Quote:
Originally Posted by originalmatthew
I'm sure we could quibble all day long wether you can audibly tell the difference between different grades of wire inside the guitar. In the electrical world, less resistance=less signal degridation. I was lucky enough to score a roll of vintage .22 ga. silver wire (not the tinned copper, but real silver wire) and maybe my ears are shot, but I "think" I can hear a difference between the silver wire and the cheap thin stuff used today. Of course, it could all just be in my mind, but silver wire has a certain cool factor. I agree that shielding is far more important to sound quality (ie, reducing hum and such) than the wire is. The pots, switches, and especially the pickups are still what they are. Maybe we could get pots and switches with silver or gold contacts?!?!?! I wonder what pups wound with fine silver wire would sound like? Might be an interesting experiment, but even then, resistance is resistance. How many Henries does it take to make a Hank? I dunno. I'm sure a fancy machine could actually measure the difference. I just know I either like a sound or I dont. :o)
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I’m glad you admit it could be in your mind. Rational thought is always welcome. One would think (as you have) better electrical conductivity of conductors in a guitar would result in fewer losses. That seems perfectly reasonable. They do. But you can’t hear it. Let’s go through an exercise.
How long is the wire in a Tele? Let’s be conservative and say it is 2 feet. Let’s say we used 22AWG copper. The resistance of 22AWG copper is 0.016 ohms per foot. So at 2 feet we have 0.032 ohms.
A typical Tele pickup has 7000 ohms of DC resistance. At 2.3 Henrys of inductance the reactance of the pickup at 1kHz is 14.4k ohms. To keep it simple let’s ignore the typical 160pF of inter-winding capacitance of a Tele pickup and the amp/cable capacitance. The complex impedance of the DC resistance plus the reactance of the pickup inductance is:
Square root (14400^2 + 7000^2) = 16011 ohms impedance
Now let’s add the copper wire resistance. 16011 + 0.032 = 16011.032
(This conservatively ignores the wire from the pots to the jack and the amp cable. Let’s use it up where it impacts the voltage divider of the pots)
Assuming the volume and tone pots are 250.000000000000kohms, and the amp impedance is like a Fender Twin or 1.00000000000 meg ohms , the total load is around:
1/(1/250k + 1/250k + 1/1000k) = 111.1111 kohms
Let’s say the voltage generated in the pickup winding is 100mV. The voltage drop due to the pots and amp as voltage dividers are:
Vout = 100mV(111.111k ohms)/(111.111kohms + 16011.032 ohms) =
0.87404990505501044854286155526526 mV
ENTER SILVER WIRE
The resistance of silver is 95% of copper resistance. So now your 2 feet of wire is:
0.032 ohms x 0.95 = 0.0304 ohms
Let plug the new low resistance of silver into the equation.
Vout = 100mV(111.111k ohms)/(111.111kohms + 16011.0304 ohms) =
0.8740499160560921940718152657826 mV
The output with silver wire is 1.26 millionths higher or 10.9 millionths of a dB !!!!
Ok. I got a little crazy with the precision of the numbers, but I did it to prove a point. Wiring resistance is completely negligible. The variation of tolerance of the potentiometers is orders of magnitude more significant than wire resistance. Notice I assumed the volume and tone pots were at 250.000000k ohms. They are never close. Most pots measure closer to 240K ohms.
I can address the capacitance part of the effect later if anyone cares.
I don’t intend to come across in a negative way here. Electronics engineering is not common knowledge, and one would expect silver wire to be better because is it lower resistance. It’s just so completely negligible in guitars, unless of course you can hear 10.8 millionths of a dB.
