18 Watt tone control?

[jeff]

 Having never built or heard a marshall 18watter I have a question about the tone control. When the tone is rolled all the way back why doesn't it kill all the volume? At first I thought there was an error in the schematic and the .005 and .01 were switched.

 When the tone is rolled all the way back the .01 goes to ground and the .005 goes to the volume pot. But if the .01 is grounded doesn't that ground all the signal? If those caps were reversed I could imagine that any signal that wouldn't pass through the .005 would go through the .01 to the volume. If a cap passes AC, and the bigger cap is grounded how is there any signal left to go to the volume?

Below is schematically equivelent to the tone being rolled all the way back.

I guess a tone stack is like that too. If the mid is off grounding one .02 cap, and a cap passes AC, how is any signal going through the other .02 cap to the bass control that's not being grounded by the mid cap?

[phsyconoodler]

Because the tone doesn't kill any signal to the grid of the preamp tube,it just alters the frequency that bleeds through.
 It's not like the bass control on a Twin Reverb.There is still a path through the caps to the grid of the preamp tube.
Or rather still a path to the PI grid I should say.

[jeff]

I'm having a problem with visualizing how a cap works.

I was told two ways to think of a cap.

1) A cap passed AC but blocks DC. If I think of it this way then it appears to me the guitar signal has a direct (AC) short to ground for all frequencies above the .01 caps rolloff.

or

2) A cap charges and discharges and keeps the DC voltage constant, removing the AC ripple. If I think of it that way, since the .01 has a lower roll off point than the .005, why doesn't the .01 to ground remove the total ripple(guitar singal) keeping the DC constant?

Either way I can't understand how is there any signal left to send through the .005 cap to the volume pot?

I know it works but realize I have a fundamental misunderstanding of the operation of caps. What am I missing?

[sluckey]

1) A cap passed AC but blocks DC. If I think of it this way then it appears to me the guitar signal has a direct (AC) short to ground for all frequencies above the .01 caps rolloff.

That's only partly true. Any capacitor has a reactance (resistance to AC) that is inversely proportional to the frequency of the AC signal, ie, reactance INCREASES as FREQUENCY decreases. For a cap to be a direct short (zero ohms reactance) the frequency would have to be infinity. The formula is ...

X_C = 1/6.28FC, where X_C is capacitive reactance in ohms, F is frequency in hertz, and C is capacitance in Farads.

So, that .01uF cap would have a reactance of 159K ohms for a 100Hz signal, or 31.8K ohms for a 500Hz signal, or 3.18K ohms for a 5000Hz signal. The reactance gets smaller as frequency increases, but it will NEVER EVER get to zero ohms (except on the chalk board).

I've annotated your 18W schematic to show the reactances involved for each cap for the above frequencies. Maybe seeing it on the schematic and thinking of the caps in terms of their AC resistance for each frequency will help you understand.

[jjasilli]

Another way to think of it is in terms of decibels.  If one cap is used, this is a first order filter.  It attenuates signal at the rate of 6dB per octave, starting at the cut-off, a/k/a corner, frequency.  If two caps are used, one series / one shunt, this is a second order filter.  It attenuates signal at the rate of 12dB per octave, and so on for third order filters, etc.  Above the corner frequency of the RC circuit, there is no attenuation of signal (for an ideal cap).  Below the corner frequency, attenuation kicks in (much like an object falling at the rate of 32ft per second, per second):  the impedance of signal increases more and more as the signal's frequency falls below the corner frequency.

[jeff]

Beautiful, More pieces to the puzzle! I think I'm starting to get the picture.

Thank you so much!

So if I think of a cap as passing frequencies like this:
            __________
           /
          /
         /

When shunted, it will ground frequencies like this:
        \
         \
          \___________
            
And even though the .005 rolloff starts at a higher frequency than the .01 there is still a point in it's responce to pass signal because it's a rolloff not a cutoff.
             ___________(.005 passes)
      \    /
       \  /
        \/
        /\
       /  \ __________(.01 grounds)

resulting in this:


         /\
        /  \___________
I think what was thorwing me was that the shunt cap is bigger than the series cap.

[jeff]

Another interesting thing is it looks like instead of just a variable resistor to ground the other end of the pot incerases/decreases resistance to ground. I imagine this compensates for loss in volume as the treble is bled off.

[sluckey]

I think what was thorwing me was that the shunt cap is bigger than the series cap.

And that's a tricky point. The thing you have to remember though is that a cap is not a switch (either on or off). Sure, for any frequency, the .01 shunt cap offers 1/2 the AC resistance that the .005 series cap offers. And that means more signal is shunted (attenuated) than is passed thru the series cap. But neither cap is ever 0 ohms nor infinity ohms. It's not a logic switching thing.

Keep it simple. The best way to get this concept (and it looks like you have it) might be to simply calculate the reactance for a particular frequency and visualize the cap as a resistor with a certain value for that particular frequency. It may be confusing to think about rolloff, cutoff, corner frequency, etc. And don't even mention phase shift!