Here I will present what several people have asked me for: a derivation of this formula. Up to now the whole thing was based on empirical observation; can it also be proven?
To take a swing at deriving the formula, I had to brush up my complex math skills; it has been 25 years since I last used them in university. Luckily, pages with basic math definitions exist (thanks!) so I was good to go.
setting up
(x: frequency in octaves; y: decibels)
What was the goal again? Ah yeah, it was to make the black curves match the orange one, for as long as that is practically possible. The orange curve is the ideal of Cpin decoupling Zpin (without taking the kathode resistor into account). The black curves are the same Cpin and Zpin with different Rk values thrown into the mix. See part three for all the details.
To achieve the goal, we look at one frequency of interest (encircled above) that is above ƒ-3dB, take the value of the orange curve at that point and figure out how much the kathode capacitance has to be increased to make each black curve match. The formula for that is:
|1 / [1 + 1/j2πCpinƒinterest]| =y is here the factor with which Cpin has to be increased to make the match. The formula is already normalised for the actual loading of the tube by Rload; Rk is expressed in terms of Zpin. We make this look slightly less unappetising by defining—
|1 / [1 + 1/(j2πyCpinƒinterest + Zpin/Rk)]|
D = 2πCpinƒinterestand by dropping the one-over part on both sides:
|1 + 1/jD]| = |1 + 1/(jyD + Zpin/Rk)|I have called it D for distance—the one between ƒ-3dB and ƒinterest. Above is the (implicit) equation I had my math grapher plot in part three:
(x: log2(Rk/Zpin); y: log2(y))
This is for different values of D: D=1 is the right-most, this means ƒinterest = ƒ-3dB. The left-most curve is for D=10, i.e. ƒinterest is one decade higher than ƒ-3dB. We can observe here (just by watching):
- each curve collapses on the left; this is where the matching we are trying to achieve is no longer practical—the shelving of the response interferes;
- every high-D curve incorporates the good parts of the low-D curves;
- to get good coverage of a practical range of kathode resistors (compared to Zpin), we need to work with a D that is 10 or higher.
D > 10Before we get going there is one more convenience definition—
x = Zpin/RkThis makes our working equation something I can handle:
|1 + 1/jD]| = |1 + 1/(jyD + x)|
proof
The job is now to solve for y. First untangle the devision on the right by multiplying top and bottom with the conjugate of the denominator:|1 + 1/jD]| = |1 + x/(x2 + y2D2) - jyD/(x2 + y2D2)|Then take the magnitude on both sides (drop the square roots immediately):
1 + 1/D2 =Drop the ones on both sides and multiply all terms by x2 + y2D2:
1 + 2x/(x2 + y2D2) + x2/(x2 + y2D2)2 + y2D2/(x2 + y2D2)2
y2 + x2/D2 = 2x + (x2 + y2D2)/(x2 + y2D2) = 2x + 1Rearranging that:
y2 = 1 + 2x - x2/D2Now we remember that D > 10. That means that even for extremely small-value kathode resistors (Rk = Zpin/20, i.e. x=20) 2x will be at least 10 times larger than x2/D2. For the more normal value of x=1, 2x is 200 times larger than x2/D2. So now we perform the trick that no pro-physics proof shall be without: drop the piddly term to get a nicer result:
y2 = 1 + 2xFilling in x and performing square root on both sides:
y = √(1 + 2Zpin/Rk)There we are, with y being the factor to multiply Cpin with to get the Rk-compensated Ck, we get:
Ck = √(1 + 2Zpin/Rk) × CpinAs they say in physics: bingo.
If you thought it a bit fast to drop the x2/D2 term just like that, remember that to investigate really small-valued kathode resistors, you need a really large D. In the second graph above, the x2/D2 term is what brings on the collapse for every curve. To properly investigate, a collapse has to be avoided, i.e. D has to be large enough so that x2/D2 is insignificant.
happy ending
It is now proven, every kathode resistor can be correctly bypassed, with smooth, predictable results:Please observe the the licence of this blog’s content, at the bottom of this page.
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