AC voltage, Multimeters, ocilloscopes, etc.

Jim Harrington posted:

stan2004 posted:

Jim Harrington posted:

...

Still doesn't add up, but it's closer...

√(8.5²+7²)≈11v

Having received the same readings with my old Radio Shack meter, I will be returning the Fluke; unless someone can convince me otherwise...

I still want to know where your "12V and 14V" RMS readings come from.  After all, you are measuring 16V (not 14V) 

12v and 14v taken from an older post...

https://ogrforum.com/t...-unit-output-voltage

 

I see I even participated in that dialog.  But being over 3 years old I have no re-collection of it.   

Anyway, in reviewing that thread my take-away is this is a topic with too many pitfalls, trap-doors, and slippery-slopes.

The smoke unit has two and only two power levels - half-power and full-power (boost).  End-of-story (in my opinion).

gunrunnerjohn posted:

7.2 watts is about right for really good smoke from the 20 ohm resistor.  Remember, it's a 3W resistor, so in free air it'll handle 3 watts without getting nearly hot enough to vaporize the smoke fluid. 

Based off a suggestion from another post, I ordered a 5W 20 Ohm Resistor.  What the result /difference in using a 3w vs. 5w resistor?

Well, if you left it in it's original form, it wouldn't work, not to mention it wouldn't fit.  However, once you crush the ceramic and expose the resistor, it's the same size and shape of the 3W resistors I use most often.

stan2004 posted:

The smoke unit has two and only two power levels - half-power and full-power (boost).  End-of-story (in my opinion).

That's where I ended up basically.  I tune the smoke unit for good smoke and sustainable current draw and leave it at that.  That turned out to be 20 ohms minimum in general for standard TMCC applications.  It's nice to know generally how you got there, but I think we get wrapped around our axles in the minutia at times.

gunrunnerjohn posted:

Well, if you left it in it's original form, it wouldn't work, not to mention it wouldn't fit.  However, once you crush the ceramic and expose the resistor, it's the same size and shape of the 3W resistors I use most often.

 

Well of course I removed the ceramic...

I bought a few of them, and few of the 22 ohm variety, in case there was too much smoke!

Why the 3w vs. 5w?  I'm assuming it not analogous to a light bulb, where one with a higher wattage will draw more current; the resistor with a higher watt rating will "resist" more current before failure.

Looking at the product data sheet, it appears that he 5w will dissipate about 30% more heat than the 3w at it's rated load.

I just want to make sure i'm not set up to cook something else (triac?).  On that note, can anyone provide a part number for the R2LC triac?  Just curios to see what it's output rated at.

Remember, I'm here to gather enough information to be dangerous...

Thanks

The wattage tells you how much the power the resistor is rated to dissipate.  Since we actually want the resistor to get hotter than it's normal operating temperature, we want one that's rated somewhat less than the power we're putting into the smoke circuit.  Typical smoke circuits put from around 5W to 8W into the resistor for normal operation.

That same ceramic resistor you crushed to create a smoke resistor would be much too cold to vaporize the smoke fluid in it's original state, even if it would fit into the smoke unit.  Removing the ceramic body also reduced it's heat dissipating qualities, thus making it a lower rated resistor and thus it gets hotter in the smoke circuit.

Thanks.

So a 3w would get hotter than a 5w, correct?

Yep, given the same applied voltage.