1/8 Watt Resistors in Guitars?

[malloryboy]

Hi,

I normally use 1/4w resistors in my guitar circuitry as they are the most commonly available and cheap.

Someone gave me some cool 1/8 Watt Carbon Composition resistors and I was wondering if there is any reason these can't be used in guitar circuitry.

Sure they are smaller which is not bad say for treble bleeds and small cavities.

Given the low voltage of a guitars circuit I can't see it as having any difference but I don't know what wattages are involved.

Could someone with Electronics knowledge say something about what wattage values for resistors are good and which to avoid?

Many many thanks

[newey]

MB-

You asked:

Could someone with Electronics knowledge say something about what wattage values for resistors are good and which to avoid?

I won't claim to have much in the way of electronics knowledge, so take this answer with a rather large grain of salt. But my wholly-off-the-cuff response would be that 1/8 watt shouldn't be a problem.

Watts = volts X amperes. Your guitar generates about 1 volt. The current level is very low, measured in milliamperes. Doing the math, only tiny fractions of a watt are involved, so 1/8 watt should cover it.

But you may want to wait for a more definitive answer before wiring anything up . . .

[ChrisK]

Could someone with Electronics knowledge say something about what wattage values for resistors are good and which to avoid?

I don't know, how hard do you play? ;D ;D

The current level is very low, measured in milliamperes.

It's even lower if'n you measure it in amps....... ;D ;D

A typical single coil pickup has an internal resistance of aboot 5 K Ohms. As per the maximum power transfer theorem, the maximum power is, uh, transfered when the load impedance equals the output impedance. This would be at 5 K Ohms. Using the 1 VAC example, this would result in 0.5 VAC across the external load of 5 K Ohms, which results in 0.5/5,000 or 100 uA (0.0001 A, or in newey's case 0.1 Ma ). This results in a power dissipation of 50 uW (0.00005 W).

This is aboot the maximum worst case that can generated by passive electric guitar circuitry. The reason why we use 1/4 or 1/8 watt components is that we tend to lose sight of those with much smaller ratings once we put them down (somewhere, dang it).

'Tain't much, aboot the magnitude of a bee's sneeze.

[newey]

Ok, so I was off by a decimal place . . . At least I was off on the safe side!

[ChrisK]

Ok, so I was off by a decimal place

Nobody was off at all!

100 uA, 0.1 Ma, or 0.0001 A is the same exact thing in an absolute sort of way.

50 uW, or 0.05 mW or 0.000,05 W is the same thing as 0.000,000,05 KW or 0.000,000,000,05 MW.

To be really safe we should go with 0.000,000,000,1 MW resistors.

[malloryboy]

You've all certainly answered my question without any doubt.

ChrisK, your comment about why we use bigger things certainly is the case. I am always misplacing things as it is.

Well a big thank you to all of you for taking the time to put my mind at ease.
Cheers

[ashcatlt]

Thought I'd post this link in this thread, since it seems to be related. R.G. is pretty well respected over there, being a "big name" pedal designer.

He says SC ~ 100mV, HB ~ 500mV.

Later in the thread, we get the info that EMG actives can be up to ~4V.

[ChrisK]

Most EMG active pickups have an internal 10K Ohm resistor in series with their output (otherwise you couldn't select two together - the outputs would short).

Yep, this means that their maximum output power transfer occurs with a 10 K Ohm external load. This is 2 VAC across 10 K Ohms or 2^⁴/10,000 = 0.0004 Watt or 400 uW.

As the external load increases or decreases, the transfered power decreases.