Please help me with my dropping resistor math

[Choan]

Hey guys, if you could spare some change and help a non-mathematically-inclined man with his math, I'd appreciate it.

I've been reading about lowering B+ voltages to get a particular sound (I have an AA270 Super) and I know the answer is to use ohm's law. I'm not sure if my math is mathin' though. Let's say I want to drop B+ off of the final power filter stage, at the 10k resistor in the doghouse, and let's assume my amp is using schematic voltages. I want to decrease the 360 vdc to 285, instead of 315. I'm calculating 3.91mA by dividing 45W by 115 wall voltage.

using R=V/I... I'm getting 19,181k, .29watt rating.

Is this math correct? Haven't taken a math class since high school so I doubt my comprehension. If it's correct, I'm assuming I could just take my actual voltages and plug them into this formula (with the understanding the resistors need to be higher wattage).
Is there any reason to progressively step down the voltages across the power supply?

Sincerely, someone who thinks math is cool but isn't good at it.

[Willabe]

Just put that 20K R in and see what happens, can't hurt anything.

Is there any reason to progressively step down the voltages across the power supply?

In part because the preamp small bottle tubes need lower plate dc voltages than the power tubes. (Unless it's an EL84 power tube.) And it just happens as you go down the B+ string.

The larger the dropping R between 2 filter caps the more ripple is reduced. You can use a bigger filter cap uF value with a smaller R or use a bigger R with a smaller filter cap value for the same ripple reduction. R's cost less than caps and years back e caps cost a lot more money than today. And it just works out dropping the B+ dcv as you go down the string.

But, there's limits here because you can only use so large a value on the 1st filter cap if you have a tube rectifier. And you can only use so high of a R or you drop too much dcv. Have to find the balance that works.

[mresistor]

It really helps if guys would include a schematic link,,, they are on this site and easy to find for you. https://el34world.com/charts/Schematics/files/Fender/Fender_super_reverb_aa270_schem.gif


Notice the Power supply C node is not used.  Why not repace the 1K 1W with a 10K 1 or 2W and see how much that drops the voltage on the D node.


Edit:  oops it's been pointed out that node c feeds the PI        more coffee..   

[Willabe]

Notice the Power supply C node is not used. 

C node feeds the PI.

[stratomaster]

You need to know the load on that final node, you can estimate it as 1mA per triode, but I prefer to calculate it.  This will give you the current through the resistor.  Remember to use Ohm's Law across the resistor you're interested in.

The schematic shows the Node you want to drop, D, is at 360v, and node C before it is at 395v.  A 4,700 ohm resistor separates them.

Ohm's law applied to the difference of 35v gives about 7.45mA through the resistor. 

You can use this to estimate the resistance needed for your desired voltage drop.  There is interaction that isn't accounted for and differences in the current drop due to the resistor values, but this is a good first estimate.

I'm having trouble parsing what your target voltage is and following your description as there isn't a 10k in the schematic for the AA270 Super Reverb, but let's assume instead of 360v you want 315v.  This means your difference is 80v instead of 35v. 

V/I=R, so a 10.7k resistance is the target (assuming the same current through the resistor).  Nearest standard value is a 10k, so use that, and measure. 

To estimate power use V²/R.  You want an 80v drop across your 10.7k resistor so you're looking at 0.6W. 

This isn't perfect, but it's enough to let you know that a 10k, 1W resistor is a good first step.

If you want to drop all the way to 285v, then the voltage across the resistor becomes 110v.  Assuming the same current gives ~14.8k.  So a 15k, 1W resistor would be a good first guess. Though a 2W might help you sleep better at night.

Bigger changes like this from stock start to really stretch the limits of this approximation as the difference in current due to resistance value becomes more noticeable.

[mresistor]

guys disregard my suggestion 

[Choan]

Thanks for thee feedback y'all! Sorry about the lack of schematic. I hesitated to post it because it's a Super Reverb but it seemed to be a bit of a bastard... as in it has some similarities to the Twin AA270, like 10k and 2.2k resistors on node D and C instead of the super values. Also had some variable resistor values and voltages when I got it. A Twin schematic for the curious...

https://www.prowessamplifiers.com/schematics/fender/Twin_Reverb_aa270-Schematic.html

[mresistor]

There is more than one Fender AA270 kinda like there is more than one AB763.  Just use the right one.

[Choan]

Thanks Stratomaster that's very helpful, didn't know that in V/R that V is the voltage drop sum. Much appreciated. Like I posted above, there are some weird inconsistencies in my amp and the super aa270, but I can take my amp's values and use your insight to get the result. Cheers