Shielding/Capacitor question

[gator]

I am going to shield my guitar (older Yamaha strat copy) following the instructions on GuitarNuts. I just have to wait on getting my new pickups and BluesKaster from Torres Engineering. My selector switch is bad so I figured that this would be a good time to put in some better electronics. My question is which capacitor to use? Will any capacitor classified as a .33uF 400v work? I mean there are Polyester, Polypropylene, etc. I am looking at getting part # 150334J400IE (Mallory) from Mouser. Will this one work?

[sumgai]

gator,

Yep, that'll work just fine. Might be a tad expensive for shipping, though.

BTW, to the forums!


sumgai

[gator]

Thanks Sumgai, so if I go to like Radio Shack or a place like that all I have to look for is a capacitor that is .33uF and 400v? Whether its Polyester or Polyporpylene really doesnt matter?

[sumgai]

gator,

Correct, it won't matter what the construction material is, for our purposes here.

However, Radidio Shabby has a poor selection of capacitors. You may need to purchase 3 each 0.1 µfd caps, and wire them up in parallel in order to obtain the value we want. 0.3 is close enough to 0.33, no problem there. And it'll still be cheaper than shipping from Mouser.


sumgai

[vonFrenchie]

You could also buy 3 1 µfd capacitors (yes 1.0) and wire them in series to achieve .33 µfd of capacitance. I love one uping sumgai

[ChrisK]

Well, I hate to one-down you.

Since one cannot control the voltage developed across each capacitor in series (unless equalizing resistor are used which defeats the purpose of the cap), each cap would have to have the full 400 VDC rating at a minimum.

Cost and size are.

[vonFrenchie]

Wait wouldnt that still be one-uping?

Anyway finding three 1 µfd non polarized 400 volt capacitors isnt very easy. I've never seen 1µfd capacitors above 250 volts. You see it's my useless knowledge. It always shows.

Anyway I recommend these
www.smallbearelec.com/Detail.bok?no=176

The sprague ones are good too but they are a buck a piece.

[gator]

Thanks for the info guys. I found out that there is an electronics store in my area that is a little more specialized than Rat Shack.

Funny story though...I have not soldered anything in like 15 years so I figured I would practice with some 12 gauge stereo cables that I have left over from my home theater. I was concentrating on heating the area and not touching the solder to the iron. Well as I am melting the jacketing and the copper is changing color the solder would still not melt (good thing this is practice). I am thinking to myself "How the hell do people get the solder to melt without touching it to the iron? I can barely get it to melt when I touch it directly to the iron." Well I finally look at my solder and it is silver bearing solder size .093 hahaha. Well I picked up some .050 63/37 solder and let me just say that it is a relief how easy soldering really is when you have the right supplies. Now all I need is for those darn pickups to arrive. ;D

[vonFrenchie]

Its called Radioshack for a reason. If there was a better selection it would be RadioApartment or RadioHouse

[sumgai]

vF,

Wait wouldnt that still be one-uping?

Anyway finding three 1 µfd non polarized 400 volt capacitors isnt very easy. I've never seen 1µfd capacitors above 250 volts. You see it's my useless knowledge. It always shows.

Anyway I recommend these
www.smallbearelec.com/Detail.bok?no=176

The sprague ones are good too but they are a buck a piece.

Sorry, but ChrisK has it right, you've been downed.

You're correct as far as you go, capacitance does indeed divide as you string units together in series. However, as Chris explained, the voltage seen by that string of units is fully present on each unit..... there is no division (or other arithmetic) at work here. This means that all 400 volts (of our example) will be present on each unit, which in turn means that your 1µfd cap must be quite sizable in order to withstand that kind of juice. And you know what that'll do to the price, I'm sure.

(Edit The next paragraph contains faulty info, see my Reply #13 below for corrections.)
OTOH, when we combine capacitors in parallel, not only do the values add together (algebraically), the voltage placed across the assembly is divided proportionally for each unit. Thus, no one unit will receive the full 400 volts (of our example), but some portion thereof. How much? Well, if each cap is the same value, then just divide the incoming voltage by the number of units, and you're done. But if the values are different, then you'll need to use the same method you did when combining the capacitor values for a series combination. Determine the total capacitance, find what percentage each capacitor is when compared to the total value, and then use the resultant percentage values as percentages of your total voltage. That'll get you more than close enough.

BTW, I liked your explanation of the name, and your extrapolation thereof. ;D But Gawd help us all if we ever have to shop at RadioCondo!


sumgai

[vonFrenchie]

I got owned to the 2nd degree

[sumgai]

vF,

I got owned to the 2nd degree

Hey, you're on your way to having your very own Black Belt of ownership, just a few more degrees to go! ;D


sumgai

[JohnH]

OTOH, when we combine capacitors in parallel, not only do the values add together (algebraically), the voltage placed across the assembly is divided proportionally for each unit. Thus, no one unit will receive the full 400 volts

sumgai

"....BEEP! BEEP! DANGER WILL ROBINSON...!!!...."

Caps in parallel all get the same full voltage across each one.

As for series, I'll believe what ChrisK says about also needing to rate the caps in a series chain at the full voltage. Resistors in a chain will divide the voltage, but caps may behave differently in practice..

John

[sumgai]

AARRGGHH!!

John, you have it correct, I was (once again) bass-ackwards.

When caps are in parallel, the full DC voltage will appear across each and every one of them. In series, they will each receive some portion of the voltage (dependent on the time constant of the cap's value). The "classic" cure for that is to run a bridging resistor in parallel with each cap. This is where the calculation I mentioned above comes into play.

The easy way to remember all this is to look at a typical Fender schematic. Therein, we see that the power supply has two 350 WVDC (Working Volts DC) wired together in series, such that the 450 or 500 volts from the rectifier will not blow up the caps.

I should have remembered this before I got into my long-winded discussion above.


sumgai