    Well-Known Member

    Nov 3, 2012
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    That happens to be typical of impedance matching used in old vacuum tube amplifier output stages. ;)
     
  20. MrChips

    Moderator

    Oct 2, 2009
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    I would expect a tube guy to know.:)
     
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