PSE9 & CS5 bit depth question 8bit / 16bit functions?

I may need to consider buying CS5...(Currently Photoshop Elements 9)
Sorry this question is probably obvious and sure most of you knew what was coming by the title alone but can't seem to find a simple answer in any of the related threads.

I've just been used to processing images from scanned transparencies which are 8bit scans. However having made the move to digital the files are now 16bit but some functions don't work in 16bit until converted to 8bit (cloning stamp for instance), is this simply the built in limited functionality of PSE9 which then work fully in CS5 as 16bit :

Thanks in advance...

Mike

Indeed it is, Mike. For earlier versions there were supposedly 'work arounds' available that would at least let one create 16-bit layers in Elements, but I could never get them to work. My policy now is to do most of the critical editing (WB, exposure, cloning/spot removal, contrast, hue and saturation) on a virtual copy of the .DNG (Adobe RGB) in LR and, where necessary, switch to 8-bit sRGB in PSE for anything else (usually just selective sharpening using layers and the high-pass filter and then resizing).

Jon

Jon

Thanks for the feedback...so put simply if I want these functions to work in 16bit I'm better upgrading to full Photoshop CS5.....yes??

Mike

Yes, that is correct.

Many thanks to you both for confirming this, much appreciated!

Mike

As a rank amateur when it comes to PP photos is there a noticeable difference in the finished quality of the photo between 8 and 16 bit format.

I use Lightroom to do all the initial adjustments to the photo and then export it as an 8bit Tiff for editing on Elements 9.

Greg

You won't notice any difference unless you are doing something very specific and then you'll probably be doing some job of work.

Everything depends on how much manipulation is being done and the final output size and purpose, but for normal use there's no problem with 8 bit.

I'll bet most users haven't even really noted that 16 bit is there.

Thanks for clarifying John.

Greg