What's Going On With This C64 PSU?
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- Опубліковано 22 сер 2024
- I decided to have a closer look at the power supply that was causing some trouble last week. Along the way I try to use the oscilloscope to measure the current drawn by the C64 on startup.
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I am thinking switch contact bounce! I've seen similar behaviour with other PSUs! Try cleaning the power switch! I think what happens is the contact closes - inrush current drawn, contacts open, PSU over compensates and the voltage goes higher, perhaps tripping over voltage protection on the PSU? Having said that - just watching towards the end now, I see using the power strip part, so not really sure.. Those power switches are pretty bad at bouncing though! The same exact behaviour happened with the C64 voltage saver prototype from bwack - and it was inrush there combined with the bounce.
Maybe somehow protection on the 9V supply is affecting the 5V output? That's supposing the 9VAC isn't just a passive transformer.
It would be helpful to have measurements of the switching transient at the 1-100us scale, I think you're likely right there's something interesting going on right as the switch turns on.
It could just plain be inrush triggering short circuit protection, forget the switch bounce. IIRC the C64 has some pretty big capacitors in the power input section, and until those are full it'll draw current like a mofo, especially if it's been recapped with modern high quality capacitors :) I would have thought the 1 ohm resistor put in place to measure resistance might have been enough to soften the inrush, but that's why I'm wondering about the 9V.
One way to soften that inrush current without just adding resistance is by adding an inductor in series -- big but not too big or it might affect regulation. Another way (harder on the switch but with less effect on regulation) would be to add a big tank capacitor on the PSU output to just supply as much inrush current as the C64 wants... brute force but 🤷
It would be interesting to crack open the IRM 10 5 and see if there is a bulk output capacitor inside it. The block diagram and test results for it don't indicate anything on the output, maybe expecting the designer to add bulk output capacitance (which appears missing).
Swap out the power switch with a new one.
The AC Power Supply is a transformer with 2x 9V AC @ 2x 5VA. There are 2 secondary windings on the transformer with 9V output in parallel and can deliver 10VA (2x 5VA) or 18V in series with 5VA. The watt (W) value indicates the real energy consumed by the consumer device, while the volt-ampere (VA) value indicates the so-called “apparent power”. The two primary windings are 115V AC in parallel in the US and in series for 230V in the EU.
The nominal 5V DC output is ratet with +/- 5% wich means that the 5V can range from 4.75 - 5.25V. So it's OK with your measurement.
The Meanwell Powerbrick is a switching power supply and can deliver 2A @ 5V DC. It also has an over current protection. So it's possible if you turn on your C64 that in 'Switch-on-mode' the Meanwell goes in Over-Current-Protection and switches on after a short time (i.e. the Capacitor in your C64 is charged). It's sometimes confusing with the markings of a transformer for the EU-Market.
Greetings from Germany.
I think your theory is right. There is an in-rush current when you turn the machine on that triggers the PSU's OCP. After half a second, whatever spurious capacitances had to charge will have charged and the OCP will turn off. What you're seeing is typical of it. Your attempt to measure the in-rush current with that little resistor will not bring you anywhere, I'm afraid. A 1Ω resistor needs to pass 1A current to have a 1V voltage drop (I=V/R), and that is assuming the resistor can take that kind of power (1W), otherwise goodness knows what kind of breakdown effects and thermal shock will be interfering with your results. In-rush currents can be caused by leaking capacitors (leaking in the electrical sense, not leaking electrolyte). If a capacitor is letting DC through, it will take longer to charge and require more current to do so. The easiest way to check is to remove the capacitor from the board and measure its capacitance (not ESR; ESR has nothing to do with DC leakage). If it measures higher than nominal, it indicates leakage (the multimeter infers capacitance based on current draw, and will interpret higher current draw as higher capacitance). With a bench power supply you can do a more reliable test: set the bench power supply to the nominal voltage of the capacitor, and apply it. Then check how fast (or slow) the voltage gets to the nominal value and the current out of the bench PSU goes to zero. If it doesn't happen pretty much instantaneously, you have a bad cap. DC leakage is an effect of EPR (as opposed to ESR), but it is dependent on the applied voltage, so you can't measure EPR with a multimeter in resistance mode. There can be other sources of spurious capacitances leading to excessive in-rush current, both on the board itself and in the ICs. In the latter case, it indicates imminent failure. Either way, you can avoid triggering OCP by using an in-rush current limiter (it's a thing), such as TDK's B57238S0709M000, in series with the 5V line. If after applying it you still have the problem, then you can rule out in-rush current issues.
Dirty contacts in the switch or plug could cause voltage to look good when no load, but act as a large resistor when any current is being called for
You could also put a bulb (5v) in series with the power supply - if it lights full on you have a big current draw.
I have got such a psu from poland and it works flawlessly. It is approx. one year old and I use it a lot. It has two connections, one for the c64 and one for the 1541. Very handy
Use several low wattage low ohm resistors in parallel - you lower the resistance and increase the wattage.
You may need to use a shorter time base to see the spike. Maybe the slow sample speed misses the spike?
Yes, 100% this. That rolling 500ms timebase is nowhere near fast enough, you're looking for protection that may kick in faster than 1ms.
Noel, my dear friend...
You're an engineer of sorts, you are great with electronics and know your stuff well. 2-thumbs up for that.
I had to check my Vic20 with the PETAcsii keyboard, and it says that its input is 117VAC @ 1amp (USA). Given that - when transformers drop voltages, amperage goes up - Law of Conservation of Energy. Since My Vic20 is a 2-prong 9VAC input, that means it is getting (rounding off numbers here) 9volats at around 10amps. 10amps?!! Of course the internals of these older Vic20s has full power regulation, conversions and dividing up power to area where it is needed. Thus when it takes 9VAC @ 10 Amps and steps it down to 5VDC, the amperage will be less because power is also divided up to 12VDC and 9VAC. So lets say it is divided up equally, so 5VDC, 12VDC abd 9VAC are each getting 1/3 of the input of 9VAC @ 10Amps; or about 3.3Amps.
You are feeding 5VDC @ 2Aamps into that C64? That is not enough power to keep the system running, especially THAT C64. If it turns on, the voltage regulator and chips will get hot because they are stressing for more power that is not there/
That Blue C64 of yours is not a Cost Reduced C64, with just a few chips and a large custom (PLA?) to take much of the logic for the 6510, same with my Vic20s (and C64, which is in storage)
I also have a couple Cost Reduction Vic20; in chip counting, they have 24 Chips compared to 30 in my PETAscii Vic20. And all the Vic20s use Static RAM, so they use less power than the C64s which uses Dynamic RAM. Using less chips, the Cost Reduction Vic20 would use less power than the PETAscii Vic20. The same is true with the C64 vs. Cost Reduced and Aldi C64s.
True, you can use a 5V USB @ 2.5amps Charger to turn on a Vic20 or C64, but it is only enough to turn it on and see it BASIC pops up on the screen. Other than that, it would be limping along, and not able to use load programs from Cassette or Disk Drive, no can you connect anything to the User Port. Your PSU is like a USB Charger, just enough to turn on the C64 to show BASIC but no more. Thus it is why you have issues with it turning on.
If you are getting a new PSU, keep in mind that you woll need at at least 5V @ 4amps or more with 9VAA @ 5amps or more. The C64 will be happy with the power increase and everything will run on it much cooler.
Great video. You always make great videos..
For a shunt resistor to measure the current you need very low resistance, 10Ohm will already limit the current to a maximum of 500mA so you aren't going to see any spike, if the fault is a semiconductor it will not be juiced enough to provoke the fault. You should use something in the range of 0.5 Ohm 20W, one of the ceramic white blocky ones, and that will already limit the current to 2.5A.
It needs to be much lower than that to capture what's really happning. The voltage drop needs to be small enough that it doesn't affect the computer, for 5V and 12V it's common to try to aim for at MOST 100mV drop over the shunt which for 2A corresponds to a maximum of 50 mOhm (0.05 Ohm!). And if you have a high resolution oscilloscope going lower (say 10mOhm, 20mV=2A) is even better as it will interfer less with the circuit and giving better results. And as a side-effect you can use small low-power shunt resistors because it's no longer burning a lot of power 🙂
@@Torbjorn.LindgrenIf you want a precise reading and a full working machine yes, but to check just a spike in what seems a problem in the power section it should be enough with 0.5Ohm.
FWIW you seemed suspicious of the Meanwell supply last stream, but to be clear they're a pretty reputable brand. The engineering firm I consulted with for a while used them in prototypes (expensive prototypes, as all prototypes are).
Obviously it's still possible for it to be faulty or not perform well in this situation for some reason! But the Polish company is doing a solid job building that module into their PSU.
Isn't for measurement voltage on resistor placed in power line differential probe needed? I think, common probe has one end grounded. Maybe it worked because C=64 is galvanic separated.
Use the adapter you made to disconnect the output of the 5V Mean Well and instead route it to your benchtop power supply (so the 9VAC is still coming from the new 64 PSU but the 5V line is being powered by your bench PSU).
Set your bench PSU to 5V/2A.
If I'm correct the 64 is sometimes drawing more amps at initial startup that's exceeding the Mean Well 5VDC PSU's max draw. My hunch is your issue will go away if you use the bench PSU. At the very least you should be able to see the true power draw the 64 is putting on the 5V rail on the display from the benchtop PSU.
If it indeed does go away, you can try replacing the LM7805 and LM7812 regulators with modern replacements. Murata and Texas Instruments make some. I can't remember the part number for the murata parts but looking around the net one person on Lemon64 used these TI part numbers:
12v: TPSM84212EAB
5v: TPSM84205EAB
Thumbs up for that idea. My bench top PSU allows for custom OCP and OVP, it shuts down output immediately if something goes even tiny bit out of the limits. Good thing to check for back EMF or something from the bouncing switch
The shape of your voltage drop-out on the oscilloscope suggests something being switched - like you'd see with a power manager chip. It limits output until it reaches a "good" state and then allows full voltage. I'd check that meanwell block to see how it is meant to behave. The mystery is still why, even if that is intended behaviour. If it is unique to that particular C64 it does still need to be doing something different. You've eliminated high current draw so there isn't too much left
I think with your 1Ohm resistor you limited the inrush current a bit so no problem manifested. You need to connect several 1Ohm resistors in paralel (the total power limit per resistor will sum up then). With 1A over the resistor you'll have a drop of 1V per 1A drawn (1A * 1Ohm). With 5 resistors in paralel, the insert resistance will be 0.2Ohm with drop of 0.2V per 1A drawn and much less power and heat losses. Also beware that the osciloscope channels sharing common ground (unless you have differential probe on input), which is commonly connected with PE wire on socket when your scope is powered directly "from the wall". This could lead to shorts. Rule of thumb is to use just single ground wire on single probe and measure everything against that potential (alt. with some math). Try to insert some coil (with high enough nominal current) into 5V supply - this could eliminate the current spike a bit. And check with much faster time base, you're searching for something really short which could be "smoothed out" by averaging of measurement samples over larger time period on the osciloscope. Sorry for "bringing trees to wood", but I think this needed to be stated for some of less experienced viewers using this method of current visualisation to prevent possible damages.
I concur. But I prefer to use things which work around 1 rather than 2. Less math, less chance of confusion. Some examples follow for understanding the principle (for other readers), not to refute your comment
5x 10ohm resistors in parallel would be 2ohms. And if they're metal film 1/8w that would be rated for 5/8 watts. Also, 1amp current would show as 2v voltage drop and you have to do math (÷2) to get the current. You'd also be feeding the device under test (DUT), the C64, with just 3v as a result. The resistors would be dissipating 2watts (well over the 5/8 watts!)
10x 10ohm 1/8watt resistors would be better, giving 1ohm resistance, with 1.25w. 1amp would show as 1v drop across the resistor, so no math.
For power handling, you have to calculate the watts using the voltage drop across the resistor, not the supply voltage, because you're calculating the power dissipated as heat.
So, if you have 5v supplied at 1amp current draw, across a 1ohm resistor, you will see a voltage drop of 1v So dissipation is 1v x 1.0a = 1w. But that gives you a brownout situation where the DUT then only gets 4v
So, we go for 10x 1ohm in parallel. That gives us 0.1ohm resistance, and 1amp current results in 0.1v drop. The DUT gets 4.9v, and you crank the 'scope vertical scale a bit so the smaller reading is still readable. Also with only 0.1v drop at 1amp, the resistor dissipates only 0.1w. With 10x metal film 1/8w resistors, you can dissipate 1.25w so 0.1w is negligible. Although this introduces math again. At 1amp, the 0.1v drop reading needs to be adjusted (x10) to equate to amps again.
0.1ohm is a probably recommended to avoid brownouts... 0.01Ohm would be less burden, dissipate less power as heat etc, but you gotta change the scale on the 'scope again, and multiply the reading by 100x. Voltage drop at 1amp is only 0.01v so the DUT would get 4.99v. Dissipated power is a mere 0.01w. Win.
It's fun to work out the best solution.
A 5W ceramic wire wound resistor would easily handle the job, if you have one at 0.1 to 0.01 ohm
10:56
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Might be something simpler, like a cold solder joint on the c64 power connector or switch? But today it was making good enough contact.
I think that the capacitor on the 5V line is aging. This can lead to currents when switching on.
Don't forget there are two 5V sources. One 5V comes from the PSU and powers the board logic and the LM7805 5V called 'CAN 5V' which is for the video section only. There's also a bunch of diodes connected to 5V so maybe one of them is suspect. Another thing, the switches are known to cause connection issues so if you are trying to check the voltage you need to bypass the switch by joining the input and output pins so that the switch is not affecting the readings. If the anomaly disappears then the switch is faulty. They are actually not that bad and I've repaired a few switches by opening them and cleaning out the old grease inside, polishing the contacts, re-lubricating it and re-assembling and it worked perfect.
Check for shorts or leakages between 9 V AC and 5 V DC inside the C64. This can be overvoltage protection, not overcurrent.
The capacitor on the right you pointed your finger at looks slightly bend out to me at the top. But I can't see it that well. If the top isn't flat anymore, but bend up, it's already a sign that the capacitor is failing/broken. Otherwise others said already, at 5Volts/2Amperes, one can estimate with that an internal resistance of 5V/2A=2.5Ohms ... So maybe taking 5% of that value as a resistor to measure over is better, not changing voltage and current too much and reducing the power at that resistor below some 0.6Watts a typical Metalfilm resistor can handle.
Hello
Thanks for all these very interesting videos.
Why not have the same test with the other psu you said did work ?
Cheers from Brittany !
The transformer probably has 2 5v taps and 2 9v taps
With the adapter you crafted I think it could be informative to check the voltage drop on the switch on the C64. you should be ok to keep your ground leads where they are at 17:13, but place one of the probes on the power led in the C64 like you did for the first test with the scope. Another thought is, if you can reproduce a persistent fault, what happens if you leave the switch on the C64 where it is, and then toggle the switch on the supply itself or the power strip you have it plugged in to. Although it is odd that the older style didn't seem to have as many or the same issue as this one.
If you want a fairly precision low value shunt resistor, and don't want to buy one, you can try using your multimeter in current mode. It will have a range of values. Probe as close to the meter as you can, use a much shorter time base and trigger on the edge. Any transients are going to be too fast for the scope to catch at 200ms devisions.
This 'soft start' that you see on some modern switching power supply chips has caused some problems in the ST scene where they can take longer to stabilise than the ancient reset circuitry is allowing for. The solution there is to replace (or parallel up) the appropriate cap in the reset circuit that provides the time delay. Of course that doesn't explain why you're seeing this problem intermittently on your supply. Perhaps there's a capacitor internal to the Meanwell that is generating the high current when it's more discharged (off for a decent time)? It got the impression you were more likely to see the slow start after a longer power off period than a short on-off-on cycle. But that could be a bit of me seeing what I want to there.
I think 2x5v means that is has a centertap and 2 5 coils and by putting them in series the (9v) is derived from that.
He didn't read it fully. It's 2x 5VA. It's incorrect to say 'watts', but if it helps you understand, think of it like that.
There are 2x 9v secondary windings (the output side) and each is rated to supply 5VA (like watts)
If you connect them in parallel, that gives you 10VA out. The C64 PSUs were rated at 1.1A or 1.2A on the 9v line, so something around 10VA is about equivalent to that.
@@CollinBaillie That makes perfect sense.
Common problem is the switch and you was measuring after the switch
The behavior is a bit weird. Maybe it is matter of the scope. Just a little maybe. I guess, you are not using an isolating transformer for it and that causes the AC/DC module to trip. Since it is a 1 Second „trip“, it should be visible from the power LED. So I would disconnect everything else (monitor, etc.) and just turn on the C64, watching the power LED coming on delayed.
I actually still wonder, what the transistor on the bottom side is doing, that I have seen in the other video.
Big chance that the elco in the meanwell lost capacity. Test with another commodore, solder an extra elco in the power supply.
If it is an inrush current problem you could try to put an 1A NTC on the 5V output of the PSU to limit that. The IMSAI guy just had a chip of the day video on these and their use in SMPS. But having had two faulty switches on vic20 and c64s my first bet would be on an dirty/bouncy switch as gadgetuk suggested.
Could be a current issue. I had a PS that worked well on some C64, but not others. Even swapping a SID or VIC could cause the C64 to not start. Replaced the PS and it worked
I would put a new 5v regulator in the computer since those are a common enough failiure point. Might be it's on it's way out and the modern power supply has some kind of protection that is sensitive enough to sense that.
Caught you
Power on reset circuit holds reset high for 0.5 seconds... Your scope showed low voltage for around 0.5 seconds. That could potentially mess up system reset.
But that doesn’t explain power LED to stay low. It is not about the CPU reset but keeping MW PSU in “safe” state
Did you check the power supply by itself? Would it behave the same if you just hooked it up to a 6v light bulb?
Power the C64 from a bench supply with output on/off capability, and do the same tests.
If it never happens, you know it's this PS.
I suspect the power supply.
Dirty contacts in the switch IMO.
My guess is a bad power switch ..
Try heating the voltage regulator with your soldering iron?
Then try freezing it.
for tests, get another C64, and PC PSU .
Good rule of thumb to know: PSU current should be double what the operating current draw is for any device. Operating at a 75% or more threshold will cause weirdness. Current draw is maximum for any electronic device (except power amplifiers if they have a protection relay) when powering them on, as all capacitors charge, and due to their ESR, put a great load on the rails while charging.
If a PSU protection circuit is at the PSU rated current, then it is far too sensitive. It should be at the surge current, else the PSU is overrated for its real rating. If that is the case for this PSU, then it's not 2 amps, it is 1 amp.