I sent a reply earlier today, but I had to answer the phone and the webpage timed out. So, when I hit the post button it just didn't save. Anyway, here goes again:

Decided to go with an older model so I could have enough money left to get a decent flash and possibly a 75-300mm lens this month. After searching around, I settled on the Minolta Dimage Scan Speed. From the reviews and everything I've seen, really does good quality scans. Good specs too:

2820 dpi (makes a 4032x2688 image from 35mm)
12bits/pixel (means you can input into PS at 16bits/channel for some sweet detail and tonal gradation)
3.6 DMax (slide film has a theoretical DMax of 3.6-3.8 I think)
All together resolution is equivalent to ~10.84 megapixels.

Picked it up factory refurbished off of E-bay. I don't want to "recommend" the guy I bought it from since I haven't gotten mine yet, but if the ad is to be believed, it is "new" refurbed from Minolta in the box with a 6 month warranty. These guys also have over 15000 feedback with under 200 negative (like a 75:1 ratio). So, not a recommendation but since they are going for a starting bid of $109 (buy it now at $119) with $22 shipping it might be worth the risk. When I get mine, I'll post to condition and whatnot (have tracking number on it now - scheduled to get here Tuesday).

That reminds me of a question that's been bugging me since I started researching film scanners. Take the hot new Canon EOS 1DS, the 11.1megapixel pro DSLR. It has an image array of 11.4megapixels and produces a max image that's 4064x2704 (just slightly larger than the image from the Scan Speed); which really comes out to 10.99megapixels. Anyway, what I keep wondering is how does a camera with an image array of 11.4mp create an image that's 11mp when it takes 3 of the array elements (1 each for R, G and B) to create 1 pixel? Is the reported array size of 11.4mp already taking this into account and the actual array has 34.2million elements?

Most film scanners don't have this problem (they don't split 1 pixel across 3 elements).

-=>Donald