Continue to Site

# Power Supply Build

Status
Not open for further replies.

#### Sundermeyer

##### New Member
I am a computer technician with some electronics understanding, but not enough to build circuits on my own. I have built a 5-node cluster computer out of old Acer Aspire One Netbooks just for fun but I wanted to make a new power supply for it to reduce some of the clutter. I want to have one power supply that feeds all five systems rather than having five AC adapters. This way I can fit it all in a small case with one power cord.

The battery for a single system outputs 10.8v
The AC Adapter outputs 19v and also lists the amperage at 1.58 - 3.42A

Each computer appears to use 14.3W

I have a car adapter for one of these and it lists its output as 19v at 2A

The idea would then to be to create an AC Power supply that fed each system with the needed voltage and amperage equally, but would not kill anything if one of the systems were disconnected. It would need to power one system the same as all five for when I need to perform maintenance or testing on it.

If there is a way to use an existing AC supply that I would not have to build and then just add on the circuitry for the five outputs, that would be ideal.

Thanks

-Sundermeyer

#### rjenkinsgb

##### Well-Known Member
Most Helpful Member
Something like this, all negatives commoned and each positive fed via eg. a 3.15A fuse to protect against short circuits?

Note that is has an output voltage trim control next to the terminals; if 18V does not work you should be able to adjust it up slightly.

Laptops with three cell batteries often use chargers around 15 - 16V, so I cannot really imagine them not operating at 18V.

#### Diver300

##### Well-Known Member
Most Helpful Member
The battery voltage is unimportant here. The computers have circuitry that converts the input voltage (19 V) to whatever is needed to charge the battery, and further circuitry to convert the battery voltage down to the 5 V, 3.3V, 1.8 V or whatever the processor runs on.

If one computer is using 14.3 W, that would be just over 0.75 A at 19 V. How did you measure the 14.3 W? If you measured the power taken by the AC adaptor, that will have some losses, so the computer would be taking less, probably 0.65 A

The computers will take very variable amounts of power. If the battery is charging, or the processor is working hard, or the display is bright, then more power will be taken. Start-up can be particularly hard as the processor often runs at full power for a short time.

I don't understand why the AC adaptor has a lower rating. It has to be able to supply no current, or it would fail when plugged into the mains without a computer connected to it, or with the computer turned off.

You should be able to take a 19 V power supply with a large enough current rating, and run all 5 computers from that. It does not matter if your power supply has a larger current rating than the computer, as long as it is 19 V. The power supply produces 19 V, and the computer takes what current it needs. *

It would be good to know what the maximum current is, because the figure of 0.65 A (from the 14.3 W) is a lot less than the 2 A for the car adaptor, which I assume works fine. The power supply figure of 3.42 A might be for larger computers, but I'm not sure, as it could be that 3.42 A is the maximum current if the battery is charging, and the computer is turning on, and the screen is bright.

Without knowing the current more exactly, you could allow for around 10 A for five computers. Here is one possible supply:-
https://www.newark.com/xp-power/ves180ps19/adapter-ac-dc-1-o-p-19v-9-47a/dp/33AH2569

I agree with rjenkinsgb that a fuse for each computer is a good idea.

* There are quite often questions on this forum about whether you can have a supply with too much current rating. Generally, you can't. You are quite OK to plug the power supply for your router into the mains socket, where that mains socket can supply a 1500 W heater. Both the heater and the router power supply are rated at 120 V, and each takes the current it wants. Similarly, a car has an interior light that takes 0.05 A from a car battery that can supply 500 A to crank the engine. That is fine because both the starter and the interior light are designed to run from 12 V.

The only consideration is fusing. If something fails it can take more current than intended. It's then a good idea to have fuses to protect the wiring. If your car's interior light shorted, with no fuse the battery could supply enough current to burn the thin wiring that supplies an interior light. That's why there's likely to be a 5 A fuse somewhere in that circuit. On the other hand, the starter has wires as thick as your finger to take the huge current.

#### Sundermeyer

##### New Member
The battery voltage is unimportant here. The computers have circuitry that converts the input voltage (19 V) to whatever is needed to charge the battery, and further circuitry to convert the battery voltage down to the 5 V, 3.3V, 1.8 V or whatever the processor runs on.

If one computer is using 14.3 W, that would be just over 0.75 A at 19 V. How did you measure the 14.3 W? If you measured the power taken by the AC adaptor, that will have some losses, so the computer would be taking less, probably 0.65 A

I pulled the power information from a Wikipedia article on these systems.

The specifics relate to the paragraph under "Power Management":
" The Intel Atom platform has a specified maximum TDP of 11.8 W. Individual figures are 2.5 W for the N270 processor, 6 W for the 945GSE chipset and 3.3 W for the 82801GBM I/O controller. The AUO B089AW01 LCD panel is rated at a maximum power consumption of 3 W. Typical read–write power consumption for the SSD is around 0.3 W, and 0.01 W when idle. The different HDDs are rated at about 1.5–2.5 W for read–write operations and around 0.7 W when idle."

In my case, all of the hard drives are mechanical and not SSD so i used the higher figure of 2.5W just to cover any extremes. Also, I have removed the LCD panels so that was not included in my initial posting's calculation. I see that I missed one of the values so it should be 6W more actually, giving 20.3W.

-Sundermeyer

Status
Not open for further replies.

Replies
8
Views
2K
Replies
16
Views
1K
Replies
41
Views
4K
Replies
20
Views
3K
Replies
11
Views
2K