Thats mostly over my head but the values shown in the example are different. They say use as small a value for the input as u can get away with. My PI has 1M where that PI shows 470k, a 1.6k cathode where mine is 470R, and tail is 10k on mine vs 47k in the example. They are saying a .01uf minimum which is twice the size of the .0047uf i liked. Would the differences in the values i listed mean i can get away with a smaller minimum than in the example?
The 10 nano-farad reference is for a turn over frequency of 20 hz. Low E on a guitar is about 80 hz. Look at the formula that was referenced.
"so
470k grid-leaks give an input impedance close to 940k.
Because the grids are at a high DC voltage, an input capacitor is required to block the DC reaching previous stages in the amp. The input coupling capacitor may be chosen in the same way as for any coupling capacitor:
C = 1 / (2 * pi * f * R)
Where:
f = the lowest -3dB to be passed.
R = the source impedance of the previous stage plus the input impedance of this stage. However, often the source impedance is small enough compared with the input impedance that it may be ignored.
For example, using 470k grid-leaks gives an input impedance of approximately 940k.
Taking f = 20Hz:
C = 1 / (2 * pi * 20 * 940000)
So, using the same formula with a turnover frequency of 80 hz. C = ~.002
C = 1 / (2 * pi * 80 * 940000)