Vol pot and cap in series Question

[tubenit]

In example C,  there are caps in series that essentially reduce the value of capacitance in that area.

What about example A .........  and example B?    Is the capacitance being reduced there when a resistor or a wiper to side terminal resistance is in play?

with respect, Tubenit

[HotBluePlates]

A and the 300pF in B are qualitatively different than example C.

In A & B there are resistive voltage dividers which reduce the level off all frequencies; except the caps bypassing them allow some range of high frequency to pass un-reduced. In the case of the volume control bypass cap (300pF), turning the volume to maximum raises all signals and bring their volume in line with that of the highs passed by the 300pF. Therefore, the amp sounds fullest at full volume and brightest at minimum volume.

C has series caps after the volume control to strip out most body from the signal. But at low settings, the first coupling cap is isolated by the resistance from wiper to the top of the pot. At maximum volume, there is no isolation, and the 1st coupling cap is in series with the other caps, lowering their effective value. This circuit is bright to start with, and is brightest at maximum volume.

[tubenit]

Circuit C capacitance is LESS than 300p because of the caps being in series.

What about circuit A or circuit B capacitance.  Would they be less than 300p also?   Or is that a misleading way of thinking about it?

Let's take circuit A  and compare it to circuit B.   IF the volume pot were set on "4",  which of those circuits would be the brightest?
Circuit A or circuit B?

With respect, Tubenit

[PRR]

The 300 and 500 caps in "C" could be replaced with a single 187.5pFd cap.

No firm prediction can be made about A B or C until you show the missing piece (the stuff to the right of where you cut-off).

[terminalgs]

Circuit C capacitance is LESS than 300p because of the caps being in series.

What about circuit A or circuit B capacitance.  Would they be less than 300p also?   Or is that a misleading way of thinking about it?

Let's take circuit A  and compare it to circuit B.   IF the volume pot were set on "4",  which of those circuits would be the brightest?
Circuit A or circuit B?

With respect, Tubenit

These networks are frequency dependent impedance networks.



(a) you need to know the frequency of the AC waveform, and (b) you need to know what follows in the circuit (impedance wise).


The effect of the value of ex.B's 500pf cap is undetermined without knowing the the resistance to ground following the cap.  Likewise for the 300pf+500pf series capacitance in Ex.C. 


Ex.A's full story can't be told without knowing what follows either. If Ex.A's following component is a 200K grid-leak resistor to ground, you could look at it as two voltage dividers in series, one after the other, each with a treble-bleed cap across the top.   any given frequency will see the entire network as having different impedances, and as such, will "pass thru" with different attenuated results.  (in Ex.A), if the pot's wiper is dead-center in the middle (250K,250K), and to the right  of the schematic is a resistor to ground (maybe the next stage's grid leak), call it Rg, Rg=200K,   you know that the lowest frequencies will be attenuated 50%X50%,  or 25%, because  those treble-bleed networks will have discouragingly large impedances (because of the resistance to ground at the right side of the cap).  The highest frequencies, (if high enough!) will see the impedances of the treble-bleed networks as being very low, and will pass with much less attenuation.


What if, in ex.A, that Rg to the right is 2M. The 500pf will have very little effect.  what frequencies see the 500pf part of the network as have an discouragingly large impedance will be attenuated 10% (Voltage divider's R1=200K, R2=2M).. So some frequencies "pass" at 100%, the  rest at 90%ish.  to the ear, if Rg=2M, the 500pf is nearly non-existent.   If Rg=10K, its a completely different story.


What if the next component to the right is a output jack?  what if there is no resistor 'Rg' in the circuit?  in that case,  whatever you plug into that jack would have some impedance to ground, so effectively Rg (or maybe better called "Zg"??) would still be there and still be important.

Is the capacitance being reduced there when a resistor or a wiper to side terminal resistance is in play?

the impedance of the cap & resistor,  or cap & pot (resistor+resistor), is frequency dependent.  for instrument AC signals "hoping to pass",   they don't see "just the cap",  they see the cap and resistors together as a network, and depending on the frequency, they see an impedance somewhere between impassibly high, and easily-passibly low...


that's my take on it!

[jojokeo]

What about circuit A or circuit B capacitance.  Would they be less than 300p also?   Or is that a misleading way of thinking about it?

Let's take circuit A  and compare it to circuit B.   IF the volume pot were set on "4",  which of those circuits would be the brightest?
Circuit A or circuit B?

With respect, Tubenit

1st part of 1st question is no, misleading. You will still get lower freq signal output from either the pot and/or resistor too along with the upper freq getting bypassed through the caps. Also, know that if the volume pots are at maximum, then the 300pf cap doesn't exist - it's shorted via the pot's lug & wiper.

2nd question - circuit B would likely sound brighter because the other signal with more lower frequencies gives the impression that it isn't as bright but a better way to put it is that it would sound "thinner". Only highs getting through the 500p cap w/ no mids or lows. Circuit A would allow lows through the 200k resistance therefore it wouldn't sound or seem as bright because of the lows being along with it BUT - technically they would both pass along the same amount of highs if measured and set exactly the same (kind of a trick question here) - the "official" answer is that they'd both have the same amount of highs.

[tubenit]

Guys,  I appreciate all the responses.

Here is what comes "after" those caps .....................

The amp actually has an OD section that switches in also, but I am focusing on the clean channel here.

With respect, Tubenit

[terminalgs]

 

Another way to draw those components is like this:





Your 500K pot is R1 and R2.  Lets put it right in the middle so R1=R2=250K.


R1 250K
R2 250K
R3 200K
C1 300pf
C2 500pf


For a given frequency at "IN", C1||R1 presents some impedance, lets say Z1. There are online calculators for determining Z1. We can generalize that for lower frequencies, Z1=R1 more or less.  Even at 1KHz, Z1~=225K.   Its not until 4KHz that Z1<125K.


So what do we do with Z1?   follow it with R2 to ground.   The impedance of R2 is 250K, call it Z2.  Take Z1 and Z2 and make a divider: (Z2/(Z1+Z2)). 


For all lower frequencies, the divider factor is about 50%. For higher frequencies the factor is higher.  (ex. 10KHz is 83%). So the voltage of the audio signal at the center point where R1,R2,R3 connect is:


Vout=Vin X 83% for 10Khz
Vout=Vin X 50% for <1Khz


 
If you had a grid-leak resistor following R3||C2, say "R4",  you could make the same calculations for C2||R3 and R4  as we did for C1||R1 and R2 and see what kind division you'll get.  However, you don't have that resistance to ground following C2||R3,  instead, you have the inter-electrode capacitances + miller capacitance of that next triode.   the impedance of a capacitance decreases with increasing frequency, so if anything, you'll have the same dividing behavior with R3||C2 as you did with R1||C1 and R2.  (I don't know if R2+R3's Z is effective at the right side of R3||C2 or not!  this is complex!)


The other aspect of of these parallel caps and resistors that is very important is the phase shift.   Go back to C1||R1.  those higher frequencies that see C1||R1 as low-Z, are subject to a higher degree of phase shift.  As Z1 drops, negative phase shift increases.  the low frequency 10Hz sees Z1=250K  but nearly 0deg. of phase shift, whereas 10Khz sees Z1~=50K but is at -78deg. phase shift.  This kind of phase shift is undesirable in general for clean audio amplification.  It is a big reason why Leo Fender invented his classic tone-stack with the 100K slope resistor. Of course, maybe you don't want clean audio amplification? In this case, I think you do, by "clean", we aren't talking about dirt/overdrive/fuzz, we talking about the ear interpreting all the frequencies from a plucked string as occurring at the same time or not, and avoiding potentially undesirable effects related to phase-shift..

You'll get compounded phase shift for those frequencies that see C2||R3 as low-Z.

In the end, for guitar audio, I don't think R3||C2 is doing much for you that C1/R1/R2 can't be designed to do.  If you want to open/close the door for certain frequencies, adjust C1.  Its not like R3||C2 is a mid-filter, its the same basic function as C1/R1/R2.
Since you have a robust tone-stack following the IFR820,  I'd leave R3 and C2 out and leave tone shaping to the tone-stack.


This is all my take on this, I'm not an engineer, so take it FWIW.

[PRR]

> Here is what comes "after" those caps ........

OK, a bare-naked grid.

"B" won't work at all. The grid has no DC path to anywhere. It will float to some random voltage. V2B will either cut-off or slam down; neither is good amplification.

Example "A" will keep the tube happy. But the 500P+200K do essentially nothing, because there is "no load" on them. Actually you can try "naked grid" = "100Meg + 100pFd". At low frequencies the 100Meg on 200K is 0.998 voltage transfer, no loss. At some high frequency response is more like 500p driving 100p or 0.83, a 1.5dB droop/shelf. The droop happens around 500pFd+200K or around 1.6KHz. It's a very fine tweak but in a critical area (where fundamentals stop and only overtones happen). Dunno if you can hear it. Sure less difference than between Tuesday and Friday in the same bar (more customers, more treble absorption).

From an engineering perspective: the 500K pot is large enough that at mid-settings you will lose some of the top of the audio band. Since the load is a tube grid, the hasty fix is a 100pFd from top of pot to wiper, to balance the 100pFd wiper to ground in the tube grid. The guitar-amp fix is to up-size that cap to give mild treble boost at mid-setting, rolling-out to flat response full-up.

> which of those circuits would be the brightest?

What do you want? BRIGHT?

Then cut bass/mids with resistor divider. Bypass the divider with a cap to shoot the treble through. You often see, from wiper to grid, 470K wiper to grid and 470K from grid to ground. This cuts everything to half (and also assures DC condition). The high resistance, against the tube capacitance, rolls off some treble. Bypass the wiper-to-grid resistor with more than 100pFd.... 220p, 470p, to taste.

[tubenit]

Gentlemen,

THANKS for the replies and explanations!  I appreciate the help and the info.

What happened is when I changed speakers recently, the clean channel now seems a little too bright to me & I wanted a dab of warmth returned.  (The OD channel is great & I am happy with the range of tones the OD can offer.)

With respect, Tubenit