Would you use these?

[leadfootdriver]

I bought these for my EL34 cathodes, so I can do the cathode current bias method. But now that I have them in my hands, I'm not to sure about them because of their thin leads.

Here's the Mouser link.
RW80U1R00FB12 Vishay Wirewound Resistors - Through Hole

[Tom Phillips]

Yes. I'd use them. The specs look more than adequate for your application.

[pdf64]

I wouldn't use them, but that's due to them costing £1.23 each!
1/2 watt MF types are fine for this application.
But as long as the leads aren't fragile and liable to fracture or something, they should be fine, it's not as if they are carrying significant current.

[MWJB]

If you use the bridge between pins 1 & 8 to take your measurement, then you won't be stressing the resistor leads.

[exclamationmark]

I wouldn't worry about the leads unless it's for structural reasons. The resistance of the leads is in the milli-ohm range compared to the actual resistor... so as you would expect, the resistor would melt from over dissipation before the leads ever did. (As a side note... long leads actually help dissipate heat and in most cases have minimum lengths specified to meet dissipation spec - especially small silicon diodes).

[diagrammatiks]

man I love those things.

When I first got a set of the newer silicon coated wirewounds I was pretty confused as well.

They are hilariously tiny, especially next too an old style wirewound with the same wattage.

They work great though. The 2 watt 1ohm is what I use.

[tubeswell]

I use 0.6W 1% 1R metal film resistors for the bias resistors. They are quite small, and the leads are quite tiny, but they only see a tiny voltage drop and a tiny amount of current, so they are fine. Your ones look fine too leadfootdriver.

[leadfootdriver]

Thanks fellas!

[R.G.]

Ohm's law: V = I*R
Power P = I*V = I*I*R

I = SQRT(P/R) = SQRT(2W/1ohm) = 1.414 amperes at rated power.

Therefore, we know the manufacturer thinks the leads will reliably conduct over 1.4A.

A conventional guitar amp power tube conducting 1.4A is in deep trouble. Therefore, the resistor will last longer than the tube will.

They're provably fine.

[leadfootdriver]

I guess I got caught in the size matters zone. I have 1 watt CF resistors for the majority of the amp, and they're big honkers with thick leads in comparison.

I will say that the 2W1R's seem to have leads made of a higher tinsel strength. You could definitely impale yourself with one of these. The spec sheet says that they have stainless steel caps, and the leads are tin plated to help with soldering.

[tedmich]

what about these?
CRCW25121R00FNEG Vishay/Dale Thick Film Resistors - SMD
1 ohm 1w 1% 500v $0.12 each

SMD 3.15 mm W x 6.3 mm L x 0.6 mm H !

[diagrammatiks]

eh there's much better thick film available.

[Enzo]

Size only partly matters.

But you have to have some idea of the size of the JOB first.

SO what RG said.

Its only 1 ohm. SO when calculating wattage, how much current can the tube conduct, and what will that do with only 1 ohm. How about 100ma? .1 x .1 x 1 = 0.01 watt.

[imaradiostar]

Ohm's law: V = I*R
Power P = I*V = I*I*R

I = SQRT(P/R) = SQRT(2W/1ohm) = 1.414 amperes at rated power.

Therefore, we know the manufacturer thinks the leads will reliably conduct over 1.4A.

A conventional guitar amp power tube conducting 1.4A is in deep trouble. Therefore, the resistor will last longer than the tube will.

They're provably fine.

Why did it take till the 9th post for someone to post the actual math? Glad R.G. did, just amazed no one else did sooner! Any time I build something for work this is one of the first steps- make sure you've picked an electrically sound value that won't die from excessive heat or exceeding the power handling of the component.

Keep in mind that small leads are rarely an issue in a tube amp. The currents we work with are teeny-tiny compared to what the average 18 to 22 awg wire is capable of carrying!

jamie

[J M Fahey]

At least *here* Math raises its ugly head often enough !!
There are a zillion other Forums where there is not the least danger of that.