Mac Portable, Dead Battery, Compatible AC Adapters

[techknight]

Circuits only draw the required amperage from a power source. Once a circuit starts to fail, more current can be drawn in a quick surge or latch-up condition, which can lead to a cascading failure if the supply was able to deliver the current. From a service standpoint, lower wattages/current availability is preferred because if the supply cant push out the amount of current the "surge" is requesting, FAR less components are damaged.

Maybe i should have been more clear. I have a tendency not to be clear sometimes.

But all this babble is moot if your circuitry is operating perfectly. you can hook up a 7.5v at 400amps if you wanted to. it doesnt matter as the machine is only going to draw a certain amount as needed. Its when a failure begins to start, is when that current matters. so if a transistor shorts or a cap starts to short, and your supply can push 10 amps, it can cause some serious damage and cascading failure as the current availability is there.

think of it like a bomb or something, if the detonator false triggers, and its yeild is only 1 ton(available power), itll do far less damage than say if it had a 10 megaton supply. (maybe a bad example). So basically if your power supply is current limited, and a transistor shorts, it might get warm or something but thats it. If you had a crapload of available current, consumption would drastically increase and cause the motherboard or other failed hardware to explode/cause a fire.

But Wattage is Voltage x Current. But i am thinking of voltage as a steady supply yes. only variable current. as current increases, wattage increases, same way with voltage. But im thinking of a rock steady voltage in my explanations. But of course theres that whole factor if your power supply is only 2 amps, and a short is pulling a 4 amp load, your voltage will drop significantly preventing further damage because its a lower wattage power supply, than say, one that provides 4 or 6 amps.

Well anyway, I am a repair technician by trade so its kinda my train of thinking on that subject. Even though it may mean nothing by what others were saying. But hey....

 

[JDW]

...all this babble is moot if your circuitry is operating perfectly. you can hook up a 7.5v at 400amps if you wanted to.

Agreed. And that's why I want to know why there is all this talk, all over the web, about users needing to exercise caution about using AC Adapters that can supply much more than 2A. It makes no sense to me because I cannot imagine that every single Mac Portable out there has failing circuitry!

Now ponder this discussion, which warns about AC Adapters and talks about burned SMD resistors in the Portable's board:

http://www.applefritter.com/node/9864

(Note my own posts there back in 2006. I've been wondering about the technical details for YEARS! Still no clear-cut answers are forthcoming.)

What are your thoughts on this?

 

[techknight]

No, All I was doing was presenting a scenario where a higher wattage/current power supply would be a bad thing in the event of a circuit failure. thats it. nothing more.

if the circuits are perfectly fine, there is no issues that I can speak of. Now of course there could be issues in the charging circuitry that a higher current power supply will allow a lead acid battery to charge faster, IF and only IF the battery is demanding for it. but if not, it dont matter.

Edit: As far as the SMD resistors, id like to see or know where and what resistors. If i knew what part of the circuit they were in, i could explain why. hehe. SMD resistors burn up because the amount of current/voltage flowing through them has increased or other factors leading to the wattage the resistor can dissipate to be exceeded. But that is just a general "fill in the blank here" statement that means nothing, but at least i said it as it is true. I could be more specific if i knew what circuit.

 

[techknight]

Now i will say that if the power supply circuitry on the logic board is designed with components that are only meant to have a certain amperage flowing through it at any given time, and you allow more amperage to flow through it than rated, it will burn components up too. Again, the cascading failure scenario.

I havent see the schematic of a portable (if one exists) but if the battery is used to run the laptop at least partially before the AC adapter takes over, that means there are 2 power sources. If you remove the battery, that causes the entire load on the DC jack supply circuitry. if there are regulators in between, that means they have a heavier load. They burn up if they cant handle it.

So for example if the DC jack circuit is now pulling 1.78 amps becuase of the battery missing, when normally it would pull 1.25Amps, it can cause problems if the circuitry is designed "cheaply"

For example most 0805 SMD resistors can only hold a max of 1/8watt, and nowadays they are making 1/4watt SMD 0805 resistors, so that statement isnt true anymore at least today...

But say if they had 1/8 watt resistors and maybe there is a 82% power dissipation across those resistors under NORMAL circumstances, there is not a problem. Remove the battery and use a higher current power supply to run the unit, you just changed the entire power configuration in a way its not designed to run, it can push those resistors from say 82% to 130% power dissipation. What happens? they burn up over time. The more you push it past 100% the faster they burn up. (hotter they get).

as far as the macintosh portable and this being the cause, thats all still debatable and means nothing because i dunno what resistors burned up or which part of the circuit they are in.

But since I am going in blindfolded, I feel this: The macintosh portable battery is required becuase it can handle surge currents that the DC circuit supply cannot. So even though the power adapter charges the battery AND runs the machine, it actually has to be at a specific charge before the machine will run. because when it is running, the DC circuit will run the machine, but any peak surge currents are handled by the battery, once the load/demand goes down, it returns to charging and running from DC circuit/power adapter.

if the battery is missing/dead/gone and you force the DC circuit to run the entire machine, the surge currents arnt handled by the battery anymore. they are now handled by the power adapter/DC circuit. So guess what, certain circuitry can exceed thier ratings just by this alone IF the circuit is built cheaply with nearly underrated parts. This is probably the reason the portable is the way it is in the first place, because if the circuit was designed otherwise, it would run by itself without a charged batt in the first place from the factory.

Without seeing a schematic, my guess is the battery and DC power circuit, there are components in between that arnt rated for high surge currents. So the battery takes over for that type of conduct.

Best way to test this scenario is grab a digital ammeter. Run the machine and do idle and intensive work with the battery + charger. monitor current draw from DC jack. Do the same with a PB100 charger and NO battery. youll probably notice a huge difference. and that difference alone can blow stuff up if its not rated for that kind of demand.

 

[techknight]

I got a backlit portable myself that i need to find time to finish restoring, maybe ill mess with it some.

 

[JDW]

...there could be issues in the charging circuitry that a higher current power supply will allow a lead acid battery to charge faster...

We can safely exclude that scenario, I think, as most Portable owners for years have been trying to get their Portables to work without a battery, since the batteries are all dead now. So it seems those SMD resistors (as discussed in that other thread) were killed without the lead acid battery being connected.

And that's what all this is about really. It's a crying shame that the Portable cannot be used with an adapter only! It's a travesty of engineering that Apple engineers forced the user to have that huge 9v lead acid battery in there.

 

[techknight]

I know i can get it to run without a battery, without blowing up SMD resistors. just might take some circuit modifications, but ill do it. not a problem ;-)

Once i get mine restored, ill figure that out.

 

[shred]

The main battery is a 6v battery consisting of 3 x 2v, "Cyclon" 5.0Ah "Gates Cell". I think Gates changed hands some years ago and they are now made by Hawker Energy... no, wrong... it's now EnerSys http://www.enersysreservepower.com/!

I repacked my battery a couple of times last century and still have the original set of Gates cells from it (its surprising that they haven't corroded and spewed their innards by now).

I recall seeing a post somewhere (possibly on 68kMLA) where someone bought a small 6v gel battery and installed it in the Mac Portable. It wasn't the right shape and so they had used a battery with considerably less capacity to make it fit, but the computer booted reliably. I'm not sure how well this would function in the long term, since the Gates cells have slightly different chemistry and charging characteristics to regular lead acid batteries. I've started my PowerBook 100 on occasions with a pair of fly leads out to a regular lead acid gel battery and I guess this technique would also work on the Portable - as long as you have the battery door in place. If the battery door is removed from the Portable, the little micro-switch changes over to the 9v backup battery and you won't get very far at all with booting up.

 

[Dark Goob]

According to Apple you can use a 9V AC adapter with all the batteries removed:

http://support.apple.com/kb/TA40602?viewlocale=en_US

It's the funniest thing I've ever read. It suggests that although technically you can power the Portable off the mains, why would you ever want to do this, since:

Any weight savings obtained by removing the battery would be negated by the size and weight of a 9-volt adapter of the appropriate capacity, and not being able to use the machine without AC power available. These size, weight, and convenience factors seem to be the primary reasons for having a portable machine.

... size! ... weight! ... convenience factors! ... portable machine! ROFL

:lol:

Ahem.

-=DG=-

 

[JDW]

According to Apple you can use a 9V AC adapter with all the batteries removed:http://support.apple.com/kb/TA40602?viewlocale=en_US

As you properly point out, that Apple article says you can power a Portable by removing the big lead acid battery, and also removing the small alkaline 9v PRAM battery, then connect a 9v AC adapter to the small 9v battery's contacts. But alas, not all the pertinent questions have yet been answered...

1) If this truly works, why then do we Google about it and find virtually no one using that method? Everyone on planet earth is instead still scratching their heads over how to rebuild lead acid batteries to get the Portable bootable again. And it's not like there is no such thing as a high-amperage 9v AC adapter either, so I am confounded to find such a dearth of end user feedback on that method. So if indeed the Apple article is correct, why then would not everyone be talking about that method? The lack of talk seems to cast doubts on the method. Or, if the method is correct, then such proves everyone on planet earth is a lazy bum for never having tried it! Agreed?

2) After some Googling, I was able to find one man on planet earth who claims to have tried a variation of Apple's advice. But in his case he claims that two AC adapters are required. Read his post here:

http://www.vintage-computer.com/vcforum/showthread.php?1957-Apple-laptop&p=30625#post30625

Therefore, would anyone care to comment on the soundness of his approach? And if it works, why then would Apple not have mentioned the need for 2 AC adapters in that TA40602 article?

3) Even in the utter absence of user feedback, let's assume for the moment the Apple article contains information on a feasible solution. What kind of 9v adapter then does one need to use in place of that tiny 9v alkaline PRAM battery? In other words, should it be a 9v AC adapter rated at 2 Amps? 3 Amps? 4 Amps? 5 Amps? And when you answer this question, please explain WHY you would recommend an adapter of such a current rating.

I look forward to your replies.

 

[techknight]

if i ever get a chance to restore mine, i could find out.

 

[JDW]

if i ever get a chance to restore mine, i could find out.

The weekend is fast approaching. Hoping you can give it a change then!

 

[techknight]

ya i know, ;-)

 

[techknight]

Looks like I am going to be placing an order from digikey for some of these caps, I ran out of stock on a couple of them. Weeee......

Anyway, as a sidenote, i can discuss a little bit about the circuit design of my board. not every board is the same, some of them have integrated power supply hybrid IC, mine does not.

It has 2 DC-DC circuits onboard, both of them Linear Technologies branded parts. LT1054CS, and LT1070C

the LT1070C appears to the 5V regulator circuit to power all the logic circuits. Has a rather large inductor. It also has a wide range of input voltages, up to 60v. But when its configured for 5V out, it has a power handling capacity of 10amps in its output. Of course, thats heatsinked and everything, plus at about 60 to 80% efficiency, you would need at least 15 or 20 amps input. lol.

Anyway, since its not heatsinked, its probably setup to deliver around an amp or two on 5V. there are alot of 5V logic circuits including the drives as well, so you need something beefy to supply 5V.

On the other hand, the LT1054CS circuit has a maximum voltage input of 15V, so that pretty much limits the portables power input handling at tops 15V. since this is a switched-capacitor based circuit, I am assuming this supplies 3.3v for the logic circuits. Maybe 12v, but im not sure what part of the motherboard would need 12V except for the drives. Then again, i could have it all backwards and the LT1054 circuit supplies the 5V and the beefy 1070C circuit supplies the 12V. Not sure, will not know until i rebuild my board to get it to fire up. the caps are so bad they are leaking through the PCB Vias. But what I do know are the facts I had stated about power input handling capabilities and what circuits they are.

Ok, now for the part you might be interested in:

the DC jack appears to feed into an IRF9Z30 P type MOSFET Switch. the jack also has a sense switch when something is plugged in, will break from ground. This goes down to another linear technology part, the Sense OP-AMP. its output goes back to the 9Z30. From there it probably turns on the 9Z30 to allow current to come in.

The main power bus from the 9Z30 when switched on, goes directly into the battery via a fuse. So the circuits ARE interconnected via only a single IRF9Z30. Thats it. this 9Z30 is NOT heatsinked. So, if run entirely off the DC jack, you could in theory allow the 9Z30 to run hotter because its RDS(on) isnt very good. But really, thats it. Also it goes through a pico fuse of unknown type, so it could in theory pop the fuse, but i doubt it.

Once through the fuse, and into the 4pin battery jack, I dont know where it goes, it branches off into an internal power plane.

the DC jack/battery power bus also feeds into Q15, another IRF9Z30 that feeds voltage to both the LT1070C, and LT1054CS circuit, to supply it power. So this MOSFET serves as a power on/off switch to this regulators, this is controlled by Q204 SMD transistor.

Also, near the battery jack, with a heatsink is Q16, and L10 which form once another DC-DC converter circuit. Not sure of output voltage. Probably since it has a large heatsink, one of the voltages that requires a high load, such as the 12V for the drives. Battery output also goes to Q20, which serves to only switch on/off the LCD display.

From looking at this, it seems the DC jack keeps the system hot all the time until the charge monitor cuts off the 9Z30 so the battery will discharge to the point where the DC jack cuts in again. A form of trickle-charging.

So there you go, To enable a portable to run without the battery safely, all you need to do is short the source and drain on Q1, IRF9Z30 at the DC jack. This will interconnect the DC jack to the same bus as the battery. To feel safer, you could remove Q1 entirely and do the same. Short Source and Drain. Reason i say this, if Q1 ever shorts internally, itll short Drain and Source, AND.. Gate together which will blow the LT op-amp. and its game over.

 

[techknight]

To add, What does this all mean? Well.. A complete paradox! Something that probably started all this confusion IMHO.

Something has to turn on that IRF9Z30 before DC jack power will get into the system, maybe the PRAM or main batt does this, still needs investigated. Maybe enough charge leaks by for it to turn on by itself. However i highly doubt it. If there is no other power available ANYWHERE inside that board, the 9Z30 cannot turn on to let DC current in. So if the main batt is completely wasted, i guess that roll is taken over by the small 9V. So basically by apple putting that transistor switch on the DC jack without some sort of trickle-back feedback, you now have a chicken and the egg syndrome.

it also kind-of puts to rest about the 1.5Amp charger not running the machine by itself. Then if this is the case, if its not enough to run the system on its own, then how do we expect it to run the system AND charge the main battery when the main battery is low? because then it would pull MORE current. not less....... I guess in this case the machine has to be OFF before the batt can charge, Which like i said, Q15 and Q20 is in charge of this, it can cut off the DC-DC circuits and the LCD just from these two transistors alone. Then you have Q18 which im not sure of its function, it also, can be turned off.

But once the battery is charged, yes the system could run, BUT, if the AC adapter cannot run the board by itself, then its going to start pulling from the battery. Therefore youll notice the battery slowly start going dead while the machine is running! So if the batt stays charged while plugged in and running, then the 1.5Amp theory is out the window, as the charger CAN run the board by itself (missing batt/batt charged 100%).

There are various smaller DC circuits around the board which probably what preserve RAM when the rest of the regulators are powered off.

So crunch on this for awhile. I have never operated or owned a portable until recently, so maybe it ONLY runs on the batt and once dead it has to shut down and charge. I dunno.

 

[JDW]

techknight, you certainly are the Knight of Tech!

But in light of your technical explanation, how do you use that information to explain what Apple clearly writes here:

"It might be possible to use an AC/DC 9-volt adapter to power the system, thus eliminating the need for the lead acid battery"

Source page: http://support.apple.com/kb/TA40602?viewlocale=en_US

Or are you going under the assumption that the word "might" used on that web page is just that -- something completely untested by Apple and nothing more than a "wild guess" by them? And yet, does Apple normally post wild guesses on their website that are not based in fact at all or otherwise could pose a danger? Would the author of that Apple page not first have consulted with Apple engineers to see if the content of that page were safe to recommend?

I would appreciate hearing your thoughts on this.

Thanks.

 

[techknight]

just depends on where the information had been obtained from inside apple. In theory, its possible. From an engineering standpoint, something needs to turn on that IRF9Z30 FET before DC power can come through. the 9V battery may do that.

But when I get mined restored, I am removing Q1 completely and jumpering it. i bet it runs fine.

as for running the system ENTIRELY off of the 9V battery jack, i dont think so. not only the gauge wire they use isnt thick enough, Im not sure where the 9V connects up into the circuit becuase it runs inside an inner layer of the system board. Also they quote a question someone asked, They state this "It might be possible to use an AC/DC 9-volt adapter to power the system,". I emphasize on the word MIGHT, with basically the same idea as you.

when someone says MIGHT, it basically means they have no f***in idea.

For me though, if i were to guess, the 9V probably connects into the same power bus is the rest of the circuit, its just when the main battery dies and AC power isnt connected, the DC-DC circuits shut down completely, 9V only keeps the PRAM and PSRAM up, assuming the sense and select circuit detects the presence of 9V but absense of main batt/AC.

 

[JDW]

when someone says MIGHT, it basically means they have no f***in idea.

Yes, but this is Apple we are talking about. Furthermore, they as an American company (in outrageously litigious US of A) have to take care when offering advice about anything. Hence, it seems illogical to me that they would flippantly post such information, especially without first having consulted with Apple Engineering.

My problem is that I don't have a Portable to test if connecting a 9-volt, 3-amp variable power supply to the 9v alkaline battery connector of the Portable (with all other adapters and batteries removed) would work in firing up the Portable and keeping it running. Therefore, if any of you Portable owners have such a power supply and are adventurous enough, would you be willing to perform the test?

 

[nvdeynde]

Hi,

I have several Mac Portables.

Since it's a valuable machine I recommand to rebuild the main battery. The battery housing is very easy to open and you can either replace the old 2 volt Lead Acid Cells with Hawker "Cyclon X" 2 Volt cells ( 3x of them ) , which are 100% identical to the original one's, or use a cheap 6 volt Lead-Acid battery of 5 Amps.

As for booting with high current power adapters only: It can kill your logic board as the circuitry is not designed to work on a power adapter only. Otherwise one would assume that Apple would have included a more powerfull PSU. The Powerbook 100 and all other powerbooks I have ( 100, Duo and 500 series ) perfectly work without the battery. The Mac Portable M5120 can work on even on the original 1.5 Amps PSU, if you disconnect the Connor Hard drive as this drive asks more than 2 Amps just to spinup.

The logic boards of the Mac portables regrattably also suffer badly of leaking capacitors: all the SMD capacitors are shot ( ESR > 20 Ohm ), also in most cases the axial and radial one's too. If you want to keep your Mac portable running in the long term: capacitor replacement is a must as soon as possible.

It will fix many issues with charging, no or whining sound, rolling barrs when adjusting screen contrast, instability, if the electronical components haven't suffered too much already...

After replacing the capacitors, you can also measure that the machine draws less power to start and also in standby/sleep.

As you will probably already know, the Mac Portables are always in Sleep mode even if you choose "Shutdown" in Mac OS: it makes no difference towards power consumption between Sleep and Shutdown. Even a fully recapped Mac portable drains a fully charged battery in about 7 days.

Good Luck,

Nico

 

[techknight]

Not going to "kill" anything.

The AC adapter and battery end up on the same exact power bus. only a MOSFET in between, and a picofuse. the reason the battery has to be in place with juice is because something has to turn on this mosfet to allow DC current to flow into the circuit from the adapter.