We are using Karl Suss MA6/BA6 and the lamp power shows 350 W and we do exposure for 10 sec. i am confused how to find power per unit area? which area to consider - mask or source area??? kindly help
Well divide the power illuminating a certain area by the area and you have the power per area. However, you should know if your lamp distributes the 350 W (not mW?) over a complete sphere or a certain cone.
Lamp power may be 350 W. As you power up your machine, you do a lamp test and see the intensity shown in the display nearby the lamp power display. It will be some mW/cm^2. You can believe this value if your machine intensity is calibrated on a regular basis. This is the source intensity. If you spin coat a photoresist for a particular thickness, find out the dosage required for the same thickness and divide the dosage with the intensity of lamp. You will get the required time of exposure.
Jose
As Jose rightly points out, you need to measure the UV intensity (MW/cm^2) so that you can calculate the required exposure time.
We currently use a 'Karl Suss UV Intensity Meter-Model 1000' with the 'P365nm probe'. Often the UV lamp in the mask aligner will require a warm up period (ours takes around 30 minutes to stabilise).
When taking measurements you'll probably find there is a slight difference in intensity between the centre and the perimeter of the chuck and this is worth noting in case you need to use the perimeter.
Use Laser power meter to check the intensity at the exposure area (at the mask). If you don't have one, borrow one then calibrate with your photo detector then use your detector to monitor any changes. Regularly check your detector for any degradation or failure. I hope this will help.
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Materials Science
- Materials Chemistry
- Polymer Chemistry