Choosing an L-Series™ UV LED Flood Lamp by Curing Area — A Model-by-Model Walkthrough
Picking a UV LED flood lamp for a large part usually starts with one question — which model's curing field actually covers the footprint — and the answer changes the intensity, the chamber, and the exposure time that follow. Q: How do you choose an L-Series™ model for a large curing area? A: Start from the largest footprint that must cure in one exposure, select the smallest L-Series™ curing field that fully contains it, then confirm that model's intensity delivers the material's required dose within the available cycle time. If it doesn't, either step to a smaller field and index the part, or move to the water-cooled W-Series™ at the same area. The Nine Curing Fields, in Square Inches Converting each model's field to area makes the comparison easier, because parts are rarely square. All intensities below are Incure's published figures at a 2-inch working distance. Model Field Area (in²) Intensity (mW/cm²) L11 1″×1″ 1 4,300 L22 2″×2″ 4 4,300 L33 3″×3″ 9 3,100 L44 4″×4″ 16 3,100 L64 6″×4″ 24 1,900 L88 8″×8″ 64 1,900 L1010 10″×10″ 100 1,900 L1212 12″×12″ 144 1,900 L1414 14″×14″ 196 950 Two step changes stand out. Moving from L44 to L64 adds 8 in² of coverage and drops intensity from 3,100 to 1,900 mW/cm². Moving from L1212 to L1414 adds 52 in² and halves intensity from 1,900 to 950 mW/cm². Between L64 and L1212, coverage grows six-fold with no change in quoted intensity, which makes that band the efficient zone for mid-to-large parts. Fit the Shape, Not Just the Area Area alone can mislead. A 5″×2″ strip has 10 in² of area, which looks like an L33 job on paper, but it will not fit inside a 3″×3″ field. The deciding dimension is the longest side of the part. L64 is the only rectangular field in the line, which suits elongated parts such as long connector housings or narrow display strips that would waste most of a square field. Converting Intensity Into Exposure Time Dose equals irradiance multiplied by time, so the exposure needed scales inversely with the model's intensity. Consider a hypothetical material that requires 3,800 mJ/cm²: Under an L1212 at 1,900 mW/cm², the exposure is 3,800 ÷ 1,900 = 2 seconds. Under an L1414 at 950 mW/cm², the same dose takes 3,800 ÷ 950 = 4 seconds. Under an L44 at 3,100 mW/cm², it takes roughly 1.2 seconds, but only a 4″×4″ area is covered. The figure of 3,800 mJ/cm² is illustrative only; use the dose from your own material data sheet and verify it with a radiometer at the production working distance. Email Us with your part outline and required dose, and Incure's engineers can run the same calculation against every L-Series™ model before you commit. When to Index Instead of Covering in One Shot Covering the full part in a single exposure is not always the right call. If a part measures 13″×13″, an L1414 covers it in one shot at 950 mW/cm². An L1212 at 1,900 mW/cm² would…