5 ms·

Depending on the PMIC and the SSD DC conditioning, even 1.2v might be enough for it to brownout/latchup without self-resetting. (or it might power up the PHY pa

Depending on the PMIC and the SSD DC conditioning, even 1.2v might be enough for it to brownout/latchup without self-resetting. (or it might power up the PHY partially or in a bad state and never link up) Try more resistors in series? (or just a bigger one if you have any -- scratch that we needed smaller ;-) ).

starslab 2y ago
12 ohms brings the rail down to 1.47 volts, still no SSD. 6 ohms is enough to finally break/trip whatever circuit is allowing this situation, bringing the rail down to 0v in power-off. Of course, that's almost 2 watts of constant draw during the power-on state, so not a long-term solution.
- bunnie 2y ago
  Hmm the usual solution in this case is to put a FET to ground that is turned on when the standby signal goes active. The reason you need such a low impedance is that a linear resistance pulls less current as the voltage goes down so the solution is less effective the stronger the leakage path is. The amount of leakage is on the high side, so it does point to some device that is wired to the standby power that is probably 'powering' the 3.3v line via some ESD network between the power rail and pads. this is usually indicative of a part that was not chosen correctly for the job, or the correct part was wired to the wrong power bus. If I were the designer I'd put the resistor in, power down the system, get a cup of coffee, come back and image the system with a thermal imager and the problem chip will be the 'hot one' (slightly warm, if you're unlucky its hot enough to just feel with your finger). 2 watts static current is nothing for a beefy PC motherboard but it will need a special heat sunk resistor which is annoying. The bigger problem is whatever is supplying the leakage is possibly an i/o pad and eventually that thing will fry if you keep pulling power from it (but from the description this feels possibly like a buck regulator that is 'off' and just leaking current, in which case it's fine to draw that amount of power). It's one of those annoying things where it's hard to say who did what wrong. On the one hand, you could say the motherboard is out of spec by not bringing 3.3v to 0; but in reality 'we all know' that at power off things always hang out around 0.7v (a diode drop above ground) for a very long time without active countermeasures. on the other hand you can say the drive is out of spec for not going into reset when 3.3v goes out of regulation. I haven't looked at the atx spec specifically so maybe this edge case is specified, but when you are doing a fully integrated design this is just one of the things you learn to check for in integration testing.
- oneplane 2y ago
  Oof, that is a giant leak somewhere. It's really sad we have to go to some shady websites to find schematics for mainboards, otherwise we could just get to the cause of this pretty quickly.
- numpad0 2y ago
  6 Ohms! Might as well just jumper it(don't) Does it sound like reverse current through SBD? They have higher reverse current and leaky I-V curve. 3.3V of drop must mean something inline.
starslab 2y ago
> scratch that we needed smaller ;-) Well... Needed smaller in terms of resistance, but needed bigger in terms of power rating, in the interests of not catching fire.

starslab 2y ago

12 ohms brings the rail down to 1.47 volts, still no SSD. 6 ohms is enough to finally break/trip whatever circuit is allowing this situation, bringing the rail down to 0v in power-off. Of course, that's almost 2 watts of constant draw during the power-on state, so not a long-term solution.

bunnie 2y ago
Hmm the usual solution in this case is to put a FET to ground that is turned on when the standby signal goes active. The reason you need such a low impedance is that a linear resistance pulls less current as the voltage goes down so the solution is less effective the stronger the leakage path is. The amount of leakage is on the high side, so it does point to some device that is wired to the standby power that is probably 'powering' the 3.3v line via some ESD network between the power rail and pads. this is usually indicative of a part that was not chosen correctly for the job, or the correct part was wired to the wrong power bus. If I were the designer I'd put the resistor in, power down the system, get a cup of coffee, come back and image the system with a thermal imager and the problem chip will be the 'hot one' (slightly warm, if you're unlucky its hot enough to just feel with your finger). 2 watts static current is nothing for a beefy PC motherboard but it will need a special heat sunk resistor which is annoying. The bigger problem is whatever is supplying the leakage is possibly an i/o pad and eventually that thing will fry if you keep pulling power from it (but from the description this feels possibly like a buck regulator that is 'off' and just leaking current, in which case it's fine to draw that amount of power). It's one of those annoying things where it's hard to say who did what wrong. On the one hand, you could say the motherboard is out of spec by not bringing 3.3v to 0; but in reality 'we all know' that at power off things always hang out around 0.7v (a diode drop above ground) for a very long time without active countermeasures. on the other hand you can say the drive is out of spec for not going into reset when 3.3v goes out of regulation. I haven't looked at the atx spec specifically so maybe this edge case is specified, but when you are doing a fully integrated design this is just one of the things you learn to check for in integration testing.
oneplane 2y ago
Oof, that is a giant leak somewhere. It's really sad we have to go to some shady websites to find schematics for mainboards, otherwise we could just get to the cause of this pretty quickly.
numpad0 2y ago
6 Ohms! Might as well just jumper it(don't) Does it sound like reverse current through SBD? They have higher reverse current and leaky I-V curve. 3.3V of drop must mean something inline.

starslab 2y ago

> scratch that we needed smaller ;-) Well... Needed smaller in terms of resistance, but needed bigger in terms of power rating, in the interests of not catching fire.