Homebrew PCBs

>Power output divided by area = 60 / 923 = 0.065005
>(using Thinktink's numbers as an example) Dry film requires 42.5 mJ/cm^2
>42.5 / 0.065005 = 653.792
>Double that = 1307.58 seconds exposure

The problem is one of units.  The first number you calculated is the lamp intensity in W/cm^2=J/(s*cm^2).  Your dry film dose is specified in mJ/cm^2= 0.001 J/cm^2.  You need to convert this to J/cm^2 to keep your units consistent.  42.5 mJ/cm^2 = 0.0425 J/cm^2.

 

Thus:

 

time = 0.0425 J/cm^2 / 0.065 J/(s*cm^2) = 0.654 s

 

doubling this gives 1.31 seconds.

 

-Marc



 


To: Homebrew_PCBs@yahoogroups.com
From: ygroups@...
Date: Tue, 2 Jun 2009 16:37:38 +0100
Subject: Re: [Homebrew_PCBs] Setting exposure with a step gauge







Zoran A. Scepanovic wrote:
> Check <http://www.thinktink.com/stack/volumes/volvi/filmimag.htm> there
> you have the explanation on using step gauge

The math for step 4 seems utterly wrong though:

# Calculate an initial exposure time by dividing the power output of your UV 
lamp (in Watts) by the total area being illuminated (in cm2). Divide the 
result into the total energy requirements of the dry-film (in miliJoules/cm2). 
Multiply the result by 2. The final number that you come up with will be a 
pretty good first estimate of the needed exposure time in seconds.

Every time I try this, my resulting exposure time is either way lower than 
seems sensible (less than a second) or far too high (1300 seconds, or over 20 
minutes)...

The UV box I'm calculating for is a Mega AZ210 -- four 15W lamps in each side 
(eight in total, 60W per PCB side, or 120W all told), and an exposure area of 
35.5x26cm (923 cm^2).

This is what I've got thus far:
Number of lamps = 4
Watts per lamp = 15
Total lamp power = (NumLamps * WattsPerLamp) = 60W

Exposure area width = 35.5cm
Exposure area height = 26.0cm
Exposure area total = (ExpAreaW * ExpAreaH) = 923 cm^2

Power output divided by area = 60 / 923 = 0.065005
(using Thinktink's numbers as an example) Dry film requires 42.5 mJ/cm^2
42.5 / 0.065005 = 653.792
Double that = 1307.58 seconds exposure

Or if I interpret it the other way:
Power output divided by area = 60 / 923 = 0.065005
0.065005 / 42.5 = 0.00153 (?!)
Double that = 0.003059 seconds

I must be doing something wrong, but I can't see what...

-- 
Phil.
ygroups@...
http://www.philpem.me.uk/








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