Delving into TCI

[Em7]

I mentioned in a previous post that all LC circuits belong to a class of circuits known as "tuned circuits;" therefore, TCI with respect to tuned circuits is nothing new.. What I believe TCI is is an effort to replicate the best examples of pickup circuits with a relatively high degree of precision. The fact that Paul has wrapped it in mystique is causing a lot of controversy. However, there is no such thing as bad publicity.

With that said, Paul mentioned that the pickup he removed from JM's Strat tested as having 6dB more bass than the pickup he had previously wound for the Silver Sky. What I can ascertain from that claim is that the resonant peak of the pickup is either lower in frequency and/or the Q of the guitar circuit is lower.

Let's look at the formula for the resonant frequency of a tuned circuit.

f0 = 1 / (2pi * SQRT(LC)), where f0 = resonant frequency, pi = 3.14, SQRT is the square-root function, L = inductance in henries, and C = capacitance in farads

If we know our desired resonant frequency and inductance, re-writing the equation to solve for C yields:

C = (1 / (2pi * f0)) ^2 / L, where C = capacitance in farads, pi = 3.14, f0 = resonant frequency, the symbol "^" means raised to the power of, and L = inductance in henries.

The average Strat pickup has an inductance in the 2H +/- a few points range with a resonant frequency of around 5K; therefore, solving for these parameters yields:

C = (1 / (6.28 * 5000) ^2 / 2 = 0.00000000050712 fareds or 507.12pF

If we want more bass with less brightness, then we have to move the resonant frequency down in frequency. Let's pick 3K. Solving for these parameters yields:

C = (1 / (6.28 * 3000)) ^2 / 2 = 0.00000000140867 farads or 1408.67pF or 1.40867nF

As we can clearly see, we need to add capacitance to this tuned circuit to lower its resonant frequency, that is, if we do not alter its inductance. What if the pickup removed from JM's Strat had more inductance than the pickup Paul had previously wound for the Silver Sky, say 2.5H? How would that difference change the equation?

C = (1 / (6.28 * 3000)) ^2 / 2.5 = 0.00000000112693 farads or 1126.93pF or 1.12693nF

As can see, an increase in inductance lowers the amount of capacitance that is needed to achieve a resonant frequency.

Now, how do we increase inductance in a pickup? For the most part, we add turns of wire to the bobbin. How do we increase capacitance? We alter the winding pattern and/or use magnet wire with thinner insulation. Paul mentioned gaining access to the original wire manufacturer used by Fender. My SWAG (Scientific Wild Arse Guess) is that this wire has a different insulation thickness, more than likely thinner. That would allow him to wind a pickup that has more turns on the same bobbin, increasing inductance. Thinner insulation would also afford him more distributed capacitance. However, I do not think that insulation thickness alone would be enough create that large of a difference in capacitance by itself because inductance increases by the square of the turns; therefore, we cannot double the amount of distributed capacitance with an increase in inductance of 25%. That realization leads us to one thing; namely, the resulting TCI pickup is less scatter wound than the original pickup that Paul wound because scatter winding increases the amount of air in a coil; thereby, reducing capacitance.

One last thing, Paul mentioned volume pot resistance and wiring a resistor across the terminals to achieve a specific resistance. The resistance of the volume pot in a guitar adds to the DC resistance of the pickup coil to alter the pickup's quality factor (Q-factor). The Q-factor is how sharp the signal rises and falls around the resonant frequency. More combined resistance means less peaky frequency response. The tone pot is also part of the equation. It adds DC resistance to the circuit until wound all of the way down, reducing the Q of the circuit along the way. However, what also happens when the tone pot is wound all of the down is that the tone capacitor adds to a pickup's distributed capacitance to lower the resonant frequency of the circuit. Reducing the Q of the circuit while shifting its resonant frequency down in frequency results in a less peaky, more bass heavy tone. The capacitance of a guitar cable also alters the circuit’s resonant frequency. It is not as much a treble bleed as it is an incremental shift in resonant frequency based on the per foot capacitance of the cable employed.

That's my SWAG and I am sticking with it.

Q.E.D.

 

[Black Plaid]

Killer post!

When talking about the cable, I assume you didn't mean 'treble bleed', but 'low pass'. The treble bleed in a guitar circuit seem to me to be a way to bleed treble "into" the circuit despite the volume being down.

Question for you about that. For a wireless adapter, can we assume the resonant peak will be unaffected, or is there still enough inductance in it to affect it?

 

[Em7]

Using treble bleed to describe the cap on a volume pot is a poor choice of terms. That circuit does not bleed anything. It merely forms a high pass filter around the volume pot. When I think about bleed, I think signal loss. Let’s think about it from a blood point of view. When we bleed, we loose blood. We do not reroute it. That is not what a treble bleed circuit does. A treble bleed circuit allows the frequencies above the RC constant to bypass the attenuating effects of the volume pot. The reason why this circuit works is because the human ear is less sensitive to treble frequencies at lower volumes due to what is known as the Fletcher-Munson curves; therefore, we need more treble at lower frequencies to sense the same tone.

 

[Black Plaid]

Using treble bleed to describe the cap on a volume pot is a poor choice of terms. That circuit does not bleed anything. It merely forms a high pass filter around the volume pot. When I think about bleed, I think signal loss. Let’s think about it from a blood point of view. When we bleed, we loose blood. We do not reroute it. That is not what a treble bleed circuit does. A treble bleed circuit allows the frequencies above the RC constant to bypass the attenuating affects of the volume pot. The reason why this circuit works is because the human ear is less sensitive to treble frequencies at lower volumes due to what is known as the Fletcher-Munson curves; therefore, we need more treble a lower frequencies to sense the same tone.

I completely agree about the 'bleed' term, always seemed odd to me.

 

[Domingo Lantigua]

Using treble bleed to describe the cap on a volume pot is a poor choice of terms. That circuit does not bleed anything. It merely forms a high pass filter around the volume pot. When I think about bleed, I think signal loss. Let’s think about it from a blood point of view. When we bleed, we loose blood. We do not reroute it. That is not what a treble bleed circuit does. A treble bleed circuit allows the frequencies above the RC constant to bypass the attenuating affects of the volume pot. The reason why this circuit works is because the human ear is less sensitive to treble frequencies at lower volumes due to what is known as the Fletcher-Munson curves; therefore, we need more treble at lower frequencies to sense the same tone.

Remember that guitar related things have a habit of being misnamed or described ie: tremolo is really vibrato and a few others that I can't think of

 

[Em7]

Remember that guitar related things have a habit of being misnamed or described ie: tremolo is really vibrato and a few others that I can't think of

Another area of confusion is the capacitor on the tone control. Without an understanding of tuned circuits, it does in fact look like a filter that shunts highs to ground. However, if that were true, there would be a noticeable drop in volume. The reality is that cap is part of an RLC circuit and any capacitance or inductance that is part of an RLC circuit affects the resonant frequency of the circuit.

 

[Em7]

Question for you about that. For a wireless adapter, can we assume the resonant peak will be unaffected, or is there still enough inductance in it to affect it?

I forgot to answer this part of the equation. A wireless system inserts a buffer into the equation that decouple’s the tuned circuit from downstream loading effects. The same kind of thing occurs with active pickups. The preamp in the active system buffers the tuned circuit from the loading effects of the guitar cable.

 

[Jazzedout]

Great post as always Em7!!

 

[Mozzi]

I mentioned in a previous post that all LC circuits belong to a class of circuits known as "tuned circuits;" therefore, TCI with respect to tuned circuits is nothing new.. What I believe TCI is is an effort to replicate the best examples of pickup circuits with a relatively high degree of precision. The fact that Paul has wrapped it in mystique is causing a lot of controversy. However, there is no such thing as bad publicity.

With that said, Paul mentioned that the pickup he removed from JM's Strat tested as having 6dB more bass than the pickup he had previously wound for the Silver Sky. What I can ascertain from that claim is that the resonant peak of the pickup is either lower in frequency and/or the Q of the guitar circuit is lower.

Let's look at the formula for the resonant frequency of a tuned circuit.

f0 = 1 / (2pi * SQRT(LC)), where f0 = resonant frequency, pi = 3.14, SQRT is the square-root function, L = inductance in henries, and C = capacitance in farads

If we know our desired resonant frequency and inductance, re-writing the equation to solve for C yields:

C = (1 / (2pi * f0)) ^2 / L, where C = capacitance in farads, pi = 3.14, f0 = resonant frequency, the symbol "^" means raised to the power of, and L = inductance in henries.

The average Strat pickup has an inductance in the 2H +/- a few points range with a resonant frequency of around 5K; therefore, solving for these parameters yields:

C = (1 / (6.28 * 5000) ^2 / 2 = 0.00000000050712 fareds or 507.12pF

If we want more bass with less brightness, then we have to move the resonant frequency down in frequency. Let's pick 3K. Solving for these parameters yields:

C = (1 / (6.28 * 3000)) ^2 / 2 = 0.00000000140867 farads or 1408.67pF or 1.40867nF

As we can clearly see, we need to add capacitance to this tuned circuit to lower its resonant frequency, that is, if we do not alter its inductance. What if the pickup removed from JM's Strat had more inductance than the pickup Paul had previously wound for the Silver Sky, say 2.5H? How would that difference change the equation?

C = (1 / (6.28 * 3000)) ^2 / 2.5 = 0.00000000112693 farads or 1126.93pF or 1.12693nF

As can see, an increase in inductance lowers the amount of capacitance that is needed to achieve a resonant frequency.

Now, how do we increase inductance in a pickup? For the most part, we add turns of wire to the bobbin. How do we increase capacitance? We alter the winding pattern and/or use magnet wire with thinner insulation. Paul mentioned gaining access to the original wire manufacturer used by Fender. My SWAG (Scientific Wild Arse Guess) is that this wire has a different insulation thickness, more than likely thinner. That would allow him to wind a pickup that has more turns on the same bobbin, increasing inductance. Thinner insulation would also afford him more distributed capacitance. However, I do not think that insulation thickness alone would be enough create that large of a difference in capacitance by itself because inductance increases by the square of the turns; therefore, we cannot double the amount of distributed capacitance with an increase in inductance of 25%. That realization leads us to one thing; namely, the resulting TCI pickup is less scatter wound than the original pickup that Paul wound because scatter winding increases the amount of air in a coil; thereby, reducing capacitance.

One last thing, Paul mentioned volume pot resistance and wiring a resistor across the terminals to achieve a specific resistance. The resistance of the volume pot in a guitar adds to the DC resistance of the pickup coil to alter the pickup's quality factor (Q-factor). The Q-factor is how sharp the signal rises and falls around the resonant frequency. More combined resistance means less peaky frequency response. The tone pot is also part of the equation. It adds DC resistance to the circuit until wound all of the way down, reducing the Q of the circuit along the way. However, what also happens when the tone pot is wound all of the down is that the tone capacitor adds to a pickup's distributed capacitance to lower the resonant frequency of the circuit. Reducing the Q of the circuit while shifting its resonant frequency down in frequency results in a less peaky, more bass heavy tone. The capacitance of a guitar cable also alters the circuit’s resonant frequency. It is not as much a treble bleed as it is an incremental shift in resonant frequency based on the per foot capacitance of the cable employed.

That's my SWAG and I am sticking with it.

Q.E.D.

Its beyond my comprehension but if I have some grasp of what you are saying, as these parts - Pick-ups, tone/volume pots etc - vary off the manufacturing process, could they achieve the Capacitance and Inductance wanted by measuring the individual parts and using the ones that fit the equation the best. We know a 500k pot for example can vary by +/-30k for example and the output of pick-ups, despite being wound by machines with the best wires etc, still vary too. Therefore, if you know the vales exactly of these from the manufacturing process, specifically picking the parts based on their value rather than just picking the 'next' by random as manufacturers have always done.

If that is the case, then it really shouldn't look any different inside. The only difference is that the electronics were selected on purpose because of their 'values' to achieve the correct inductance and capacitance. You aren't going to see a radical new wiring system or some new part in the chain you haven't seen before. The tone, volume and maybe even the resistor (I don't know how much they vary myself) could all of been hand picked to go inside that specific guitar because of the exact values of the Pick-up. If you analysed a LOT of the same models, and used the calculations, maybe you would find that the inductance and impedance would be remarkably similar despite the minor variations of the individual parts and a LOT more consistent than manufacturers who still use random selection when fitting electronics...

As I said, this is beyond my comprehension but if you cannot get the parts to come off the line perfectly the same, then the next option is to hand pick the values to get the most consistent output. I don't know if you need a higher value tone/volume pot or resistor for a the PU's that are hotter to compensate and give the same result as the PU's that are a bit lower than average but if you know the equation, then you know what value would be needed to achieve the most consistent capacitance and impedance across the range...