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I have a number of early-80s era microcomputers that are missing their external power supplies, or in some cases have internal power supplies that are dodgy or broken. These take various (and sometimes unusual or even unknown) external voltages from the PSUs for which they were designed and almost invariably regulate them down to no more than four standard voltages: +5, +12, -12 and -5. Most often this is done with linear regulators, typically 7805 and suchlike, which are frequently on the motherboard itself.

I'm thinking that rather than trying to replicate the various missing external power supplies (or fix the internal ones), it would be convenient just to use a modern +5/+12/-5/-12 switching power supply to directly supply the voltage rails on the motherboard, ignoring (as much as possible) the original power supply circuitry. For machines where the original PSU can be completely disconnected from the motherboard there's obviously no problem: just unplug the original PSU's cables and in its place connect my modern PSU.

But for many machines only part of the PSU can be easily disconnected. (This part being the external "wall wart" or internal transformer, often including the rectifier and other related circuitry.) Thus, if I clip my modern PSU to the power rails, the remaining part of the PSU circuitry on the board, such as the regulator output, would be seeing the rail voltage with no voltage at its input.

What I'd like to know is what sort of issues I should be looking at to try to do this safely (i.e., without damage to the microcomputer) and minimally intrusively. From questions like this and threads like this I gather that it's sometimes safe to leave the input of a voltage regulator unconnected when supplying power to a circuit via alternate means, but in other cases a protection diode is needed to ensure that the regulator input is at the same level as the output.

Obviously I could simply cut off the regulator's output pin, but I'd rather avoid making such drastic changes to vintage equipment, especially if it would make it unusable with the original PSU. Soldering in a diode would be a lot less drastic, but still a modification, which I'd like to avoid if possible.

So how do I tell how safe it is to supply any particular board in this way, what should I be looking for that might cause problems with doing this, and what can I do to mitigate them?

(I'm aware that some computers use their PSUs for more than just power: for example the 9 VAC from the Commodore 64 external PSU is used not only to generate +12 VDC but also as a frequency source for timers. To keep things simple, you can ignore any considerations of this nature in your answer.)

In case it helps, here's one example of an external power supply (which I do not have) and the regulator circuitry on the microcomputer's motherboard that the external PSU feeds. I am guessing that pin 2 is actually -8 VDC, as marked on the diagram, not +8 VDC as it might appear from the text at the top. These are from a National (Panasonic) JR-100, released in 1981.

JR-100 external PSU JR-100 motherboard - regulators

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    \$\begingroup\$ Also keep in mind that depending on the device, a specific power-on sequence may be necessary. Such as +5v first, then +/-12v. The more complex the device, the likelier this is. There are even power "supervisor" chips to monitor this, and prevent the device from starting if the power is not 100% correct. \$\endgroup\$
    – rdtsc
    Commented Mar 12, 2021 at 12:51
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    \$\begingroup\$ @rdtsc A good point, but this seems pretty rare in early 80s microcomputers, isn't it? \$\endgroup\$
    – cjs
    Commented Mar 12, 2021 at 12:55
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    \$\begingroup\$ Just remove the regulators. That doesn't help with the power-on sequence, but it prevents any backfeeds. \$\endgroup\$ Commented Mar 12, 2021 at 13:30
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    \$\begingroup\$ As electrolytic capacitors age they can dry out. When that happens, the voltage is not as "smooth" (i.e., increased ripple). Those would be the first components I would check and replace before powering up something that spent 40 years in the attic. \$\endgroup\$
    – spuck
    Commented Mar 12, 2021 at 22:48
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    \$\begingroup\$ @spuck Retrocomputing Stack Exchange has advice along similar lines. (Here's a detailed, though low-score, answer explaining a procedure in detail.) \$\endgroup\$
    – wizzwizz4
    Commented Mar 12, 2021 at 22:59

4 Answers 4

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In general it is not safe since most of the time regulator properties are not guaranteed when the input side isn't powered on. Some regulators are specified to be powered 'into the load', as in, with an already present voltage on the output.

The diode trick for the 78xx regulator (which was IIRC actually in the datasheet) is the minimum to keep the part biased but there could be other problems too depending on the internal topology. For example most regulator can only supply power but not sink: this is not universally true, however.

I think it should be decided case by case

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  • \$\begingroup\$ +1 Yes, and the schematic would probably be hard to find or draw from the board to be sure. If I understood correctly, the "diode trick" is to discharge an output capacitor in case of input short. We can't guarantee it will survive constant power from the output. \$\endgroup\$
    – devnull
    Commented Mar 12, 2021 at 12:27
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    \$\begingroup\$ Ah. This was not discussed in the Fairchild LM78XX datasheet I have, but in the TI datasheet §8.1.1 shows that diode configuration, and §8.1.2 "Raising the Output Voltage Above the Input Voltage" reads to me as if powering the rail with the input unconnected could destroy the regulator. Is that correct? And as for "case by case," that would really involve working out the full schematic of the PSU section and getting hold of all the datasheets, right? \$\endgroup\$
    – cjs
    Commented Mar 12, 2021 at 12:35
  • \$\begingroup\$ That's exact. Every regulator has its own quirks. Maybe not all the schematics but at least check around the regulator ICs \$\endgroup\$ Commented Mar 12, 2021 at 14:52
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All Linear Regulators only pull up current while the loads pull down towards 0V.

All linear regulators have a max reverse bias voltage where it is either clamped to input by a protect diode or damage may occur if the input is short circuited abruptly with a stored higher voltage on the output. If the input is floating, it must be pulled up by the output. For example Veb on most transistors is 5V in reverse where it will begin to breakdown and possibly fail at low power levels. The same is true for LEDs (-5V max safe limit), but not deeply diffused junctions like diodes where they are designed for reverse blocking. So adding this is your best safety net.

Yet it is safe to parallel an active linear regulator with the same or slightly higher external supply with shared gnd. But not safe for a Buck Boost regulator or any half bridge that also pulls down.

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As already noted, it's not in general safe to backfeed power to the output of a linear regulator. One option is to use an adjustable external power supply (or supplies) connected to the external power socket of the computer and set to give the voltages that the original external power supply would have provided or that the internal regulators need. So a 7805, for example, will need an 8 V supply.

If you need a negative rail, as with your pictured power supply, and you don't have a supply with a suitable bipolar output make sure any separate supply has a floating output (i.e check the negative output isn't connected to ground) otherwise you may short-circuit the supply when you connect its positive output to the 0 V rail of the other supplies.

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    \$\begingroup\$ One option is to use an adjustable external power supply (or supplies).... Yeah. It probably wasn't clear from my question that I'd considered this, but it's not my preferred option. Not only is it moderately costly (I currently have only a single-output bench PSU), but doing frequent readjustment as I switch between machines is prone to error. Not to mention that it also requires knowing what the inputs are supposed to be. (It was months before I lucked across a readable picture of that JR-100 external PSU on an auction site.) \$\endgroup\$
    – cjs
    Commented Mar 12, 2021 at 14:41
  • \$\begingroup\$ Oh no, your wishes were perfectly clear in the original post but, as Mick Jagger noted, you can't always get what you want. As I agree with you that it is undesirable to modify vintage computers, where you can't disconnect the internal regulators the next option is to provide their supply side with the voltages they expect. If you look for surplus or cheaper imported power supplies you might find you get what you need. And try not to blow any fuses, 50 A or otherwise. \$\endgroup\$
    – Graham Nye
    Commented Mar 12, 2021 at 18:26
  • \$\begingroup\$ Yes, I agree that the JR-100 situation is looking like a Mick Jagger situation. At the very least I will probably need to temporarily rig up a three-voltage external supply to confirm that the on-board PSU circuitry's design is what it appears to be from looking at it. Thanks. \$\endgroup\$
    – cjs
    Commented Mar 14, 2021 at 7:25
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As pointed out in the other answers, typically you must not let the input of a regulator see a voltage lower (or more than slightly lower) than that at the output, or you risk damaging the regulator. Thus, even using a diode to back-feed the input from the output can be risky.

This answer suggests another solution: connect external PSU to both the input and output of the regulator. There is still some risk here; not only should the regulator data sheet be checked, if possible, to confirm that equal voltage at input and ouput is allowable, but also the rest of the circuitry on the input should be checked to confirm that it will be ok when fed from its "output" side. (Possibly the external PSU should be feeding that circuitry as early in the chain as possible, even as far back as the microcomputer's external power connector.)

N.B.: I'm posting this answer to get the idea out here, but I don't have the technical knowledge to know how generally safe this really is or in what circumstances this is likely to go wrong. If someone can write up a better version of this answer that goes into more detail about this, I'd appreciate it.

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