The rabbit hole opens and now the heat is on the find that elusive tech note!
PowerBook power adapter compatibility?
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Great archive resource on old Apple tech notes here:
TIL16168 - PowerBook: Battery, Recharger & AC Adapter Identification
Slightly different versions:
TIL16168 - PowerBook: Battery, Recharger & AC Adapter Identification
Slightly different versions:
@Fizzbinn
Thank you for the links. No scary warning that I could find (yet) about AC Adapters rated for more Amps than the stock Adapters, but the search is still on for that.
I see in some of the documents you linked that NiCd batteries were mentioned. I've seen people replacing those with NiMH cells even though the PowerBook's charging circuit was designed specifically for NiCd. I guess this is yet another case of people not really knowing what's best simply because many people have tried it (NiMH) replacements and they seem to work. Arguably, they are a vastly better choice than NiCd, but the concerns about the charging circuit remain. You don't want to overheat or overcharge your batteries.
Thank you for the links. No scary warning that I could find (yet) about AC Adapters rated for more Amps than the stock Adapters, but the search is still on for that.
I see in some of the documents you linked that NiCd batteries were mentioned. I've seen people replacing those with NiMH cells even though the PowerBook's charging circuit was designed specifically for NiCd. I guess this is yet another case of people not really knowing what's best simply because many people have tried it (NiMH) replacements and they seem to work. Arguably, they are a vastly better choice than NiCd, but the concerns about the charging circuit remain. You don't want to overheat or overcharge your batteries.
The only scary note like that I see is for the APS-17adp AC Adapter:
The fourth AC adapter, 17 W, shipped with the PowerBook 150, and works with the PowerBook 100 or 150.
Note: AC adapters producing more than 19 watts can damage the PowerBook 100 and 150.
You're saying that Apple put current limiting features in the power adapter instead of inside the Mac. You can't use third party power adapters if that's true. You would have to test the adapter to verify how much current you can draw from it. What happens to the adapter when that current limit is exceeded? A fuse would make it inoperable until it is replaced.
This note is almost useless. Watts = Volts x Amps. Which quantity causes the damage? Can you use an adapter that's 19V 1A? 1V 19A?
Adapters have Volts. There's no current or power until something is connected to it. If you connect a wire (small resistance) between + and -, then you have a short and you get lots of current/power until something melts. Apple adapters have something other than a fuse that makes a short not happen?
Actually, you can. I know people who have safely used modern AC Adapters with their Portable but those Adapters match the Apple Adapter in terms of output being 7.5V @1.5A. Voltage and Current output must be the same, and such Adapters do exist. Tolerances of those modern Adapters (as reported to me by others — I've not tried them) seem to be tight enough on the modern adapters such that very slight variation of voltage and current doesn't cause damage. Of course, the issue comes down to how slight, but I guess 100mV and maybe 200mA.
My Macintosh Portable uses a Battery Eliminator which means I can use any AC Adapter I like (even a 5A adapter), so long as the voltage is right. That's because the Battery Eliminator is an add-on card that charges up super capacitors and eliminates the need for a battery. It bypasses the plug you normally use for the AC Adapter entirely. Instead, you plug your Adapter into the Battery Eliminator card. I think it's great because batteries are a pain, and the other benefit is you don't need to worry about having an Adapter that outputs "too much current."
The motherboard burns and heatsinks change color, as @techknight clearly shows in this video (an absolute MUST WATCH):
I cannot give you an exact answer because I do not know, but I can tell you that both VOLTAGE and CURRENT are likely going to be an issue if either way off the nominal values, just as they are on the Mac Portable. Put 7.5V @ 2.0A into a PowerBook 100 and it works great. Use the same Adapter on a Portable, and you'll see what you saw in TechKnight's video above. (Also see my spreadsheet for specs.) But put 9v @2.0A into a PB100? You're likely going to fry something. Put 7.5V @ 3.0A into a PB100? I'm guessing (haven't tested, so I don't know), that the same thing that happens to a Mac Portable will happen to the PB100; namely, that something will overheat or fry. The Apple Tech Note seems to indicate that to be true.
In other words, your output voltage and output current will clearly have certain allowable tolerances for the machine to run without issue. Exceed the tolerances, and sadly things will fry.
What are the tolerances? I do not know. And like you, I don't like "not knowing." But for now, we simply do not know.
Yes you can use them. I meant to say you can't trust them. Usually the current rating is a minimum that the adapter can supply without damage to the adapter. It does not mean the adapter can't supply more current - at least for a short time. If the Mac can force an amount of current from the adapter that damages the Mac then that remains a problem.
Interesting video. Need an electrical engineer to understand this failure mode.
The battery is bad, so it has too high voltage or too low voltage? I would guess the latter?
The battery is used for the reference sense voltage of the 5.2V voltage regulation.
A voltage regulator tries to output a voltage that matches the reference in some way?
If the reference voltage is too low, how does this cause high current or high voltage (probably the latter but you all say the former)?
Q16 and Q1 that were pointed to in the video are transistors.
A voltage regulator VR1 (near the ADB port) is a LT1070 which can take input voltage 3V to 60V...
https://www.analog.com/media/en/technical-documentation/data-sheets/10701fe.pdf
The Macintosh Portable Schematic shows the LT1070 being used as a Boost Converter (5V to 12V).
Found this thread with more description of the results of the failure (but not really why the failure happens):
https://68kmla.org/bb/threads/macintosh-portable-charger-1500ma-max-aftermarket-options.28386
The V1M HYBRID module has a LT1054 which is a Switched-Capacitor Voltage Converter with Regulator.
https://www.analog.com/media/en/technical-documentation/data-sheets/LT1054-1054L.pdf
It might be fun to simulate the circuits involved to see how they can cause damage.
Let's talk through this using @techknight's video as an example. What could have caused this particular board to be destroyed? It is a 5126 (backlight) with bluing of the power transistors and holes blown in the AC logic chips. (We will assume this is non-abuse unintentional damage. That is, someone did not power on the bare board on a metal table or connect the 12V rail to the 5V rail. Also, my analysis does NOT consider the 5120, as I don't have a schematic for the hybrid module so I don't know how it controls Q16.)
For the 74AC logic chips to be destroyed like that, they would need to exceed their absolute voltage rating of 7V. How?
All of the power adapters we're discussing provide > 7V. However, a healthy lead acid battery only produces 6.3V. The battery will 'fight' the power adapter's 7.5V such that the combined voltage is likely < 7V. But if the battery is missing or 'dead', then the input voltage can be the full 7.5V. So, we seem to have our first conditions:
1. The Portable must be supplied power from the power adapter to exceed 7V and...
2. The battery must be weak or missing to allow the final input voltage to exceed 7V. Or... the power adapter must have sufficient current to 'win' against a healthy battery to exceed 7V.
I argue the above conditions must be true to have an input voltage >7V. Yet, at this point, that doesn't mean this voltage reaches the 74AC logic chips.
Walking backwards from the 74AC logic chips, the voltage they are supplied must first pass through Q16. This transistor is used as a buck regulator. That is, it turns on to allow a greater input voltage (say 7.5V) to flow in until it detects the 5V rail is high enough and then switches off. There is a big capacitor (C16) and an inductor (L10?) and the total resistance/capacitance/inductance of the rest of the motherboard that smooths this out. So, it's not like the 74AC logic chips are suddenly seeing 7.5V and then 0V. Instead, the 5V rail stays around 5V with tiny up/down oscillations.
If Q16 should fail short when the input voltage >7V and the current is sufficient (9V batteries are too weak), then the Portable motherboard will be destroyed. The 74AC logic chips could pop and the power transistors would change color due to too much current.
If Q16's gate floats or is driven low constantly (input >7V, etc), then the Portable motherboard will also be destroyed for the same reason.
This is a bad design. Besides the buck regulator, there isn't any secondary overvoltage or overcurrent protection. You could argue that the fuse (F1) is overcurrent protection, but the 5A rating is too high.
Let's get back to Q16's gate (on/off enable signal). It is first driven through a transistor/op amp combination that provides undervoltage protection. That is, the Q16 'enable' signal will be unable to reach Q16 if input voltage goes too low. Instead, a pull-up resistor will tell Q16 to shut off. This is the hardware shutdown mode to prevent data corruption if the battery suddenly dies (power adaptor unplugged with low battery).
The actual Q16 enable signal comes from the LTC1179 (U1M) which compares a LT1004 (U2L) 1.2V precision voltage reference to a 0.244 voltage divider (100K/(200K+100K+9.1K+100K)) of the 5V supply voltage. That is, it multiplies the 5V output * 0.244 (which would equal 1.22V when the 5V output is actually 5V) and compares that to 1.2V. If the divided 5V rail is higher than 1.2V, it turns off Q16. If it is lower, it turns it on. Thus, the 5V output stays around 5V.
Therefore, if something goes wrong with the LTC1179, the resistor divider, or the precision voltage reference, then Q16 could be turned on too long to allow the 5V rail to rise to >7V. For example, if leakage from capacitor C26 causes the nearby pin 8 of the LT1004 to disconnect, then the '1.2 voltage reference' would rise, causing LTC1179 to think the 5V output was too low in comparison, causing the Q16 gate to be turned on too long, cause >7V to hit the logic chips.
Aside: The bluing on the power transistors would just be a side effect of the gate being left on while the logic chips short and burn out.
I have not come up with an ordinary scenario where the maximum aperage of the power supply alone would be the cause of a motherboard failure. But instead, something else must fail first and a healthy battery must not be installed.
I am really interested in anyone who can come up with a different point of failure.
For the 74AC logic chips to be destroyed like that, they would need to exceed their absolute voltage rating of 7V. How?
All of the power adapters we're discussing provide > 7V. However, a healthy lead acid battery only produces 6.3V. The battery will 'fight' the power adapter's 7.5V such that the combined voltage is likely < 7V. But if the battery is missing or 'dead', then the input voltage can be the full 7.5V. So, we seem to have our first conditions:
1. The Portable must be supplied power from the power adapter to exceed 7V and...
2. The battery must be weak or missing to allow the final input voltage to exceed 7V. Or... the power adapter must have sufficient current to 'win' against a healthy battery to exceed 7V.
I argue the above conditions must be true to have an input voltage >7V. Yet, at this point, that doesn't mean this voltage reaches the 74AC logic chips.
Walking backwards from the 74AC logic chips, the voltage they are supplied must first pass through Q16. This transistor is used as a buck regulator. That is, it turns on to allow a greater input voltage (say 7.5V) to flow in until it detects the 5V rail is high enough and then switches off. There is a big capacitor (C16) and an inductor (L10?) and the total resistance/capacitance/inductance of the rest of the motherboard that smooths this out. So, it's not like the 74AC logic chips are suddenly seeing 7.5V and then 0V. Instead, the 5V rail stays around 5V with tiny up/down oscillations.
If Q16 should fail short when the input voltage >7V and the current is sufficient (9V batteries are too weak), then the Portable motherboard will be destroyed. The 74AC logic chips could pop and the power transistors would change color due to too much current.
If Q16's gate floats or is driven low constantly (input >7V, etc), then the Portable motherboard will also be destroyed for the same reason.
This is a bad design. Besides the buck regulator, there isn't any secondary overvoltage or overcurrent protection. You could argue that the fuse (F1) is overcurrent protection, but the 5A rating is too high.
Let's get back to Q16's gate (on/off enable signal). It is first driven through a transistor/op amp combination that provides undervoltage protection. That is, the Q16 'enable' signal will be unable to reach Q16 if input voltage goes too low. Instead, a pull-up resistor will tell Q16 to shut off. This is the hardware shutdown mode to prevent data corruption if the battery suddenly dies (power adaptor unplugged with low battery).
The actual Q16 enable signal comes from the LTC1179 (U1M) which compares a LT1004 (U2L) 1.2V precision voltage reference to a 0.244 voltage divider (100K/(200K+100K+9.1K+100K)) of the 5V supply voltage. That is, it multiplies the 5V output * 0.244 (which would equal 1.22V when the 5V output is actually 5V) and compares that to 1.2V. If the divided 5V rail is higher than 1.2V, it turns off Q16. If it is lower, it turns it on. Thus, the 5V output stays around 5V.
Therefore, if something goes wrong with the LTC1179, the resistor divider, or the precision voltage reference, then Q16 could be turned on too long to allow the 5V rail to rise to >7V. For example, if leakage from capacitor C26 causes the nearby pin 8 of the LT1004 to disconnect, then the '1.2 voltage reference' would rise, causing LTC1179 to think the 5V output was too low in comparison, causing the Q16 gate to be turned on too long, cause >7V to hit the logic chips.
Aside: The bluing on the power transistors would just be a side effect of the gate being left on while the logic chips short and burn out.
I have not come up with an ordinary scenario where the maximum aperage of the power supply alone would be the cause of a motherboard failure. But instead, something else must fail first and a healthy battery must not be installed.
I am really interested in anyone who can come up with a different point of failure.
Schematics at
https://www.macdat.net/repair/apple_schematics.html
https://tinkerdifferent.com/resources/macintosh-portable-5120-schematic.133/
but doesn't show the V1M hybrid module. For that there's this:
https://github.com/ppieczul/macintosh-portable/releases
Thank you. BTW, I based my 5126 analysis largely on the PowerBook 100 schematic that appears on your macdat.net link. I then compared it to my 5126 motherboard to be sure.
The 5120 recreated hybrid schematic largely reflects the 5126 analysis. Input voltage (6.3B or 7.5V) comes in through RC3 to voltage reference DC1. I assume this is a 1.2V reference. Capacitor CC1 keeps that reference steady. RC3 limits the amount of current that DC1 has to handle so it doesn't heat up and change value.
The 5V rail output voltage comes in V1M-11 through the voltage divider RC5/(RC5+RC7)=0.238 which is about the same as the 5126. So, if the 5V rail is 5.0V, then OP20 sees 0.238*5.0V=1.19V on input pin 3. OP20 sees 1.2V (from DC1) on pin 2. It compares these two and turns on Q16 if the 5V rail is too low, or off if the rail is too high.
What this means is, it doesn't matter whether the input voltage is 6.3V or 7.5V, 1.5 amps or 3 amps. The divider and voltage reference will still generate a valid 5V output to the chips.
They improved this slightly for the 5126 and PowerBook 100 by adding hysteresis through a feedback resistor back into their version of the OP20.
Let's say corrosion on the hybrid module causes damage or disconnect to DC1. Then, CC1 will rise to the input voltage (let's say 7.5V). OP20 will see input pin 3 is low in comparison and turn on Q16 constantly, destroying the Portable logic chips. Likewise, if RC7 is corroded/aged to a higher resistance or cap juice conducts across RC5, then pin 3 voltage will drop, causing Q16 to turn on constantly, destroying the Portable logic chips. Or, more simply, what if OP20 becomes damaged by cap juice and stuck 'on'.
This last paragraph explains why an old crusty and juicy hybrid module is so dangerous to the 5120 portable. There isn't any protection if V1M-8 -> Q16 is constantly enabled. A simple crowbar circuit would fix this.