Best way to get a 100:1 arc voltage out of Powermax 85

[SteveJustSteve]

I'm surprised I can't find anywhere this has been mentioned before here, but I have searched up and down and can't find it, so...

What's the simplest and/or best way to get a 100:1 signal from a Powermax 85 divider board?

I realize a 100:1 signal is not the best to use but I'd still like to know.

Thanks

[djreiswig]

The internal divider doesn't provide 100:1 ratio. You'll have to build your own divider to get that ratio.
Google voltage divider. You just need some resistors in parallel.

[SteveJustSteve]

The internal divider doesn't provide 100:1 ratio. You'll have to build your own divider to get that ratio.
Google voltage divider. You just need some resistors in parallel.

Thanks for taking the time to answer. Right, I was wondering if there was a combination of DIP switches on the Hypertherm that could possibly make 100:1 by combining somehow. I've tried a simple voltage divider with two 1K resistors to cut the 50:1 signal in half and it didn't yield me the results I expected but if that's the best option I'll try it again. Thanks

[adbuch]

The internal divider doesn't provide 100:1 ratio. You'll have to build your own divider to get that ratio.
Google voltage divider. You just need some resistors in parallel.

Thanks for taking the time to answer. Right, I was wondering if there was a combination of DIP switches on the Hypertherm that could possibly make 100:1 by combining somehow. I've tried a simple voltage divider with two 1K resistors to cut the 50:1 signal in half and it didn't yield me the results I expected but if that's the best option I'll try it again. Thanks

How did you have your two 1K resistors connected? Perhaps share your wiring diagram.
David

[tcaudle]

You probably ned to use much higher value resistors. With 1K you are loading the pee-waddly out of the signal. That 50:1 is already high impedence (meaning you can't load it down with an external divider without skewing the voltage way off. I don't remember the values they use for the 50:1 ratio but when you place two resistors in series (1k each) you put a 2K load on the 50:1 output and it drags it down. Its not going to be a real simple divider . I would start with at least two 10K resistors or even two 49.9K 1% precision so you 100K of load. Then you need to worry about what you are feeding.
Your external divider is in parallel with the bottom resistor of the internal divider so it drops the bottom resistance of that by a large factor. Lets say the internal bottom resistor is 2K. At a 50:1 ration the upper value is like close to 100K. Its a ratio. Change to bottom value and it changes the ratio by a bigger delta than just the difference in value. Then hope the load you put on it is high impedance since it becomes part of the equation too
Also keep in mind the SNR (signal to noise ratio) wiil be poor) it says each..010 volt (10mv) of change (or noise) translates to 1Volt of reading change. Full scale for most plasma would be 2.5V.

Look up running resistors in parallel and its always lower than the lowest of the two .

.

[adbuch]

You probably ned to use much higher value resistors. With 1K you are loading the pee-waddly out of the signal. That 50:1 is already high impedence (meaning you can't load it down with an external divider without skewing the voltage way off. I don't remember the values they use for the 50:1 ratio but when you place two resistors in series (1k each) you put a 2K load on the 50:1 output and it drags it down. Its not going to be a real simple divider . I would start with at least two 10K resistors or even two 49.9K 1% precision so you 100K of load. Then you need to worry about what you are feeding.
Your external divider is in parallel with the bottom resistor of the internal divider so it drops the bottom resistance of that by a large factor. Lets say the internal bottom resistor is 2K. At a 50:1 ration the upper value is like close to 100K. Its a ratio. Change to bottom value and it changes the ratio by a bigger delta than just the difference in value. Then hope the load you put on it is high impedance since it becomes part of the equation too
Also keep in mind the SNR (signal to noise ratio) wiil be poor) it says each..010 volt (10mv) of change (or noise) translates to 1Volt of reading change. Full scale for most plasma would be 2.5V.

Look up running resistors in parallel and its always lower than the lowest of the two .

.

This is great info for fellow EEs such as myself. Nice explanation. But for most folks most likely way too technical.
David