I saw an discussion at:
Claiming an rebuilt scanner with 40 UV leds will properly expose an pcb with the speed of 2cm/minute.
Known: 2 cm/minute 40 UV LED in line.
Assumed: Scan area = A4 width = 210 x 297 mm P_led = 3.8V*.04A (mentioned in the text) t = 297/20*60 lambda = 400 nm
Math: Q = 3.8 * .04 * t * 40 = 5417.28 Joule A = .210*.297 = .06237 m²
PCB needs 86857 Joule/m² to be properly exposed. Ignoring inefficiency of the LED, distance to pcb and it's beam angle.
If a board with each LED responsible for an 30 mm² area on a 100x160 mm board is built it will need to shine for: (.100*.160*86857)/(3.8*.040*60*((100*160)/(30**2))) = 8.57 minutes
Correct?
Especially how much energy is actually needed to properly expose an pcb board photoresist..? I'm interested in knowing practical numbers for this type of application. And if the boards gets sharp? Would pulsed UV led be more efficient?
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