Getting total power requirements

ddt · 02-23-2003, 12:14 AM

I'm trying to revamp my install a bit because I'd REALLY like to use an Opus PSU because of all of it's intergrated circuitry. I'm trying to figure out power requirements for a Celeron-based system and this is what I've come up with so far.

Asus CUV4X-C 30W 3.3V and 5V
Celeron 1GHz 30W 5V
TV Tuner 7W 5V
Lucent PCI Adapter+WiFi Card10W 5V
Geoforce 4 64MB 30W 3.3V
512 MB PC133 RAM 40W 3.3V
Maxtor 80GB 10W 5V and 12V
Slim DVD ROM 10W 5V and 12V
Deluo USB GPS ??? ???
Touch Controller (USB) ??? ???
LCD Controllers (2) ??? ???
IRMAN ??? ???

Help me out here, does anyone have a guess about the power requirements of the last 4 components?

Also when I'm converting watts to amps, for use in an automobile, should I use 12V or 14.4V???

digitallexus · 02-23-2003, 12:30 AM

Not sure how you came up with all of those numbers.

You should be measuring how many Amps each devices uses and at what voltage.

Most devices will list their power requirements. Harddrives, cdroms, dvd's, etc should show on the label.

Lets say I have two components:

my DVD drive requires:
12v 0.8A
5v 1A

My laptop harddrive requires:
5v 1A

Now add it up. Let's start with 12v:
Dvd(0.8A) + Harddrive (0A) = 0.8A

Now 5v:
Dvd(1A) + Harddrive (1A) = 2A.

So the power supply must supply at least:

12V .8A
5V 2A

The wattage is more of a generalization on power supplies since it lumps all of the different voltages together.

to find Watts, you multiple Current(Amps) times Voltage.
so 12v * .8A = 9.6W
5v * 2A = 10A
for a total power consumption of 19.6W

If the device doesn't list it, then you can just measure it with an Ammeter. Most meters will measure Amperage up to 10A.

You simply put the meter in-line with the device. Power up the computer and watch the needle to find the peak.

ddt · 02-23-2003, 12:47 AM

Quote:

Originally posted by digitallexus
Not sure how you came up with all of those numbers.

Online guide to power requirements for PC Components.

Quote:

You should be measuring how many Amps each devices uses and at what voltage.

Look at my question at the end. I realize that Amps = Watts / Volts. I need to know if I should use 12V or 14.4V in the calculation. Then and only then would I be able to get an accurate representation of what I need.

Quote:

Most devices will list their power requirements. Harddrives, cdroms, dvd's, etc should show on the label.

I have not purchased the components yet. I would like to know if it works prior to purchasing them. Saves money!

Quote:

Lets say I have two components:

I have two components.

Quote:

my DVD drive requires:
12v 0.8A
5v 1A

My laptop harddrive requires:
5v 1A

Now add it up. Let's start with 12v:
Dvd(0.8A) + Harddrive (0A) = 0.8A

Now 5v:
Dvd(1A) + Harddrive (1A) = 2A.

So the power supply must supply at least:

12V .8A
5V 2A

The wattage is more of a generalization on power supplies since it lumps all of the different voltages together.

to find Watts, you multiple Current(Amps) times Voltage.
so 12v * .8A = 9.6W
5v * 2A = 10A
for a total power consumption of 19.6W

If the device doesn't list it, then you can just measure it with an Ammeter. Most meters will measure Amperage up to 10A.

You simply put the meter in-line with the device. Power up the computer and watch the needle to find the peak.

Once again I realize exactly just how the math portion works and I know the specs of most components AND the PSU. My questions were:

does anyone have a guess about the power requirements of the last 4 components?

AND

when I'm converting watts to amps, for use in an automobile, should I use 12V or 14.4V???

MP3DUB · 02-23-2003, 01:09 AM

Anything usb cant greater than 500ma x 5vdc, and id guess the irman uses next to nothing. The lcd controllers im not sure about, but I know the specs for my 5" ntsc lcd + backlight + inverter + controller say its less than 750ma@12vdc. hth.

digitallexus · 02-23-2003, 01:10 AM

If you are trying to convert the Wattage listed for each component back to how many amps, to match up with the OPUS, then you divide the list watts by the device voltage. Has nothing to do with being in a car, or 12v or 14v.

Asus CUV4X-C 30W 3.3V and 5V
no way to calc this without knowing how many watts for each voltage.

Celeron 1GHz 30W 5V
this requires 6A at 5V.

TV Tuner 7W 5V
1.4A at 5V
(this seems high)

Lucent PCI Adapter+WiFi Card10W 5V
2A at 5V

Geoforce 4 64MB 30W 3.3V
9.1A at 3.3V

512 MB PC133 RAM 40W 3.3V
12A at 3.3V
(this seems high to me)

Maxtor 80GB 10W 5V and 12V
Can't calc.

Slim DVD ROM 10W 5V and 12V
Can't calc

digitallexus · 02-23-2003, 01:14 AM

Google is your friend:

From
http://www.pocketaprs.com/gps.html

Deluo/EverMore GM-305

Power requirements. Rated voltage is 3.8 to 8 VDC; current, by measured testing, 140 mA during acquisition, 125 mA with satellite lock (manufacturer rating is 105 mA). This GPS comes configured to mooch power from the PS/2 (keyboard) port of a laptop PC. Variable voltage testing revealed that internal regulation was linear, so I would not recommend connecting this GPS to straight 13.8V mobile power (lest it go "poof!").

ddt · 02-23-2003, 01:32 AM

Quote:

Originally posted by digitallexus
If you are trying to convert the Wattage listed for each component back to how many amps, to match up with the OPUS, then you divide the list watts by the device voltage. Has nothing to do with being in a car, or 12v or 14v.

okay gotcha. THAT was screwing me up.

Quote:

Asus CUV4X-C 30W 3.3V and 5V
no way to calc this without knowing how many watts for each voltage.

Celeron 1GHz 30W 5V
this requires 6A at 5V.

TV Tuner 7W 5V
1.4A at 5V
(this seems high)

Lucent PCI Adapter+WiFi Card10W 5V
2A at 5V

Geoforce 4 64MB 30W 3.3V
9.1A at 3.3V

512 MB PC133 RAM 40W 3.3V
12A at 3.3V
(this seems high to me)

The site I looked @ quoted 10W per 128MB.. now it might be for newer faster DDR RAM. I dunno.

Quote:

Maxtor 80GB 10W 5V and 12V
Can't calc.

Slim DVD ROM 10W 5V and 12V
Can't calc

Thanks for doing the math for me

makes my slightly intoxicated life easier!

ddt · 02-23-2003, 01:34 AM

Quote:

Originally posted by digitallexus
Google is your friend:

Rated voltage is 3.8 to 8 VDC

I'm unaware that the Opus has a 3.8 to 8V DC line

. Google isn't my friend it seems...

I'll go with MP3DUB's answer, being that the GPS was of the USB flavor and doesn't get power off of the PS2.

and just a little FYI... I'm the LAST person you need to tell to search.

digitallexus · 02-23-2003, 01:36 AM

the device itself probably needs 5v

MP3DUB · 02-23-2003, 02:08 AM

Also, im sure you have your reasons, but isnt 512 a bit much for a carpc? And doing a bit of googling, it seems that pc133 has a maximum power draw of 10w per module, with average consumption being lower.

ddt · 02-23-2003, 03:13 AM

Quote:

Originally posted by MP3DUB
Also, im sure you have your reasons, but isnt 512 a bit much for a carpc?

Not for my application.. and I have one 512MB stick of PC133 RAM sitting around so I'll probably use it if I go this route.

zootjeff · 02-23-2003, 10:38 PM

512 MB PC133 RAM 40W 3.3V

Does your SDRAM have a massive heatsink?
Thats 12 Amps! You must be using 16, 32 meg modules....

ddt · 02-23-2003, 11:35 PM

Quote:

Originally posted by zootjeff
512 MB PC133 RAM 40W 3.3V

Does your SDRAM have a massive heatsink?
Thats 12 Amps! You must be using 16, 32 meg modules....

Read the posts above in which I stated "The site I looked @ quoted 10W per 128MB.. now it might be for newer faster DDR RAM. I dunno.."

digitallexus · 02-24-2003, 03:37 AM

In device manager, look under Universial Serial Bus Controllers,
then under the USB Root Hub (there may be more than one)
Double click USB Root Hub and then select the Power tab.

A list of all attached devices is shown with a power requirement in mA for each. I suspect this is a hard coded value set by the device manufacture in the driver. For this reason, they may not always be 100% accurate, but it's probably at least a decent estimation of power required.

As was already pointed out, you can have up to 500mA on each port.

zootjeff · 02-24-2003, 10:37 AM

Quote:

Originally posted by ddt
Read the posts above in which I stated "The site I looked @ quoted 10W per 128MB.. now it might be for newer faster DDR RAM. I dunno.."

Your safer doing a chip calculation or going to the manufacturers website.

10Watt/ 128 still seems a little high. I just looked up a 512 megabit (key word BIT) SDRAM part and the absolute max power dissapation is 2 watts. (which is typically a conservative estimate)

How many chips is your SDRAM? I would guess that a safe estimate is 1.5 watts per SDRAM chip.

Here is a data sheet for some samsung memory.

http://www.samsungelectronics.com/se...k4s510632d.pdf

Giving their absolute max ratings, that would put a 512M byte module at 16 watts. I would guess if you measured it, it would come out around 8 watts..