Originally posted by Old Tele man
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Where does AC signal go that is blocked by coupling capacitor value?
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"Take two placebos, works twice as well." Enzo
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Originally posted by Helmholtz View PostIn other words, the effective input impedance of a LTP is about twice the input grid resistance. For the example circuit shown, this means a resistive input impedance of about 1M and a (second) corner frequency of 1.7Hz.
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That's without output stage negative feedback isn't it?
Input signal current only flows in the grid leak resistor. Input impedance is given by input signal voltage divided by input signal current. So if the current through the resistor is reduced by bootstrapping, input impedance is increased proportionally. The closer the cathode signal voltage gets to the input signal voltage, the higher the input impedance.
A positive voltage applied to the PI input produces a positive feedback voltage from the output of the transformer, thereby raising the cathode signal voltage. This in turn increases input impedance with closed NFB loop. The input impedance without global NFB thus is a lower bound.
Instead of doing a complete circuit analysis I suggest to find the actual input impedance by measuring the differential signal voltage across the grid leak resistor.
To those interested in circuit analysis and calculus I highly recommend the books by Richard Kuehnel, especially "The Fender Bassman 5F6-A".- Own Opinions Only -
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