Blown 1 OHM current-sense resistors

[R.G.]

The sheet says that for the resistors I ordered that they can withstand 5x their rated power for 5s, which is about 3.9A.

The reason that some of this is confusing is that the answer to how much they can take are tied up in things that are not easy to measure and not something our normal experiences tell us about. The thing that kills parts on overcurrents is the internal temperature it gets to. That varies hugely with how fast the heat gets out.

Here's a heating-up analogy I just thought of. Imagine that temperature is like the pressure inside a balloon. You pump up the temperature by blowing air into the balloon. However, the balloon has a tiny, constant pinhole leak. So if you blow air in at some rate, the pressure (temperature) goes up. But the pressure across the pinhole leak goes up, too, and more air leaks out. At low rates of air coming in, the pinhole can reach a place where it just balances the air coming in, and the pressure (temperature) levels off at some higher-than-zero pressure in the balloon. If you quit blowing into it, the pressure declines back to zero.

The speed at which you blow air in matters, as does how long you blow. If you blow with low flow rate over a long time, you are counting on the pinhole leak (the cooling rate) to keep up. But you can blow in short puffs too. So as long as the puffs average out to less than the pinhole can take care of, you're OK.

But what if you are pressurizing the balloon from a scuba diver's air tank? If you just turn the valve on, the balloon pops almost instantly. But what if you had a microprocessor controlled valve that could turn on and off in a microsecond? You could ping the valve for a few microseconds and a puff of 2200psi air goes in, but the puff quits before the pressure/temperature pops the balloon. As long as the puff doesn't pop it, the pinhole has time to drain the pressure back down.

The total amount of air the balloon can stand in one puff is its surge rating. It can take microseconds-puffs from high pressure, or a second from 200psi, or 10 seconds from 20 psi, or forever at 2psi.

It's like that with heat. You pump in a slug of heat with current. There is some value the thing will stand forever, because it's cooling mechanism can get rid of that amount of heat on a constant basis. Over that level, constant heating will kill it, but you can put in short slugs of current and if the single slug doesn't kill it, and it has time to cool off before the next one, it still lives. It's the pressure times the number of seconds it's on, compared to the leak-down rate that matters.

Your resistors can take 3.9A for 5 seconds. The power into the resistor is I²R. The energy pumped in is I²R times the time it acts. So your resistors can stand (3.9)*(3.9)*R*5seconds, or 76.05*R. They will take 8.7A for one second, and 27.5A for a tenth of a second without overheating inside. But they'll only take 2.76A for 10 seconds, and so on. (assuming I got my math right)

That's the actual surge rating.

I have a feeling I'm missing something about the 1N4007's 1A rating. I mean, if there is a short to the cathode for any amount of time that would cause the 1W to blow at 1A, shouldn't I have the same concern about the diode? I understand that it can tolerate a very large surge for , but what about a sustained surge?

Yes, you have exactly the same concern about the diode. It works the same way - it has an I²T rating which represents the rate at which it can get rid of pulsed heating. For diodes they generally express this differently, though, so it's harder to compare this apple to the resistor being an orange.

1N400x diodes are usually set up as power line frequency rectifiers. The worst thing that happens to them normally is when the power switch is turned on and the diode starts letting current into a completely unfilled filter cap. Assuming it starts at the zero-crossing of the AC wave, it conducts the incoming voltage into a completely flat cap, so the current is limited only by the imponderables of the ESR of the cap, the resistance of the wires, etc., so the diodes are usually rated in terms of the peak current reached by a half-sine wave of current for one half cycle of the AC power line. This amounts to the same thing as the resistor, but to determine the total I²T that goes into the diode, you have to integrate the square of a sine wave for a half-cycle over some time. If you do a square pulse of current as we did with the resistor, the math comes out the same as for the resistor. Diode makers just quote the half-sine surge rating as being more useful to power supply designers.

[Gaz]

Just wanted to say thanks to everyone, and big thanks to Enzo and RG for sticking with me in this big boring self-centered thread I have the 3W WW resistors and cathode fuses in place. Time will tell if they were worth the effort

[Gaz]

I wanted to give an update to what I've done with this amp as well as ask another question.

Before adding the beefier 3W wirewound resistors in place of the 1W metal films, I wanted to try out the cathode fuses with LED and limiting resistors. This is how I did it:

I was running the amp with the old tubes (which I though were maybe okay), then the sound got funny, then hmmmmmm. The amp still made sound, but I shut it down (I'm pretty sure it was a runaway tube).

It turned out it blew the new cathode fuse (Yea!), but it also blew the 1W resistor just like before! I couldn't see in what order they blew, but my fuse was a 750mA fast blow, so I figured that the fuse blew first.

I was really puzzled by this, but then realized that even with the fuse blown the 10k resistor and LED still provided a path to ground, and that the 1R resistor was still in series with the 10K and LED. My thinking is that when the fuse blew, the same fault current was still flowing through the 1R resistor and caused it to blow

The only way I could think to solve this problem would be removing the LED circuit, or removing the bias resistors. Is that correct, or would there be a way to keep both?

Thanks!

[Steve Conner]

Ohm's law dude. To exceed the power rating of a 1R, 1W resistor, you need more than 1 amp. To force 1 amp through a 10k resistor, you need 10,000 volts.

I've yet to see a tube guitar amp with a 10,000 volt supply so I have to assume the current through the 10k resistor was most probably not what blew the 1R one. You just need some heftier 1 ohm resistors or some protection diodes across them as we discussed.

There is an outside chance that the 10k resistor might have exceeded its voltage rating and arced over. If this was the case, there'll be evidence, a destroyed LED.

[Gaz]

Thanks for the reply, and I think I get it now.

Haha, just pretend you don't understand electronics for a second...

I was originally thinking that when the fuse blows, the fault current stills flows through the 1R and series 10K/LED, causing the 1R to take the heat and blow.

BUT! what you're saying is that when the fuse blows the series 10K resistor limits the fault current through both resistors, protecting the 1R, correct?

If that's correct, then I was initially (and stupidly) thinking that for some reason the fault current would just keep on flowin' even with the 10K resistor in place.