what is the wattage of r1?

[georage]

so i actually dont know anyone else i can ask about this.. im trying to assemble a gibson ga5 clone and im trying to calculate the wattage value of the resistors in the preamp, especially r1 in my diagram using ohms law. the schematic gives only resistor values so i used the 12ax7 data sheet to fill in the plate resistance and ma current in the circuit. if you feel like it could you glance at my work and comment whether I am "on track" or "way off"? any help is appreciated.

[catalin gramada]

You have 2x 12ax7 sections conected to those nodes as shows is loaded draws, say 1mA each. So you have 2 mA draw through 22k resistor meant 0.002A x 22000ohm = 44V drop. So 44V x 0.002A = 0.088W, say 0.1W. Any 0.5W resistor will work. These is a large margin calculation.

[georage]

that makes sense, so im off by a factor of 10+

[catalin gramada]

Use 1 w if you can, they have longer leads and more tolerant at heating when solder on heavy terminals. And don't forget the voltage ratings.

[georage]

i think i mistook wattage for amperage in the final calculation. thatnks for helping!

[catalin gramada]

i think i mistook wattage for amperage in the final calculation. thatnks for helping!

You're talking about power ("wattage") and current ("amperage"). Speak English damn' it ! joking.

[pdf64]

To accommodate any voltage surge at power up, it’s a good guideline to have the voltage rating of all the HT droppers at least as high as the full unloaded HT, eg with no valves other than the rectifier fitted.
I acknowledge that may leave an unnecessary degree of voltage ’headroom’ for droppers at the end of the chain, to the preamp node.

[Helmholtz]

Resistor power calculation:

When R is not known, use P = V * I,

when I is not known, use P = V²/R,

when V is not known, use P= I² * R.

With I = 2*1.2mA, this would give P= 0.13W.
But actual tube current with a 2.2k cathode resistor shoud be only around 0.6mA.
The 1.2mA from the datasheet is just a reference value.

You can't use the internal plate resistance rp for DC calculations as rp is a dynamic (or differential) resistance.
It works with AC signal only.

[georage]

i was going to ask why the plate resistance values made no sense at all. i tried a few times to make sense of it in the drawing. im really trying to understand how to guess the right resistors instead if just changing them willy nilly and seeing what happens.

[Helmholtz]

As said, the internal plate resistance rp is not a DC resistance, so can't be used with calculations using a DC current.

[FrusOG]

IMO, I would assume 2 shorted tubes (worst case) which gives about 220K Ω + 2K Ω = 222K Ω per tube. Two of these in parallel = 111K Ω. Add that to the 22K Ω that's in series to 350 Volts of B+ = 133K Ω. 350 Volts ÷ 133K Ω ≈ .00263 Amps. .00263 Amps X 350 Volts = .921 Watts. A 1 Watt resistor should survive a worst case scenario and be well within "normal" operating parameters.

[The Dude]

I'll preface by saying that I don't disagree with anything said in this thread. This is just a side point. In many circumstances it's best not to overate parts too far planning for "worst case scenarios". It's often better to have a cheap resistor burn up in case of a short than more expensive alternatives.

[FrusOG]

You make a valid point.

In this case, the tubes maximum current specification is .008 Amps per tube. This is much more current (≈ 3 times as much per tube and ≈ 6 times as much total current) than the resistors will allow, assuming 350 Volts B+.

I just thought, by eliminating the tubes characteristics, it made it easier to figure out.


I did make an error, in that, I did not figure the Wattage for the single resistor. I only computed the total Wattage spread across all of the resistors.

Assuming .00263 Amps through 22K Ω ≈ 58 Volts across the resistor. 58 Volts x .00263 Amps ≈ .15 Watts. Even a 1/4 Watt resistor should work for worst case. A 1 Watt resistor would be a waste.

Let's figure the 220K Ω resistor Wattages, too:

.00263 Amps divided by 2 ≈ .0013 Amps per resistor (and per tube). .0013 Amps x 220K Ω = 286 Volts across the resistor. 286 Volts x .0013 Amps ≈ .37 Watts. A 1/2 Watt resistor should work for worst case. (The 2.2K Ω resistors will dissipate even less Wattage than the 22K Ω resistor does, so I won't bother to figure out that value.)

[georage]

interesting senerios Frusog. that is definetly what i was trying to understand. its a lot for me to comprehend, but it makes sense to think of the tube as a short circuit for ease of calculation and for safety sake. as an electrician im always designing circuits that can conduct enough current to blow a fuse or breaker during a short circuit. anyway the amp is built now and so i should state some actual values.

[georage]

i did some measurements. the wattage...i mean power of r1 ended up being .29 watts, just over one quarter. 1/2 watt would have been ok, but 1 watt is definitely a good common sense rating.
its really interesting how v1 a and b played out as v1a conducted more current than v1b even though it has a higher resistance cathode resistor. i really dont know if that is good or bad or why that happened.