I have decided not to have a pot in the crack at all. Instead, I am using a Submissive pasive preamp to control volume (gain?). BTW, how id gain defined? How is attenuation defined? How is impedence defined?
Man, you have a whole thread worth of questions here, I'll try to do my best! The Submissive idea is perfect, your implementation is exactly why I wanted to put the submissive out in the first place. Just be sure to put a resistor from pin to shell on each RCA jack on the Crack. Any value between 100K and 500K should be OK.
Gain can be defined in terms of voltage, current, power, etc. In our discussions on the forum, it is generally assumed that we are only talking about voltage gain. Grainger stumbled on to why, as when you feed 1V into a 100K pot and 1V into a 50K pot, there is a difference of power in those situations, and that difference is not useful to us. So, we define gain as 20log(Vin/Vout).
Impedance is defined as either a resistance, or as something that behaves like a resistance. For example, if we have a rectifier that drops 0.7V at 1A of current, then we can say that the rectifier has an impedance of 0.7 Ohms. (Even though our voltmeter wouldn't measure a resistance across it)
1. Put a ~100K Ohm resistor between the red signal cable where it comes from the RCA input and the input tube
and
2. Put a ~100K Ohm resistor between the white signal cable where it comes from the RCA input and the input tube
If you have 1V coming into the pot, and 1V coming out of the pot at full level, a 100K resistor between the pot wiper and the tube will not decrease the level. It will, however, have a negative impact on the high frequency response of the amplifier.
What is the specific relationship between resistance, current and output volume in DB? (I assume that this would be expressed as a percentage, not a unit as such)
Resistance divides voltage, change in voltage goes into the 20log(Vin/Vout) equation, then you get dB. Current isn't generally considered important in this particular discussion.
1. Why does he want to see 75 volts on the plate?
It allows about half of the available B+ to appear from plate to cathode on the output tube. It also helps define the maximum amount of heat that can be generated by whatever is under the cathode.
2. Is there any reason why I could not change the 1.5V diodes for biasing?
There are some other options, other LED's and other resistors. I would try to keep the bias voltage above 1V if possible.
3. Is optimal performace gained by having tube perform at a linear point on the Vg (bias?) curve?
Yes, this is very important, but you need to pick the operating point *and* draw the load line.
From the little reading I have done it would be best to have the tube working at a linear point on this curve. With a 75V plate voltage and 1.5V diodes for biasing, this point is not particularly linear on the ECC32.
It actually looks very linear with a 50K load.
However, if I reduced the bias to 0.5 V and maintained a plate load of 75V, a more linear point in this plot would be obtained. Using Ohm's law (if I understand Paul correctly), I would require a plate load of about 23 K Ohms, which is remarkably close to the 22.1K of the stock circuit.
If you reduce the bias voltage to 0.5V, you'll limit how much voltage you can feed into the amp.
Would changing the diodes for biasing cause other unwanted problems with the circuit?
Yes, a lot of diodes have a voltage/current curve that you have to be careful of. You may find a diode with a 0.5V forward drop, but only at 1A. If you go down to a few mA, that voltage may wander all over the place. (Some parts may also be a tad noisy)
Assuming my reasoning is correct, if I took the path of changing the bias would I achieve this by changing the diodes? Would I need to change anything else in the circuit to achieve the change in the bias? Would the change in the bias make it necessary to change anything else in the circuit?
I think you'll get lower distortion and better overall performance with the 50K plate load resistors.
