Curse

ubrikkean · May 21, 2006, 02:38 PM // 14:38

Well, I think I'm as far to getting this thread solved as I ever will. Unfortunately, I have no idea where to find a small form-factor graphics card for you, cosypief (and I really don't mind the hijacking, as far as I can tell there wasn't any more help anyway

). I think I'll just buy the x1600 and be done with it.

sigried · May 22, 2006, 06:01 PM // 18:01

if you goin to buy the radeon 1600 pro you will need a 450 watts for normal sytm operations. I have one X1600 pro

Lurid · May 22, 2006, 09:58 PM // 21:58

For the record wattage does not matter

Wattage = Volts x Amps, the important thing here is amperage. A PSU with a large amounts of amperage on a rail that sees less use, and less amperage on one that sees more use will result in your precious wattage meaning next to nothing. Stop spreading this rumor people!

Matsumi · May 22, 2006, 10:44 PM // 22:44

Regardless, that's how psu's are identified. You won't notice the amount of ampere distribution unless you take a meter to it. You can do some simple math and figure out how many amps it's capable of distributing, but not what it's actually distributing to each device without a meter or doing some more math by looking at how much voltage/resistance = current (measured in amperes). You could also do some calculations to figure out power (measured in watts) = voltage*current. If a 100 watt light bulb operates on 120 volts it's operating at .833~ Amps, or 833 mili amps (mA). To figure this out you simply calculate current = power (watts)/ voltage.

Lurid · May 22, 2006, 11:07 PM // 23:07

Your logic is intriguing. We'll bump heads alittle, in order to perhaps learn alittle more. As i'm no electrician.

"Regardless, that's hwo psu's are identified."

Thats not the only way, thus the side sticker with all the other information, amperage for each rail being one of them.

"You won't notice the amount of ampere distribution unless you take a meter to it.You can do some simple math and figure out how many amps it's capable of distributing, but not what it's actually distributing to each device without a meter or doing some more math by looking at how much voltage/resistance = current (measured in amperes). You could also do some calculations to figure out power (measured in watts) = voltage*current. If a 100 watt light bulb operates on 120 volts it's operating at .833~ Amps, or 833 mili amps (mA). To figure this out you simply calculate current = power (watts)/ voltage."

I'm not sure how you come to this conclusion, I understand where your coming from though. Your saying that the amperage charts show what its capable of distributing, though thats not always the amount that is being distributed.

Which i'm fairly sure is well and good, for all but a few things. PSU's use three rails for power distribution (sp?) meaning that (atleast from everything i've read and seen) a specific portion of the PSU powers certain devices. The most powerful of which being the 12v rail, which supplies most of the power to the mobo / cpu.

If i'm understanding your logic correctly, then the fact that the distribution isn't always certain, doesn't really have much to do with what I mean. That is to say that i'm saying, for example:

12v - Weak
5.5v - Medium
3.3.v - Strong

In theory you'd get most of the components that draw a large portion of the amperage drawing from a weak rail, meaning that possible instability issues could ensue. Even though the distribution here isn't certain, we can assume that certain elements of the system are getting more power than others. Meaning that my formula of having rails that power more power hungry components, be more powerful still makes more sense. Otherwise i'm confused here, as like I said, i'm still learning myself.

Matsumi · May 22, 2006, 11:29 PM // 23:29

Well, these are just simple basic calculations one can do to figure things out with math concerning power ditribution, and measurements in electricity. I'm an electronics graduate, and was just simply introducing some of these simple calculations and ohm's law from begining knowledge. Power fluctuations and distributions don't always contain a set value. Having set amounts of power for ditribution whether it be measured in amps, volts, watts, etc. is fairly common in electronics, having more devices absorbing more of that set amount of power will result in instability and may require a larger set amount. I'm not really challenging you to anything, just providing information. Saying "wattage does not matter" in relation to this though is not really correct, seeing how watts or power is a mathematical representation used to calculate current or amperes as well in a lot of cases. Also heat can be measured (in joules) as well, but that's beside the point.

Lurid · May 23, 2006, 12:07 AM // 00:07

Oh no, not a challenge of who is right or wrong. I'm more or less challenging myself to understand what you are saying, as it could prove useful later.

I was under the impression though from everything that i've read that the wattage as a whole is not as important as the distribution of the amperages that govern the amount of end wattage. As in having a higher amperage on a rail that is used less would result in a higher end wattage with an overall lessened amount of power to the components that use most of the power.

Matsumi · May 23, 2006, 12:43 AM // 00:43

Although it's true that current is the main reason why some things operate, in a DC circuit or in digital, some things are more reliant on voltage signals as well. Like say a digital clock or other things, when all the numbers reset to 0 as a result a carrier is sent to a part on the chip like say, 5 volts for a second, to indicate that a number needs to be displayed if it was 0999 it turns to 1000.

I'm sort of getting off topic here, but basically what I was trying to point out is that you'll notice amps fluctuate when distributed to a device, similar to the voltage (although less noticable). Power (watts) in this equation does matter, given the power source. If watts/volts = current you will have less current or amps in the equation if a lesser watt power source is introduced. So, not really disagreeing with you, but just trying to show how they're related to eachother.

ubrikkean · May 23, 2006, 12:50 AM // 00:50

Guys, you can take it to the PSU thread if you'd like, I just picked that PSU beacuse I found a review that said it was one of the best ones for under $30. From the sound of it, no matter how you measure it, I will need a pretty expensive PSU if I get the X1600 Pro, so I'll just stick with the X1300 and hope for the best, possibly buy that $30 PSU if I need it for the card.

Matsumi · May 23, 2006, 12:55 AM // 00:55

lol, ya sorry, glad you found a solution or what you wanted to buy.

Lurid · May 23, 2006, 01:40 AM // 01:40

Yeah, I apologize aswell. Grats to you for finding a solution that works for you

Nah, I wasn't arguing as in "omg your wrong, lemme show you the way it is" rather as in, "well didn't know that... elaborate", lol.