I need to replace an old graphics card, but not sure what to replace it with.
I'm currently running on Win XP Pro, 1.23 Gig Ram, 80 Gig HD.
Can anyone help please.
There are a couple of major brands most here use. I have a ATI Radeon 9200 card and it works fine, and mine isn't the latest and greatest card they have out.
The first question you want to answer is how much dinero in your pocket?
I am looking to spend between £80 to £100.
This is just a suggestion (also happens to be what I use) it is I guess roughly 20 pounds over the 100, but would be well worth the extra money over a standard 6600.
You didn't mention what your current card is, I'm assuming you have an AGP slot on your motherboard..
There are many people here who know much more about this than I do so hang on and you will get some more suggestions.
Hi Picard have a look at this site
The price your willing to pay depends on what your mobo is. (AGP or PCI-E)
This store does some good deals from time to time.
Sorry , I currently have a nVidia T4200, which has given up the ghost, in an AGP slot.
Thanks for the info so far guys.
I have the last on this list and find it does an excellent job.
ATI Radeon 9800 Pro 128MB (AGP)
Im running with Nvidia 6800 ultra and im more than satisfy with its performance.
I think it depends on your CPU/mobo. I don't believe you want to spend the cash for a hi-end expensive card unless youhave a rig that is already fast, say a P4-2.5 Ghz +. Otherwise get a middle-of-the-roader that will get you thru til you get a new rig, and save as much as you can toward that end. I too use the 6800 Ultra and love it, but my rig is P4/3.4 Ghz. I know Radarman has always been happy with his card, altho I lean toward Nvidia. Maybe if people suggest cards, they should also inc what CPU/Mobo/Ram they are running with that card. I use the 6800 Ultra with a Asus P4C800E-Deluxe mobo/1 G Ram....self-built/XP Home. Good Luck, rob
the Radeon 9800 pro is a great card for the money
Thanks for all the info guys.
I will take a look at what has been suggested.