Are you familiar with reducing a pot value by adding a resistor? For instance, a 1M pot with a 1M resistor soldered to terminals 1 & 3, will result in a pot with ~500k resistance. Essentially, it will function as a 500k pot.ultimately I decided it was redundant to have the 47nF coupling cap, followed by another 22nF coupling cap, especially with this 'lead channel always on' design where there isn't any switching.
so i dropped that 22nF cap completely. ... maybe it does matter? i don't really know.
By removing the 22nF cap, the previous 100k resistor is now acting as a parallel resistance with the 1M pot (of course there is the 680k and 470k resistors involved). Let's ignore the 470k for now and look at the 680k in series with the 1M pot. When you parallel the 100k resistance, the total resistance will be ~94k. When we figure in the 470k the total resistance can get down to around 90k.
So, removing the 22nF cap may not have much affect on the low frequency cutoff but the circuit will lose highs. IDK if this preamp would be better with less highs. Subjective.
When AC is full wave rectified into DC there is a strong ripple current. It is double the 50Hz frequency from your power source (Wall voltage). 100Hz ripple will become *signal* and amplified by this preamp if measures are not taken to reduce the ripple. This will sound like a buzz or hum.yep, first dropping resistor, 10K, was dropped to *2.2K (just checked the layout)
and yes, the following B+ dropping resistors were changed to 3.3K (from 1K), as per a suggestion made my by a wiser poster, earlier on in this thread.
im not going to claim that i can explain why 3.3K is a better choice over 1K, but apparently it's for better filtering.
The 47uF reservoir cap smooths the ripple by charging and discharging 100 times per second (100Hz). When you get rid of the (10k or 2.2k) resistance after the reservoir cap the B+2 cap (22uF) is in parallel with the reservoir cap. These two caps in parallel make a capacitance of 69uF. This 69uF capacitance is now the reservoir cap. The potential problem that is created is that the B+2 components are now being powered with DC that contains the strong ripple current. There is no additional filtering at B+2 which would have been there if the resistance after the 47uF were still in the circuit.
The dropping resistors, drop voltage of course but they play another role. They form 1st Order Low Pass RC filters with the filter caps. If you like math you can calculate the desired frequencies to filter out the 100Hz content in the DC supply. I tend to just use an online low pass filter calculator to get the numbers.
A 1st Order Low Pass RC filter rolls off frequencies at 6dB per octave. The cutoff (or corner) frequency starts at 3dB below the cutoff frq. If we choose a cutoff frequency of one octave below 100Hz (=50Hz), the 100Hz content will be 9dB (6dB+3dB) quieter. (An octave lower is half the Hz.)
Using a low pass filter calculator with the B+2 original values. R=10k C=10uF the cutoff frequency is 1.7Hz. How many octaves is that below 100Hz?
100/2 = 50
50/2 = 25
25/2 = 12.5
12.5/2 = 6.25
6.25/2 = 3.12
3.125/2 = 1.6
It looks like about 6 octaves.
At 6dB per octave, plus that 3dB... it is close to a 39dB reduction of the 100Hz content at the B+2 node.
On this 12.6V heater supply, it appears he is using the impedance of the filament of V3 to replace the two 100R resistors. As you know the filament is connected from pin9 to pin4 and to pin5 on a 12A_7.Just don't understand what he means by his channel version with only pin9 of V3.
But I can give it a try as described here.
Note that he has lowered the voltage a bit from 70V to 48V.
Elevating heaters. The 70V or 48V is not critical. This preamp has a cathode follower (V3b). The cathode voltage is comparatively high on this cathode compared to the other cathodes in this amp. The tube manufacturers have a rating stating they can handle around 100v difference (depending on the tube). To hopefully keep heater hum low, elevating the heaters so the difference is not near 100v is the goal.