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47.136.5.103
Given the arrangement below(Fig. 3). If the 47K needs to be replaced with a volume pot, which arrangement is correct between Fig. 1 and 2. Why, and does the grid resistance of the tube the basis of the calculation for the value of the pot? Will 100k log tapered be fine here or 250K or higher?Thanks!
Edits: 12/03/16Follow Ups:
-Otherwise you may encounter weird effects, like more hiss in the middle of the control.
300-1000 ohms should serve nicely.
Have fun!
You want to wire it like Fig 2.If you use fig 1 the stage driving the pot would have to drive a very low impedance as you turn the volume down.
There are two reasons why you don't want to use too large a value for the pot.
All tubes have grid leakage current.
The higher the grid leakage current to lower the Max permissible value for the grid resistor.
The other reason is Miller capacitance. The higher the pot value the higher it's output impedance. The higher the output impedance the lower the -3db point of the low pass filter that is created between the output impedance and the Miller capacitance.
What tube?
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 12/03/16 12/03/16
Thanks Tre'!
The tube is 201A.
Edit, I made a mistake. The datasheet I looked up must not be for a ux 201a.Please disregard the info below WRT the gain and the Miller capacitance of a ux 201a.
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Depending on operating point, a 201A stage has a gain of 25.
The grid to plate capacitance is 5pF.
5 X 25 is 125pF plus 2.9pF grid to filament for a total Miller capacitance of 127.9pF.
A pot has a worse case output impedance of 1/4 it's resistive value at the -6db point.
A 250k pot would have an output impedance at the half voltage point (-6db) of 62.5k ohms.
62.5k ohms driving 127.9pF of capacitance creates a low pass filter with a -3db point of 19.9kHz
I think if your source was happy driving the 47k grid resistor you should stay with that. Replace the 47k resistor with a 50k pot.
A 50k pot has a worse case output impedance of 12k ohms.
12k ohms driving 127.9pF gives a -3db point of 103kHz.
The amplitude will be down only 1db at 51.3kHz and the phase will be shifted all the way down to 10.3kHz but I doubt you will notice.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 12/03/16
A 201A has a mu of 8, is this the same as gain?
I must have looked up the wrong tube.I'm very sorry. I try hard not to make this kind of a mistake.
Thank you for pointing out my mistake. I have edited my post above.
But to answer your question, The mu of the tube is the most gain you can get from it in a circuit. It would be the gain of the stage, when and only when, the load line is totally horizontal.
You would have to load it with a plate choke or a CCS to get close that much gain. The gain of a stage is always lower than the mu of the tube.
Can you give me a link to the ux 201a datasheet?
Edit, I found the datasheet for the ux201a. Sure enough, mu of 8.
I have no idea what data sheet I was looking at earlier. Again, my mistake.
BTW If the mu of a ux 201a is only 8 then everything I said about the Miller capacitance is wrong and driving that Miller will be much easier and will not require a low drive impedance.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 12/03/16 12/03/16
Clearly a different tube.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
"A 50k pot has a worse case output impedance of 12k ohms."
Did I miss something here? Even if the source is zero ohms, the pot when set to its electrical midpoint will look like 25k to the grid. I'm not saying that's a bad thing, just wondering what you meant.
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Buy Chinese. Bury freedom.
Assuming a zero ohm source driving the pot, 25k ohms from the wiper to ground (the bottom of the pot) and that 25k ohms would be in parallel with 25k ohms from the wiper to the top of the pot (a zero ohm source).That's 12.5k ohms WRT AC ground and thus a 12.5k ohm source impedance to drive the 201a grid....no?
Note...even if the source impedance driving the pot is 5k ohms that doesn't change the number much...25k in parallel with 30k* (*the other half (25k) of the 50k pot plus the 5k driving source) is 13.6k
BTW Sorry for the typo, I meant to type 12.5k in my original post.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 12/03/16 12/03/16 12/03/16
"Assuming a zero ohm source driving the pot, 25k ohms from the wiper to ground (the bottom of the pot) and that 25k ohms would be in parallel with 25k ohms from the wiper to the top of the pot (a zero ohm source)."
Hey, I was just approximating. :) :)
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Buy Chinese. Bury freedom.
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