is amd radeon r9 270 the same graphic card with asus r9 270?

[gamer4life253]

it maybe a silly question but i have this question for some time and i was hoping that someone could enlight me.

 

[Math Geek]

yes they are the same more or less. a graphics card is like a mini computer. it has it's own processor, ram, controller chips and cooling. a "reference" card will be what amd suggests as a model for the card. other brands may make reference cards to match what amd released. they also have the option of making custom cards using the graphics chip from amd/nvidia. this will have custom cooling, power delivery, ram, outputs and other possible changes to the card. it will still use the same processor but can change many other things to make it different from other cards on the market.

the r9-270 is the graphics processor on the card for both examples you asked about. you'll have to read the specs on the asus card you're looking at to see if it is "reference" or not. more than likely it is a custom version with better cooling and such. the amd one will be reference since that is all amd releases for themselves.

 

[gamer4life253]

yes they are the same more or less. a graphics card is like a mini computer. it has it's own processor, ram, controller chips and cooling. a "reference" card will be what amd suggests as a model for the card. other brands may make reference cards to match what amd released. they also have the option of making custom cards using the graphics chip from amd/nvidia. this will have custom cooling, power delivery, ram, outputs and other possible changes to the card. it will still use the same processor but can change many other things to make it different from other cards on the market.

the r9-270 is the graphics processor on the card for both examples you asked about. you'll have to read the specs on the asus card you're looking at to see if it is "reference" or not. more than likely it is a custom version with better cooling and such. the amd one will be reference since that is all amd releases for themselves.

thanks for your answer!!

 

[Math Geek]

glad i could help