VOGONS


Gigabyte GA-586HX (Rev. 1.53)

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Reply 20 of 71, by snufkin

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Ok, I think I mostly understand that now, although I'm unsure about the voltages on pin 12&13.

I'm going to assume that pin 13 connects directly to R151 (2.2k) and then R182 (10k), R183 (100k) and SW1. The other side of SW1 connects to Ground. R151 and R182 connect to a +ve supply (possibly different supplies). R183 connects to the Base of Q9. Emitter of Q9 goes to Ground.

So when SW1 is open/off then pin 13 is pulled high (but only to 2.1V?) through R182 and R151. Current also flows through R183 in to the base of Q9, so current can flow from C to E. This pulls pin 12 to Ground. So the inputs are FSEL[1..0]=01=66MHz.

When SW1 is closed/on then pin 13 is connected directly to Ground (FSEL0=0) and no current flows from B to E so Q9 is off. The Collector of Q9 connects to R184 (470k) and R152 (2.2k). They probably connect to a +ve supply, so with no current flowing through Q9 they will pull pin 12 high (FSEL1=1), so FSEL[1..0]=10=60MHz. But they only pull it up to 1.7V, which is technically below the level needed to guarantee it's read as a '1' (2V).

So there's direct control of pin 13 (FSEL0) through SW1, with the slightly trickier thing to find an easy way to control pin 12 (FSEL1). Shorting B-E of Q9 will force it off and hold FSEL1 at 1. Shorting Collector to Emitter will hold it at 0. So maybe a 3 pin header connected to Q9, middle pin on Emitter, side pins on Base and Collector. Jumper CE for 0, BE for 1, remove to give control back to SW1.

It all seems a bit more complicated than necessary. The PLL52C59-14A has internal 150k pull ups on the FSEL pins, so I'd expect they'll be there on the -14L, so in theory it'll run at 55MHz with nothing connected and only needs external pull downs. So all the extra external resistors (R182, R184, R151, R152) seem odd

Things that would be good to know though, particularly since I don't understand those input voltages:
Voltages on the vias connecting to R182, R184 and R151.
Voltages on pins 1, 8 and 14 of the clock gen (those should be the Vdd pins, may not all be the same)
There's an odd stub and a via from the Base of Q9. It can't (I think) connect to a power plane, so is there a trace on the other side of the board? I'm curious where it goes because maybe something else can control Q9.
Probably best to check that I'm right about the connections from the clock gen to SW1. So with the board off check for shorts between pin 13 and R151 / R182 / R183, and pin 12 and R152, Q9 pin 3 / R184.

[edit: I drew a circuit of how I think it's connected up, and a possible mod to give independent control for FSEL1 without having to remove any components or bend any pins (suggested method on that previous mod page for the HX2). I just stuck the supply voltage to 2V for the moment:
https://www.falstad.com/circuit/circuitjs.htm … ImvWzpioapK8EAA
]

[further edit: Ah, and re-reading that other mod page it looks like they already tried the pin 5 mod on the -14L and it is an OE control, so no hidden frequencies there. So it is, as they said, change the clock gen or change the crystal]

Reply 21 of 71, by majestyk

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Thanks for the research!
I tested pin 5 in the meantime - everybody was right, it´s an output enable input for the PLL52C59-14LSC. As soon as it´s pulled to ground there´s no output.

I´m currently trying to get my hands on some 75 or 83 MHz clock-gen. As soon as I can get one I´m going to continue this issue here.

In the meantime I returned to revision 1.53 for a second and modded it´s VRM cooling:

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When the RTC needs to be replaced I´ll have to disassebmle the heatsink first 😉

It´s now safe to use a K6-III+ @ 400 MHz and with 2V core voltage without burning the VRM. The new "pimped" heatsink won´t get any hotter than 40°C.
With the original heatsink - due to the high voltage drop of 3V it´s useless if you want to use a K6-3+ or K6-2+, only a Tillamook has a low enough power consumption.
The original VRM circuits were designed for core voltages around 3V so the voltage drop of the regulator would be around 2V.

Reply 22 of 71, by pentiumspeed

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Good modification; back in the day, when I had Asus motherboard with 430fx chipset, that low profile heatsink for regulator is hot but I found pointing a fan at this kept regulator cold. Believe me that was easiest way.

After that, I don't like linear regulators for high current stuff anymore.

Cheers,

Great Northern aka Canada.

Reply 24 of 71, by majestyk

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Compared to the original Gigabyte heatsink it´s a real "heathunk". 😉

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I´m no fan of linear regulators either. So I searched for some ready-made step-down replacement everywhere. But all the offers I found are either limited to about 3A or, if they are capable to handle 5-10A, they are even bigger than my heathunk.

An AMD k6-III+ 450 has a maximum power consumption of 16W (12W typical). At 2V core voltage (I think I/O supply is neglectable here) it draws a current of 8A / 6A. This current has to be handled by the VRM also.

In case it´s a linear regulator hooked to the 5V rail this means a voltage drop of 3V and a power dissipation of 24W / 18W! Hence the need for a heathunk.

The "AS2880" on this board has a maximum rating of 8A, so it´s good for the job:

https://datasheetspdf.com/pdf-file/866433/Alp … ductor/AS2880/1

Reply 25 of 71, by snufkin

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I'm looking at that datasheet, and looking at the Thermal Resistance for Junction-Ambient, and looking at 0.65C/W, and blinking a lot, and wondering if they meant 65C/W for no heatsink, which I think sounds more believable, if I remember how thermal resistance works. If correct then without a heatsink at 24W it'd hit 1,560C. When do ceramics melt? I suppose there's a chance it might fail first.

This is quite cute: https://www.farnell.com/datasheets/2239371.pdf
Adjustable output 12A DC-DC buck converter on a ~2cm x 1cm x 1cm board. Looks like it hits about 93% efficiency with Vin=4.5V and Vout=1.8V @ 8A. So probably around 90% with Vin=5V and Vout=2V@8Amp. Which I think means it'd burn under 2W.

Doesn't have quite the same gravitas, or gravity, of a big hunk o' metal.

Reply 26 of 71, by pentiumspeed

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These chinese dc to dc converters are trailing end technology and too low margins using under rated IC and most does not use true PWM controller and toroid and open bobbin inductors are not the best. Most are analog PWM designs. The well designed dc to dc and VRM used current designs that can handle stuff properly and many have closed inductor or few turns or band inductors to keep magnetic field contained and less noise.

Cheers,

Great Northern aka Canada.

Reply 27 of 71, by snufkin

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I think Murata are Japanese? My first bit was looking at that AS2880 which is an ancient (as things go) LDO, second bit was about the DC-DC converter which I thought looked quite nice, although I haven't closely read the datasheet so it might have some gotchas.

Reply 28 of 71, by majestyk

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Murata is a renowned Japanese company with a very good reputation when it comes th inductors and ceramic capacitors. They probably produce some / most components in China or other Asian countries today as most Japanese companies do. Panasonic electrolytics are made in Malaysia for example.
Personally I don´t care a bit as long as the quality is as expected.

The form factor would be perfect for my purpose so I`m planning to give it a try. Mouser is expecting some stocks for February (Farnell and Digikey demand a minimum order of 250 units).

If all else fails I can live with the heathump 😉

Edit: just ordered a couple ot this 20A version (33x14x9mm), and it´s Made In Japan:
https://www.mouser.de/datasheet/2/281/okl2-t2 … -w12-472031.pdf

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It would be nice to solder it on some small pcb with the necessary resistors and jumpers for let´s say 1.6 1.8 2.0 2.2 2.4 2.8 V

Reply 29 of 71, by snufkin

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Ah, the more power version. It's huge though. 3cm long!

I thought to myself whether a rotary switch would be nice, and there are a few that have binary like outputs, e.g. https://docs.rs-online.com/c640/0900766b816e2fc3.pdf . So I wondered if it was possible to make a resistor network that would give the correct voltages by selecting multiple different branches by the rotary switch. I couldn't see that it was obviously impossible, sketched out a 3 branch (so 7 positions + all off) circuit. Realised I only had 5 resistors for 7 equations, so probably not precisely solvable. Thought for a moment (very short moment) of trying to work out a way to get a reasonable approximation, then threw it at the LibreOffice solver.
Behold:https://www.falstad.com/circuit/circuitjs.htm … nvXdlOdz9X-0nEA

Then I thought, maybe rather than an off state, I could set that to a sensible value and get one more voltage level, so added a resistor that can't be switched out. That seemed to confuse the solver, so I started fiddling by hand. Then I looked for rotary type dip switches and found that 8 isn't common, but 16 is. So thought I could add a 4th branch as a quick fudge. Carried on playing and found that a couple of the resistors could actually be set to 0 without doing much harm. Eventually ended up here:
https://www.falstad.com/circuit/circuitjs.htm … xh-vvG-dFfbaWIA

That should give:

0 - 7.58k, 1.60V  |  8 - 4.58k, 2.20V
1 - 6.21k, 1.80V | 9 - 4.05k, 2.40V
2 - 5.26k, 2.00V | A - 3.62k, 2.60V
3 - 4.57k, 2.20V | B - 3.28k, 2.80V
4 - 3.79k, 2.51V | C - 2.86k, 3.11V
5 - 3.41k, 2.71V | D - 2.64k, 3.31V
6 - 3.11k, 2.91V | E - 2.45k, 3.51V
7 - 2.85k, 3.11V | F - 2.29k, 3.71V

There's a fair bit of overlapping between the first and second half, so there's probably a better arrangement, but I thought that was ok for 5 resistors (2 of the same value even). I might have another go at it aiming for smaller steps and less overlap. Precision of resistors would be important to the final voltage output.

Of course, a normal dip switch with resistors will be far easier. But where the fun in that?

[edit: I realise resistor value could be an issue, I went to 3 sig. fig. for them, so the ones given as 7.6k are actually 7.58k]

[edit: values for 1.8V - 3.3V in 0.1V steps, using EIA96: 6.19k (always on), 69.8k (1s), 34.8k (2s), 17.4k (4s), 8.66k (8s) Looks like for even steps you first set the '0' state which fixed the always on resistor. Then you set the '1' state, which is just two resistors in parallel, with the resistor connected to the 1 position on the switch. That resistor sets the step size. After that you just keep halving the resistor value for positions 2,4 and 8. Given the resistors are all in parallel, it's probably easier to think in conductances which add in parallel. So the always on resistor has a certain value of conductance, and then the first resistor adds a certain amount more (x), the next adds twice as much, then 4x, then 8x. And the Voltage out is linear with conductance, 6.9*S+.69]

[last one, 10.5k, 46.4k, 23.2k, 11.5k, 5.76k for 1.35V - 3.6V in 0.15V steps. Datasheet mentions a sense connection, but doesn't say anything about it. Tempting to make the starting resistor 11.5k to match the other one. That would make it roughly 1.3V to 3.55V in 0.15V steps. Allow for a bit of voltage drop, call it 1.25V -3.5V. Feels like a good range.]

[I wonder if a carrier PCB could get away with just the TO220 mounting pins. Ah, no, there's no direct ground on 3 pin LDOs. Pity. Those 5 pin regulator pinouts would work though. They've got Ground and the feedback voltage. Or could just go with ATX CPU 12V connector on the carrier? Could then still mount in place of a TO220. Maybe]

Reply 30 of 71, by majestyk

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This will be a perfect concept for the final layout for all common voltages!
Today I started with a first test. Just set the output voltage to 1,9V with 2 x 2K7 and connected 3 wires of the tiny thing to the mainboard. It´s so light-weight, it doesn´t even need any additional fixture.

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It´s working perfectly so far, I can´t feel the slightest heat dissipation. It´s go time for Heathunk...

Reply 31 of 71, by majestyk

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While at it, my good old Gigabyte GA-586DX got the same treatment for it´s 2 processors. I´ll always be running 2 x Pentium MMX 233 in it, so the voltage can be fixed. With a 3K3 resistor you get 2.78V at the output - perfect for these CPUs.
The old 7.5A linear regulators had roasted the PCB for years.

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This system is my oldest still active server, running NT 4.0 Server. It serves as a FTP-Server in my local network. 😀

Reply 32 of 71, by snufkin

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Regulate all the things? Power-on-Load is a fun thing to search on. I think these modules are good for drop in without modding the motherboard, what with the built in input and output capacitors, and accessible input/output connections.

I think if space is a constraint then there are some that come without input/output capacitors (effectively just the main control+FET chip with a big inductor glued on the back, e.g. https://www.farnell.com/datasheets/2820765.pdf , 6.4mm x 10.5mm, price/availability could be a problem) that could maybe be used without extra caps because the motherboard will almost certainly have an output capacitor for the original LDO. Might need a bit of motherboard modding like connecting Adjust pin to Ground and replacing any smoothing caps with very low ESR poly caps. That type would definitely need a carrier PCB to get the chip connections out. But it's pretty close to fitting within a TO220 footprint. Think there was a brief thread a few months ago about this sort of thing.

Reply 33 of 71, by majestyk

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It´s amazing how small components like this have become!

On the GA-586DX there´s an input-capacitor (330µF 25V) at the input of each regulator:

http://blog-imgs-15.fc2.com/d/u/a/dualsocketw … ld/DSC01617.jpg

For every CPU there are 4 x 330µF 25V Rubycon ZLH. The total ESR of 4 parallel capactors might get close to or even below 0,01 R so maybe we´re already exceeding the 1000µ limit (a little). We have 0.056 R for a single capacitor @ 100 KHz - so 0.014 R total ESR. Not sure what ESR applies @ 400 KHz...

Haven´t checked with the GA-586HX so far, since the regulator will be modyfied anyway.

Reply 34 of 71, by snufkin

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majestyk wrote on 2022-01-20, 21:19:

For every CPU there are 4 x 330µF 25V Rubycon ZLH. The total ESR of 4 parallel capactors might get close to or even below 0,01 R so maybe we´re already exceeding the 1000µ limit (a little). We have 0.056 R for a single capacitor @ 100 KHz - so 0.014 R total ESR. Not sure what ESR applies @ 400 KHz...

Hmm, true. I was assuming that any caps off the carrier PCB would be far enough away that trace inductance would push up the impedance. But maybe not enough. I don't have the skills or tools to model that. Looking at the technical notes at the end of the datasheet I think the ESR numbers are still based around 100kHz, even thought it's switching at 400kHz.

There's also a possible issue around ripple increasing at low load. Any chance you can get a scope and compare the output on the converter with the output measured as close to the CPU as possible and check what the actual ripple is?

Reply 35 of 71, by majestyk

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I did this today. The ripple signal has the same amplitude at both regulators. There´s no difference between the values measured directly at the regulator´s output and at the Vcore CPU pins.
Ripple at the regulator input (taken at the input filter cap) is a little lower - about 75%.

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The Y-scale is at 50 mV/cm so ripple is about 40 mVpp.

Timebase is set to 2 µS so the frequency is 1/([1.25 x 2]/1.000.000) = 400 KHz.

40 mV seem reasonable, the regulators are hooked to the 5V rail here btw.

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Reply 36 of 71, by snufkin

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majestyk wrote on 2022-01-21, 16:35:
I did this today. The ripple signal has the same amplitude at both regulators. There´s no difference between the values measured […]
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I did this today. The ripple signal has the same amplitude at both regulators. There´s no difference between the values measured directly at the regulator´s output and at the Vcore CPU pins.
Ripple at the regulator input )taken at the input filter cap) is a little lower - about 75%.

ripple_DX.JPG

The Y-scale is at 50 mV/cm so ripple is about 40 mVpp.

Timebase is set to 2 µS so the frequency is 1/([1.25 x 2]/1.000.000) = 400 KHz.

40 mV seem reasonable, the regulators are hooked to the 5V rail here.

ripple_datasheet.JPG

And that' s with a nominal 2.78V output? So that's less than +/-1% ripple? Well that sounds ok. Any obvious spikes going from no to high load (like starting and stopping prime95 or something)?

Reply 37 of 71, by majestyk

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Yes, it´s at 2.78V. When CPU load changes (both CPUs at 100% 2 x Memtest) , the ripple amplituide remains unchanged, only the small harmonic changes a little bit rhythmically.
No spikes, not even at startup. The ripple appears the second the system is turned on.

Reply 38 of 71, by snufkin

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What do you feel about components on the back or not? Keeping it all on one side means is can lay flat on a surface, but components on the back keeps it smaller. I'm leaning slightly toward having it stand upright with components on the back. Also gives a chance for some silkscreen text. With much less heat being generated, shouldn't need to heatsink to the motherboard PCB.

Reply 39 of 71, by Sphere478

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Is it possible to make a version of these that looks more like a oem part? Maybe a small pcb that has these ICs on it that fits where the old regulators went and bolts to the heatsink? (Even if it didn’t need the heatsink)

No offense, these look tad rigged. Loving the idea and results though, these could help achieve higher overclocks on many boards.🤷‍♂️Maybe.

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