I'd love to see an answer to this, this mobo is almost the same as the 486SLE I have ... at least the 3.3v missing parts and area as identical... 😀

Robin4 wrote:

So is it correct that the voltage regulator is always an 5 volt type on an 486 motherboard??

Very unlikely. If the component in question really is a regulator, it would be an adjustable regulator, with jumpers configuring the resistors for correct voltage. You can't adjust a fixed 5V output regulator to any lower voltage, but you can cheat it a bit to get higher output voltage.

And with a linear regulator, it would be very unlikely that it would use 12V input to make 3.3V or 4V or 5V, as it would not require a low drop-out model and it would waste a huge amount of power as heat. So most likely 5V is used for regulator input and 3.3V or 4V or whatever out, and when only 5V is required there just is a jumper.

So i guess i need an transistor instead?? (dont know if it would be an mosfet?)

On the other hand i thought that the lower voltage would be possible by a set of resistors.. So when you set the right jumper you just set the resistor configuration so it lowers the voltage output on the transisor/mosfet)

I measured some pins on the motherboard.. With the jumpers on i measure 5 volts (4.98 or so) on the shimmed jumpers)

Hardwired bridge on JP6 = 3.11 volts.
Hardwired bridge on JP8 = 4.98 volts.

The highest pin of TR1 = 4.98 volts (when shunts connected for 5 volts setting)
The middle pin is also 4.98volts (when shunts connected for 5 volt setting
The lowest pin is 3.11 volts (when shunts connected for 5 volt setting)

All plusses of the capacitors i measure also 4.98 volts.. On the negative side i cant get any read out.

Now remove the shunts so i would be set as 4 volt i guess (because the hardwire is bridged on the right side)

( i think it would be easier to first remove that bridge and soldered an single row 3 pins jumper header in, and see what it does) Later i will do if i have me desolder equipment.

I removed the stunts, so the board is now set for 3.3volts..

I dont get any current read-outs from JP6 anymore (i guess this is because not have an voltage regulator)
Hardwired bridge on JP8 = 4.98 (still again)

The highest pin of TR1 = 4.98 volts (when shunts removed for 3.3 volts setting)
The middle pin is also 0 volts (no read out) (when shunts removed for 3.3 volt setting
The lowest pin is 0 volts (no read out (when shunts removed for 3.3 volt setting)

Conclusion for now:

The board uses an 5 volt source (no higher voltage then that)
The pins for the setting of the shunts: The lower pins seems to been 5 volt positive), the upper are the negative

I think that the upper pin on TR1 is our V-in pin for the voltage regulator.. So the middle pin could be the V-out pin (because of thicker electronic trace) I red that the how lower the voltage how more current (or amps) there is) And the lower pin could be the ground or V adjust pin (i guess the last)

I think the following now:

JP6 -> need to remove the hardwire bridge and solder an single row 3 pins header in (you need that to select 4volt as well for cyrix processors)
JP8 -> only need to remove the hardwire bridge, and nothing more. (my SLD / SLP with voltage regulator doesnt have one, so i guess it wouldnt need anymore)

Remove that black jumper shunt, and solder an LT1085 voltage regulator in..

http://cds.linear.com/docs/en/datasheet/108345fg.pdf

(Vin-Vout) = 5V (the board supply 4.98v = 5 volt current) MIN = 3.2 volt TYP = 4.0 MAX= no value.

So i guess JP6 is needed to set it from 3.2v (i guess actually 3.28volt = 3.3volt) to 4.0volt.

so you have actually found the missing part? an LT1085?
I just checked 2 486 mobos from my dead pile, one has a 1085CT next to to the voltage pins (sis based) and the other is hardwired to 5v with a missing spot that is labeled "LT1085CT" 😈

I havent tested it yet.. Need a desolder station so i can do it whithout damage things, and could do it quick..
If its installed, then need to measure again, if the voltages are you right to have.. If it is, it should be working.

I think you could use others as well, but i guess, the LT1085 is the better options, and it more power efficient.. Because its an LDO one.

^ who needs a desoldering station for this? wick is your friend! 😀

OK, you have some good info there. Let's do some mental work, bear with me and keep LT1085 datasheet open to see if you agree or not.

The missing part in the picture looks like a place for a part TO-220 case. LT1085 is available in TO-220 case.

You measured 4.98 volts on topmost pin, and that is pin number 3 in datasheet, labeled VIN. So I think you guessed right.
You also measured 4.98 volts on middle pin, and that is pin number 2, labeled VOUT. When the regulator is missing, there must be a jumper to short VIN to VOUT so CPU gets 5V. I think you guessed right again. And the middle pin is also connected to the TO-220 case metal tab which rests against the bare shiny metal plane on circuit board, so can you verify the shiny metal plane is either disconnected from everything or connected to pin 2 ? If this is not the case, then you need to have some electrical isolation there so that the tab does not touch the plane, and you even have to isolate the screw and nutyou are going to use to bolt the regulator to motherboard. I believe you can find suitable parts from inside an old broken PC power supply perhaps if you don't want to buy new ones.

This leaves the bottom pin as pin 1, labeled VADJ, which adjusts the voltage.

Now, if for a moment we analyze how voltages read when regulator would be there, making 3.3V to CPU. So the regulator needs resistors to set the output voltage so that there is 1.25V between VOUT and VADJ pins. VADJ=VOUT-1.25V. As VOUT is 3.3, VADJ=3.3V-1.25V. You would measure 2.05V at VADJ pin (in reference to ground).

But since the regulator is omitted, and VOUT is also 4.98V instead of 3.3V, the VADJ voltage should be proportionally higher as well as the resistors are the same. VADJ without regulator and 4.98V on output would be 2.05*4.98/3.3=3.09V.

Now, as you measured 3.11V which differs less than 1% I will have to assume the adjustment resistors are there and would correctly adjust to 3.3V on output.

So far everything seems like you could drop in a LT1085 there, disconnect the jumper between VIN and VOUT and it would work.
I have no idea about 4V CPU support, but it needs the regulator and some jumper to select resistors differently so you get approximately 2.75V on VADJ pin.

Please note that your multimeter may read 0V because either the wires are shorted together or they are completely open so there is no connection and nothing to measure. You need to measure with the resistance mode to see what is really connected to where.

And in this case, you need an LDO because it can make 3.3V from 5V, while a non-LDO regulator can not, it needs more than 5V to make 3.3. It has nothing to do with efficiency here!

well, I have a very similar board, a spare LT1085 , a multimeter, wick, pins, a 3.3v cpu and a sense of adventure... 🤣
I might try that tomorrow evening, restore the pins + LT and measure the cpu voltage... what would be a good point to measure what goes in the 486cpu other than the output from the 1085?

keropi wrote:

^ who needs a desoldering station for this? wick is your friend! :) […]
Show full quote

^ who needs a desoldering station for this? wick is your friend! 😀

Iam just very careful with my old motherboard, dont want to wreck them.
Yes wick is your friend, but it would still be a hassle to get the solder out of the holes.
I know there is also an tin sucker pump, but that isnt the best method to work with.. You wont want to heat te board up to much.. With solder wick you will heat it up..Because you need to heat the wick up so it can suck the solder out.. For hole trough components is much easier with an desolder station, because the suck pump is already on the solder iron, it will much be more efficient. An hand solder sucker pump doesnt have the right angle, so you need to heat it more, on more chances.. An de-solder heats it up, and sucks the whole solder tin out of the holes..

I had this LT1085 regulator few weeks ago also mounted on one of my other board.. In the PCB reads; LT1085.. So i knew there need to be mounting an LT1085 there.. Iam also measured that board too and the whole regulator.. And yes there is current on the outer tab.. I hope i also can solve the problems with my other boards i have.. It seems the regulator circuit could be a little bit different.. But i guess there is always an capacitor on the V-in line, and one on the V-out line.. and also an tantalum capacitor.. ( i think that last one is for fine filtering of the current)

Jepael wrote:

. […]
Show full quote

.

This leaves the bottom pin as pin 1, labeled VADJ, which adjusts the voltage.

Now, if for a moment we analyze how voltages read when regulator would be there, making 3.3V to CPU. So the regulator needs resistors to set the output voltage so that there is 1.25V between VOUT and VADJ pins. VADJ=VOUT-1.25V. As VOUT is 3.3, VADJ=3.3V-1.25V. You would measure 2.05V at VADJ pin (in reference to ground).

But since the regulator is omitted, and VOUT is also 4.98V instead of 3.3V, the VADJ voltage should be proportionally higher as well as the resistors are the same. VADJ without regulator and 4.98V on output would be 2.05*4.98/3.3=3.09V.

Now, as you measured 3.11V which differs less than 1% I will have to assume the adjustment resistors are there and would correctly adjust to 3.3V on output.

So far everything seems like you could drop in a LT1085 there, disconnect the jumper between VIN and VOUT and it would work.
I have no idea about 4V CPU support, but it needs the regulator and some jumper to select resistors differently so you get approximately 2.75V on VADJ pin.

Can you explain me how to calculate the VADJ voltage?? I really dont know how to calculate it, and how that works..

About 1.25volt Vadj for 3.3volt could be absolutly right!.. I measured also a third board i have where the regulators is on the board, but that was only a 4 pin design. But i really like to know where you get that Vadj voltage value from?? Can you explain by an calculation??

If there wasnt the smd resistors on the board how can you calculate that resistor(s) iam missing..

Can you comfirm that does regulator are all have fixed voltage setting, or does it differ?

@Robin4

even if you manage to heat the mobo so much , there is nothing important nearby... not even an IC to damage 😉 if you use some extra paste as wee then you are all set! just my 2 cents...

I'm gonna add that 1085 tomorrow and see what happens...

Robin4 wrote:

Can you explain me how to calculate the VADJ voltage?? I really dont know how to calculate it, and how that works.. […]
Show full quote

Can you explain me how to calculate the VADJ voltage?? I really dont know how to calculate it, and how that works..

About 1.25volt Vadj for 3.3volt could be absolutly right!.. I measured also a third board i have where the regulators is on the board, but that was only a 4 pin design. But i really like to know where you get that Vadj voltage value from?? Can you explain by an calculation??

If there wasnt the smd resistors on the board how can you calculate that resistor(s) iam missing..

Can you comfirm that does regulator are all have fixed voltage setting, or does it differ?

It's all in the LT1085 datasheet, it will take you less time to read it than me to write here. Page 11 has a picture, formula and description. Page 15 has a typical adjustable output circuit so just select fixed resistances instead of variable pot.

Totally different regulators could have totally different output setting mechanisms, also found in their datasheets.

Ok, i though you had calculated that value.. But every information is just in de data sheet.. funny.

OK , I just added the missing parts: the LT1085 , the tantalum cap and proper pins (btw that thing soldered on the TR1 spot is just sockets, there is no connection between them, one can just insert there an LT1085 to test) :

I powered up the mobo without a CPU , used the following image for the jumper settings that Robin4 provided in another thread and got these readings from the LT1085:

5v setting:
Vin = 5.10v
Vout = 5.10v
Adj = 3.17v

3v setting:
Vin = 5.10v
Vout = 1.245v
Adj = 0v

4v setting:
Vin = 5.10v
Vout = 3.30v
Adj = 2.05v

Apparently my 486SLE mobo is very similar but not the same as the 4SLE200 we used the jumper settings from. There are other differences in jumper locations and there is no silkscreening on the 486SLE to indicate any kind of 4v support. The way I see it I have a 5v and a 3.3v setting.
Now . before I go plug a CPU on it to test 3.3v , anyone knows what pin on the socket3 is the cpu voltage? I want to measure it before I add a cpu, just to be safer 🤣

edit: I went ahead and tested an AMD486/100 cpu with the 4v setting above, it works , it gets detected as an "486DX2 100mhz" and using a small sized heatsink on it (no fan) it doesn't get much warm... but I do need some sort of BIOS upgrade IMHO

also what about TC29 and TC30 ? do you think they need populating as well? both their 4 points are connected to the LT1085 through some capacitor... I assume they are extra filters, certainly a good thing? 😕

R52 and R53 seem like voltage setting resistors. It is a bit blurry but I think the other reads "121" and other reads "201", meaning 120 ohms and 200 ohms, in theory providing 3.33V VOUT. No other resistors nearby. And it seems the other jumper position sets the VADJ pin to ground, so VOUT is 1.25 V.

It means, if the jumper has VADJ in the middle, one pin going to 200 ohms resistor and third pin going to ground, you can select a suitable resistor between VADJ and ground to set 4.0V output.

Just a warning, if there is ever a loose connection on that jumper, it will let VADJ pin float and nearly 5V can get to the CPU, frying it. Don't remove it while a non-5V CPU is socketed.

Does your board have the option on `JP8` left near your cpu socket?? Mine have an hardwire on that JP8 section.. ( on the lower two pins of that connection) the upper single pad is just open) Maybe JP8 is to enable the resistor line to adjust the VR..

TC29 and TC30 has named TC4 and TC3 on mine two board.. But they are also empty on does two boards.. So i think its normal you missing them..

That your board displays DX2 100Mhz is because:

1. The cpu configuration on your board doesnt match the processor or:
2. Bios doesnt support the processor officially, so you need one that does support it..

I also get one SLE board here soon.. I saw it had a newer bios.. So i can help you on that one.. I can recommend you to use an other eprom cmos chip, so you always can fall back to the old one.

And nice work on the soldering skills, looks like is standard.

The voltage might be low because there is no current flowing?

You could test the voltage again while the machine is running.

thanks for the replies 😀

@Jepael

I left my workplace now so I can't have a better look at the mobo anymore... what jumper are you referring with
"It means, if the jumper has VADJ in the middle, one pin going to 200 ohms resistor and third pin going to ground, you can select a suitable resistor between VADJ and ground to set 4.0V output." , do you mean to try a combo on the JP1/JP2/JP4 block?

@robin4
will check the mobo tomorrow, can't remember about JP8 at all...
About the CPU config, it is silkscreened on the edge that the DX2/DX4 share the same exact jumper settings , here is a pic:

Any idea what M7 and AMD-SL are for? And is P24D the 5x86?

I think I just need a BIOS update, hopefully your upcoming mobo will provide that 😊
The soldering is a quick one atm, just wanted to test things... I plan to bend the pins on the LT1085 and screw (or solder it) on the ground plane it has behind to act as a heatsink... for now I have a small fan blowing air at it 🤣 🤣

@Mau1wurf1977

Voltage measuring was done with the machine running without a CPU... so it's pretty much accurate AFAIK

M7 is a Cyrix kind processor.. -> http://www.chipdb.org/data/media/1389/DSC02257.jpg
I guess the AMD-SL processor is an AMD 486 processor but in an different package (correct if iam wrong) (like this: http://www.google.nl/url?sa=i&rct=j&q=&esrc=s … 388982081209754 )
P24D = 486 DX4 VS8B -> Write Back version of the DX4 processor..
P24T = Overdrive processor.

For 5x86 you need to config it as P24D.. System bus speed to 33Mhz so the total would be 66mhz.. Then you also need to set the DX4 multiplier to 2x..then its 66x2= 132Mhz, but sees as 133Mhz.. There is only one downside.. Your bios need to regonize the processor system code the let the cpu show up as 5x86 - P75 -133 mhz.