19 Jul '16 17:56>
I am designing a vacuum chuck, our substrates are about 210 mm X 120 mm alumina. We have a spinner/coater (photoresist) that we spead the liquid on the substrate and it spins at various rates, up to about 3000 RPM's. We use a poor design now and my chuck will use an o ring instead of the flat design we now use which has a number of problems.
My question is: Call it 8 inches by 4 inches for grins, and with full vacuum on the backside you would get around 350 pounds of force on the substrate.
The thing is, with the force we see now for the inferior chuck arrangment, we can pull the substrate off the chuck with fingers.
Now if it was a perfect seal and a perfect vacuum, say 1 E-5 torr or some such, wouldn't it take 300 pounds to pull off the substrate?
If so, and we pull upwards with say 5 pounds of force, would that equate to the inches of mercury used in the vacuum part of the system?
That is to say, the actual inches of mercury say with perfect vacuum causing a downwards clamping force of say 300 pounds but the vacuum was instead of 28 inches of mercury, was instead 2.8 inches of mercury, wouldn't that equate to a force needed to pull the substrate free, now being 30 pounds?
But we are clearly not anywhere near 30 pounds of force to free the substrate while under vacuum.
So if 3 pounds of upwards force is needed to pull the substrate loose while under vacuum, does that equate to around 0.28 inches of mercury? or 1% vacuum, or 7 odd torr? I don't have a gauge to measure the vacuum right now, could do that but have to modify present plumbing.
So if we have a vacuum pump capable of say 20 inches of mercury vacuum, wouldn't that require at the size of our substrates to need around 2/3 of perfect vacuum force or say around 200 pounds upwards force to pull off the substrate? Which would surely break the fragile alumina, its only 1 mm thick.
Sorry for the use of inches, just relating that to STP units, PSI and so forth.
My question is: Call it 8 inches by 4 inches for grins, and with full vacuum on the backside you would get around 350 pounds of force on the substrate.
The thing is, with the force we see now for the inferior chuck arrangment, we can pull the substrate off the chuck with fingers.
Now if it was a perfect seal and a perfect vacuum, say 1 E-5 torr or some such, wouldn't it take 300 pounds to pull off the substrate?
If so, and we pull upwards with say 5 pounds of force, would that equate to the inches of mercury used in the vacuum part of the system?
That is to say, the actual inches of mercury say with perfect vacuum causing a downwards clamping force of say 300 pounds but the vacuum was instead of 28 inches of mercury, was instead 2.8 inches of mercury, wouldn't that equate to a force needed to pull the substrate free, now being 30 pounds?
But we are clearly not anywhere near 30 pounds of force to free the substrate while under vacuum.
So if 3 pounds of upwards force is needed to pull the substrate loose while under vacuum, does that equate to around 0.28 inches of mercury? or 1% vacuum, or 7 odd torr? I don't have a gauge to measure the vacuum right now, could do that but have to modify present plumbing.
So if we have a vacuum pump capable of say 20 inches of mercury vacuum, wouldn't that require at the size of our substrates to need around 2/3 of perfect vacuum force or say around 200 pounds upwards force to pull off the substrate? Which would surely break the fragile alumina, its only 1 mm thick.
Sorry for the use of inches, just relating that to STP units, PSI and so forth.