Is 2 Watts enough for power resistors?

[Natman]

I know a lot of power resistors are 3W metal oxide, 5-15W ceramic and even wirewound. Will 2W be enough for a typical 2x6v6 build (20-40 uF filter caps)?

[Bruce / Mission Amps]

I know a lot of power resistors are 3W metal oxide, 5-15W ceramic and even wirewound. Will 2W be enough for a typical 2x6v6 build (20-40 uF filter caps)?

I think we need a little more info here.
What do you want the resistors to do?
Are these cathode biasing resistors or screen current limiting resistors or the screen or preamp node B+ dropping resistors? etc.

[Natman]

Ah -this is where my newbishness comes out. I might not be using the right term, but I'm referring to the resistors on the B+ rail among the filter caps. I went and ordered 2W and then realized I maybe should have gone with 3W or higher. Thanks for chiming in Bruce!

[Enzo]

Even then it matter which ones in the circuit, their value, the amount of voltage dropped across each, which is proportional to the current through them.

Nothing shameful about being a novice, we all started somewhere. if you learn anything in electronics, learn Ohm's Law. it is a very simple little formula that describes the relationship between voltage, current, and resistance. If you know two of those three factors, you can simply calculate the third. So if you have some resistor, and you find it has 100 volts end to end, then Ohm's Law will tell you the current flowing through it. And current (in amperes) times voltage is power in watts. If a resistor calculates that it will be dissipating (having to deal with) 1 watt, then we want the resistor to be rated for at least 2 watts, and to me preferably more.

Example:
10k resistor (10,000 ohms) with measured voltage end to end of 100v. Ohm's Law tells us that Volts = AMperes x resistance (in ohms). Amperes, amps, usually goes by the letter I. So
V = I x R

Which we can massage to find current:
I = V/R = volts divided by ohms.

So current in the example is I = 100/10,000 = .01A = 10milliamps

Power (in watts) is simply current times voltage:
P = I x V

So in the example:
P = .01 x 100 = 1 watt.

SO I would use a 2 or 3 watt resistor for that.

I have been soldering now for about 55 years, and I can honestly say I get out my little pocket calculator and figure some little result with Ohm's Law every single day. You won;t regret learning it.

Look at this Fender Deluxe schematic: http://www.schematicheaven.com/fende..._6g3_schem.pdf

The voltages along the B+ rail are given.

375v at the rectifier, and 365 at the screens, and 1000 ohms between.

I = V/R = 10/1000 = 10ma again = .01A

P = I x V = .01 x 10 = 1 watt. SO you see they used a 2W.

Now look to the left, the 27k resistor has 270v and 325v at its ends - 55v.

I = V/R = 55/27000 = .002A = 2ma

P = I x V = .002 x 55 = .11 watt. So a 1 watt resistor is plenty large enough.


There is a short cut for this too:
solving the power formula in the terms of the earler formula, we can say:
Power = voltage squared divided by resistance = V^2/R
or
Power = current squared x resistance = I^2xR.

So for the 55v across 27k ohms, we get
P = (55x55)/27000 = .11 watt


SO the bottom line is, there is no set power rating for all those resistors, some need to be heftier than others. But the calculation really is pretty simple.

And it is not just for resistor wattage. For example in the same schematic, look at the 12AX7 section top center. See it has 1.4v on the cathode with 1500 ohms (1.5k) to ground, so 1.4v across 1500 ohms. How much current is flowing through that tube then. Try it. And note for that same tube that there is 165v on the plate, while there is 270v on the B+ node, and 100k+15k (115,000 ohm) between them. So 105v across 115,000 ohms. What current do you get there? How does it compare to the current on the other end of the tube. it is after all the same flow of current going in one end of the tube and coming out the other.

Why is that usefull knowledge? If I know how much current is flowing through each tube, and I know all that current comes through my B+ supply, then at each step I can add up the current through each tube to know how much current is flowing through those resistors you asked about.

[Natman]

Once again, great info. Thanks!

I am well versed in Ohm's Law but usually don't know how to apply (anything) when it comes to tube amps. Heck I'm still learning nomenclature and all the bits & pieces.