Homebrew PCBs

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 !
>  
>


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