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-   -   Tone caps in series or parallel (http://www.talkbass.com/forum/f38/tone-caps-series-parallel-1016436/)

 wild4oldcars 09-18-2013 05:54 PM

Tone caps in series or parallel

so I was sitting in calculus class today, and of course, I stated daydreaming about tone pots :p

so I ask you, how would tow tone caps on the same pot behave if they were wired in parallel on the pot itself (one leg of each cap connecting to ground, and to the third terminal). How about series (ground->cap->cap->third terminal)? just a thought, feel free to completely shoot me down.

 GlennW 09-18-2013 06:00 PM

The values add when wired in parallel.

 wild4oldcars 09-18-2013 06:21 PM

cool, thanks

 line6man 09-18-2013 06:24 PM

Series: CTotal=1/([1/C1]+[1/C2]+...[1/Cn])
This is because series wiring increases plate separation.

Parallel: CTotal=C1+C2+...Cn
This is because parallel wiring increases plate area.

 Hopkins 09-18-2013 06:43 PM

Quote:
 Originally Posted by line6man (Post 14886283) Series: CTotal=1/([1/C1]+[1/C2]+...[1/Cn]) This is because series wiring increases plate separation. Parallel: CTotal=C1+C2+...Cn This is because parallel wiring increases plate area.

My brain just shut down.:eyebrow:

 ExaltBass 09-18-2013 06:45 PM

Quote:
 Originally Posted by Hopkins (Post 14886354) My brain just shut down.:eyebrow:
Me too!

 line6man 09-18-2013 07:13 PM

It's only basic algebra. Division and addition.

 giorob815 09-18-2013 07:29 PM

Quote:
 Originally Posted by Hopkins (Post 14886354) My brain just shut down.:eyebrow:
LOL. When I first saw this, I heard Jeff Foxworthy in my head going: "Letters? Since when did they start putting letters in MATH?!" :p

Rob
:bassist:

 line6man 09-18-2013 08:01 PM

Quote:
 Originally Posted by giorob815 (Post 14886495) LOL. When I first saw this, I heard Jeff Foxworthy in my head going: "Letters? Since when did they start putting letters in MATH?!" :p Rob :bassist:
Sometimes Calculus formulas are almost entirely letters.
If h(x)=f(x)/g(x) then h'(x)=(f'(x)g(x)-f(x)g'(x))/g(x)^2
Or dy/dx= dy/du*du/dv*dv/dx.
What the hell!? It's hard stuff! :bawl:

Lol.

 giorob815 09-18-2013 08:05 PM

Oh, I remember. Once upon a time, I used to be very smart. ;) :p

Rob
:bassist:

 wild4oldcars 09-18-2013 08:27 PM

so calculating total capacitance, is kinda the inverse way of calculating total resistance? (high school physics, don't fail me now lol)

 line6man 09-18-2013 08:45 PM

Quote:
 Originally Posted by wild4oldcars (Post 14886690) so calculating total capacitance, is kinda the inverse way of calculating total resistance? (high school physics, don't fail me now lol)
Parallel resistance, yes. It's the same formula (series capacitance) for parallel resistance, parallel impedance, and parallel inductance.

 megafiddle 09-18-2013 09:06 PM

When you only have two values, just use product over sum -

C total = ( C1 x C2 ) / (C1 + C2) for caps in series

R total = ( R1 x R2 ) / (R1 + R2) for resistors in parallel

...

 Hopkins 09-18-2013 09:09 PM

This thread makes me realize how uneducated I really am.

 Jensby design 09-19-2013 04:21 AM

Quote:
 Originally Posted by line6man (Post 14886283) Series: CTotal=1/([1/C1]+[1/C2]+...[1/Cn]) This is because series wiring increases plate separation. Parallel: CTotal=C1+C2+...Cn This is because parallel wiring increases plate area.
:rolleyes:Duh
.022u + .047u parallel = .069u
.022u + .047u series = .015ishu
Simply math really:smug:

 bassbenj 09-20-2013 01:33 AM

Quote:
 Originally Posted by Jensby design (Post 14887569) :rolleyes:Duh .022u + .047u parallel = .069u .022u + .047u series = .015ishu Simply math really:smug:
The confusing part of the series formula is when you try to do the one over the one over thing. What I mean is that in series the values would be: 1/C total = 1/C1 + 1/C2 + 1/C3 + etc.

Then to get the actual total value you have take one over the sum. It's a simple thing but when you write it all out in one step in a formula it messes with your mind.

By the way the same thing is true for Ohms only the formulas are reversed. Resistances in series simply add and resistances in parallel use the "one over" formula above. Hence the above formula also works for finding the Ohms of cabs in parallel (the usual connection when plugged into the back of an amp)

Example: 4 Ohm cab and 8 ohm cab: 1/4 + 1/8 = .375 and 1/.375 = 2 and 2/3 ohms. Which is an important question to see if your amp can handle the load. Answer: A head with a 2 Ohm minimum would handle it, an amp with a 4 Ohm minimum would blow up.

We are all old enough here to have not been subjected to modern public education, right?

Quote:
 Originally Posted by Jensby design (Post 14887569) :rolleyes:Duh .022u + .047u series = .015ishu Simply math really:smug:
1/22 + 1/47 = 47/1034 + 22/1034 = 69/1034 => 1034/69 = 14.986

14.986 nF > 0.015 µF q.e.d. ... :)

 Jensby design 09-20-2013 11:07 AM

Quote:
 Originally Posted by Cadfael (Post 14892372) 1/22 + 1/47 = 47/1034 + 22/1034 = 69/1034 => 1034/69 = 14.986 14.986 nF > 0.015 µF q.e.d. ... :)
Precisely :D
__________________
damn kids and their math

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