Random arc signal lost

snowgoer540 wrote:

I’d think the same 100mA would apply, the reading I did suggested they pick a standard amperage value. So for 24V I’d need a 240ohm resistor... somewhere... to make the same 100mA draw. I do know my board is rated at 800mA max (7i76). I planned to measure the current amperage draw tomorrow out of curiosity.

I’m sure oxide does form, just not sure how much of an issue it is. If Jim Colt has advised these same relays have been used for years and years, again, you’d think the manual would bring it up so they didn’t see as many field issues.

Ah yes, but in the same post Tom Caudle from CandCNC said that Arc OK circuits usually have some resistance in them (but never said how much). I think the 4.7k resistance is better than the 7i76e but in my calcs (if I'm right), the circuit has about 1 amp and it needs 1.2 amps. I think the minimum current will vary with voltage (higher voltage has more power)

I think that my 1K resistor would put about 4.5 amps through the relay without affecting the Mesa card. That is well within the 10 amp relay spec.

So its possible that the manufacturing tolerances may mean that most machines will work OK but maybe you won the jackpot and were closer to the quoted spec than many others!

7i76e on page 18

The 7I76E field inputs have a nominal input resistance of 20K Ohms to field power ground.

Ok, so 60 mA comes from the power formula P = I * V
I = P/V
I = 0.3/5
I = .06 A (60 mA)

So lets design a circuit that has about 200 mA of current after I have dinner!
We don't want to go over 5 watts of power as it will be harder to find an off the shelf resistor
We know it will have 2 resistors in parallel, the 4.7k via the 7i96 and the unknown pull down resistor where V = 24.

See what you come up with while I am off the air.

I think you can forget the 4.7k so 24/0.2 = 120ohms that should give about 200mA
But then you would need about at least a 5w resistor

Clive wrote:

I think you can forget the 4.7k so 24/0.2 = 120ohms that should give about 200mA
But then you would need about at least a 5w resistor

Yes, I agree, including the internal resistance which reduces the circuit resistance to about 117 ohms, I make it will generate 4.92 Watts. the resistor specs are typically +- 5% so I think you need to allow at least 20% headroom. RS components list some 6.5 W resistors but its probably cheaper to go for 10 W like this one.
au.rs-online.com/web/p/through-hole-fixed-resistors/1991675/

So there you go, after 6 pages going down the debouncing rabbit hole after the issue was correctly identified, we have a clear solution backed by the laws of physics.

This should be adopted on all builds. It should still work for a 7i76e as the 20 k resistance on that card won't stress a 10 W resistor.

I would use a wire wound R.

Clive wrote:

I would use a wire wound R.

Well, you'd drop back to a 7 W resistor but that is enough headroom.
au.rs-online.com/web/p/through-hole-fixed-resistors/2060953/
or
au.rs-online.com/web/p/through-hole-fixed-resistors/4852811/

Here, we can't just go and walk into a local store to buy stuff like that so if I wanted to fix something in a hurry, I'd probably get 2 of these
www.jaycar.com.au/68-ohm-5-watt-wire-wound-resistor/p/RR3270

But mount these resistors proud of the circuit board for heat dissipation. They will get hot!

Actually by this this definition , the first option I suggested is probably better.