Need some tutoring on Ohm's Law for determining resistor power rating

[PharmRock]

Trying to educate myself on Ohm's law and the necessary calculations needed to determine the power rating of a voltage dropping resistor.

This is for a JTM45 build with KT66s and GZ34 rectifier.  345-0-345 PT.  7H Choke.

Although the schematics already have the values, I wanted to do the numbers to see what I came up with. 

Attached is the data table for the tubes for Class AB1, Push-Pull, Pentode operation, Fixed Bias.

While I haven't yet got the amp up and running, I'm guesstimating a rectified B+ of 450V. 

My first question in determining the screen resistor value is, do I use idle or max signal values?  The KT66 data table states at idle Vg2 = 395V, Ig2 = 3mA.  With max signal, Vg2 = 360V, Ig2 = 19mA. 

So the resistor value at idle would be (450-395)/0.003 = 18.3K screen resistor.  That doesn't seem right as its way higher than I've seen on most schematics. 

At max signal we have (450-360)/0.019 = 4.7K.  That's more in line with what I've seen, but still a little on the high side (schematics have anywhere from no voltage dropping resistor to an 8.2K coming off the choke, and 470R on the screens at the sockets).

I currently have a 470R/5W on pin 4 of each KT66, and a single 1K/2W coming off the choke to feed both of the 470R resistors. I suppose this would result in a series resistance to each tube of 1.47K.  If I stick with this, then my voltage drop at idle would be: V = I*R = 0.003*1,470 = 4.41V.  So if the voltage coming off the choke is in the neighborhood of 450V, I will have 445V at the screens.  That doesn't sound good.
A max signal: V = I*R = 0.019*1,470 = 27.9V, putting my estimated screen voltage around 420V. 

If I replace the 1K voltage dropping resistor with a 5K, at idle I will have a voltage drop of 0.003*5,470 = 16.5V, putting the screens around 430-435.  Still a little high.  At max signal I will have a drop of 0.019*5,470 = 104V, which puts me on target for 360V per the data sheet.

Ok, so let's say I go with a 5K dropping resistor.  What power rating do I need?
P = I^2 x R = (0.019)^2 x 5,000 = 1.8W.  I've read that is always a good idea to go with 2-times the required power rating...so a 5K/5W resistor?

For the 470R on the sockets: (0.019)^2  x 470 = 0.17W.   I have 5-watt resistors on there right now, which is way overkill as I see it.

Anyways...first time doing this.  Thanks for your patience as I rambled, and if you can provided any guidance/insight, it is much appreciated.  I suspect I am overlooking something major here, or did a calculation wrong. 

[SILVERGUN]

I'm about to drift off, but read this whole page and make sure you make it all the way to the bottom:
http://www.valvewizard.co.uk/smoothing.html

[PharmRock]

Thanks SG...
I think I've read and reread everything on Valve Wizard's site.  I also bought his book on preamp (his power amp book is no longer in print).  Its just a matter of getting all of this to sink in, which is why I tried to apply some of what I've learned to the screen power supply. 
Reading up some more, I found a post on this forum by PRR related to the topic. https://el34world.com/Forum/index.php?topic=18985.0
Basically he says that while the datasheet put the screens substantially lower than plate voltage, the reality is that they typically run 90-98% of the plate voltage and we set the bias accordingly.  So if I have 450 on the plates, I'm looking at 405-440 on the screens.
I worked out the calculation for the choke and arrived at 5H (using 32uF and target of below 10H resonant frequency).  I have a 7H choke so that's good.

Valve Wizard utilizes an EL34 in the example you provided, he states, "The data sheet suggests a screen-to-anode current ratio of 6.5, so we can expect the screen currents to amount to 125 / 6.5 = 19mA for the pair."
I can't seem to find the ratio for a KT66 tube, but assuming the ratio is about the same, I arrive at about 17mA for the screens if plate voltage is 450V.

Thus, if the plates are at 450, and I shoot for a idle screen voltage of around let's say 425 (voltage drop of 25V coming off the choke) and screen current of 17mA, I end up with a resistor value of 1,470R.  Which is exactly what I have (1K + 470R in series).
Since P = I^2 x R, the power dissipation of the 1K resistor would be 0.3W, and the 470R resistors would each have about a quarter watt.  I know that current and voltage would change substantially during playing conditions, so the 2W and 5W values I have (in series) should be plenty.

In other words, my set up is good.

Am I on the right track here?

[pdf64]

The 1k is typically shared, carries the g2 current of both KT66. Therefore the current through it will be doubled.
My finding is that to accommodate heavy overdrive, the 1k needs to be 5W, whilst the 470 is fine at 1W.
Beam pentodes (eg KT66) tend to have a lower ratio of g2 to a current than suppressor grid pentodes (eg EL34).

[PharmRock]

Thanks PDF...
Regarding the doubled current through the shared 1K, when I calculated the 17mA current draw of the screens that included both tubes.

The closest I have in a 5W resistor is an 820R.  Seems like this would be "close enough", but can't hurt to order some. 

[HotBluePlates]

... calculations needed to determine the power rating of a voltage dropping resistor. ... in determining the screen resistor value is, do I ...

For where you're at, you're not ready to determine the power rating of a screen resistor.  Instead, focus on your original question of a "voltage dropping resistor in the power supply."

JTM45 Schematic

V3 is fed from a 16µF filter cap that is annotated "310v."  This cap receives voltage from a dropping resistor of 10kΩ that itself is fed by a 16µF filter cap with 350v present.

    -  We can see how many volts are across the 10kΩ:  350v - 310v = 40v
    -  We can use Ohm's Law to calculate the current through the resistor:  40v / 10kΩ = 0.004A = 4mA
    -  Power dissipated is Volts (dropped) x Current (through):  40v x 0.004A = 0.16 watt
    -  The resistor should be rated for at least 2x the dissipated power:  1/2w would work, but the marked 1w is better.

We could double-check the resistor current a different way, by looking at the current pulled by downstream gain stages.

    - V1 shares an 820Ω cathode resistor for both sections; the voltage chart says there is 1.6v across this resistor:
      1.6v / 820Ω = 1.95mA
    - V3 has 1.1v across an 820Ω cathode resistor:  1.1v / 820Ω = 1.34mA
    - The other V3 section has 190v across the 100kΩ cathode follower load:  190v / 100kΩ = 1.9mA

    - Total Current = 1.95mA + 1.34mA + 1.9mA = 5.19mA
    - It's normal that the numbers don't add up exactly, as the drop of "350v to 310v" was obviously rounded.
    - Different form of Power:  Current x Current x Resistance = 0.00519A x 0.00519A x 10kΩ = 0.27 watt
    - Now 1/2w seems marginal, but 1w has plenty of margin.

[HotBluePlates]

... the necessary calculations needed to determine the power rating of a voltage dropping resistor.

... determining the screen resistor value is, do I use idle or max signal values?  ...

The reason you're not ready for going after the screen resistor is it is a complex calculation.  We don't use the screen resistor for a static voltage-drop, but to keep the screen below its rated dissipation limit over the signal cycle.

The condition considered is the output section driven to full-power.  The data sheet figures are average values, as the screen current is changing over different points in the signal cycle.

    -  You appear to be referencing the condition on Page 3 of this KT66 data sheet.
    -  Multiply the average screen current by the screen voltage, perhaps the idle 395v shown.
    -  Screen Current shown is ]per pair so 0.019A / 2 = 9.5mA.
    -  0.0095A x 395v = ~3.8 watts
    -  The first page says the screen is rated for 3.5 watts maximum.

We need a resistor that drops the screen voltage so the dissipation is no more than 3.5w.  The Max Signal column on that Page 3 condition shows that at maximum output, the screen voltage falls from 395v to 360v.

     -  360v x 0.0095A = 3.42 watts
     -  Let's assume each tube has an individual dropping resistor:  (395v - 360v) / 0.0095A = ~3.7kΩ
     -  Resistor Dissipation = 35v x 0.0095A = ~0.33 watt ----> Double to 1 watt

     -  Assume a shared dropping resistor:  (395v - 360v) / 0.018A = 1,944Ω ----> Round up to 2kΩ
     -  Calculate power using resistance & current:  4kΩ x 0.018A x 0.018A = ~1.3 watt ---->. Round up to 3 watt

Some of the calculated resistance above may come from elsewhere in the power supply.  That is, if the supply naturally sags 15v without any added screen resistor, we only need to cause an additional 20v of screen voltage-drop.  And so we'd use a smaller resistor than calculated above.

____________________________________________

The part above was "simple."  The "complex calculation" is that we normally don't have a figure for "average screen current over the signal cycle, at maximum output power."  That means it's up to us to:

    -  Plot a loadline for our output stage
    -  Break up a complete signal cycle into many parts
    -  Plot the screen current at each point in the signal cycle
    -  Calculate an average-value from the individual currents found.

Many folks won't go through all that, and simply copy a setup they've seen in other amps.

[PharmRock]

HBP,
First, let me say "thanks" for taking the time to break it all down for me. I really appreciate that. I'm going to read through this a few times while looking at the schematic and follow along.

I think the schematic you were looking at is the tremolo version of the amp...I'm doing the typical JTM45 plexi without the tremolo.  However, I fully appreciate that I can still use this and your answer to learn, which is what I plan on doing.

Thanks again! 

[HotBluePlates]

... let me say "thanks" for taking the time to break it all down for me. ...

You're welcome!

... I think the schematic you were looking at is the tremolo version of the amp...I'm doing the typical JTM45 plexi without the tremolo.  ...

That doesn't matter:  we needed a schematic with voltages listed, and that was the one JTM45 schematic I could find quickly with voltage figures shown.