Inkjet Throughput Computation

January 14, 2016

Steve DiBartolomeo
Applications Manager

There is a large difference between using an inkjet for R&D work and using one in production. Production engineers like to think in terms of panels per hour. So a machine that is quite suitable for the lab might not fit into a production line.

Let's do some basic computations and see what kind of times we can achieve.

Here are the parameters of our ink jet equipment.

    panel size - 460 x 610 mm

    scan speed - 6000 mm/minute

    width of inkjet head - 36.1 mm

    number of nozzles - 512

    nozzle pitch - 70.5 um

    printing DPI - 1441


Each head scan (along the 460 mm width) should take (in seconds)

[[460 mm] / [6000 mm/min]] x 60 sec/min = 4.6 seconds

How many scans needed to cover the panel?

610mm/36.1 mm = 16.89  = 17 scans

However our nozzles are separated by 70.5 um and our actual DPI is 1441 which is 17.625. So in order to get the resolution we need to image our lines we need to make 4 passes of the head; each pass is shifted by 17.625 um

. throughput2.gif

So we actually will need 17 x 4 scans which is 68. At 4.6 seconds per pass you can see that this will take 312 seconds or slightly more than 5 minutes. In practice we get considerably less performance because most printing is done uni-directionally and this requires a measurable amount of time to scan back across the panel. A bidirectional printing scheme needs very little time to make the shift.

To reduce the time to paint a panel, most inkjet systems use multiple heads. While this adds complexity due to alignment and timing control, it is the only practical way to get a board covered in one minute or less.