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.