Christian & Adam,
Where I'm having an issue understanding is where the time comes from.
It looks like the 100mJ/cm^2 is really 100 mW∗Seconds/cm^2 as 1 Joule =
1 Watt Second.
To get 100mJ/cm^2, you need 100 mWSeconds/cm^2, or (25 mW ∗ 4 Seconds)/cm^2.
With 25mW from the LED, concentrate it onto 1 cm^2, & it takes 4 seconds
to expose.
Since the area of exposure is a 0.9mm diameter (?) circle, the area is:
pi ∗ r^2 = 3.14159∗(0.45mm)^2 = 0.64 mm^2 = 0.0064 cm^2.
The power required to expose this area = 0.0064 cm^2 ∗ 100 mW Seconds/cm^2
= 0.64 mW∗Seconds
The time required = 0.64 mW ∗ seconds / 25 mW = 0.0256 seconds.
Since the area is 0.64 mm^2, you can "pretend" that is the spot is a
square 0.8mm on each side. What that buys you is the ability to
calculate a nominal velocity
for the spot to expose the board. if the spot is moving at a constant
speed, to expose a point,
it must be within the spot for 0.0256 seconds,
V = 0.8mm / 0.0256 seconds = 31.25 mm/second
Now comes the magic:
I assumed different things in different parts of this equation for a reason.
I assumed the spot was circular to determine the area. This is probably
right.
I assumed the spot was a rectangle to calculate the speed. This is
patently wrong.
But the 0.8 mm sides provides us with a better working velocity than
using the 0.9mm diameter.
This is because using the 0.9mm diameter means only the exact center of
the beam provides a
full 0.0256 second exposure to the board. Further, the energy
distribution of the LED is
probably gaussian (I don't know that, but when you don't know, gaussian
is a good guess).
That just means the center of the beam is 'brighter' then the edges, and
to expose the edges,
you have to expose them for longer.
By varying the velocity of the 'print head', you vary the exposed trace
width. I am sure that
if I were smarter (or more stubborn) I could tell you what velocity
would give you what trace
width, but I expect there are too many variables to know it well: what
is the actual energy curve,
what is the thickness of the photo resist, how 'active' is this
particular batch, etc. My gut reaction
is to try a series of traces in all 8 cardinal directions. Vary the
velocity from about 1.5 cm/sec
to 3.5 cm per second, develop it and check the trace width in each
direction. While I like the
idea, I am a little worried that you will be generating inconsistent
widths, but as long as it works,
then it's 'wicked good'.
Good luck, Christian
Richard
PS, sorry, I can't help it, but you two together have a seriously
religious naming convention ;)
> From: Adam Seychell <a_seychell@...>
>Subject: Re: UV laser exposure
>
><snip>
>
>
>Where did you get 4 seconds from ? Shouldn't the time to expose 1cm2 be
>equal to: 100mJ/cm2 / 25mW ∗ (1cm / (0.09cm ∗ 0.09cm)) = 493 secons =
>8 minutes. A typical 100 x 60mm single sided PCB with 50% track usage
>will take at least 4.1 hours !
>
>
[Non-text portions of this message have been removed]