Josh and Humbuckers and 250k Pots

I met a nice man named Josh today. He brought in a couple of Tele’s and was interested in getting quotes for wiring. He knew how to wire them himself for the most part. He had a good question though and one a lot of people ask. He asked me if you can you use 250K pots with a humbucker? And the answer is yes you can! I gave Josh a very quick intro to Ohms law but if you are not familiar with it, then it usually makes one’s eyes glaze over, it is quickly forgotten and conversation moves on to something else. So I’ll explain it here as short as I can.

Ohms law is a simple math formula that describes the relationship of Voltage to Current and Resistance. It is fundamental to ALL electrical theory. It applies to your house, your car, your stereo and your guitar. Here it is —- V = (C x R) . This means the following:

V = Voltage – Voltage is measured in volts or smaller pieces of volts, don’t worry about it. Just know volts.
C = Current – Current is measured in amps or smaller pieces of amps, don’t worry about it. Just know amps.
R= Resistance (your pot) – Resistance is measured in ohms, usually thousands of ohms, don’t worry about it, just know ohms.

Voltage divided by Current equals Resistance.

Voltage divided by Resistance equals Current.

Current times Resistance equals Voltage.

All very simple math. But how it is applied can be confusing!

Your pickup is a voltage generator and it has it’s own internal resistance and your Pot is obviously a resistance. Let’s make up some simple phony numbers so you can see how it works and why a 250k pot or 500k pot will work with any passive pickup and what the effects will be. It is the relationship of the Voltage, Current and Resistance that you should understand. The phony numbers I use here are not anything like the real numbers and that is on purpose to keep this simple.

Let’s just say that the voltage output of your pickup is 20 volts and it’s own internal resistance is 10 ohms.
The highest current your pickup can generate when a string is plucked by Ohms law is 20 volts divided by 10 ohms equals 2 amps.

Let’s say that your pot is 10 ohms and we connect that to your pickup. Now the total resistance is the internal pickup resistance of 10 ohms plus your pot resistance of 10 ohms and that equals 20 ohms.

So now your pickup can only make 1 amp of current because Ohms law says 20 volts divided by 20 ohms equals 1 amp.

Now if we have 1 amp traveling through two resistors and each one is 10 ohms we need to know the voltage at the point between the two resistors. Ohms law again says 10 ohms times 1 amp equals 1 volt.

So this 1 volt is the voltage on the pot that goes to your amp when the pot is at maximum.

Here we go, now let’s say we change your pot to 5 ohms and we do all the math again but with just this pot changed.

The maximum current your pickup can generate by Ohms law is 20 volts divided by its internal 10 ohms and that equals 2 amps. Same as before.

We connect the 5 ohm pot to your pickup and now the maximum current your pickup can make by Ohms law is 20 volts divided by 15 ohms (10 ohms internal resistance plus the 5 ohm pot equals 15 ohms) and that equals 1.3 amps.

Now we want to know what the voltage is at the top of our 5 ohm pot and from Ohms law it is 5 ohms times 1.3 amps equals 6.5 volts. This would be the highest voltage your humbucker will give your amp with the 5 ohm pot at maximum.

From this simple example your 500k pot is the 10 ohm pot in the first example and the 250k pot is the 5 ohm pot in the second example.

So you can see from these numbers that by using a 250k pot instead of 500k you will have lower output from your humbucker but it will still work and you might like it!

It does get much more complex than this because there are other parameters that can be considered like capacitance and inductance. These are used in complex AC analysis to determine frequency response but they are not necessary to show you how different pots will work. Study electronics if you want to know this.

You might find that using a 250k pot with a humbucker reduces the high frequency sound of your pickup in addition to reducing the output. This would be because of the other parameters mentioned, capacitance and inductance. If it does and you don’t like it then just change the pot to a 500k.

I hope this helps those who are interested in learning how pots work in guitars.

By the way, 250k or 500k really means 250,000 ohms and 500,000 ohms respectively, the ‘k’ means thousands. And pickups generate 100’s of millivolts. A millivolt is 1 thousandth of a volt. So don’t go around telling people your pickup puts out 20 volts at 2 amps because people that know won’t invite you to any of their parties. Just don’t worry about real world numbers as they are not important to understanding how the circuit actually works.

One more note. Guitar pots vary in resistance by up to 20%. If you are NASA and spend taxpayer money then you can get them in 1% tolerance. But a guitar player cannot generally get them that way.  When you buy a 500K pot it might be 400k ohms or 390k ohms or 480k ohms and the same for 250k pots! They can be all over the place. Really picky people will measure them to get the closest to the desired value but in real world application it makes so little difference it truly is not worth the trouble. Spend your time playing your guitar. If you tell an engineer you spent all night matching pots for a guitar you won’t be invited to his or her party either. That is for another article sometime on debunking the myths of pots (big ones versus little ones), capacitors(some real poopoo in here), wiring, wood, cryogenic treatment and a host of other mojo unexplainable type things.

Note – For all you electrical snobs out there I did not use EIR on purpose because non engineers think in terms of V for voltage, C for current, and R for resistance because it makes sense. Besides, who would know that E means voltage and I means current? What genius thought that up anyway? 🙂