Placement of clipping cap in BMP

[WhenBoredomPeaks]

There were changes in the placement of the clipping cap in the BMP.


Which one of them filters the clean signal first so the diodes can clip the filtered, less bassy signal?

I am asking this because i think distorting a bass filtered signal would cause less harmonic content in the lower freqs, so maybe it could make chords less "chaotic".

[nocentelli]

Yeah, I've never known which is correct/most common/standard, or even understood fully the difference this makes: I've seen schematics both way rounds: I always assumed the pisotones website was consistent (i recognise the schems), but there you've got Violet Ram's head style on the left and Russian on the right (which looks standard to me) now I look the Russian is  - Can anyone gives us a run-down?

[WhenBoredomPeaks]

Yeah, I've never known which is correct/most common/standard, or even understood fully the difference this makes: I've seen schematics both way rounds: I always assumed the pisotones website was consistent (i recognise the schems), but there you've got Violet Ram's head style on the left and Russian on the right (which looks standard to me) now I look the Russian is  - Can anyone gives us a run-down?

If they make a difference than it should be this: one of them clips the full signal and then removes some bass from the clipped signal. (diodes first) This would make the higher harmonics of the clipped bass frequencies untouched.
The other one should remove the bass from the clean signal first so they won't get clipped and they won't leave their harmonics in the signal. (cap first) This should be a less complex signal than the other.
But i am not even sure that there is a difference between them.

[CynicalMan]

There is no electrical difference between the two circuits. In a series circuit, the current flowing through each component is equal, and is only related to the total impedance of the circuit. This means that the order of components in series is irrelevant, as long as the components or circuit blocks only have two external connections each.

[WhenBoredomPeaks]

There is no electrical difference between the two circuits. In a series circuit, the current flowing through each component is equal, and is only related to the total impedance of the circuit. This means that the order of components in series is irrelevant, as long as the components or circuit blocks only have two external connections each.

And if they are equal, what happens first, the filtering or the clipping? /chicken or the egg? /

[petemoore]

  And if they are equal, what happens first, the filtering or the clipping? /chicken or the egg?
   Not that it helps, but figure out which way the electrons are flowing, calculate the distance-time it takes for the electron to move the <1/4'' or so at somewhere near light speed and have an infanitecimally small, insignifigant number.
   Then of course try it both ways in order to determine what the differences really are.
  At least the first blurb could produce a calculated answer...much smaller than what could produce a detectable phase non-linearity.
   

[asatbluesboy]

There is no electrical difference between the two circuits. In a series circuit, the current flowing through each component is equal, and is only related to the total impedance of the circuit. This means that the order of components in series is irrelevant, as long as the components or circuit blocks only have two external connections each.

God damn it... I went through serious pains to reverse their order in the Tonepad layout... *facepalm*

[WhenBoredomPeaks]

  And if they are equal, what happens first, the filtering or the clipping? /chicken or the egg?
   Not that it helps, but figure out which way the electrons are flowing, calculate the distance-time it takes for the electron to move the <1/4'' or so at somewhere near light speed and have an infanitecimally small, insignifigant number.
   Then of course try it both ways in order to determine what the differences really are.
  At least the first blurb could produce a calculated answer...much smaller than what could produce a detectable phase non-linearity.
   

I have it on a breadboard so i gonna try it (changing the order of the diodes/caps) after my pots arrived. But it looks like that i should not expect anything from it.

[EATyourGuitar]

I'm a complete noob but I understand that when your talking about an AC signal, it will see the cap first or the diodes first on each half of the waveform. flipping it around is the same as inverting the waveform first and running it through the same old circuit. as long as the waveform going in is symmetrical to begin with. with a symmetrical diode pair and a non-polar cap, it will still be a symmetrical signal after it is clipped and filtered. even if you found asymmetrical artifacts on the scope, your ear cant tell the difference what side of the waveform its on.

[Electron Tornado]

I doubt you'll hear any difference by changing the location of the 0.05uf cap. Try different values for the cap to change the sound. You may find that you'll need to make changes in more than one of the stages.

As I understand it, removing the 520p caps (or whatever value is there, they vary) from all three stages will brighten the sound, which is what the "Wicker" switch does. See the videos:

http://www.youtube.com/watch?v=AZTN2gaVGjY

http://www.youtube.com/watch?v=pE5LwNvt6BA

[CynicalMan]

I'm a complete noob but I understand that when your talking about an AC signal, it will see the cap first or the diodes first on each half of the waveform. flipping it around is the same as inverting the waveform first and running it through the same old circuit. as long as the waveform going in is symmetrical to begin with. with a symmetrical diode pair and a non-polar cap, it will still be a symmetrical signal after it is clipped and filtered. even if you found asymmetrical artifacts on the scope, your ear cant tell the difference what side of the waveform its on.

No, there is no difference at all between the two setups. The results are identical. In a series circuit, the order of components with two connections each is irrelevant. We can treat the diodes as one component with two connections.

Think of it this way. The DC resistance of the 50n capacitor is infinite. That means that there's no DC current flowing through the capacitor or diodes, and no DC voltage can develop over the diodes. The AC current flows through fine, which means that the voltage over the diodes is the difference between the AC signals at the transistor's collector and base. This happens whatever the order of the components is.

[blooze_man]

I tried both configurations in a muff and I could swear that there was an extremely audible difference. But I very well could've made a mistake somewhere.

[WhenBoredomPeaks]

Is it possible to filter some bass from the signal being clipped without changing the input and coupling caps? I don't want to loose the bass of the whole circuit, i just want to clip a thinner signal.

And i know the role of the 560pf on my pic but i need some highpassing not lowpassing. If it is possible at all.

[CynicalMan]

If you reduce the value of the 50n capacitors, then less bass will be clipped.

[stringsthings]

If you reduce the value of the 50n capacitors, then less bass will be clipped.

+1

... and for those BMP lovers that want more bass frequencies to be clipped, simply raise the value of the 50n capacitors ...

[WhenBoredomPeaks]

If you reduce the value of the 50n capacitors, then less bass will be clipped.

+1

... and for those BMP lovers that want more bass frequencies to be clipped, simply raise the value of the 50n capacitors ...

So that cap filters the signal before clipping? (this is the original question of the topic)

[CynicalMan]

The order is irrelevant. It's better to think of it as a matter of impedance. If the AC voltage is under 0.7V, then the impedance of the diodes is very high. Once the voltage hits 0.7V, then the diodes clip and their impedance becomes very low. Then, the total impedance becomes more or less the reactance of the capacitor, equal to 1/(2*pi*c*f), which decreases as the frequency gets higher. Because the circuit is a negative feedback loop, the gain decreases as the impedance decreases. So, when it's not clipping, the impedance is high, and the gain is high across the frequency range. When it is clipping, then the impedance decreases as frequency increases, and there's more bass than treble. The overall effect is that treble is clipped more than bass is.

[EATyourGuitar]

The order is irrelevant. It's better to think of it as a matter of impedance. If the AC voltage is under 0.7V, then the impedance of the diodes is very high. Once the voltage hits 0.7V, then the diodes clip and their impedance becomes very low. Then, the total impedance becomes more or less the reactance of the capacitor, equal to 1/(2*pi*c*f), which decreases as the frequency gets higher. Because the circuit is a negative feedback loop, the gain decreases as the impedance decreases. So, when it's not clipping, the impedance is high, and the gain is high across the frequency range. When it is clipping, then the impedance decreases as frequency increases, and there's more bass than treble. The overall effect is that treble is clipped more than bass is.

very helpful, thank you for this. you make it so easy to understand whats happening here.