Melting Pot Pains

[Harrylee]

As I was casting some bullets the other day I was having a hard time trying to keep the temperature at my desired point. Now on the Lee pot I have just has a dial on the top not the most high tech method of controlling the temperature but I keep my thermometer in the lead and keep an eye on it, the problem is was you adjust the dial you can’t tell when the power is on. So that’s when I thought of putting a light in the circuit to tell me as I adjust the dial when the power goes on and off too help fine tune the pot. The problem I had was my pot is 220 volt and I couldn’t find a 220 volt light that easy. So off to Radio Shack the only lights they had where 120 volt so by add a 470k ohm resistor to one of the leads to the light dropped the voltage so that light will work. The one thing I’m not sure of and I’ll have to wait and see if the unit gets to hot for the light where it is mounted, if it’s to hot I’ll mount the light outside. The whole thing wasn’t bad the light $2.79 resistors $1.89. I thought that was a cheap fix if your pot is 120 volt no need for a resistor

[Bawanna]

Glad your on our side. You'd be hell on IED's I think. I never could grasp electricity myself, just totally foreign to me and I hate getting shocked.

In my remodel days I'd shut down the whole block rather than risk getting shocked.

Sparkies are crazy. I'd ask one if a certain line was hot, rather than grab a tester they'd just grab it and if they got jolted, yup it's hot.

Happened several times, it's like they enjoy it or something.

Nice effort on your part.

[Harrylee]

Gee thought I was going to get that was a Bright idea or something like that and after the numbness in you arm goes away you are good to go again. But really after I got the light hooked up I needed to try it and it helped a lot being able to see the temperature on the thermometer and if I wanted it to go up or down by seeing the light I knew if the power was on or off. Still get a bit of a temperature swing but not as bad, so all in all this did help. There are some other things I may try but that’s down the road

[CJB]

For those who may be inclinded......

Keep in mind that the 470 ohm value is not universal.

You need to consider the voltage AND amperage used by the light you'll be installing.

The formula would be V1-V2 divided by A equals R

Not the correct nomenclature for amps and volts, (I and E) but you can remember it.

That is, if the bulb you use is a 120volt bulb and you're running it on 240, then V1-V2 is 240-120 or 120.

If that bulb uses 20 milliams that would be diveded by .020 or be 6000 ohms. A resistor of that value is not easily had, so you could use the next biggest value, being 6200 or the more commonly found 6800 ohms, in our example.

You also need to consider the wattage of the resistor, with a little headroom. In this case the voltage of the bulb is 120v, and the amperage is .020. A simple 120 x .020 is 2.4watts The 120 is derived from the volts operational voltage of the bulb, not the volts dropped, but they're the same value in this example. A 2.4 watt resistor is not easily found, but you might find a 3watt, or more commonly a 5 watt resistor.

[Bawanna]

Ok, that's it. Get the Advil, I got a headache.

I just plug stuff in, if it don't blow up or catch on fire we're good to go.

[100percent]

You talked about a 470 ohm resistor but the package shows a 470,000 (470K) ohms. Which is right?

My Lee pot is usually good at about 7 on the dial.

[kwh]

For those who may be inclinded......

Keep in mind that the 470 ohm value is not universal.

You need to consider the voltage AND amperage used by the light you'll be installing.

The formula would be V1-V2 divided by A equals R

Not the correct nomenclature for amps and volts, (I and E) but you can remember it.

That is, if the bulb you use is a 120volt bulb and you're running it on 240, then V1-V2 is 240-120 or 120.

If that bulb uses 20 milliams that would be diveded by .020 or be 6000 ohms. A resistor of that value is not easily had, so you could use the next biggest value, being 6200 or the more commonly found 6800 ohms, in our example.

You also need to consider the wattage of the resistor, with a little headroom. In this case the voltage of the bulb is 120v, and the amperage is .020. A simple 120 x .020 is 2.4watts The 120 is derived from the volts operational voltage of the bulb, not the volts dropped, but they're the same value in this example. A 2.4 watt resistor is not easily found, but you might find a 3watt, or more commonly a 5 watt resistor.

What??? Where is Jocko when we need him to clear things up.

[100percent]

BTW a 470K might work but I bet that the neons that you got already have a resistor built in the fixture. If it doesn't light or is too dim try it with less resistance. Maybe none would be ok. Neons are pretty tolerant of this. It isn't like a LED.

Neons will drop about 70 volts the resistor takes up the rest of the voltage.

[Harrylee]

OK my typo should have been 470k for the resistor the neon light does have a built in resistor. On 120 volt it is brighter than when it’s on 240 volt with resistor. Checked the light and is pulling 2 mA so that would be .002x120=.25 watt by the formula that I see here. I have a ½ watt resistor in the circuit that should be good. I think I must have dropped the k on the floor and forgot to pick it up and put it on the 470. Such as life, but you did see the package and it did say 470k. Bawanna now I think I have a headache
I did correct the original thread

[downtownv]

For those who may be inclinded......

Keep in mind that the 470 ohm value is not universal.

You need to consider the voltage AND amperage used by the light you'll be installing.

The formula would be V1-V2 divided by A equals R

Not the correct nomenclature for amps and volts, (I and E) but you can remember it.

That is, if the bulb you use is a 120volt bulb and you're running it on 240, then V1-V2 is 240-120 or 120.

If that bulb uses 20 milliams that would be diveded by .020 or be 6000 ohms. A resistor of that value is not easily had, so you could use the next biggest value, being 6200 or the more commonly found 6800 ohms, in our example.

You also need to consider the wattage of the resistor, with a little headroom. In this case the voltage of the bulb is 120v, and the amperage is .020. A simple 120 x .020 is 2.4watts The 120 is derived from the volts operational voltage of the bulb, not the volts dropped, but they're the same value in this example. A 2.4 watt resistor is not easily found, but you might find a 3watt, or more commonly a 5 watt resistor.

Gotcha!

[b4uqzme]

Interesting thread Harrylee. Beats the heck out of talking about gay marriage. THANKS!

[garyb]

Good idea Harry. Has the light improved your process?

[muggsy]

I didn't have to be an electrical engineer when I melted lead on mom's gas stove. Oh for the good old days.

[CJB]

That last post explains a lot.