Simulcast capacitor voltage ratings

C3 & C10 need to be 35V or more.
C11 & C16 should be 25V or more.
The other electrolytic caps should be 15V or more.
The film & ceramic caps will all be at least 50V because you'd have a hard time finding any with a lower rating, so no need to worry about those.
 
Okay thanks. I have few questions. I apologize in advance for the mess of capacitor references to follow. A diagram may be a better way to communicate this...

But first, by C16, did you mean C100 or C6? There's no C16 on the schematic or BOM.

I'm mostly concerned about fitting C3 and C10 (the 120u electrolytic caps). I didn't drop below 25V for any of my tolerances. Currently on my build the bottom row of 10u caps (C12, C13, C14, C15, C1, and C4) are all rated for 35V and have a diameter of 5mm. No problem there. The top row of caps is where things get tight. C11, C8, and C6 are 100u caps. I have them rated at 25V with a diameter of 6.3mm and they fit fine. C3 and C10 (120u 35V) finish out that row. I can't find 120u 35V caps with a diameter of 6.3mm, though. The only ones I could find are 8mm, and that is too big. Is my best option to replace C6 and C8 with 5mm caps and use the freed space to fit the 8mm 120u caps in?

EDIT: Since there's space to the side of C8, I guess an easy solution would be to bend leads and shift everything over to make room for the 120u caps that I have now. So, C3 (120u), C10 (120u), and C8 (100u) would all be shifted a bit.
 
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On mine, for the larger caps I just used heat shrink as insurance on the exposed leads and didn't insert them all the way down then kinda bent em over. Sorta like this...
IMG_20190920_090305.jpg
 
On mine, for the larger caps I just used heat shrink as insurance on the exposed leads and didn't insert them all the way down then kinda bent em over. Sorta like this...
View attachment 1476

Ah okay. So you shifted C3, C10, and C8 (in the edit to the reply above)? I looked at your build report after I posted the first reply and noticed that your caps in the top row didn't match with the silk screen. I figured that that is what you had done. Thanks for the picture and explanation! It's odd there aren't 6.3mm 120u 35V caps. Mouser has some listed as 'New!', but they are non-stocked.
 
So I must have been using an older build doc that didn't specify voltages on the capacitors. For the 120uF caps, what voltage is actually running through them assuming a 9v power supply? How much danger am I in by using 25v caps there?
 
Ah okay. So you shifted C3, C10, and C8 (in the edit to the reply above)? I looked at your build report after I posted the first reply and noticed that your caps in the top row didn't match with the silk screen. I figured that that is what you had done. Thanks for the picture and explanation! It's odd there aren't 6.3mm 120u 35V caps. Mouser has some listed as 'New!', but they are non-stocked.

Also, Digikey has those caps actually in stock.

 
Sorry, typo, I meant C6. You could sub 100uF for the 120uF caps and not hear the difference. Here's the deal with capacitor voltage ratings: you want some margin. You don't want to run 25V caps at 25V. I prefer at least 20% margin. Vcc will run at about 26V if the input is 9V. Fresh batteries and most power supplies deliver more like 9.2V. That will make Vcc closer to 27V. A 35V cap will have 30% margin.
 
Right on with the ratings...I just wasn't sure exactly where in the circuit the voltage divided from Vcc, so I just stuck with 25V minimum for everything that wasn't specified 35V.

I can't really tell what the purpose of the two 120u caps is? Could you unpack that?
 
Sorry, typo, I meant C6. You could sub 100uF for the 120uF caps and not hear the difference. Here's the deal with capacitor voltage ratings: you want some margin. You don't want to run 25V caps at 25V. I prefer at least 20% margin. Vcc will run at about 26V if the input is 9V. Fresh batteries and most power supplies deliver more like 9.2V. That will make Vcc closer to 27V. A 35V cap will have 30% margin.

Thanks for response! Curious how you got to the 26v calculation. Any insight on how the charge pump dishes out so much voltage? I think I'm only familiar with charge pumps as voltage doublers so I thought it was at 18v operation.
 
C3 & C10 are power supply bypass caps. Any AC noise that tries to get on the power bus gets shunted to ground thru those caps. What's so special about the 120uF value? Nothing. Anything in the ballpark will work.
Cheers. That makes sense. I didn’t really consider that since there’s power filtering before the charge pump.
 
The TC1044, LT1054 and ICL7660 charge pumps are capable of voltage inversion (making -9V from +9V) and voltage multiplication. They all do pretty much the same thing and are more-or-less pin-compatible. In the case of voltage multiplication, the charge pump switches pin 2 between pin 8 (9V in) and pin 3 (ground). C12 & C14 ride on pin 2. When pin 2 is connected to ground, C12 and C14 are charged by D1 & D3, respectively. When pin 2 is connected to +9V, the charges on C12 & C14 are dumped thru D2 & D4 into C13 and C15. For the moment, we'll ignore the diode drops for simplicity. Since C12 is charged to 9V when pin 2 is low, when pin 2 goes high, the + side of C12 is now at 18V and C12 dumps its charge into C13. C14 gets its charge from C13, so it gets charged to 18V when pin 2 is low and when pin 2 goes high, the + side of C14 goes to 27V and dumps it's charge into C15. It's similar to a bucket brigade. Here's how you calculate the output voltage from a multiplier like this:
1) Count the number of capacitors connected to pin 2 and add 1. In this case, that number is 3.
2) Multiply that number by the input voltage, usually 9.0V to 9.2V. 3 * 9.2V = 27.6V
3) Subtract out the diode drops. Schottkys like 1N5817 drop about 0.2V each and there are 4 of them, so in this case we subtract 0.8V. 27.6V - 0.8V = 26.8V. That's the approximate output voltage of the multiplier.

In some pedals the diodes are 1N400x, which have a drop of about 0.65V each.

You can multiply the voltage as high as you like by adding more diodes and capacitors. The only limitations are the voltage ratings of the capacitors and the amount of power you need to deliver.

And that concludes tonight's lesson on charge pump voltage multipliers.
 
The TC1044, LT1054 and ICL7660 charge pumps are capable of voltage inversion (making -9V from +9V) and voltage multiplication. They all do pretty much the same thing and are more-or-less pin-compatible. In the case of voltage multiplication, the charge pump switches pin 2 between pin 8 (9V in) and pin 3 (ground). C12 & C14 ride on pin 2. When pin 2 is connected to ground, C12 and C14 are charged by D1 & D3, respectively. When pin 2 is connected to +9V, the charges on C12 & C14 are dumped thru D2 & D4 into C13 and C15. For the moment, we'll ignore the diode drops for simplicity. Since C12 is charged to 9V when pin 2 is low, when pin 2 goes high, the + side of C12 is now at 18V and C12 dumps its charge into C13. C14 gets its charge from C13, so it gets charged to 18V when pin 2 is low and when pin 2 goes high, the + side of C14 goes to 27V and dumps it's charge into C15. It's similar to a bucket brigade. Here's how you calculate the output voltage from a multiplier like this:
1) Count the number of capacitors connected to pin 2 and add 1. In this case, that number is 3.
2) Multiply that number by the input voltage, usually 9.0V to 9.2V. 3 * 9.2V = 27.6V
3) Subtract out the diode drops. Schottkys like 1N5817 drop about 0.2V each and there are 4 of them, so in this case we subtract 0.8V. 27.6V - 0.8V = 26.8V. That's the approximate output voltage of the multiplier.

In some pedals the diodes are 1N400x, which have a drop of about 0.65V each.

You can multiply the voltage as high as you like by adding more diodes and capacitors. The only limitations are the voltage ratings of the capacitors and the amount of power you need to deliver.

And that concludes tonight's lesson on charge pump voltage multipliers.

This is fantastic!!! Seriously thank you so much for the thorough explanation. I so appreciate it!
 
The TC1044, LT1054 and ICL7660 charge pumps are capable of voltage inversion (making -9V from +9V) and voltage multiplication. They all do pretty much the same thing and are more-or-less pin-compatible. In the case of voltage multiplication, the charge pump switches pin 2 between pin 8 (9V in) and pin 3 (ground). C12 & C14 ride on pin 2. When pin 2 is connected to ground, C12 and C14 are charged by D1 & D3, respectively. When pin 2 is connected to +9V, the charges on C12 & C14 are dumped thru D2 & D4 into C13 and C15. For the moment, we'll ignore the diode drops for simplicity. Since C12 is charged to 9V when pin 2 is low, when pin 2 goes high, the + side of C12 is now at 18V and C12 dumps its charge into C13. C14 gets its charge from C13, so it gets charged to 18V when pin 2 is low and when pin 2 goes high, the + side of C14 goes to 27V and dumps it's charge into C15. It's similar to a bucket brigade. Here's how you calculate the output voltage from a multiplier like this:
1) Count the number of capacitors connected to pin 2 and add 1. In this case, that number is 3.
2) Multiply that number by the input voltage, usually 9.0V to 9.2V. 3 * 9.2V = 27.6V
3) Subtract out the diode drops. Schottkys like 1N5817 drop about 0.2V each and there are 4 of them, so in this case we subtract 0.8V. 27.6V - 0.8V = 26.8V. That's the approximate output voltage of the multiplier.

In some pedals the diodes are 1N400x, which have a drop of about 0.65V each.

You can multiply the voltage as high as you like by adding more diodes and capacitors. The only limitations are the voltage ratings of the capacitors and the amount of power you need to deliver.

And that concludes tonight's lesson on charge pump voltage multipliers.

Any idea how the Siumlcast would sound with only one 120u cap?
 
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