What purpose do these resistors serve?

[SILVERGUN]

In this PPIMV circuit, what is the intended purpose of the resistors that I have highlighted in the attached pic?

[Leevi]

I don't see any good reason for them.
I would rather see them connected from wiper to ground.
/Leevi

[jjasilli]

Per KOC TUT1, a resistor of that value in that location, following a MV, helps restore low end frequency response, which may suffer when a MV is used.

[alerich]

Specifically its purpose is to restore low end when the MV is turned to minimum levels to keep the tone from being too fizzy at low MV settings.

[SILVERGUN]

Thanks for the responses guys,,,,I had them in there and then took them out and didn't really notice any difference...

Then again I never had the MV below 5, because I'm trying to voice the amp aroud true stage volume.

I'll try 'em again at some point, but for now they're just in the way  

[Leevi]

Thanks for the responses guys,,,,I had them in there and then took them out and didn't really notice any difference...

I agree. I have built several PPIMVs and never used these resistors. Unfortunately I don't see the point theoretically either.
/Leevi

[HotBluePlates]

You've used 1/(2*pi*R*C) to calculate the bass roll-off of a coupling cap?

Well, the "R" isn't just the grid resistor following the cap (220k in your case); it's all the resistance from one leg of the cap to the other leg.

In a typical master-volume Marshall, they don't have the same arrangement your schematic fragment shows, and there's no second cap isolating the master volume control. When you have the master volume control at maximum, the resistance from leg-to-leg is at a maximum and the bass roll-off frequency is at its lowest.

As you turn the master down, there's less resistance from leg-to-leg and the bass roll-off keeps becoming a higher and higher frequency. The extra resistor in series with the wiper is added to have the roll-off frequency at a minimum master volume setting (slightly above off) at/below the desired cutoff so the amp doesn't seem crazy-bright at low master volume settings.

This assumes you chose the cap value to sound right when the master was at full volume. If you picked the value of the cap at a low volume setting, or if you made the cap value big enough that is was well below the guitar's range, then you might not need the isolating resistor in series with the wiper.

Try listening to a JCM800. They're much brighter with the preamp volume up and the master low, than if you switch volume settings and have the master volume at max.

Your diagram has a second related issue apparently not mentioned by the source of the schematic: at the full master volume setting the 2 coupling caps used are in series and their total value is half the individual cap values. Again, that's fine if you designed the amp with that minimum value in mind; if you add a PPIMV to an existing amp, you will probably want to make the new added cap 10 times the size of the other, original cap so that it doesn't reduce the total capacitance at full volume (and again, shave the bass but for the opposite reason addressed by the series resistor above).

[SILVERGUN]

You've used 1/(2*pi*R*C) to calculate the bass roll-off of a coupling cap?

No,,,unforunately, I'm just not that good...

BUT, Thanks again for taking the time to give me all the answer I need...
Basically and honestly what I AM doing is using known good(?) schematic snippets, to experiment and see what sounds good or not.....

This assumes you chose the cap value to sound right when the master was at full volume.

Yes, I've tried to keep that consistent, and made most of the "voicing" choices before putting the MV in there...
Since I put it in, I always keep it set at 8 to make modifications.......it's loud and my ears are paying the price, but I don't see another way...
I never bothered trying to get the PPIMV to sound good at low volume, because it's only there as a practice tool for giving my ears a break (it just doesn't sound all that good and I'm OK with that).......I just needed some control over full output and I was actually considering going down to 100K on the dual gang pots,,,in hopes that would give me a broader usable range near full output..

Here's the way it looks in my amp, purely from trial and error, and tweaking by ear...

[SILVERGUN]

You've used 1/(2*pi*R*C) to calculate the bass roll-off of a coupling cap?

I would like to undestand this.... (and I think I should before I continue on)
Do you have a couple minutes to teach a dummy how to wrap his head around that equation?
This is how I have to break it down in my head to start to even look at it :
1 divided by the sum of 2 X pi X resistance X capacitance = what?   (the exact frequency?)

I guess I should've paid attention in school (25 years ago).....I was too busy reading "Guitar For The Practicing Musician"

[HotBluePlates]

The equation is:

f = 1/(2*pi*R*C)

If you swap f and R, you get the equation for capacitive reactance:

R = 1/(2*pi*f*C)   --->  Xc = 1/(2*pi*f*C)

The formula for capacitive reactance tells you how much opposition to a.c. a cap has at a given frequency (akin to resistance at d.c.; "impedance" is the combination of reactance and resistance to describe the total opposition to a.c.)

Back to the original formula... By swapping the places of f and R, you're finding the frequency at which the cap has a reactance equal to the resistance in the circuit, R. When the two are equal, signal voltage is split equally between the cap's reactance and the circuit resistance, like a voltage divider.

As a result, this is voltage split equally between the resistor and the cap's reactance, a 45-degree phase shoft, and results in an output of 0.7071 times the input voltage. We describe that in decibels as -3dB, so the rearrangement gives the -3dB point of bass roll-off for a resistor/cap circuit.

Do you have a couple minutes to teach a dummy how to wrap his head around that equation?

"2*pi*f" is also called ω (lower-case omega), and is a mathematical shortcut to relate other parts of the equation to a.c. It makes the math come out right, and relates frequency to a complete a.c. signal cycle.

R and C are self-explanatory. C is always in farads (so you have to work with a bunch of decimals or scientific notation).

1/....  occurs mainly because the cap's reactance gets smaller as frequency goes up; inductance uses the same formula as capacitance, but has no 1/.... because the reactance of an inductor gets bigger as frequency gets bigger.

[sluckey]

As a result, this is the same as a volume control set to half, and results in half-voltage (and half-current) output. We describe that in decibels as -3dB, so the rearrangement gives the -3dB point of bass roll-off for a resistor/cap circuit.

-3dB relates to 1/2 power. -6dB relates to 1/2 voltage or current.

[SILVERGUN]

Thank you*1,000,000

I dont get it yet, but as with your other posts, I'll have to read it 10 or so times and try to apply it before I can say "I got it"

I can say that this is what separates the men from the boys on these subjects....I'm sure there are many guitar players here who cant get past the math,,,and I'm having a bit of a block with it myself

I think the guy who writes the book "Tube Amp Circuits for Guitar Players" with simplified math, and over-simple explanations of how they apply to specific circuits (with big words and pretty pictures) could make a billion dollars

Most of us get to the equations at the end of Merlin's first chapter and say to ourselves "what did I just read?"  "that was chapter one??" 

[SILVERGUN]

Thank you as well Sluckey....

I obviously have a lot more reading to do.......I can only describe it like this:
"I feel like the kid who missed the first year of Algebra, and just enrolled in Algebra 2"......way too many holes in my understanding

As a result, this is the same as a volume control set to half, and results in half-voltage (and half-current) output. We describe that in decibels as -3dB, so the rearrangement gives the -3dB point of bass roll-off for a resistor/cap circuit.

-3dB relates to 1/2 power. -6dB relates to 1/2 voltage or current.

I understand that explanation,,,but still can't wrap my head around how you got there.....where did -(3)db come from....is that a constant?

[sluckey]

frequency at which the cap has a reactance equal to the resistance in the circuit, R. When the two are equal, signal voltage is split equally between the cap's reactance and the circuit resistance, like a voltage divider.

where did -(3)db come from....is that a constant?

No, it's not a constant. -3db = 1/2 power. It's a logrithmic expression of the ratio of one power level to another power level. Specifically...

-3db = 10log(P1/P2), when P1/P2=1/2, so -3db = 10log(0.5)  (10log is used for power expressions)

-6db = 20log(V1/V2), when V1/V2=1/2, so -6db = 20log(0.5) (20log is used for voltage expressions)

Crank it on your calculator. Don't ask where 10log and 20log came from yet.

[12AX7]

This is apparently the same as the mod many implement with a pre PI master of putting a resistor off the wiper. What i found with my homebrew is at low settings without the resistor the amp was so dark i had to make the amp overall very bright in order to use it at low settings which is the only way i use it these days. But i didn't like the fact that as i turned it up it became intolerably bright and harsh and whole different and cr@ppy sounding amp. Putting the resistor there made the amp sound that bright even at low settings requiring me to re voice the amp and start using a lot of NFB where before i used none or very little. The result was great however because the amp now sounds better overall. The thing about this i don't understand is that it was mentioned above that a JCM sounds bright at low master settings, but why ?  Mine was the opposite. Without the resistor it was pure mud, which makes more sense to me because it's the same thing that happens with a guitar volume that has no treble bleed.

[SILVERGUN]

-3db = 1/2 power. It's a logrithmic expression of the ratio of one power level to another power level. Specifically...

-3db = 10log(P1/P2), when P1/P2=1/2, so -3db = 10log(0.5)  (10log is used for power expressions)

-6db = 20log(V1/V2), when V1/V2=1/2, so -6db = 20log(0.5) (20log is used for voltage expressions)

Crank it on your calculator. Don't ask where 10log and 20log came from yet.

I'm trying to stay with you Sluck,,,My brain just seized up  ......
I don't think there is a problem with the way you explained that,,,just with the way I see it
10log is the same as (P1/P2)?..or should I be multiplying 10log X P1/P2 ?....what dictates the value of P?............(the questions only get dumber from here)
(I'm over at Amazon.com looking for a simplified algebra book....I can't go on feeling this dumb anymore...it's embarassing)

I do think I can get it.....just need to stare at it for a while 

Very sorry that you guys have to babysit on this  ........I guess you don't have to do anything,,,so I really do appreciate you're time and help,,,,,just feel like I'm having trouble holding up my end

[sluckey]

Maybe a simple example... Amp 1 puts out 18 watts (call it P1). Amp 2 puts out 36 watts (call it P2). So, amp 1 only puts out one half the power as amp 2 (18/36=0.5). 18/36 is a ratio that equates to 1/2. I chose these numbers purposely. Another way to express this power ratio is with a logarithm similar to a bell curve. The unit would be Bell, but it's more convenient to use 1/10th Bell, or decibel, or dB. So, if you wanted to express that original power ratio in dB format, the conversion would be...

Power ratio (dB) = 10(log(P1/P2)) = 10*(log(18/36)) = 10*(log(.5)) = 10*(-0.3) = -3dB

Using Windows calculator, key in 18/36=log*10=, and the answer is -3.0103-----

...so, amp 1 is -3dB relative to amp 2. Another way to say it is amp 1 is 3dB DOWN from amp 2. And still another way to say it is amp 2 is 3dB UP (hotter than) from amp 1.

dB is just another way to express ratios. The advantage is more apparent when dealing with very small or very large ratios.  But all this talk is about POWER. When talking about voltage or current ratios, the conversion will be 20log(V1/V2). It's OK if you roll your eyes!

EDIT...strike thru my silly comment.

[SILVERGUN]

Thank you for breaking it down into smaller pieces for me  

I had to switch my calculator to scientific to see the log function  

At least the grey matter stopped dripping out of my ears,,, for now!

By walking me through the calculator entries, it really helped me grasp the equation..... \

The unit would be Bell, but it's more convenient to use 1/10th Bell, or decibel,

Learned something else in the process ......it's really nice to have a definition for a word I throw around pretty loosely

[Willabe]

Another way to express this power ratio is with a logarithm similar to a bell curve. The unit would be Bell, but it's more convenient to use 1/10th Bell, or decibel, or dB.

So that's where dB comes from as in;  .1 Bell? I'm thinking Bell cure is named after the man who came up with it?


            Brad    

[sluckey]

So that's where dB comes from as in;  .1 Bell? I'm thinking Bell cure is named after the man who came up with it?

I may have made up part of that! Sure seemed real though. I'm having trouble separating my realities! 

quote from wiki...

The decibel (dB) is a logarithmic unit that indicates the ratio of a physical quantity (usually power or intensity) relative to a specified or implied reference level. A ratio in decibels is ten times the logarithm to base 10 of the ratio of two power quantities.[1] A decibel is one tenth of a bel, a seldom-used unit named in honor of Alexander Graham Bell.

[Willabe]

Well look at that it's Alexander Graham Bell.

Decimal point Bell = 1/10 of a Bell = .1 Bell = Decibel = dB.


          Brad    

[SILVERGUN]

The equation is:
f = 1/(2*pi*R*C)

So, now that I'm a math wiz  ,,,I'd like to try to apply this equation to my coupling caps, to understand what I've done there

Sluckey or HBP,,,maybe you could walk me through this one because I want to make sure I understand where to get the resistance values from....

Equation will be 1/(2*pi*R*20000)= frequency....

20000 farad entry is based on .02 * 1,000,000.....correct?

Lets start with a dual-potentiomter setting of 100K to ground

[HotBluePlates]

Sluckey or HBP,,,maybe you could walk me through this one because I want to make sure I understand where to get the resistance values from....

Don't use that circuit as a starting point. Use a simple non-master volume amp. You pick the circuit and we'll show you.

I say that because it's no the value of 1 resistor in the circuit... it's the value of all resistance, including parallel paths, from one leg of the cap to the other.

[SILVERGUN]

OK, that sounds good....

How about this one,,,,this is how my build started: (pre-PPIMV)

[HotBluePlates]

Alright. We're gonna pretend a section of the 12AX7 looks like 62kΩ (I don't want to go through the torture of calculating it in this type of circuit).

Let's pick the upper coupling cap. On the left side, we have an 82kΩ plate load to B+ and the 62kΩ internal resistance of our tube. Since each of these goes to "ground" as far as a.c. is concerned, they are parallel resistances. 82k ll 62k = 35.3kΩ

If you started at the left lead of the cap, you might imagine the plate load and tube replaced by an equivalent resistor of 35.3kΩ to ground. To get to the right lead of the cap, you'd have to come up through the 220kΩ grid reference resistor. Ground is irrelevant, and traveling from one leg to the other, these resistors are in series. 35.3k + 220k = 255.3kΩ. Round that off to 255kΩ.

Now we'll use f = 1/(2*pi*R*C)

0.022uF = 0.000000022F  (or 22 * 10^-9)

f = 1/(2 * pi * 255,000Ω * 0.000000022F) = 1 / 0.03524867 = ~28Hz

The response will be 3dB down at 28Hz. Take it on faith right now that there will be roughly no roll-off at 10 times this frequency (at 280Hz). This arrangement therefore passes much of the guitar range with little or no effect.

[SILVERGUN]

0.022uF = 0.000000022F  (or 22 * 10^-9)

I went the wrong way with the zeros on my uf to f conversion earlier,,,so that's a big help

Thanks again, so much...I can see that!

I'll have to try to apply it myself to say "I got it" , but at least I CAN see it......and I see why the PPIMV circuit can make the calculations much more difficult 

[HotBluePlates]

You should also see why the setting of the master volume will change the apparent resistance to ground on one side of the coupling cap, and therefore change the bass roll-off frequency.

If the bass roll-off is very low to begin with, you might not even notice until the master volume is set very low. Funny thing is, at low acoustic volume, your ears perceive less bass as well so there's a double-whammy.

Aside from people saying an attenuator seems to result in treble loss, the reduction in volume contributes to why they say it doesn't feel like it goes as deep. I guess I shouldn't mention how your ears also are less sensitive to very high frequencies at low volume, too... Google "Fletcher-Munson Curve" if you want to see more on this topic.

[SILVERGUN]

You should also see why the setting of the master volume will change the apparent resistance to ground on one side of the coupling cap, and therefore change the bass roll-off frequency.

Yes,,,and that takes care of my initial question.....now I truly understand what those resistors are for, and I can go back over them later and see if I need them in there without just assuming I don't..........

Aside from people saying an attenuator seems to result in treble loss, the reduction in volume contributes to why they say it doesn't feel like it goes as deep. I guess I shouldn't mention how your ears also are less sensitive to very high frequencies at low volume, too... Google "Fletcher-Munson Curve" if you want to see more on this topic.

I am experiencing that first hand here, because the amp sounds dead and lifeless with whatever attenuator I try on it.....and I believe that it's the way my ears are hearing less high end at the reduced volume........and highs cover the whole spectrum, so the bass notes are missing the harmonic overtones of the highs (thats how i see it)........so I understand why someone would describe it as playing with a blanket over the speaker

My amp sounds OK from 1-6 (muffled) and then "BAM" 8 is soooo much better.....it's like the whole big round, crisp, detailed tone just jumps out of nowhere

[PRR]

-3dB is really the same as a stick slanted 45 degrees.

Figure the "normal" (midrange) gain of an amp is a stick straight up.

WAY-WAY outside the audio band, the stick has fallen flat, near-zero output.

It is mathematically convenient, and psychoacoustically useful, to define the fall-off as the point where the stick has fallen-over 45 degrees, its height is 0.7 times full height (any carpenter knows this), which in the log-scaled named for Bell is -3dB.

You can actually, barely, hear a -1dB drop. This happens about an octave inside the -3dB point.

For guitar, you want the core amplifier (before tone-tweak) to be -3dB below 80Hz and above 6KHz. For hi-fi we normally demand -3dB better than 20Hz-20KHz. (Which incidentally gives -1dB 40Hz-10KHz, which is about as much as speakers and listeners will manage _flat_.)

When you have many stages, each one must be scaled even wider. It is not unusual for "82Hz guitar amps" to have a lot of 10Hz-20Hz coupling networks.

dB cheat-sheet:
Measure in Voltage:
-1dB is 90%
-3dB is 70%
-6dB is 50%
-12dB is 25%
-20dB is 10%
-40dB is 1%

______________________________________________________

If you really can't do equations with pie....

0.01uFd into 1Meg gives 17Hz roll-off.

Then transpose by decades.

0.1uFd into 1Meg is 1.7Hz

0.1uFd into 100K is 17Hz

10uFd against 1K is 17Hz.

So 1uFd against 1K is 170Hz.
And 0.68uFd against 1K is about 250Hz. (Marshall bass-cutting cathode cap.)

Or find a Reactance Chart which mostly covers the audio band.

_____________________________________________________

> I went the wrong way with the zeros on my uf to f conversion

Starting with a slide-rule FORCES you to think about which way the zeros go.

In electronics, where resistor and capacitor values can vary over a million to one each, the zeros (decimal place) are often more important than the numbers. 0.1uFd and 0.5uFd will *both* work. 0.1pFd won't work for crap (and you can't buy one). 0.1 *Farad* is much-too-good (and costly).

It's like the US budget. Who cares if it's 1.23 or 1.47? The real thing is the zeros: billions or trillions?

[SILVERGUN]

PRR, thanks so much for taking the time to write that out for me.....I'm a sponge right now, and your "rules of thumb" can be absorbed quickly and are very valuable at this stage in the learning process.....

The last time you gave me a rule of thumb, I was immediately able to apply it to what I was looking at in my own schematic, while trying to gauge values for my plate and cathode resistors.....you said :
> When you have a significant plate resistor, the cathode bias resistor is *about* Mu times less. 12AX7 with 100K in the plate, try a 1K cathode resistor. 12AU7 with 47K in plate, try 2.2K.
(I had never looked at it that SIMPLY before)
And even though that is not an exact science, it "inspired' me to learn more about where that rule of thumb came from,,,and gave me a baseline to work off of...

So when you type, I read,,,,and then go back and read some more,,,,and I seriously thank all of you guys that take YOUR time to do that!!

[HotBluePlates]

-3dB is really the same as a stick slanted 45 degrees.

...

It is mathematically convenient, and psychoacoustically useful, to define the fall-off as the point where the stick has fallen-over 45 degrees, its height is 0.7 times full height (any carpenter knows this), which in the log-scaled named for Bell is -3dB. ...

Thanks for posting that! This clears up something I've half-known for a long time, but said wrong above (where Sluckey had to correct me).

45 degrees -> Π/√2 radians -> 0.7071 -> voltage split between cap and resistor (but not 0.5 voltage due to phase angle)

Thanks for the lightbulb moment!

[HotBluePlates]

About which part?

Figuring coupling cap rolloff frequency?
R-C Coupled Triodes, Chapter 12.2 pp 483-484; Filters, Chapter 4.8 pp.172-175

R-C high-pass/low-pass filters?
ibid; Pretty much all of Chapter 15 in various ways

Voltage division between resistance and reactance (the piece Sluckey and PRR corrected me on)?
Vectors and j-notation, Chapter 6.5-6.6 pp. 283-285; Condensers in a.c. circuits, Chapter 4.4 pp. 137-140

I probably should have gone back and looked, and it would have saved me time correcting wrong statements...