Random arc signal lost

snowgoer540 wrote: Honestly I'm on the fence about it. I do believe that oxidation exists, contact wetting prevents it, etc. I'm just not convinced that it's a big enough issue to warrant a solution. I also worry about pitting of the contacts from the sparks associated with switching a load. Again, maybe not a big issue.

Well you could do a Halscope plot and if you don't see an oxidisation issue then you needn't worry.

phillc54 wrote:

snowgoer540 wrote: Honestly I'm on the fence about it. I do believe that oxidation exists, contact wetting prevents it, etc. I'm just not convinced that it's a big enough issue to warrant a solution. I also worry about pitting of the contacts from the sparks associated with switching a load. Again, maybe not a big issue.

Well you could do a Halscope plot and if you don't see an oxidisation issue then you needn't worry.

At the moment she's happily chooching along consistently with 5ms of debounce (Cody would also agree).

I know I know, Halscope is on the to-do list

phillc54 wrote:

rodw wrote: The jury also noted that the Hypertherm relay used for ArcOK takes 10 ms to turn on and 4 ms to turn off so it is unlikely that debouncing will be required.

Could the jury explain how this works, it doesn't seem to gel with data from relay manufacturers.

Well, lets start with the 7i96 which has internal resistance to ground of 4.7k. Mesa says it draws about 5 mA. More precisely, it is 24/4700 = 0.005106 Amps (or 5.1 mA) just to confirm PCW's maths. (Hint, google Ohms Law).

So according to the jury, it does not matter that the spec says 5 volts, the minimum current draw remains at 100 mA regardless of the voltage. I specifically asked the same question Phill asked. Technically, you should calculate the resistance of the total circuit (eg the pull down and the internal resistance)
eg 1/Rt = 1/R1 + 1/R2 (where Rt = total resistance)
but the jury said there is a 10 times rule you can apply where if one resistance is 10x greater than the other, you can in this case ignore the impact of the internal resistance. So what current should we shoot for?

Well, the feeling was we want to be well past the minimum for certainty. So we shot for 200 mA. If you play with the maths it becomes apparent there is a tradeoff between the current through the relay and the power that needs to be dissipated. Plus the jury wanted to provide a cost effective solution (a mere $2.04 Aussie pesos. (< $1.50 dino Dollars) So 90 watts was easy and cheap to achieve so we end up with:
24/90 = 0.2666 A or 267 mA

So what about the power? well P = V * I or:
P = 24 *.267 = 6.408 Watts!

Phew! I just took the Jury's advice at face value! I'm glad that worked out when I did it!

So in answer to people's concerns about blowing up their 7i96 et al, I need to share a a circuit diagram the Jury made me draw up so they understood what I was trying to achieve. It clearly shows that the 7i96 input is unaffected by the pull down

Typing takes a long time while thinking is much faster, so while typing the brain has time to process other things, hence errors during typing.
But this is exactly the beauty of public forums, even when there are errors there is always someone to correct them.

snowgoer540 wrote: Honestly I'm on the fence about it. I do believe that oxidation exists, contact wetting prevents it, etc. I'm just not convinced that it's a big enough issue to warrant a solution. I also worry about pitting of the contacts from the sparks associated with switching a load. Again, maybe not a big issue.

Well ignore the Jury's recommendation at your peril. You are inviting Murphy's Law. There is no load being switched here. The relay can handle 10 amps and all we ask of it is is 267 mA. Trust me, when it happens it will be at the worst possible time. So don't be a tightass and spend your $1.50 dino dollars!

It really is foolish not to cover off the risk if you want to build a reliable machine.

rodw wrote:

phillc54 wrote:

rodw wrote: The jury also noted that the Hypertherm relay used for ArcOK takes 10 ms to turn on and 4 ms to turn off so it is unlikely that debouncing will be required.

Could the jury explain how this works, it doesn't seem to gel with data from relay manufacturers.

Well, lets start with the 7i96 which has internal resistance to ground of 4.7k. Mesa says it draws about 5 mA. More precisely, it is 24/4700 = 0.005106 Amps (or 5.1 mA) just to confirm PCW's maths. (Hint, google Ohms Law).

So according to the jury, it does not matter that the spec says 5 volts, the minimum current draw remains at 100 mA regardless of the voltage. I specifically asked the same question Phill asked. Technically, you should calculate the resistance of the total circuit (eg the pull down and the internal resistance)
eg 1/Rt = 1/R1 + 1/R2 (where Rt = total resistance)
but the jury said there is a 10 times rule you can apply where if one resistance is 10x greater than the other, you can in this case ignore the impact of the internal resistance. So what current should we shoot for?

Well, the feeling was we want to be well past the minimum for certainty. So we shot for 200 mA. If you play with the maths it becomes apparent there is a tradeoff between the current through the relay and the power that needs to be dissipated. Plus the jury wanted to provide a cost effective solution (a mere $2.04 Aussie pesos. (< $1.50 dino Dollars) So 90 watts was easy and cheap to achieve so we end up with:
24/90 = 0.2666 A or 267 mA

So what about the power? well P = V * I or:
P = 24 *.267 = 6.408 Watts!

Phew! I just took the Jury's advice at face value! I'm glad that worked out when I did it!

So in answer to people's concerns about blowing up their 7i96 et al, I need to share a a circuit diagram the Jury made me draw up so they understood what I was trying to achieve. It clearly shows that the 7i96 input is unaffected by the pull down

Sorry but this dissertation has no relevance to the question I asked.

So here is the circuit diagram. You can see that the 7i96 is unaffected. It still draws its 5 mA and the 3 x 270 ohm resistors take 267 mA


So this becomes my weekend project. To bring my existing circuit in line with the Jury's recommendations.

Just thought to point out where the extra current is going:



Now I am not saying everybody should do this. But just to show that the Mesa card input is not affected by this. As long as you do it right, of course.

phillc54 wrote: Sorry but this dissertation has no relevance to the question I asked.

Sorry, This is the data sheet I matched the part number to.
www.te.com/commerce/DocumentDelivery/DDE...e=DS&DocLang=English

Operate/Release time: 10 ms / 4 ms
So debounce is of no value if its < 10ms
The relays I use trigger in 250 usec so this is a very slow relay. Perhaps by design?

Aciera wrote: Just thought to point out where the extra current is going:

Now I am not saying everybody should do this. But just to show that the Mesa card input is not affected by this. As long as you do it right, of course.

But I am saying everybody should do this. Why would you ignore Ohms Law and basic electronic design? I learnt the hard way and had no idea how to start cutting again.

And absolutely correct. The Mesa card is unaffected.