Hi Ralph. Im curious if you use an inverse RIAA network to check your EQ. If so where did you get the values? I built mine from the advice of Peter Moncrieff, Mitch Cotter and Dick Sequerra who were the first to discover that many RIAA EQ curves were off. Of course I used precision parts.Of course! Its been a really long time and I do not remember where I got the values, but when used I got flat response. FWIW it also plays flat when I cut a lacquer on my lathe and play it back on the preamp. The Westerex has pretty tight curves in this regard; the electronics are matched to the cutter head.
5. What is the best and cost effective way to separate power supply between output tube B+ and driver tube B+?@alexberger
You can of course make sure that the timing constants in the power supply are low enough that the driver and output section can’t talk to each other. Here’s the math:
F = 1/RC times Pi squared.
That formula results in some hard to work with numbers, since F is frequency, R is resistance and C is capacitance in Farads, which is really inconvenient. So I usually useF=1,000,000/RC * 6.28
(6.28 being Pi squared). This formula results in F in cycles per second (Hertz), R in ohms and C in uf (microfarads), which is more real-world.
In your power supply, C is the power supply bypass capacitor, which is probably an electrolytic device. R is the resistor between the power tube B+ power supply filter cap and the filter cap for the driver. The thing is, there will be a certain amount of current that the 6SN7 needs, so you have to make sure the resistor is large enough wattage to survive that.
For that you need Ohm’s Law which is R=V/I
R is resistance in ohms, V is volts, I is current (C was taken already so I is the convention for current).
To calculate wattage (of the resistor) W=V x I
So for example, if you use a 40uf filter cap, and the driver tube is drawing 9mA (0.009amps) between the two sections, then for the power supply to have a cutoff at 0.5Hz the resistor value will be 8,000 ohms. There will be a 72 volt drop across the resistor and it should be a 2 watt device.
Now the trick here is to make sure that the coupling capacitor has a cutoff frequency higher than the power supply cutoff! Otherwise, the circuit can become unstable, prone to ’motorboating’ (a low frequency oscillation) and IMD will be higher. Use the same formula to calculate the value of the coupling capacitor; R will the resistor in the grid circuit of the power tube that the capacitor is driving. To get 0 phase shift at 20Hz, the coupling cap should go 10 octaves lower, or to 2 Hz; our power supply cutoff is safely below that (although it would not hurt to go an octave lower by doubling the value of the filter capacitance). The problem here is that SET output transformers often don’t have good low frequency bandwidth, and the 300b is right on the cusp of where 20-20Hz is actually sort of possible with a good output transformer. Even though your speaker may not go that low, its a good idea to get as much bandwidth as you can to preserve phase relationships that the ear uses to detect soundstage width and depth. So get a good output transformer.
If the manufacturer has issues with low frequencies saturating the transformer, you can reduce the value of the coupling capacitor, but I would be hesitant to do much of that as phase shift in the bass robs the amp of perceived bass impact, even though its flat on the bench.
A less cost effective way to do the power supply for the driver is to use a separate power transformer and power supply for it. In this way no matter what signal condition exists in the output section, no amount of noise in the power supply of the output tube (such as a general voltage sag) can talk to the driver section. However in SETs, this usually isn’t a problem unless you run the power tube in class A2 or class A3. The latter classes of operation can draw more power from the power supply as output power is increased; class A1 does not do that.
