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ACK Radio Supply - 1/8w resistor in place of 1/4w?

Started by AwesomeTyler, January 08, 2017, 03:17:59 PM

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AwesomeTyler

I recently made a trip to an absolutely awesome place in Atlanta called ACK Radio Supply. It's run by a couple of gentlemen and lady, all of which SERIOUSLY know their stuff. They also happen to carry just about every electric component under the sun, including different brands. I Definitely suggest it if you happen to be in the area, even if just for the relief that such a place still exists. They all hangout by the parts cabinets in the back while you callout your parts list to them. You call it out and go down the list (literally did this with the Moodring shopping list up on my phone) and they fill little baggies with what you're calling for and label it.

The downside of this is that you don't get to checkout the components yourself, aside from the cool stuff they have hanging up in their storefront, which is rather small. After I was done with the parts, I checked out and went on my merry way. After I'd left, I remember thinking that $30 was a helluva lot for a handful of resistors and a few caps, so I take a look and notice that almost all the resistors are metal film Vishay-Dale RN55D (BADASS!), a few of them Xicon, and the caps are ceramic dipped or multi-layered..which I probably won't use, seeing as that I can't be certain of the material.

ANYWAYS..tl;dr - It looks like all the resistors I got were 1/8 watt instead of 1/4 watt, and I'm unsure of whether or not they can still be used with the Moodring board.

Would there be any discernible/noticeable differences if I use these resistors, if I can even use them at all? I've recently read that the power rating of a resistor is directly related to it's potential contact noise.

Thanks in advance!  ;D

-Tyler


jimilee

Atlanta isn't too far from me, I will check it out!
Pedal building is like the opposite of sex.  All the fun stuff happens before you get in the box.

reddesert

Power dissipated in a resistor is P = V^2/R.  So R = V^2/P.  For a pedal with a 9V power supply (and no charge pump/voltage doubler), the max voltage a resistor would ever see is 9V.  To dissipate 1/8 W, you would have a resistance of R = (9^2) / (1/8) = 648 ohms. Any higher resistance will dissipate less power.

To leave a small safety factor, any resistor over 1K in a 9V pedal will always be dissipating under 1/8 W.  In the Moodring, the only small resistor is a 470 ohm that is from opamp output to coupling cap (so it won't see a lot of voltage), and most of the circuit runs off 5V from a regulator anyway. I would not see a problem using 1/8 W resistors; you could use a 1/4 W for the 470R if you wanted to be extra cautious.

I would just go ahead and use the capacitors regardless of knowing what kind of ceramic cap they were. No reason to waste them.

AwesomeTyler

Quote from: reddesert on January 08, 2017, 10:49:13 PM
Power dissipated in a resistor is P = V^2/R.  So R = V^2/P.  For a pedal with a 9V power supply (and no charge pump/voltage doubler), the max voltage a resistor would ever see is 9V.  To dissipate 1/8 W, you would have a resistance of R = (9^2) / (1/8) = 648 ohms. Any higher resistance will dissipate less power.

To leave a small safety factor, any resistor over 1K in a 9V pedal will always be dissipating under 1/8 W.  In the Moodring, the only small resistor is a 470 ohm that is from opamp output to coupling cap (so it won't see a lot of voltage), and most of the circuit runs off 5V from a regulator anyway. I would not see a problem using 1/8 W resistors; you could use a 1/4 W for the 470R if you wanted to be extra cautious.

I would just go ahead and use the capacitors regardless of knowing what kind of ceramic cap they were. No reason to waste them.

Thank you so much! But will the additional 'work' on the 1/8w cause any problems with noise?

Regards,
Tyler

reddesert

The power rating is just a "do not exceed" rating. Two 1K resistors of, say, 1/4 W and 1/8 W function exactly the same in a circuit, and one doesn't cause any more loading or power dissipation than the other.

It's possible that they could generate slightly different amounts of internal noise, but, speaking in generalities, many effects circuits will be totally insensitive to that, and in those that are sensitive to noise, it will probably only be certain parts of the circuit that are, and shielding / crosstalk between adjacent wires is likely a bigger effect, etc etc. It shouldn't be a high priority issue to worry about when starting to build pedals.