It still looks like nothing gets back in to the battery except via D3. I think safe charging voltage is around 1.4V/cell (nominal voltage 1.2V/cell), so 4.2V for the battery. So maybe D3 was a ~1V zener diode, which would drop the board 5V down to 4V, with the 10 ohm R26 acting as a current limiter. When the battery reached 4V then current would stop.

But I think the better option here is to just use an external battery pack. I'm not happy about making guesses about how the charge circuit works, plus (and this is the more important point) I don't think the battery should be replaced. They leak and cause problems, so in 20 years either you or someone else will be cursing the person who fitted a battery to the board. D3 is apparently already blocking any current that would try to charge the battery, so there doesn't seem to be any risk of accidentally charging non-rechargeable batteries.

I don't know how a transistor works but I'm curious, why Q1 has 4.22V on the collector and 0.70V at the base and 0V at the emitter.
I only want to repair the board the best I can, like I said I soldered terminals to the battery soldering points, I'm not going to solder a battery there, but I'd like to see the battery charging. I know that I can use an external battery and even solder a CR2032 battery holder because the charging circuit is not working, but, is the feeling that the board has something wrong what annoys me.
Tomorrow I'll recheck the connections to Q1, I think I didn't do it right.

Hoping wrote on 2021-12-04, 22:46:

I don't know how a transistor works but I'm curious, why Q1 has 4.22V on the collector and 0.70V at the base and 0V at the emitter.

I think Q1 and Q2 are about connecting the board supply to the CMOS storage, and stuff powered by the battery when the board is off. I'm assuming that the nearest ends of both R22 and R23 go to the base of Q1, and that the far ends go to something like on the PSU Ground and +5. Q1 is an NPN transistor, so it conducts from Collector (pin 3) to Emitter (pin 1) when current flows in to the Base (pin 3). Not much current is going to flow since R22 and R23 are quite large (10k and 51k). So less than a mA. The gain of both transistors (how much more current flows from Collector to Emitter) is around 100. So the maximum current flowing from collector to emitter for Q1 is going to be a few mA.

Q1 base-to-emitter is effectively a diode, so the maximum voltage drop is 0.7V. Assuming that the emitter is connected to ground then the maximum voltage on Q1 Base is 0.7V, as you've measured.

Next look at Q2. That's a PNP transistor, so current flows from Emitter to Collector when current flows out of the base. In this case than we can assume from your measured voltages that the base of Q2 (pin 2) is connected to the Collector of Q1, and we know that Q1 will allow a few mA to pass. So a few mA are flowing out of the base of Q2. Similar to the NPN transistor, on the PNP then there's a diode from Emitter-to-Base, so the voltage drop will be about 0.7V (in this case a bit more, you've measured 0.8V). But the voltage isn't actually important here, it's more about what current is flowing. Q2 has the same sort of gain as Q1, about 100. So the few mA flowing out of the base will allow a few hundred mA to flow from the Emitter to the Collector.

Finally, the Collector of Q2 connects to the supply to the things powered by the battery when the board is off.

So, when a very small current (microamps) is supplied to the base of Q1, then Q2 is turned strongly on, and the board +5 supply is connected to the battery CMOS supply. When the board is off then there's no current flowing in to Q1, so Q2 can't conduct, so the CMOS supply is disconnected from the rest of the board.

I think. Never really got to grips with analog stuff. Could be interesting to check what the far ends of R22 and R23 connect to. I'm assuming they connect to something on the PSU connector.

ASCII diagram time:

1                                              +5V (5.09V)                            
2                                               |                                     
3                                              /E                                     
4                     |------------- 4.22V--B-|    Q2 (PNP)                           
5                    /C                        \C                                     
6R22&R23-- 0.7V---B-|    Q1 (NPN)               |                                     
7                    \E                       4.87V                                   
8                     |                 (CMOS supply,                                 
9                    GND (0V)            also goes through 10ohm R26, diode D3??, to battery)
10
11Emitter - Pin 1, Base - Pin 2, Collector - Pin 3

So a very small Q1 Base-Emitter current (Ibe) allows a larger current to flow through Q2 Emitter-Base current (Ieb) and then Q1 Collector-Emitter, which in turn allows a much larger Q2 Emitter-Collector current to flow, allowing board +5 supply to connect to CMOS without dropping much voltage. Of course these days you could use a mosfet with roughly 0 gate current and roughly 0 on resistance to achieve the same thing, without even dropping the 0.2V that are dropped here.

Still can't see how current gets through to the battery to charge it. And as far as I know there's no way to visually identify if D3 is supposed to be a zener diode or not.

[Actually, scratch this zener idea, they don't go down low enough. I was imagining something with a breakdown voltage of around 1V, and they only seem to go down to 1.8V, which would be too low. So I'm stumped for the moment]
[Slightly daft question, but have you checked that the -ve terminal of the battery goes to Ground? There's a bit of damage to the board around there and I'm wondering if the battery connections in to the motherboard have cracked]

Yes, the negative is connected to ground, it could have been the same as with the MIC, but no.

I've been reading things about the zener diodes and maybe D3 is something like this https://www.mouser.es/ProductDetail/Nexperia/ … jR4HAUWpw%3D%3D
a 3,6V zener diode, If I understood well what I have been reading, that should drop from 4,90v to 4.2v, maybe less, and the resistor before the diode would be R26, and R26 will only have to hold 0.70V.
Can this be the problem??, I'm not sure, but I'm going to continue to study it.

Quick explanation: normal diodes will only conduct passing current forward. If you try and push current the other way (reverse bias) then it won't conduct. If you keep increasing the voltage to try and force some current to flow then eventually the diode will just break. Zener diodes work the same way as normal diodes when passing current forward, but are a bit different for reverse bias. At low voltages they don't conduct, but as the voltage is increased, rather than just breaking they reach a certain voltage and then start conducting, but without breaking. The voltage won't rise any further as more current flows. That voltage is the zener voltage, and tells you what the voltage drop across the device will be.

So in this case we've got something a bit under 5V supplied to one side of D3 and the battery on the other side. We want the battery voltage to be around 4V when charged, so we want to drop a bit less than 1V across D3. A 3.6V zener (I think the one linked is a 4.3V zener) would reduce the 5V by 3.6V, leaving 1.4V on the battery side, so too much voltage is lost across D3. If I had been right about D3 being a zener (I'm pretty sure I was wrong) then it'd have to be something like a 0.8V zener. Which don't seem to exist. Mouser gives one device with a zener voltage of 1V, but I'm pretty sure that's a typo and it should be 10V.

You measurements for R22 and R23 make sense I think. The 51k (R22) goes to +5V and the 10k (R23) goes the GND. The point between them is then held at 0.7V because of the diode voltage across the Base-Emitter of Q1. Net result is that about 0.01mA flows in to the base of Q1, so about 1mA (gain of 100) flow out of the base of Q2, so about 100mA can flow through Q2.

So the problem is that we still need to find a way for current to flow in to the +ve side of the battery, and it can't come via D3. Couple of things to try. Does you meter have a current measuring mode? If you haven't already, with the board on try measuring any current flowing from +ve to -ve on the battery terminals. Also try disconnecting the board, put the black probe on the battery +ve terminal, then use the red probe to measure resistances and see if you can find anything that connects to the battery (with the battery source jumper on 2-3).

What's the red 0.637 reading?

It is clear that you have much more knowledge of electronics than me, in fact I only knew zener diodes by name.
But I think you are mixing the operation of a normal diode and that of a zener.
I understand that a zener in the opposite direction begins to conduct when it reaches the voltage for which it is rated, and only allows to pass up to that voltage. The normal diodes are the ones that drop voltage when conducting. That is, a 3.6 volt zener does not start to conduct up to a voltage higher than 3.6v and only allows 3.6v to pass.

https://theorycircuit.com/zener-diode-voltage … ulator-circuit/

That would mean that if D3 were a 3.6v zener. (1N4728A) it would only let 3.6v pass.
All the information that I have been reading and that I have found in electronic manuals about the zener make me think that unless, no matter how much I review it, I still misunderstand it.
The way to know that zener is D3, would be to apply the formulas, knowing the value of the resistance before the zener. But I'm very bad at math, it was going to be difficult for me.

With all due respect, don't be mad at me.
The 0.632 is the reading between the ve+ side of D3 and the odder points. And the 0.01v-0.01v is the voltage at the ve+ side of D3

Don't assume I'm any good at this, think I said that I'm pretty bad with analog stuff. I'm going to go on a bit here, but I'm trying to explain my thinking in case you or anyone can spot If I'm going wrong somewhere.

So, I think we both agree about the function of the zener diode: when reverse biased then a zener won't conduct any current until the threshold is reached, at which point it will start to conduct however much current is required to keep the voltage from rising any higher, until it overheats. So in your example, the 3.6V zener can have anything from 0V to 3.6V across it, but nothing higher.

In plumbing terms, it'd be a sort of pressure activated valve (a bit like a relief valve). Think of something like those ketchup sauce bottles with the rubber seal, where you have to squeeze the bottle quite hard before anything starts to come out, and then will suddenly stop and seal up when the pressure drops (doesn't quite work as an analogy because the pressure to start the flow is higher than the pressure to maintain the flow, but I don't think it's too far off).

That's what I mean by the zener dropping 3.6V. If the voltage/pressure across it is less then the threshold, then nothing goes through. Once the voltage/pressure is above the threshold then potentially lots can come through. But if there's some blockage/obstacle/resistance further along, then the pressure will start to build there. That decreases the pressure difference/voltage across the zener. If the pressure across the zener drops too far then it will stop conducting.

Now, I think where things get a bit different is exactly how the zener fits in to the circuit. In your example the battery provides 9V: that's the pressure. The zener threshold is 5.1V. So without the resistor then lots of current would flow through the zener and it would break. So there's the 330ohm resistor. Current flowing through the resistor will lose voltage. What will happen is that enough current will flow through the resistor until the resistor output is at 5.1V. If any more current flowed then the output would drop and the zener would stop conducting. If any less current flows then the voltage across the zener would try to rise, which would mean more current would flow through the zener. So that means there's a fixed 5.1V across the zener, which can be used as a 5.1V supply to other things (up to about 15mA).

In that example the zener is in parallel with the load (the load goes either side of the zener). But on your motherboard, the zener is in series. I've tried to sketch it here:

On the right you can see the parallel version, similar to your example, but showing the battery to be charged. Voltage and current (Power) come from the top and through a 10ohm resistor, and there's then a zener in parallel with the battery. When the battery is empty then it will have 0V across it, so current will flow through it, but not the zener. As the battery charges up then the voltage will increase. Eventually the voltage will reach the zener threshold, at which point current will flow through the zener instead, and the battery will stop charging.

But that's not how the motherboard is wired up. It's connected like the left hand side, in series. Assume the motherboard supplies 5V and the zener threshold is 3.6V. Power comes from the motherboard, through the resistor, through the zener and then through the battery to ground. If we assume the battery is empty then it's at 0V on both sides. If we assume that there's no current flowing then there's no voltage drop across the resistor. That means all of the voltage will be across the zener. The motherboard supplies 5V, which is above the 3.6V threshold, so current will flow. As more current flows, the voltage across the resistor will increase, so there will be less voltage across the zener. Too much current will mean too much voltage is lost across the resistor, which might reduce the voltage across the zener below the threshold, and conduction would stop. So what will happen is there will be a steady state of just enough current flowing to keep the voltage across the zener at 3.6V.

But, over time the battery will charge up, and the voltage on the +ve side of the battery will increase, which increases the low side of the zener. To keep conduction happening then the high side must also increase. Which means the voltage across the resistor must decrease. Which means less current flows and the charge rate slows down. Eventually (after infinite time...) the battery will be charged up so that the voltage across it is 1.4V. The zener will have 3.6V across it. Which leaves 0V across the resistor, so no current will flow through the resistor. If the battery could charge any more then the voltage across the zener would drop beneath 3.6V and it would stop conducting. So no current flows and the battery stops charging.

For the battery to charge any higher then the zener needs a lower threshold. If the zener had a threshold of, say, 1.4V, then the battery could charge to 3.6V. But the lowest common zener threshold voltage I can see is 1.8V. Which leaves just 3.2V across the battery, which isn't enough.

So I think we're missing something. That diode D3 is almost certainly not a zener diode. And yet it appears to be the only way for current to flow in or out of the battery. But to charge the battery that means current has to flow backward through the diode, which isn't possible unless it's a zener diode, in which case it needs a threshold of at most 1.4V, which don't seem to exist.

First, thank you very much for the explanations, and I think I have fully understood the explanation after reading it several times.
I have had much worse teachers in my life.
I keep looking for another point where the + ve is connected, but I can't find it, so then your theory is overwhelming and irrefutable unless it was a 1.8v zener, as you say, and they didn't try to fully charge the battery to preserve his life. Furthermore, I do not remember that the battery was damaged, and I do not see traces of a possible battery loss, however, I remember a 386 motherboard that someone gave me at that time and the battery was damaged and had lost liquid.
It may also be that the motherboard did not charge the battery at all, however I would swear it had a typical green NICD battery, and nothing seems to indicate that it had anything else.
In the end, in the absence of any opinion from other members, I can't think of anything to object to your conclusion and I guess I'll end up soldering a socket for a CR2032, with all the pain in the world knowing that something's wrong. Although if I understand the explanation well, if it were the already repeated zener diode, this would also act as a control so as not to overcharge the battery, and would charge it when it was low in voltage, less than 3,2V.
One final question, what harm could it do to try to install a 1.8v diode in that place?, a CR2032 is 3v, so not far from 3,2v.
I'm very curious, after all, I'm learning a lot with this, but obviously I don't want to damage anything.
I am very grateful for your help, and your explanations. I hope I'm not being too stubborn about it.

Hoping wrote on 2021-12-06, 22:01:

I hope I'm not being too stubborn about it.

Oh, it's bugging me too. There must have been some way for that battery to be charged to around 4V, I just can't see how. There doesn't seem to be any way for current to leave the battery without passing through that diode in the forward direction, so there'll be a voltage drop that way. If the battery was charged to 3.2V (1.8V zener) then that would leave less than 3V to run the real time clock oscillator and maintain the CMOS settings, which wouldn't be enough. So the battery has to be charged to a higher voltage.

That said, it is possible (but I think unlikely) that the battery didn't charge. There are 3 diode locations marked, of which one's been replaced with a 0 ohm link and the other isn't fitted. I've no idea what Chicony's reputation was, but maybe they decided to save a few pennies on the parts cost and leave the battery uncharged. Looking at some picture of other Chicony motherboards then they do seem to have more diode around the battery.

A CR2302 voltage might be too low, after the voltage drop going forward through D3. I think the usual approach is to connect a 3x AA battery holder, which will provide 4.5V (so a bit under 4V after D3) and should last for a few years without being charged, and the battery holder can be placed away from the motherboard so if batteries leak then there's no risk to the board.

I know there's a general ongoing debate about how much to stick to original equipment, or how much to keep the core parts original and make the peripherals reliable (e.g. stick with spinning hard drives or replace with various solid state storage). I tend toward having original hardware, but when the original hardware causes a risk of physical damage (e.g. battery leaks) then there should be an exception.

Although I ended up using an external battery, I would like to know that the internal battery charge works, which I will not use surely, but still.
Ok, I found two resistors cut by the legs, R4 and R3 near the cristal between the memory and the first ISA slot.
This validates the theory that they wanted to cut costs, or had design problems.
So, following what I learned from your explanations about the zener, and the fact that what the position of D2 is bridged, I decided to remove the bridge and now I find a possible solution, to my little understanding of course.
Following what I understood from your zener diode classes.
D1, it could have been a 7.5v zener since on one side it receives 11.48v coming from R47, (I don't see that R47 goes to another place), which in turn receives 12v directly from the AT connector and, the other side of D1 goes directly to The battery, but it also goes to one side of D3 and it would pass through it after reducing the voltage, this is where I don't understand, what could happen when D3 receives voltage from two sources on one side, what voltage would there be on that side of D3 ??. This part is the one that I have less clear.
On the other hand D2 is connected on one side to the positive of the external battery and on the other to D3, this could be to prevent the external battery from charging.
It would be necessary to know the approximate values ​​of D1 and D2.
What do you think?

So, my guess as to how it should have worked is below. If you want to try this, make sure the battery is on some long leads and away from anything flamable...

D2 should be a diode to stop an external battery receiving any current and being charged up. The current 0 ohm link ends up connecting pins 1&3 on the battery header; a diode at D2 would mean any current entering the board battery via a jumper on 2&3 couldn't also get to pin 1. That would just be an ordinary diode.

I think a 7.5V zener at D1 would work to charge the battery, except it'd have to be backward from the board markings. That way, current from the 12V board supply would go through the 1.2k R47, then enter the zener at D1 nearest the board edge, lose 7.5V, and then be able to enter the battery. With an empty battery the voltage across R47 would be 4.5V (12V, minus the 7.5V across the zener), so the current flowing would be 4.5V/1.2k = 3.75mA. If the battery capacity was 20mAh, then that's initially charging at an absolute maximum of ~0.2C (3.75/20), which is too high for what should be a trickle charge. But the current wouldn't actually ever be that high. As the battery charged up then the voltage across it would rise, so the voltage across R47 would drop, so the current would drop, so the battery charge rate would slow down. Eventually (after infinite time) the battery would reach 4.5V. Actually it would reach whatever the PSU 12V supply was, less the zener voltage. So if the 12V supply was 5% high (12.6) then the battery would (extremely slowly) end up with 12.6-7.5=5.1V across it. Which would be bad (1.7V/cell).

But there's a snag when the board is off. Current can flow both ways through a zener diode. They're normally used reversed biased so that they can be used to control voltage. But current will also flow through them when forward biased, just with a different (lower) voltage drop. When the board is off the the 12V supply will drift down to ground. So now the battery voltage will push current out through both D3 (for the oscillator and CMOS stuff), but also through the zener at D1, and then on through R47 and on to the inactive 12V supply, and eventually down to ground. That current would probably drain the battery pretty quick. Let's assume the battery is at 4.5V and the forward voltage drop over the zener is 0.9V (to make the numbers easy). That would leave 3.6V over the 1.2k R47, so up to 3mA flowing out of the battery. Even if the battery were 80mAh, that would drain the battery in a day. So somewhere there would need to be another diode that would allow current to come in to the battery via D1, but would stop current flowing out of the battery and through D1.

With a blocking diode in series with the zener diode then that would mean when the board is off then current can only leave the battery through D3, out to the oscillator and CMOS. If the board is on then the battery would very slowly be charged up to around 4.5V (depending on the PSU 12V level), but no higher. So D3 would have 4.5V on the battery side, and 4.9V on the board side. That means that current would try to flow from the board to the battery (current always tries to move from higher voltage to lower voltage), but would be stopped by D3. So the battery and charge voltage wouldn't affect the board 5V supply.

It's all a bit odd really. It looks like they nearly had a working charge circuit, but forgot they had to drop the 12V voltage, and then just stuck in components to let both the board and external battery run the CMOS stuff, but not get charged up at all. R47 is there, which limits the current rate, which is good. D1 (if a normal diode, fitted as marked) would stop the battery draining out through the inactive 12V when the computer was off, which is good. D2 stops the external pack charging up, which is good. D3 makes sure that nothing on the battery side interferes with the PSU 5V supply when the computer is on, which is good. The only snag if components are fitted as marked is that they end up putting 12V across the battery, which is bad.

So. I think it might work by:

1. Fitting a diode in place of D2. It looks like they've used small signal diodes elsewhere, possibly 1N4148. It's all (or should be) low current and power, so they should be ok.
2. Fit the Anode of a 1N4148 to the D1 pad nearest the board edge. 12V will enter there and drop about 0.5V.
3. Fit the Anode of a 7.5V zener (1N5236 maybe?) to the other D1 pad.
4. Connect the cathodes together.

The charge voltage should end up a bit less than 4.5V because of the extra blocking diode at D1, but that's no bad thing for a test. But... I don't really know and don't have a board to test this on. I can't see how it would break anything, but it's not my board. If your PSU 12V ended up on the high side then the battery may go over voltage, so at the very least (as I said before typing all this) make sure the battery is well clear of anything.

Or, and I've only just thought of this, but one of your example zener circuits shows using the zener in parallel. So maybe fit stand 1N4148 diode as marked for D1 and D2, and then put a 4.3V zener across the battery. That way the battery would reach 4.3V, and then any extra current would go through the zener diode. It would mean a small amount of current always flowing when the computer was on, but it would only be a few mA.

The first solution looks great, apart from the lack of protection for an overcharging battery, thanks for mentioning the worst case scenario, prepare for the worst and aim for the good.
The second solution, it seems simple and elegant, since I already soldered terminals to the solder points of the battery, it is easy to solder a diode on the back of the board in parallel with the battery.
I don't know what to do, both solutions look good, maybe flip a coin.
Thank you very much, now the board would be fully functional.
And I have learned a few things from this, and your explanations are very good, even for someone with such a low level of electronics as me.

Hoping wrote on 2021-12-08, 16:44:

The first solution looks great, apart from the lack of protection for an overcharging battery, thanks for mentioning the worst c […]
Show full quote

The first solution looks great, apart from the lack of protection for an overcharging battery, thanks for mentioning the worst case scenario, prepare for the worst and aim for the good.
The second solution, it seems simple and elegant, since I already soldered terminals to the solder points of the battery, it is easy to solder a diode on the back of the board in parallel with the battery.
I don't know what to do, both solutions look good, maybe flip a coin.
Thank you very much, now the board would be fully functional.
And I have learned a few things from this, and your explanations are very good, even for someone with such a low level of electronics as me.

If you do do this, might be better to go with putting the zener in parallel with the battery. It does mean the battery will only charge to 4.3V, so it won't be at full capacity, but at least the charge voltage won't be affected by changes in the PSU supply. Charge rate for a completely empty battery would be around 10mA (12V/1.2k), which for an 80mAh battery would be a charge rate of .13C. That should be ok since it'll stop charging once it reaches 4.3V, by which time the charge current will have dropped to about 6.5mA ( (12V-4.3V)/1.2k).

Note that this isn't a good charge circuit (this isn't how NiMH should be charged), but I don't think it's actually dangerous. Before attaching the battery use your meter to check the voltage across the zener (should be 4.3V) and then measure the current when placing the probes where the battery will go (should be around 10mA). Keep checking the battery temperature and voltage for the first few days. Don't fit the battery to the board until you're sure it's not overheating or leaking, although, personally, I wouldn't attach the battery to the board at all. You'll need to experiment a bit to find out how long the battery will keep the CMOS settings intact.

I have already prepared a battery with long cables and connectors, I plan to glue it to the bottom of the box with double-sided tape por hot glue. I also have a 1N4731A zener diode on the way and the other two diodes already on the way, so this should work. Now it's time to start with another project similar to this one but with a different approach.
Thanks a lot for the help again.

Hello, I'm sorry to revive the thread, but I have acquired a Chichony badge, and when I remove the 486DX2 at 66 MHz and replace it with the 486DX4 at 100 MHz, it boots up but gets stuck after reading the memory. When I try to enter the BIOS, it freezes and won't go in; this doesn't happen with the DX2. After reading all your comments, nobody mentions how it reacts when it doesn't get its 3.3 voltage. Does it freeze? Black screen? Speaker beeps? Does it get damaged? Thanks, I love this forum. Regards.

Did you change the jumpers to match the new CPU?

I had the DX4 100 running at 5v for a while and I didn't realize it was 3.3v because it had a heatsink on it and at that time I didn't have much experience with that platform and also, I have to say that it didn't give any problems to 5V.
On the first page I already mentioned having used the processor at 5V and even having done overclocking setting the FSB to 40mhz instead of 33mhz.
Currently, I don't do this modification the way I did it, I use another simpler and more efficient method.

devius wrote on 2023-07-28, 12:45:

Did you change the jumpers to match the new CPU?

but it crashes in bios when counting ram memory

My motherboard is a REV1, maybe I have to update the bios? In the manual it says how to put the jumpers... I don't understand anything.