taking small exception to a R value.
I would replace the old capacitor at the same time. Use two 47µF 450VDC electrolytics,
with a 22 Ω 5W resistor between them.
Not bad advice, but the
“corner” (–3 dB point) of the R and C
2 is
F = 1,000,000 / (2 π RC) … Ω, µF
F = 1,000,000 / (6.28 × 22 × 47)
F = 153 Hz
Better would be to use a somewhat larger resistor. If one shoots for 50, or 60 Hz divided by
e (2.71828), it is pretty optimal. Luckily, it fits right back into the same formula:
if…
F = 1,000,000 / (2 π R C)
then by moving variables around
R = 1,000,000 / (2 π F C)
where
F = 50 Hz ÷ 2.71828
F = 18.4 Hz
C = 47 µF
so, now subbing in
R = 1,000,000 / (6.28 × 18.4 × 47)
R = 185 Ω
This will have a substantially stronger hum-suppressing effect, but at the same time not likely to drop B+ very much. An
“old amp” isn't going to draw more than what, 50 mA?
E = I R
ΔE = 0.050 A × 185 Ω
ΔE = 9.2 V
So, 220 - 9 … is about 210 volts.
Plenty still, to
“get 'er done”.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡
GoatGuy ✓ ≡=-⋅
________________________________________
PS: alternatively, one
could use a larger capacitor, since larger-value capacitors are
brilliantly cheap these days!. If you still want the very modest ΔV of the 22 Ω resistor, then just use a proportionately larger capacitor. For example
if
R = 1,000,000 / (6.28 F C) … then with rearrangement
C = 1,000,000 / (6.28 F R) … and subbing in from above
C ≈ 160,000 ÷ 18 Hz × 22 Ω
C ≈ 400 µF
And since capacitors-that-are-cheap-and-good are the ones that come in commercially competitive E–6 value sequences ( 10→15→22→33→47→68 …], then just go for the nearest
larger value
C = 470 µF
(
edit…I said 'tada', but realistically, the 22 Ω resistor's backward facing current draw will exceed the vacuum-rectifier's nominal handling.
I take it back. Better to use an even larger resistor. 330 Ω to 470 Ω, again in E–6 [10→15→22→33→47→68] values, for easy access and low cost.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡
GoatGuy ✓ ≡=-⋅
________________________________________
PPS: the output power will be impacted only by
dB = 20 log10((B+ -ΔV) / ( B+ … without CRC ))
dB = 20 log10(220 - (0.050 × 330 Ω)) ÷ 220)
dB = 20 log10( 0.925 )
dB = –0.6 dB
Hardly enough to even slightly worry about. … G
G