SSS Continuously Hum-cancelling Two-Pickup Parallel Blender

[Yogi B]

Using an MN-taper dual-gang blend pot (what ChrisK termed "a true blend pot"), this combines the three hum-cancelling two-pickup parallel combinations of a typical SSS pickup set with RWRP middle pickup. Namely these positions are: neck + middle; bridge + neck out-of-phase; and bridge + middle. The result is nominally hum-cancelling throughout the entire sweep of the blend pot.

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Proof of hum-cancellation can be derived from analysing half the above circuit and noting in order that two parallel pickups are hum-cancelling the ratio of their hum output voltage (V) to Thevenin equivalent impedance (Z) must be equal and opposite.

Assuming V & Z are equal for all pickups (noting that for a RWRP pickup V is negated, therefore has its orientation reversed below) the following diagram may represent either the neck & middle or middle & bridge pickups as connected one gang of the blend pot. As this is then placed in parallel with the remaining pickup (B connected to V positive), V_AB/Z_AB must equal V/Z in order that the overall circuit is hum-cancelling.

\begin{aligned} V_\text{AB} &= {RwV \over Rw + Z} + {R(1-w)V \over R(1-w) + Z} \\[1.5em] &= V \left( {Rw \over Rw + Z} + {R(1-w) \over R(1-w) + Z}\right) \\[3em] Z_\text{AB} &= \Bigl(Rw \parallel Z\Bigr) + \Bigl(R(1-w) \parallel Z\Bigr) \\[1.5em] &= {RwZ \over Rw+Z} + {R(1-w)Z \over R(1-w) + Z} \\[1.5em] &= Z \left( {Rw \over Rw+Z} + {R(1-w) \over R(1-w) + Z} \right) \\[3em] \therefore {V_\text{AB} \over Z_\text{AB}} &= {    V \cancel{\left( {Rw \over Rw + Z} + {R(1-w) \over R(1-w) + Z}\right)} \over    Z \cancel{\left( {Rw \over Rw + Z} + {R(1-w) \over R(1-w) + Z}\right)} } = {V \over Z} \end{aligned}

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Well some thing I need to read up on:)