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The Asus EEE PC comes with a small power supply. It is very transportable, but I'd like a second power supply I can leave at home. There are lots of multivoltage power adapters out there, but none that meet the finicky specifications of the EEE PC:

 Voltage: 9.5V
 Current: 2.315A

The voltage, in particular, is a decidely old choice. Most multivoltage adapters can be switched to 9V or 12V. The 12V setting, I fear, would fry the little laptop. Meanwhile, the 9V setting might just not quite supply enough juice to it. Moreover, many of these adapter just aren't up to delivering the current the EEE PC needs. The other problem with these multivoltage adapters is that they cost money!

One thought I have had is to reuse an old laptop power brick. These typically can deliver ample current. The only problem is that they typically deliver between 15 and 20 volts. Far too much for the little EEE PC. 

But this doesn't seem to be an insurmountable problem. By interposing a voltage regulator, it should be possible to get the voltage down to 9.5V. To me, this approach sounds the best. It saves yet another perfectly good piece of electronics (the power adapter) from going to landfill, and its a good excuse for getting the soldering iron out. The cost, too, can be kept down to a minimum.

I haven't had a chance to verify this, but the EEE PC apparently has a DC power socket with OD 4.75mm and ID 1.75mm.

==Linear Regulator==

The first thought was to use a linear regulator to bring the voltage down. The Silicon Chip Magazine [http://www.siliconchip.com.au/cms/A_110498/article.html Low Voltage Adjustable Regulator], available as a kit from [http://www.jaycar.com.au/ Jaycar Electronics] looked suitable. It has impressive currently handling for a linear regulator. By adjusting the value of a couple of feedback resistors, it was reasoned that the regulator could be made to output 9.5V at high currents.

Whilst it is indeed possible that the linear regulator could deliver this voltage at around the current required by the EEE PC, further thought indicated that the linear regulator would have to dissipate a hell of a lot of heat to do so. 

Take, for argument's sake a Dell power supply at 19V. If this is used to supply the linear regulator, then the regulator needs to drop

 19 - 9.5 = 9.5V

between its input and output terminals. This means it has to dissipate as much power as the EEE PC does at peak load, ie:
 
 P = I x V = 9.5 x 2.315 = ~22W

This a big ask for a linear regulator, even with a pass transistor taking some of the burden. It necessitates a big heatsink. It's also not a win for the environment - the power supply is under 50% efficient: under half the power drawn from the power point is going to the EEE PC...

==Switchmode Regulator==

So a linear regulator seems unsuited to the task. The better approach, it would seem, is a switchmode regulator. A switchmode regulator should be able to handle the same currents and voltage drop with only a small heatsink (if any at all).

A suitable candidate would seem to be the National Semiconductor [http://www.futurlec.com/Linear/LM2576T-ADJ.shtml LM2576T-ADJ]. It can handle up to 3A and input voltages up to 40V. Moreover, the output is adjustable using two resistors as a voltage divider. 

All that would seem to be required, in addition to this regulator is:
* An inductor
* Some low ESR filter capacitors
* A Schottky Diode (perhaps an 1N5822)
* Resistors for the voltage divider
* A DC Plug to suit the EEE PC.

The hardest part to find seems to be the inductor. I have opted to use a 150uH 3A inductor from [http://www.altronics.com.au/ Altronics]. Another possibility is to wind a DIY inductor around a powdered iron toroid, for instance the Jaycar [http://www.jaycar.com.au/productView.asp?ID=LO1244&CATID=&keywords=lo%2D1244&SPECIAL=&form=KEYWORD&ProdCodeOnly=&Keyword1=&Keyword2=&pageNumber=&priceMin=&priceMax=&SUBCATID= L0-1244] toroid (dimensions 28x14x11). As a guide, I know (from [http://www.siliconchip.com.au Silicon Chip Magazine] projects that use this toroid) that 50 turns of 0.8mm enamelled copper wire results in ~145uH inductance.

==The Design==

Whilst waiting for the LM2576T-ADJ to arrive in the mail, I thought I'd nut out the design. All this involves, really, is following the datasheet and selecting components that satisfy the requirements laid out therein. Here is my working. Note, however, that I haven't built a prototype yet - so use this information at your own risk!

===Output Voltage===

The output voltage is given by:

 Vout = Vref (1 + (R2/R1)) where Vref = 1.23V

We can express this as:

 R2 = R1 * ((Vout/Vref) - 1)

Assuming R1 of 1 Kohm:

 R2 = 1000 * ((9.5/1.23) - 1)
    = 6.723 Kohm
    = ~6.8Kohm (nearest practical resistor value)

So, R1 will be 1K and R2 will be 6.8K.

===Inductor===
E*T is given by (where F is a fixed frequency of 52 kHz):

 E*T = (Vin - Vout)*(Vout/Vin)*(1000/F)  V*uS
     = (19 - 9.5)*(9.5/19)*(1000/52) V*uS
     = 91.35  V*uS

Looking up Figure 4. of the datasheet, this gives an inductor value of 150uH for 2.5A maxium load current.

So, L1 will be 150uH.

===Output Capacitor===
Cout is given by (where L is the inductance of the inductor in uH):

 Cout >= 13300 * (Vin / (Vout * L))  uF
      >= 13300 * (19 / (9.5 * 150))  uF
      >= 177.34 uF

However, as the datasheet notes, to keep output ripple down, the capacitor should be several times larger than this figure. So, I have settled 1000uF for C2. This should be a low-ESR type capacitor.

==Diode==

The diode needs to be a Schottky or Fast Recovery diode, with a current rating of 1.2 times the maximum load current:

 Idiode = 1.2 * 2.315 A
        = 2.778 A

For our purposes, a diode with a rating of 3A should be fine.

So, following figure 3 of the datasheet, I have selected an 1N5822 for D1, which is rated at 3A and suited for a maximum input voltage of 40V.

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That's it for now. More to come when the regulator arrives in the mail.