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The article mentions 13.5nm. Is 19nm the wavelength used to blast the tin that re-emits as the the 13.5nm EUV source? Is that where the power is going to result
by Already__Taken 9y ago
The article mentions 13.5nm. Is 19nm the wavelength used to blast the tin that re-emits as the the 13.5nm EUV source? Is that where the power is going to result in a few watts at the printing surface?
- no_flags 9y agoFirst question, no, they use CO2 lasers which emit in the um range, not nm. Probably a typo. Second question, the loss of power is due to two factors: 1) Conversion efficiency of CO2 laser pulse to EUV light pulse. This is in the single digit range as a percentage. 2) Transmission loss of EUV from light source to wafer. EUV mirrors reflect in the ballpark of 50% of the light, and there are many of them inside the scanner, so you have .5 * .5 * .5...