Relating tone cap value to frequency range

[ratherdashing]

I was reconditioning a couple of old speakers I have, and in the process I learned about the relationship between the value of a capacitor, the load on a circuit, and the audio frequency rolled off by that cap. I thought it would be helpful to share this info as it relates to guitar tone caps.

The crossover frequency, as it is called in speaker circuits, is the point in the frequency spectrum at which the capacitor acts as a high pass filter (only frequencies above a certain point get through). When you have two speakers in parallel where one speaker is designed to handle higher frequencies than the other (like in a home stereo or PA speaker cab), a capacitor is used to pass the highs to the smaller speaker. What's left is collected by the larger speaker.

In a guitar's tone circuit the same principle applies, but the "high pass" is actually passing the highs to ground leaving the lows behind, so it more or less acts as a low pass filter. With the tone knob all the way "on", you are hearing the full effect of the low pass circuit.

The frequency at which the crossover occurs is determined by the capacitor value, and the load on the circuit. In the case of an electric guitar, the load is the pickup. The formula to calculate the crossover frequency is:

0.159/(C x Rh) = F​

C is the capacitance in farads. Caps for guitars are usually expressed in microfarads; to convert to farads, multiply the value by 0.001.

Rh is the total load on the circuit in Ohms. This will be the load of your pickup. Example: a Duncan Custom has a 14,100 Ohm load.

F is the frequency in Hz.

Here's an example: let's calculate the crossover of a Custom with a .022 uf capacitor, tone control cranked full on:

0.159/(C x Rh) = 0.159/(14,100 Ohm x 0.00022 f) = 512.6 Hz​

With your tone control on 1, all frequencies above 512.6 Hz are passed to ground.

So how much difference does the cap value make? Let's change it to a 0.47 uf cap.

0.159/(C x Rh) = 0.159/(14,100 Ohm x 0.00047 f) = 239.9 Hz​

Big difference! The .047 uf cap will pass way more of the frequency spectrum to ground, resulting in a much bassier tone.

This, of course, is all with the tone knob as far down as it will go. With the tone knob on 10, we're introducing a much bigger load to the circuit: the pot.

"Hold on a minute," you say. "According to that formula, increasing the load will pass MORE highs to ground, not less! We all know that's not what happens when you turn up the tone knob. What's up?"

Stay tuned for the answer ...

 

[ratherdashing]

Re: Relating tone cap value to frequency range

The answer is actually really simple: the pot has nothing to do with the crossover formula.

A pickup, like a speaker, is an inductor coil. A potentiometer is a resistor. Both have a load in Ohms, but they behave completely differently.

Think of electricity as a group of long distance runners. The pickup is a running track - it makes the runners "work out", but does nothing explicit to stop them. Increasing the pickup's load is like increasing the length of the track. The pot is an old forest full of fallen trees, vines, and nasty critters - some of the runners will trip and fall on their way through it.

The load placed on the circuit by the pot prevents current from getting to the cap, but it doesn't figure into our equation. As you turn the pot down, you are making it easier for current to get to the cap. You aren't changing the crossover frequency by turning the pot, though it may sound like you are. All you're doing is increasing the effectiveness of the crossover. With the tone on 10, the crossover is very mild and weak - on 1 it is very strong.

What about 250k vs. 500k pots? The main difference is that a 250k on 10 will let more current through to the crossover than a 500k on 10. Basically, if you want the crossover to be more apparent when the tone is at 10, use a 250k pot. This is why they are used in trebly guitars like Strats and Teles - to bleed off a little bit of treble.

So here's a summary:

Changing the cap value changes the crossover frequency

Turning the tone knob from 10 to 1 increases the effectiveness of the crossover

Using a lower value pot makes the crossover more effective in the 10 position.

 

[Kivitel]

Re: Relating tone cap value to frequency range

in. the. vault. now.

This is the only time I've read this on a non hifi website. Excellent.

 

[Sludgenutz]

Re: Relating tone cap value to frequency range

Not a bad write up ratherdashing!

I often deal with the mysterious world of electricity by equating electrical systems to water/plumbing systems!

Voltage = "Potential". The higher one lifts a water source above ground level, the more potential it has.

Amps = Actual water. One might have a small current (as in a piss stream), or a large current (as large as water supply, to a city).

Watts = WORK DONE! The equation of Amps (current flow) AND the Voltage (the water pressure) behind it determine the Watts (Power Consumption). It is rather obvious that one may take a wizz off the top of the Hoover Dam, but the supply of liquid is a fraction of a larger pipe...no matter how high the pipe is above ground level.

A person turning on a garden hose (the whiz! :fingersx at the top of the Hoover Dam cannot accomplish much "WORK".

Car batteries accomplish a lot of "WORK", like turning the engine's starter with only 12volts!...because it acts as supply of huge amount of current (a huge water pipe, full of water), but at a lower potential (height above ground).

I'm getting to caps....bear with my edits before my ISP drops this....

 

[Sludgenutz]

Re: Relating tone cap value to frequency range

Yikes! I give up! The water analogy is fine. AC theory is not particularly quantum physics...but is beyond my will to type. If anyone has particular questions please ask, I'll give it my best shot.:eek13: I'm sure other guys here can help too!

 

[nebucanazza]

Re: Relating tone cap value to frequency range

So, a low DC resistance pickups retains much more highs with the tone knob at 1.I guess tone controls should be much more useful with low DC pickups, as for the high DC models, tone knob at 1 or 2 might be too muddy and bassy.

Alternatively, it might be a taste thing.For a low DC pickup, a tone knob might only serve to tame high end shrillness over its range, and some people might need just that.Or one could try a 0.047 instead of 0.022 for getting heavier tones.

Man, am I rambling today:eek13:

 

[ratherdashing]

Re: Relating tone cap value to frequency range

Bump for the weekend readers.

 

[Stevo]

Re: Relating tone cap value to frequency range

Actaully, to calculate the resonant frequency for a bandpass filter, you need to factor with inductance, capacitance and DC resistance. The value if inductance changes with the value of the frequency.

Pickups are inductors which superimpose a variable AC signal over a DC voltage average - therefore they have special rules because you are using the actual inductor to create sine waves (in conjunction with a vibrating string), rather than applying inductance to another signal, as in an audio amplifier.

 

[Stevo]

Re: Relating tone cap value to frequency range

"According to that formula, increasing the load will pass MORE highs to ground, not less!

Actually you are increasing resistance, which opposes current. More current is more load.

 

[LJ King]

Re: Relating tone cap value to frequency range

Yes, the output impedance of the pickup (which is rarely stated as a specification) and NOT the resistance of the coil determines the frequency cutoff of a cap placed across the pickup. The impedance can be many times higher than the resistance measurement.

 

[ratherdashing]

Re: Relating tone cap value to frequency range

Actually you are increasing resistance, which opposes current. More current is more load.

I know that. Read the second post.

Yes, the output impedance of the pickup (which is rarely stated as a specification) and NOT the resistance of the coil determines the frequency cutoff of a cap placed across the pickup. The impedance can be many times higher than the resistance measurement.

Ah. So is there a way to determine the inductance of a humbucker? Or are they all more or less the same? Ballpark estimate even?

In any case, the theory is still valid as long as we're talking about different cap values on the same pickup.

 

[Sniper1]

Re: Relating tone cap value to frequency range

I was reconditioning a couple of old speakers I have, and in the process I learned about the relationship between the value of a capacitor, the load on a circuit, and the audio frequency rolled off by that cap. I thought it would be helpful to share this info as it relates to guitar tone caps.

The crossover frequency, as it is called in speaker circuits, is the point in the frequency spectrum at which the capacitor acts as a high pass filter (only frequencies above a certain point get through). When you have two speakers in parallel where one speaker is designed to handle higher frequencies than the other (like in a home stereo or PA speaker cab), a capacitor is used to pass the highs to the smaller speaker. What's left is collected by the larger speaker.

In a guitar's tone circuit the same principle applies, but the "high pass" is actually passing the highs to ground leaving the lows behind, so it more or less acts as a low pass filter. With the tone knob all the way "on", you are hearing the full effect of the low pass circuit.

The frequency at which the crossover occurs is determined by the capacitor value, and the load on the circuit. In the case of an electric guitar, the load is the pickup. The formula to calculate the crossover frequency is:

0.159/(C x Rh) = F​

C is the capacitance in farads. Caps for guitars are usually expressed in microfarads; to convert to farads, multiply the value by 0.001.

Rh is the total load on the circuit in Ohms. This will be the load of your pickup. Example: a Duncan Custom has a 14,100 Ohm load.

F is the frequency in Hz.

Here's an example: let's calculate the crossover of a Custom with a .022 uf capacitor, tone control cranked full on:

0.159/(C x Rh) = 0.159/(14,100 Ohm x 0.00022 f) = 512.6 Hz​

With your tone control on 1, all frequencies above 512.6 Hz are passed to ground.

So how much difference does the cap value make? Let's change it to a 0.47 uf cap.

0.159/(C x Rh) = 0.159/(14,100 Ohm x 0.00047 f) = 239.9 Hz​

Big difference! The .047 uf cap will pass way more of the frequency spectrum to ground, resulting in a much bassier tone.

This, of course, is all with the tone knob as far down as it will go. With the tone knob on 10, we're introducing a much bigger load to the circuit: the pot.

"Hold on a minute," you say. "According to that formula, increasing the load will pass MORE highs to ground, not less! We all know that's not what happens when you turn up the tone knob. What's up?"

Stay tuned for the answer .

I didn't see this in my Mel Bay Play Guitar In 30 Seconds Book.

 

[JacksonMIA]

Re: Relating tone cap value to frequency range

0.159/(C x Rh) = F

When I first saw this, I was trying to figure out where you came up with that formula when it's f = 1 / (2 x PI x R x C). It took me a second to see that you'd just simplified it. :smack:

Pretty good write up, though. You could have a bunch of engineers nit-pick everything for a couple of weeks to throw in the quantum mechanics and everything else that's truely involved in it, but what you've got is close enough for any guitarist to use.

 

[ratherdashing]

Re: Relating tone cap value to frequency range

When I first saw this, I was trying to figure out where you came up with that formula when it's f = 1 / (2 x PI x R x C). It took me a second to see that you'd just simplified it. :smack:

Pretty good write up, though. You could have a bunch of engineers nit-pick everything for a couple of weeks to throw in the quantum mechanics and everything else that's truely involved in it, but what you've got is close enough for any guitarist to use.

Thanks. I'd still like to know the inductor impedence of a pickup though, just to get an accurate crossover number.

 

[LJ King]

Re: Relating tone cap value to frequency range

EMG is the only maker I know of off hand that provides inductance and impedance values in addition to DC resistance. They have the info in PDF files on their website for their passive pickups.

In one example, a humbucker with 13.75 ohms of DC resistance works out to have an impedance of 74K - almost 5 times as much as the DC resistance value.

 

[JacksonMIA]

Re: Relating tone cap value to frequency range

The impedance of an inductor is time dependant.

IIRC, it's Z = V / I = (dI/dt) / I, so you'll have a hard time using that in your equation. That's the only equation I can remember, or at least the only one that keeps the equation real (no pun intended).

 

[LJ King]

Re: Relating tone cap value to frequency range

Time being another word for frequency.

The formula you are quoting happens to be ohms law at least for the first equality.

I do not know about the rest as I have no clue as to what dI and dT are supposed to mean.

Someone told me that if you take the voltage of a unloaded source, and then load the source until it produces 1/2 of the unloaded voltage, then the load will be the same as the impedance of the source. I am just repeating here but it makes sense to me because the source and the load form a voltage divider and when the voltage is 1/2 of the original voltage then the source and load have to be equal.

 

[JacksonMIA]

Re: Relating tone cap value to frequency range

Actually, frequency is the number of cycles in a given amount of time. Or did you just mean that the impedance of an inductor is frequency dependant? You're right about that.

dI/dt is just the voltage of an inductor or, mathematically, the derivative of current with respect to time.

Your solution works with a purely resistive load, but it gets a bit more complicated when you involve inductance and capacitance. It's the same principle, but you'll need more than just a multimeter to do it.