Dropping string resistor calculations question

[jcsifu]

Regarding dropping string resistor calculations for my PI tube through V1

Voltage (E) = Current (I) * Resistance (R)

So if I'm right, I use 2.2ma per point(both triodes of a 12ax7) and the voltage that I want to drop right?

Example:
I want to drop an extra 15 volts from V3 (pi) so I get 6818 ohms. So I would take the value of the resistor already there (3300) and add the 6818 to it for a new resistor value of 10100, right? So I would just sub a 10k for the 3.3k that was there previously.

That seems really high compared to other peoples amp values. Am I missing something?

[tele_player]

Looks like you're ignoring the current draw of the preamp tubes.

Robert

[jcsifu]

Looks like you're ignoring the current draw of the preamp tubes.

Robert

I used 2.2ma for the 12ax7...isn't that the current draw?

[tubeswell]

Depends on how the gain stage is set up, but good rule of thumb is

1mA per 12AX7 triode.
2mA for a 6V6 screen at idle.
5mA for a 6L6 screen at idle.

[jcsifu]

Depends on how the gain stage is set up, but good rule of thumb is

1mA per 12AX7 triode.
2mA for a 6V6 screen at idle.
5mA for a 6L6 screen at idle.

Ok.
Amp is low plate.
So you are saying that my math is correct other than maybe subbing 2mA for my estimated 2.2mA right?

I would still need something around the 10k?

[tubeswell]

Depends on how the gain stage is set up, but good rule of thumb is

1mA per 12AX7 triode.
2mA for a 6V6 screen at idle.
5mA for a 6L6 screen at idle.

Ok. So you are saying that my match is correct other than maybe subbing 2mA for my estimated 2.2mA right?

I would need something around the 10k?

Depends on the voltage you have to start with and the voltage you want to end up with. 10k with 2mA current with result in a 20V drop.

[jcsifu]

Starting with 305 and dropping to 290, it's my pi tube I want to drop. I"ll do the V2 V1 after I make sure I get the math right from ya.

[martin manning]

As tele_player alluded to above, there is more than just the PI current flowing through that resistor. What are the voltages on each end of the 3k3? You can get the total current from that voltage drop.

Be aware that there is always some guessing in this process since when you change the resistance the current everywhere will move a bit.

[jcsifu]

As tele_player alluded to above, there is more than just the PI current flowing through that resistor. What are the voltages on each end of the 3k3? You can get the total current from that voltage drop.

Be aware that there is always some guessing in this process since when you change the resistance the current everywhere will move a bit.

Hey Martin!
I think I see where your going with the voltages. I would take the voltage difference between the nodes on each side of the resistor and divide by the value of the resistor to get my estimated mA...right?

So I have 462,430= 32v difference
A=32/3900 (not 3300 as I falsely remembered)
A=.0082
so 8.2mA

So that means I need another 1829.26 ohms added to the original 3900.
So I should need something in the ballpark of 5730 total...I should be able to use either 5k6 to 6k2 as a good starting point. Is this what you were suggesting?

[Synchu]

The math seems good, but as far as I understand what Martin is saying - the whole balance will somehow shift, hence it is difficult to calculate the exact effect, but these values seem as a good starting point, to me at least.
Niki

[martin manning]

Around 6k would be a good place to start. Re things shifting a bit, as you lower the voltage the current draw from the tubes downstream will go down a little, and the voltage upstream of the 3k9 resistor will go up a little (due to the reduced current draw) so you might have to iterate to get what you want.

[tubeswell]

Starting with 305 and dropping to 290, it's my pi tube I want to drop. I"ll do the V2 V1 after I make sure I get the math right from ya.

So there's three dual triodes going through that resistor? Time to see a schematic. See this one from Merlin Blencowe's site (albeit with 12AT7s) to illustrate what the others were saying about summing the current through each subsequent supply resistor.

[img:763:351]http://www.valvewizard.co.uk/smoothing5.jpg[/img]

[jcsifu]

Synchu- Thanks for taking time to post to me bro. I don't understand how to figure the shift other than to just redo calculations after each resistor change, adjusting for the new numbers for each resistor to replace. I imagine that is prob what everyone does to hit the numbers anyway.

Martin- Thanks for verifying my starting point. I will redo calculations after each resistor change, adjusting for the new numbers for each resistor to replace as you suggest.

tubewell- I built to same specs as #124 low plate skyline, only some pot and trimmer different values. I don't know how to cut out the sections of the schematic to show so here's the link


https://tubeamparchive.com/files/124_schematic_182.pdf


HAPPY HOLIDAY TO ALL YOU BROS!!!

[tubeswell]

https://tubeamparchive.com/files/124_schematic_182.pdf

Assuming the choke is like any other choke in a showman or twin, there's probably about a 1V drop across the choke (so B+2 will be 1V less than B+1).

Let's assume each 12AX7 is set up to draw 2.2mA at idle. (I haven't worked this out - I'm just using your 2.2mA figure from earlier). So the calculation would go something like the following:

There's three 12AX7 at B+3 and a 2k2 resistor between B+2 and B+3, so that resistor would see a (2.2ma x 3)/2k2 = 14.5V drop between B+2 and B+3. The 2k2 resistor is dissipating 14.5V x 0.0066A = 0.1W

There's two 12AX7 at B+4, and a 22k resistor between B+3 and B+4, so that resistor would see a (2.2mA x 2)/22k = 96.8V drop between B+3 and B+4. The 22k resistor is dissipating 96.8V x 0.0044A = 0.43W

There's one 12AX7 at B+5, and a 2k2 resistor between B+4 and B+5, so that resistor would see a 2.2mA/2k2 = 4.8V drop between B+4 and B+5. This 2k2 resistor is dissipating 4.8V x 0.0022A = 0.01W.

So:

B+2 is 1V less than B+
B+3 is 15.5V less than B+
B+4 is 112.3V less than B+
B+5 is 117.1V less than B+

Therefore, once you know what B+ is, the above ballpark figures give you a model for starting to work out the load lines for each stage

[jcsifu]

Ok, got it.
This might be a non crucial item for me to bring up but after calculating out a new B+ node using your given formula, do I then have to calculate out any voltage consideration to include influence (if any significant amount) by the 100k plate load resistor?