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  • Coupling caps

    I'd like to learn a bit more about how the values of the various coupling
    caps used in an amp are arrived at. I know that in general the lower the
    value the less bass that gets through. What I don't understand is why
    you use different values in different places in the same amp.

    For example, in a Fender AB763 Deluxe Reverb the normal channel preamp
    outputs through a .047uf cap and the vibrato channel outputs through a
    .02uf cap. To me this means that Fender wanted the vibrato channel to
    have less bass (be more bright) than the normal channel. But then both
    these channels go into the PI through a .001uf coupling cap. Won't this
    throw away much more bass than either the .047 or the .02 caps ? Why
    not use .001's everywhere if you're not going to be keeping the bass
    anyway ? Then the PI goes to the power tubes through 0.1uf caps. Why
    so big if there isn't much bass left ?

    Paul P

  • #2
    My theory

    There are no-doubt others with far more knowldge than me who can explain it in a more articulate way. But if tried to I stop myself from thinking, it would be a futile exercise, so in my simple terms...

    The amount of frequency bandwidth you are passing through each gain stage varies, depending on what comes into the stage, and on what you do inside each stage to change it. So you aren't necessarily going to start with x-amount of bass and keep subtracting it to zero-infinity as you go through the signal path.

    For example, a distortion gain stage can sound too muffly/muddy if you have a larger value coupling cap from the previous gain stage as opposed to a smaller value cap. There is a certain amount of bandwidth of signal already in the previous gain stage, that has been modified/filtered/tuned/whatever in the amplification process and you don't necessarily want to take it all through to the following stage. By the same token, once you have gone through two or three or four gain stages and EQ'd the signal with a tone stack etc, you might be wanting to preserve whatever bass is left in the signal by maxing some/all of the caps thereafter etc etc.

    I discovered alot of this the hard way doing and undoing and redoing ad-infinitum, a truck load amount of modding to the pre-amp of my classic 30 (which is the worst kind of amp to try this with BTW).
    Building a better world (one tube amp at a time)

    "I have never had to invoke a formula to fight oscillation in a guitar amp."- Enzo

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    • #3
      The coupling caps only produce a bass rolloff in combination with the following resistance.

      You can find RC calculators on the internet that'll give you the corner frequency of the 6dB/oct rolloff a cap driving a resistor produces. In the case of a phase inverter the negative feedback "bootstraps" the input giving an effect increase in impedance- this means a puny little 1n cap is still full range or close to it when driving a phase inverter with NFB.

      jamie

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      • #4
        There is a thing to bass roll-off known as the "slope". A single capacitor into a resistor produces a LF rolloff that reduces the signal in level by 6dB for every octave below the cutoff frequency. If you cascade two of these 6dB roll-offs at the same frequency you get a 12dB/Oct bass roll-off, 3 of them= 18dB/Oct, etc., etc.

        The faster the bass rolls off the less "natural" it tends to sound. That is one, simple, way to view why Fender would "preserve" the bass and then roll it off later.

        It is quite possible that Fender is not rolling off the bass as much as you think. As others have mentioned, the cap itself doesn't do anything. The breqk frequency is relative to the resistor (or load) that the capacitor "sees". While the input to a typical Fender phase inverter may look to you like it has a 1Meg input resistance (it has a 1Meg resistor after all right?), it really has an input impedance many times that due to the electrical nature of the circuit. A typical phase inverter input is a type of a "bootstrap" circuit. The impedance of these is usually many times that of the grid leak resistor itself and depends on the circuit configuration. So that .001 capacitor in that place in the circuit may actually produce an LF roll-off similar to a .02uF cap somewhere else in the circuit (or any other value you want to pull out of a hat).

        The main thing to remeber is the slope ofthe roll-off though. I think that is the real answer to your question.

        Chris

        Comment


        • #5
          Originally posted by Paul P View Post
          For example, in a Fender AB763 Deluxe Reverb the normal channel preamp
          outputs through a .047uf cap and the vibrato channel outputs through a
          .02uf cap. To me this means that Fender wanted the vibrato channel to
          have less bass (be more bright) than the normal channel.

          But then both
          these channels go into the PI through a .001uf coupling cap. Won't this
          throw away much more bass than either the .047 or the .02 caps ? Why
          not use .001's everywhere if you're not going to be keeping the bass
          anyway ? Then the PI goes to the power tubes through 0.1uf caps. Why
          so big if there isn't much bass left ?
          Tremolo oscillators can induce some low frequency hum into the signal. Therefore, tremolo channels often need to cut bass to get rid of this. (Checkout the Vox AC30 where you have 5 lowpass filters in series at the end of the vibrato channel!).

          Also, in the Fender the tremolo channel was originally intended for guitar, while the normal channel was intended for bass and other instruments. Therefore, the tremolo channel is brighter.

          The PI input is boot-strapped (the 1M resistors are not connected to ground but to the PI tail). Therefore, the impedance is much greater than it first appears (much bigger than 1M). I'm not sure about the exact math for this, but I think the .001 cap is enough to lett full audio frequencies through in this context.

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