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Please help me understand this amp circuit

Discussion in 'Electronic Design' started by Peabody, Apr 19, 2009.

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  1. Peabody

    Peabody Guest

    The amplifier is a speaker dock for MP3 players. It's just
    a cheap single-chip amplifier with two cheesey speakers from
    Ebay, and is of no real consequence, but I would like to
    understand why the designer may have made certain changes
    from the example schematic shown in the datasheet, and what
    the consequences would be of changing back to the datasheet
    version. The as-built amp produces very little output
    volume, and appears to be capable of much more, and I'm
    curious why they elected not to let it do that.

    The datasheet and a circuit drawing of the amp as received
    are here:

    It appears the designer went way out of his way to reduce
    the input signal and the amp gain. I don't really know why.
    The actual circuit differs from the drawing in the datasheet
    in three ways:

    1. The input signal first goes through a 50K pot to ground,
    and the wiper off that then goes through another divider
    circuit of 39K/4.7K to ground, with the chip's input pin
    connected at the junction. I really don't understand this
    at all. I even wonder if the 39K resistor was intended to
    be 3.9K. (It is indeed 39K, on both channels.) In any
    event, I don't see why either of these resistors need to be
    there. It appears to be the Homoepathic version of the
    input signal - diluted down so much that there's hardly
    anything left. I tried jumpering around the 39K, and got
    nice loud volume out of the speakers.

    2. The resistor labeled Rf in my drawing is the feedback
    resistor. It's referred to in the datasheet as reducing amp
    gain. Again, it's not clear this is really needed, but I
    assume the intent is to reduce distortion.

    3. Capacitor C8 in the datasheet is not present in the
    actual circuit. I don't know what this does. I don't know
    what "bootstrap" does here at all.

    Obviously I'm not an EE, and analog isn't exactly my strong
    point. Well, at least it's not RF. But I would like to
    understand this if I can. Any help would be appreciated.
  2. Phil Allison

    Phil Allison Guest


    ** Your schem and the datasheet are both too small to be readable.

    ** The datasheet tells you - try reading it.

    ** Bootstrapping in an audio amplifier is a way of improving the available
    voltage swing from the driver stage and hence improves overall linearity.
    Usually, the load resistor for the class A driver stage is split into two
    and the mid point connected to the speaker output via an electro capacitor.

    ...... Phil
  3. Guest

    What is the source that gives the input signal ?
    In some cases it may be that the input amplitude voltage is
    high enough that it should be reduced to avoid distortion.
    This resistor fixes the gain : G = 10K / (50 + Rf)
    Have a look at the data sheet, there are
    equations for the gain that includes Rf.
    C8 at the output is a first-order low-pass (high-cut) to avoid
    oscillations of the AOP at high frequencies when the gain is high.
    I Hope This Helps.
  4. Peabody

    Peabody Guest

    This amp just plugs into the earphone jack of an MP3 player
    such as an iPod, Sansa, or whatever. So it's not really a
    typical line input. I wouldn't think the voltage would be
    all that high, but in any case, the pot is there to take
    care of that if it starts to clip. I'm testing it with a
    Sandisk Clip player, and at maximum volume on both the
    player and the amp, it's still not very loud at all. So I
    think I'm going to try playing with the 39K and 4.7K
    resistors and see what effect they have.
    Yes, I found it, but didn't understand what "JWC1" is.
    Here's the formula:

    VOUT/VIN =


    Rf + R2 + __________


    Anyway, if R2 is 50 ohms, then a Rf of 100 ohms will cut the
    voltage gain by as much as two-thirds, depending on what
    JWC1 is.
    Ok, so maybe they left it out because the gain is low. And
    I probably out to add it back if I increase the gain.
    Yes it does. Thanks very much.
  5. Yes : adjust these resistors with the pot to maximum,
    to get the max volume you need for your purpose.
    "j" is the symbol for imaginary, i.e sqrt(-1)
    "w" is omega (e.i 2*PI*F) and F is the frequency
    "C1" is the value of the condensator C1 in Farad.

    Then the expression (1 / j*w*C1) is the value of
    the resistor that is equivalent to C1 at frequency F.
    The value of C1 should be such that its equivalent resistance
    is low in regard to the serial resistor at minimum frequency.
    Two-thirds of what ?
    Just use: G = 10K / (50 + Rf)
    Sure, maybe you won't even need it, but keep in mind that
    for a high gain, if the AOP oscillates you may not notice
    it at all, because the frequency could be above 20 KHz.
    Don't mention it.
  6. Peabody

    Peabody Guest

    Jean-Christophe says...
    Well, I just meant that with Rf = zero as shown in the
    datasheet circuit, the gain is 200, but if Rf is 100 ohms as
    in my amp, the gain is 67, ignoring in both cases the effect
    of C1. I'm just saying that the 100 ohms makes a big
    difference. I assume this was done because the amp produces
    too much distortion when operated at maximum possible gain.
    If that's the case, then I'm better off if I can get where I
    need to be solely by adjusting the input resistors and
    leaving Rf alone.

    Thanks again for your help.
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