Dropping string resistor calculations question

[tubeswell]

Yeah the B+ node is stabilised to a relatively steady DC level by the filter cap at that node, and this constitutes the High Tension (HT) voltage for the gain stage(s) sourced from that B+ node.

On a static anode characteristics graph, the HT voltage is where the load line intersects the x-axis.

The plate resistor determines the load line.

To get the point where the load line crosses the y-axis, you subdivide the HT voltage by the load resistance. Hence if the HT voltage at B+5 was say 250V on the x-axis, then the point on the y-axis for stage V1a (100k plate resistor) is 250V/100k = 2.5mA.

The load line then connects between the 250V on the x-axis and the 2.5mA on the y-axis. There you go

[jcsifu]

Yeah the B+ node is stabilised to a relatively steady DC level by the filter cap at that node, and this constitutes the High Tension (HT) voltage for the gain stage(s) sourced from that B+ node.

On a static anode characteristics graph, the HT voltage is where the load line intersects the x-axis.

The plate resistor determines the load line.

To get the point where the load line crosses the y-axis, you subdivide the HT voltage by the load resistance. Hence if the HT voltage at B+5 was say 250V on the x-axis, then the point on the y-axis for stage V1a (100k plate resistor) is 250V/100k = 2.5mA.

The load line then connects between the 250V on the x-axis and the 2.5mA on the y-axis. There you go

Awesome! Thanks bro...thanks for something new to focus on. Took it to a new level didn't ya?.....I did ask.....I just spent the last hour looking at plate characteristic charts and it makes me realize that I am not one of the people who understands circuits enough to design a dropping string accurately using a complete mathematic approach.

I will continue to study up on this now that you pointed it out to me.

I think in the mean time, I will have to go with an educated guess for a starting point for the values required for my target voltages.

Very cool

[martin manning]

Starting with 305 and dropping to 290, it's my pi tube I want to drop.

I gave this some more thought, and came up with something that I believe will be accurate and generally useful.

You want to reduce the PI plate voltage by a factor 290/305 or 0.951. To do that you will need to reduce the voltage at the PI node by that same factor. Right now the PI node is at 430 (call that V2), so you want to reduce it to 409. This node is fed by a 3k9 series resistor (R) and the upstream voltage (at the previous node, call that V1) is 462. You want to find a new R (Rn) such that the new V2 (V2n) will be 409.

Rn will be found as: Rn = R x (V1/V2n - 1)/(V1/V2 - 1)

The value of the new dropping resistor Rn is then 6791, or 6k8.

[jcsifu]

Starting with 305 and dropping to 290, it's my pi tube I want to drop.

I gave this some more thought, and came up with something that I believe will be generally useful.

You want to reduce the PI plate voltage by a factor 290/305 or 0.951. To do that you will need to reduce the voltage at the PI node by that same factor. Right now the PI node is at 430 (call that V2), so you want to reduce it to 409. This node is fed by a 3k9 series resistor (R) and the upstream voltage (at the previous node, call that V1) is 462. You want to find a new R (Rn) such that the new V2 (V2n) will be 409.

Rn will be found as: Rn = R x (V1/V2n - 1)/(V1/V2 - 1)

The value of the new dropping resistor Rn is then 6791, or 6k8.

Well that sure is an easy to apply equation! That is empowering like I just got a super power bro! I feel like I can accurately calc out my values now.

Brilliant to use a ratio calculation of the plate difference to apply to the node voltage...all I have to do now is sneak away from the family while they're recovering from over eating turkey and switch some resistors out tomorrow.

I bet a lot of people could benefit from this equation, as it is so simple to apply.

Thanks Martin!

[martin manning]

You're welcome; let us know how it works out.

[jcsifu]

You're welcome; let us know how it works out.

Will do bro...I need to order some 2 watts as I don't have 6k8 on hand, but I will follow up when I receive them and put them in so we can verify this formula of yours.

HAPPY THANKSGIVING

[jcsifu]

I'm experiencing what you guys were saying about many factors involved in figuring out the dropping string values.

Since we last spoke I...
*got impatient (surprise surprise) and used the closest value I had in a 2 watt, which is 5k1, and threw it in there.

*this put me at 296/281, which of course is close enough to the desired 290/280 in #124....yeah

*was so pleased with mucking about that I didn't want to stop, so I decided I wanted to roll some tubes. I took out the black plate raytheon in the pi and put a telefunkin long smooth in there, which I like the results of, and then re-checked readings and had gained 2v/2.7v

then put 2 GE long plates in v1 and v2...because that's what I heard Dogears does and of course who doesn't want to sound like that dude.

I then re-checked readings and now have 299.7/284.6 on the pi

the reading at the nodes of each side of the 5k1 are 462,418

It would seem that anything I do anywhere in the circuit affect everything else to the extent that I now wonder if it is even possible to set this amp up and be able to swap a pre tube out without having to change the string values.

Is this just how it goes?
What do you guys do?

[tubeswell]

Each tube has slight variations in tube current, even between the same type of tube. I wouldn't sweat it. Its a ballpark thing.

[jcsifu]

Each tube has slight variations in tube current, even between the same type of tube. I wouldn't sweat it. Its a ballpark thing.

Ok bro, thanks for the confirmation.

[Structo]

What I have done in the past is find a schematic that is very similar to the build you are working on.

Some schematics have voltage notations while others don't.

I know that is cheating a bit but calculating dropping strings can get complicated depending on the amp.

And a big +1 on the differences between tubes of the same designation, even the same brand, will draw different amounts of current.
Thus different voltages as well.