[shifuimam]

I'm working on upgrading my I-Opener to a 550MHz AMD K6-2 CPU. The biggest issue is that the socket on the IO's motherboard is single-voltage, while any CPU worth using is dual-voltage.

I found some information on changing a couple resistors on the board and adding some 0 ohm resistors (or jumpers) to a couple other lines. The second part I understand well enough, but I'm not sure how to figure out what resistors I need for the first part.

The page that describes the mod (rather briefly, but with pics) is here:

Linux-Hacker.net CPU upgrade

This guy put on 7.6k and 1.4k resistors in order to cut the line down to 2.8V for the 166MHz Pentium MMX he installed. The processor I'm wanting to put in is only 2.3V, so I can't follow this to the letter, which brings me to my question:

How do I figure out what two resistors are needed to bring the voltage down to what I need (2.3V)? A clearer picture of the two resistors in question can be seen below:



The other part of the mod shouldn't be an issue; it's this part that I need some assistance on so I can complete this project (sort of).

Any help is really appreciated - I've used up all my IRL resources who might have a clue how to go about doing this.

[reader50]

Resistors are a lousy way to cut voltage. They can work for a constant-current application, like an LED. But a CPU has wildly variable current draw.

If you needed another voltage, for small current applications you could use a resistor bridge. It would be inefficient and voltage would vary somewhat under load. For any significant current draw, you would need a voltage regulator. Probably a mini board with regulator IC, capacitor to stabilize things a bit, along with a fuse + zener diode to protect the CPU if the regulator IC blew.

[ghporter]

The math for a voltage divider is almost trivial, but you need to know a couple of things to set up what you have to start with. First, what is the voltage at the top of the divider (the supplied voltage). Then what resistors of what values are in place before and after the ones you want to change.

Here's how a voltage divider works:
As long as all the values are identical, you simply divide the supply voltage by the number of separate resistors between supply and ground, and each step will be identical (i.e. 10 resistors between 10v and ground will have a 1v difference between each). Mixing things up can give you simple fractions without having to have a bunch of identical resistors.

In your existing example, the divider is (apparently) made up of a total of 9.0k ohms, with 1.4k ohms giving you the 2.8v. It's simple cross-multiplication to find what the original voltage at the top of the 7.6k resistor is (in this case, with the above assumption, it's 18v). If those are the only resistors, then you do basically the reverse of the cross multiplication, substituting 2.3v as the desired fraction, and that should tell you the value of the smaller resistor. Knowing that, you subtract the new value (1.15k) from 9k to get the larger resistor's value: 7.85k. But these are not values that resistors come in, so you'll have to kluge a couple different resistors for the small end and a couple resistors for the big end.

A practicing electrical engineer may make the numbers prettier, but that's the basics. And I made a HUGE assumption in going with those two resistors as the only components in the voltage divider... triple check to make sure before you change anything!

Reader beat me to posting (I'm long-winded) but even if the design isn't the smartest, it looks like that's how the final voltage is determined. It could easily be a zener-based source, with the Vcc (or more likely Vsource) being the only current draw on the ladder-which would be pretty clean. Again, I gotta know the rest of the schematic in order to say anything definitive about this.

[shifuimam]

Unfortunately, there is no schematic information available online for the motherboard on this thing, since I'm pretty sure it was a custom board designed specifically for this device. I believe Foxconn made the board, so I'm going to contact them on the unlikely chance that they can provide me with some information.

I think it would make sense that the original voltage is 18v - the power brick is 19v.

I'm only going off what this guy's done and says works, so I'm assuming that the replacement of these two resistors is what's necessary. It makes sense to me from a non-technical standpoint - AMD's Socket 7 processors follow the same standard as Intel's, so everything else should work as expected. What I think this mod is doing is (a) cutting the voltage to the CPU core (3.3v for the stock WinChip CPU), and (b) redirecting a 3.3v current from one location on the socket to another in order to supply the I/O voltage (which is always 3.3v).

That being said, do I have any chance of doing this mod sucessfully? If it does require more than two resistors, how should I go about putting them on the motherboard so that they do what they're supposed to do?

Boyfriend has done a little EE work and will be handling all the actual desoldering and soldering of the components, but this particular mod isn't exactly his field of expertise.

[Dork.]

Do you have the manual for the 550MHz AMD K6-2 CPU? If not, post the part number and we'll be able to find it.

There will probably be certain pins that have to be pulled high or low depending on what type of voltage you are providing to it. On a normal motherboard, you would probably be able to set the voltage by setting jumpers on the motherboard. You have a motherboard that is meant to be used with only one CPU, so the designers used jumpers to hard-set these pins, so you would have to manipulate resistors to get it to do what you want. 0-ohm resistors are common to use for the purpose of hard-wiring jumpers.

It's also possible that those two resistors are not a voltage divider at all, but are meant to be set to certain values depending on what the operating voltage of the part is. There might be a voltage regulator inside the part that relies on external resistors, for instance. If this is the case, the manual ought to tell you this.
(One way to confirm whether or not this is a voltage divider is to get a multimeter, and start probing to see if the voltage is actually being divided.....)

The manual will tell you what pins on the CPU need to be dealt with, but you're not going to be able to map that with resistors on the board. This is where your friends on the Internet help: if you find a chart which says that there are two pins that need to be connected to 7.6k and 1.4k to get 2.8V, you can assume that the chart goes with the resistors he found, and it will likely have a line for 2.3V as well.

Like Glenn said, though, this is all pure speculation on our part. Without the schematic, you're just guessing, and guessing wrong can make the blue smoke leave the chip. Blue Smoke is what gives chips their mystical powers -- once the blue smoke leaves, it won't work anymore!

[shifuimam]

Everything you could possibly want to know about the AMD K6-2 line can be found in this technical document:

http://www.amd.com/us-en/assets/cont...docs/20695.pdf

It's a 550AGR (550MHz K6-2). Pinout of the CPU (which is the same for any socket 7 CPU) starts on page 285.

I have a multimeter, but being the disorganized mess that my home usually is, I haven't the faintest idea where it is. My next goal is to either find it or give up and buy a new one.

[The Godfather]

It is possible, reader50, that the voltage divider feeds an voltage-follower amp that feeds the CPU with an exactly defined voltage.

Without the schematic, it will be guesswork. Get a tweezer style multimer (with resistometer and capacitometer) and figure out the schematic yourself.

[Dork.]

Thanks for the link to the datasheet. Your CPU runs off only one voltage, so you're going to have to figure out how to change that voltage on the motherboard.

I'm not a power guy, but I was thinking about this on the drive into work this morning, and those are teeny, teeny resistors. 0603's, maybe? (I think the board is too old for them to be smaller than that.) They're probably rated for 1/6 W of power at most? Even if it is a voltage divider, there's no way that's actually powering the CPU.

Those resistors are probably going to a voltage regulator which is (hopefully) only there to provide power to the CPU. (I say hopefully because if it's powering other stuff, then when you reduce the voltage to the CPU that other stuff may not work.) That voltage regulator is taking the power from the power brick and regulating it down to the Vcc of the CPU. But voltage regulators can output a range of voltage, which is set by external components (resistors, capacitors, and inductors). If you can find which chip the voltage regulator is, you can look at the datasheet and possibly find out which parts to change to get the voltage you want.

I would guess that voltage regulator (or possibly an op-amp, ad The Godfather suggests) is in the area near the picture you took. Big honkin' capacitors and parts whose labels begin with L (inductors) are a clue you've stumbled on a power-related area. If you have a multimeter which can measure resistance (preferably one with the buzzer/"diode mode" that beeps when you have a short), you might be able to probe and find out which IC's those two resistors go to.

Is i-Opener still in business? Maybe someone bought the schematic or the Bill of Materials at the bankrupcy auction?

[ghporter]

Originally Posted by Dork. 

I'm not a power guy, but I was thinking about this on the drive into work this morning, and those are teeny, teeny resistors. 0603's, maybe? (I think the board is too old for them to be smaller than that.) They're probably rated for 1/6 W of power at most? Even if it is a voltage divider, there's no way that's actually powering the CPU.

That's another reason I hedged as strongly as I did. I think you're right about the resistors being 0603s, and 1/6 or 1/8W resistors aren't really robust enough for much more than biasing elements...which sort of says to me "these are setting up a control circuit for the real CPU power supply component." Like maybe one of those two 8-lead, "stealth marked" ICs in the picture? You can also see a big honkin' electrolytic cap in the left of the picture and an "L" component label (though for a very small part) near that, so I think this part of the board is at least near the power handling circuitry.

Looking at the overall picture of the board, it seems that the detail photo comes from just above the SODIMM socket...

[Dork.]

Originally Posted by ghporter 

"these are setting up a control circuit for the real CPU power supply component." Like maybe one of those two 8-lead, "stealth marked" ICs in the picture?

Could be. One of the problem with IC's once they get small enough is that you can't put the full part number on them, and you only get a code that the manufacturer relates to the full part number. Which means that if you don't know who the manufacturer is, you may never find out what that code means, since Google searches on random 5-digit numbers are next to worthless....

(which is why I wonder if someone has the BOM somewhere on the Internet, or knew a guy at i-Opener who had some clues on reverse-engineering one....)

[Laminar]

Originally Posted by shifuimam 

My next goal is to either find it or give up and buy a new one.

$5 at Harbor Freight. Mine works great.

[Dakar V]

Originally Posted by Laminar 

$5 at Harbor Freight. Mine works great.

Love the post. To it, a toast!

[shifuimam]

Originally Posted by Dork. 

Thanks for the link to the datasheet. Your CPU runs off only one voltage, so you're going to have to figure out how to change that voltage on the motherboard.

I'm not a power guy, but I was thinking about this on the drive into work this morning, and those are teeny, teeny resistors. 0603's, maybe? (I think the board is too old for them to be smaller than that.) They're probably rated for 1/6 W of power at most? Even if it is a voltage divider, there's no way that's actually powering the CPU.

Those resistors are probably going to a voltage regulator which is (hopefully) only there to provide power to the CPU. (I say hopefully because if it's powering other stuff, then when you reduce the voltage to the CPU that other stuff may not work.) That voltage regulator is taking the power from the power brick and regulating it down to the Vcc of the CPU. But voltage regulators can output a range of voltage, which is set by external components (resistors, capacitors, and inductors). If you can find which chip the voltage regulator is, you can look at the datasheet and possibly find out which parts to change to get the voltage you want.

I would guess that voltage regulator (or possibly an op-amp, ad The Godfather suggests) is in the area near the picture you took. Big honkin' capacitors and parts whose labels begin with L (inductors) are a clue you've stumbled on a power-related area. If you have a multimeter which can measure resistance (preferably one with the buzzer/"diode mode" that beeps when you have a short), you might be able to probe and find out which IC's those two resistors go to.

Is i-Opener still in business? Maybe someone bought the schematic or the Bill of Materials at the bankrupcy auction?

I'll take a look at the board again tonight and see what I can find. FWIW, the CPU is definitely dual-voltage - the movement of the two 0-ohm resistors in the second part of the mod, I'm fairly sure, direct the 3.3.v power to the proper pins on the CPU for the I/O voltage requirement.

According to boyfriend, the two resistors in question are 0402s. I don't know if that's relevant to the discussion; just thought I'd mention it.

Netpliance, unfortunately, went out of business nearly a decade ago (they folded only a year after the I-Opener was introduced - they lost a massive amount of money on it once people started hacking them). I found a Chinese BBS with a lot of posts about the I-Opener...unfortunately, Google's results have the URL slightly wrong, so I'm not sure if I'm going to be able to find the posts on that site. I've Googled around quite a bit looking for the schematic for the motherboard, but have come up empty thus far.

I'll see what I can do about finding a good multimeter for cheap, so that I can figure out the numbers accurately.

I've found some possibly interesting information on the original Linux-Hacker BBS:

I-Appliance BBS - CPU Upgrade K6

I-Appliance BBS - K62P Mobile Processor Hack (this is with a mobile K6-2P, but the information may or may not still be relevant...)

I-Appliance BBS - V1/V2 mod for 2.2v Vcore - this one may have the right information, although now I've discovered that the CPU I purchased doesn't have any L2 cache. I could get a K6-2+ with 128k L2 cache, but that has a 2.0v core voltage rating, so that brings us back to "how to get the voltage down to the right number"...

http://www.linux-hacker.net/cgi-bin/...=&Session=

[Laminar]

Originally Posted by Dakar V 

Love the post. To it, a toast!

[Princess Bride Quote]

[Railroader]

Anyone see the moat?

[Laminar]

[ibookuser2]

Originally Posted by shifuimam 

According to boyfriend, the two resistors in question are 0402s. I don't know if that's relevant to the discussion; just thought I'd mention it.

I've never seen 0402s that are labeled. And judging by their size relative to the two SOICs in that picture, they're definitely 0603s. They're wider than the pin spacing on the SOICs, which an 0402 definitely isn't.

[Railroader]

Originally Posted by Laminar 

Is that the new non-hackable Apple iAppliance?

[shifuimam]

I'll post on the Linux Hacker forum and see if anyone there can provide some help. We'll see how this goes.