RMustakos wrote:
> 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 ;)
>
Well, to be pedantic we should be strictly following scientific
notation, with SI units and use standard errors were applicable :)
Your 4 seconds sounds completely reasonable. I must of had too much
coffee in the system. I agree its impossible to get accurate
calculations using such simple assumptions, but I guess we only need
ball park numbers to know if what Christian is trying to achieve is a
waste of time.
So a typical 10 x 6 cm single sided PCB with 50% track usage will take
on the order of a few minutes using a 25mW 355nm light source.
At 10 watts input power, thats 0.25% efficiency ! Now I see why mercury
discharge lamps are the choice for photopolymerization.
Adam