two stage amplifier

Discussion in 'The Projects Forum' started by andrebc, Jul 13, 2009.

  1. andrebc

    Thread Starter New Member

    Jul 12, 2009
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    This Circuit is supposed to give an amplified signal to a speaker of 25 Ohms, the common emitter gain Av is about 95, Rout is around 1.7K and the input impedance of the Follower is about 5K and the output is 50 Ohms, the issue is mis-match impedance I am having between the 1.7K CE output impedance and 5K input impedance of the buffer. It’s like a puzzle I could not make my buffer input resistance at least 10 times higher than my CE output impedance without affecting the output impedance of the follower which is supposed to be around 50 ohms to drive the speaker. I also tried using a voltage divider technique instead of self-biasing to overcome this mismatch impedance issue but I could not get it to work. I used R1=60K and R2= 47K, and RE=300 ohms for the buffer, instead of R(B)=215K. I basically get nothing out to the speaker. Any body can think of how this can be resolved. I know there are other ways of designing an amplifier among which is the use of MOSFETS instead or Darlington configuration.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The 10uF capacitor C15 appears to be in the wrong place - it is blocking the DC bias from the 215k to Q6 base. C15 should probably go between Q1 collector and Q6 base with the 215k also going to the base of Q6.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Try the attached circuit instead.
     
  4. millwood

    Guest

    tnk is right. C15 is in the wrong place.
     
  5. millwood

    Guest

    you can greatly simplify the amplifier, reduce its part account and improve its performance by going with a direct coupled design, like the one below.

    there really is no reason to try single stage amplifiers in serial.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Millwood,
    Did you see the terrific clipping of the top of your output waveform with a 1v input?
    It works OK with a .1v input, but your gain is fixed at 9.6dB, and you have an input impedance around 500 Ohms.

    In the circuit I posted, I mistakenly showed 5v for Vcc; the circuit was biased for operation at 9v. With the volume control set to 90% providing 10k input impedance, gain is 25dB with very little crossover distortion. It's not my design, Audioguru posted it awhile back.
     
  7. millwood

    Guest

    yeah, because the resistor (R7) is running out of current. you can solve that by a) running the output stage at higher current (lower R7's value); or b) using a CCS in place of R7, or if you want to retain the existing topology, c) you can lower the DC output by download adjusting R2.

    here is the waveform with R7=22ohm (a halogen bulb of the right wattage, :)), idling at about 300ma.
     
  8. millwood

    Guest

    the gain is determined by 1+R8/R9, 3x in my example.

    lower R9 will yield a higher gain. increasing R8 is not recommended because this is a current feedback topology.

    the beauty of this amp is also in its CFB topology: it has extremely wide open loop bandwidth. I would point it to like 20khz, vs. <10hz for a typical VFB amp. plot our your amp's open loop bandwidth and compare that with this amp and you will know.

    you didn't do it correctly. the input impedance would be dominated by R3 across the audio band. higher input impedance at lower frequencies (because of C2; and lower input impedance at higher frequencies (because Q1 loses its gain).
     
  9. millwood

    Guest

    the input impedance of your circuit, right of the volume pot, is much lower than that. so had you used a log pot, you will find that the actual volume doesn't follow the pot's characteristic, because the pot is being overloaded by the amp's input impedance. Now, that CAN be an advantage: you use use a linear pot to achieve a near-log chacteristic.

    the gain, and the DC working points of the amp also depend on the supply voltage, or the devices used, etc.

    and whether the output stage works in class A or B also depends on the relative characteristics between the diodes and the transistors. so I would definitely add two Re resistors on the output transistors to help define idle current and help maintain thermal stability and to protect the ops from being damaged by an output short.

    what you have is essentially the VAS (voltage amplification stage) and OPS (output amplification stage) in a typical Linn 3-stage amplifier. you can greatly improve its performance by adding a different input pair (thus making it a typical opamp), or a SE input pair, like Q1 used in my design.
     
  10. millwood

    Guest

    here is the amp, redesigned to deliver 25db gain (26db actually), by reducing R9 to 47ohm (you do pay a price for that).

    here is the waveform, amplifying a .1v to 2.1v Vp output.
     
  11. millwood

    Guest

    here is the same amp's frequency response.

    ltspice is calculating it to 1v Vac, to yield a 6db gain. but the actual input is .1v Vac (as I circled in green for you), so the actual gain is 26db.
     
  12. millwood

    Guest

    as to your amp's inability to deal with device variations, or supply variations, or thermal stability, you can just plug in different devices , supply voltages, or add Vbe disturbances and observe its behaviors.
     
  13. millwood

    Guest

    this is a comparison of those two amps, frequency only. I changed your input cap to 0.47u - it is just practically impossible to find a high quality 100uf capacitor.

    1) your amp: has a -3db frequency response of roughly 40hz - 400khz. at 10hz, its gain is -12db (vs. .1v Vac input).

    2) my amp: has a -3db frequency response of roughly 25h z- 1mhz. at 10hz, its gain is -4db.

    I also added V3 to both of our amps to help analyze thermal stability.

    with a -20mv change to the Vbe (about 8 - 10 degree warm-up), your idle current moved about 50%, and mine is basically unaffected.
     
  14. millwood

    Guest

    one important thing about making a successful amp is that its characteristics should be determined as much by resistors or capacitors as possible, and as little by active devices: because variations among active devices are far bigger than those among passive devices.

    so if an amp sounds good with 2n5550, it should sound good with 2n3904, within reason of course.

    now, take a look at yours.

    two identical amps, one with 2n3904 as the front end and another with 2n5550 as the front end.

    we are just looking at their frequencie responses. what do you see?

    the one using 3904 has a much higher gain than that with 5550.

    so if you are making a stereo amp with this topology, and you used identical prts other than those two 3904: one with higher gain than the other - pretty normal, right?

    well, one channel will be noticeably louder than the other.

    why? because your amp's close-loop gain depends on the active devices used.
     
  15. millwood

    Guest

    here is what it looks like if you add a front-end to it - a SE front end in this particular example but you can add a differential pair as well - may be simpler.
     
  16. millwood

    Guest

    now, we take the last amp (yours + a front end), and replace the 2n5550 with a 2n3904.

    again, just frequency responses at this point.

    what do you see?

    identical frequency responses, even though we used different transistors. exactly what you want to have in an amp.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    There's something rather non-trivial you left out.
    Yours is a class A amplifier, where AudioGuru's amp is a class AB amplifier.
    Even with no input, your load (R7) will be dissipating roughly 2.2W; 7.1v x 320mA.
    In Audioguru's circuit, the DC is blocked. Sure, there's a slight amount of crossover distortion, but that's to be expected in a class AB.
     
  18. millwood

    Guest

    it is not about distortion. it is about thermal stability: a small temperature rise will cause your Ic to go up significantly, which by itself will heat up the transistors significantly, which will cause the temperature to go up further. . .

    Bang! your amp goes to amp heaven in no time.

    for thermal variations, you have zero means to control it because you don't have a feedback loop for it.

    that's precisely why a typical amp don't use diodes for biasing - they use Vbe multipliers.

    and for those that do use diode for biasing, they typically use significant Re resistors to establish that negative feedback.

    that is a properly designed amp. yours is anything but.
     
  19. millwood

    Guest

    the bias current in your design is basically undefined: it depends on the devices (diodes and transistors) you use.

    and you can bias your amp in Class A and the same problem would still exist. because it is topology driven in your design.
     
  20. millwood

    Guest

    we kept talking about how we can eliminate device variation from your amp by adding a front end to it. we compared one where we used a 2n3904 as VAS vs. the same amp using a 2n5550 as VAS and they looked identical.

    but maybe the difference between a 2n5550 and a 2n3904 isn't big enough. how what if we replace a 2n3904 with a darlington formed by two 2n3904? their betas would be hundreds of times different, and their Vbe would double.

    will the two amps act differenent?

    here are their frequency responses.

    tell me if using a transistor whose beta is 100x bigger than that other made any different on their ac performance.

    now, try the same experiment on your original design and see if you can generate the same ac performance without altering the circuit or device parameters.

    :)
     
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