Product

• Brisque^® 4.x
• Brisque 5
• Lotem^® 800
• Lotem 400

Platform

all Brisque platforms

Symptoms

When an expose job using certain screen set parameters is sent to a Lotem 400 or Lotem 800, the number of beams and the throughput are reduced. For example, this occurs when the selected screen set is VC175R10C.

Cause

To prevent plate bending, some separations of the expose job are performed with a reduced number of beams. This depends on the screen angle and the screen frequency.
For example, the cyan and black separations are exposed by 48 beams, but yellow is exposed by 42 beams and magenta is exposed by 40 beams. This information appears in the Expose Info window in the Lotem controller application, and no error appears in the Lotem controller.

Solution

This variation in number of beams is normal behavior. However, it is helpful to understand when it occurs.

The optimization process that is used during an expose is aimed to increase the tolerance of the plate imaging and developing process to variations in the device setup, imaging and chemistry. The overall system tolerance is significantly increased while the overall job preparation time is not affected.

However, the optimization process affects the number of beams for specific plate screening parameters.
The resulting beam reduction can be calculated as follows:
Suppose that the number of beams is N.
If (resolution/frequency)*COS (angle value) = integer or integer and a half, then the number of beams changes:

1. If (resolution/frequency)*COS (angle value) is n, then we look for the largest integer K where K*n <=N. If K*n >= 0.8*N, then the number of beams is reduced to K*n; otherwise the number of beams does not change.
   For example, if N = 24 and (resolution/frequency)*COS (angle) is 10, then K = 2, K*n = 20 and the number of beams is reduced to 20.
   If N = 24 and (resolution/frequency)*COS (angle value) is 9, then K is 2, K*n = 18 and the number of beams does not change, since 18/24 = 0.75 and the throughput reduction would be too large.
2. Similarly, if (resolution/frequency)*COS (angle value) is integer and a half (that is, n+.5), we look for the largest integer K where K*(2n+1) <=N.
   For example, if N = 24 and (resolution/frequency)*COS (angle value) is 10.5, then K = 1, K*(2n+1) = 21 and the number of beams is reduced to 21.

Technical contact(s)

Dov Vainer

Submitted by

Frederic Mayot

Date submitted

02 Feb 2004