3 ms·

That only solves the problem if you're protecting against backward batteries, not the "two sources" problem

simcop2387 8y ago
Two sources will look surprisingly the same, this still will require both sides to have the protection but once they do it's exactly the same. The one with the lower voltage will end up looking like it's got a small negative voltage being put on it, shutting off the mosfet. You'd still want over current protection, just like any power source, for devices that don't have protection or a broken cable that's shorting, etc.
- hatsunearu 8y ago
  No. Try simulating it. It won't work. I mean, if you did exactly what that circuit shows, you're not gonna go anywhere, so you need some modifications (like changing the zener diode to a schottky, etc). The reason why that thing works is the fact that the body diode puts the input voltage onto the source voltage (sans Vf) which establishes a nice and high Vgs (negative, but you get the point). Let's assume you have a similar circuit on both sides. The bus voltage is 3.3V, because if it's not, then you don't have a bus. Somehow you need the gate voltage on one to be nearly equal to the bus voltage on one side, but on the other, near to ground. How are you going to do that? That circuit (and any trivial variation thereof) is not going to cut it. Like, any sort of diode that connects the bus voltage to the gate on either side isn't going to work asymmetrically that pulls one low and one high, so it's not going to work. You need active monitoring of some sort. I've been seeing a lot of hilariously bad comments regarding electronics that seem to think any of this shit is easy--it's not.
  - foodevl 8y ago
    It's actually far easier than you're making it out to be, because you seem to be assuming an ideal voltage source. Most linear regulators can only source current, not sink it. So it is perfectly fine to simply put a LDO at both ends and tie them together. The bus voltage will be the higher of the two. No extra diodes needed, certainly no active monitoring, and no extra current will follow even if the regulators are putting out slightly different voltages. Maybe add a small series resistance to limit initial inrush current as capacitors balance, but that's it.
    - simcop2387 8y ago
      That's a really interesting and novel way to address it. It does leave a potential issue though, it could lead to double the current being available to a device in the middle which could mean it could overdraw for it's own specs and melt or start a fire. Unlikely though. I'd also say that with an LDO, the LDO is providing the active monitoring since it'll need to monitor the output anyway in order to regulate properly. You'd definitely need a resistor following it, not as much to limit inrush or anything, but to prevent the two regulators from causing oscillations (one turns off because the voltage looks good, which causes the other to cycle too).

Two sources will look surprisingly the same, this still will require both sides to have the protection but once they do it's exactly the same. The one with the lower voltage will end up looking like it's got a small negative voltage being put on it, shutting off the mosfet. You'd still want over current protection, just like any power source, for devices that don't have protection or a broken cable that's shorting, etc.

hatsunearu 8y ago
No. Try simulating it. It won't work. I mean, if you did exactly what that circuit shows, you're not gonna go anywhere, so you need some modifications (like changing the zener diode to a schottky, etc). The reason why that thing works is the fact that the body diode puts the input voltage onto the source voltage (sans Vf) which establishes a nice and high Vgs (negative, but you get the point). Let's assume you have a similar circuit on both sides. The bus voltage is 3.3V, because if it's not, then you don't have a bus. Somehow you need the gate voltage on one to be nearly equal to the bus voltage on one side, but on the other, near to ground. How are you going to do that? That circuit (and any trivial variation thereof) is not going to cut it. Like, any sort of diode that connects the bus voltage to the gate on either side isn't going to work asymmetrically that pulls one low and one high, so it's not going to work. You need active monitoring of some sort. I've been seeing a lot of hilariously bad comments regarding electronics that seem to think any of this shit is easy--it's not.
- foodevl 8y ago
  It's actually far easier than you're making it out to be, because you seem to be assuming an ideal voltage source. Most linear regulators can only source current, not sink it. So it is perfectly fine to simply put a LDO at both ends and tie them together. The bus voltage will be the higher of the two. No extra diodes needed, certainly no active monitoring, and no extra current will follow even if the regulators are putting out slightly different voltages. Maybe add a small series resistance to limit initial inrush current as capacitors balance, but that's it.
  - simcop2387 8y ago
    That's a really interesting and novel way to address it. It does leave a potential issue though, it could lead to double the current being available to a device in the middle which could mean it could overdraw for it's own specs and melt or start a fire. Unlikely though. I'd also say that with an LDO, the LDO is providing the active monitoring since it'll need to monitor the output anyway in order to regulate properly. You'd definitely need a resistor following it, not as much to limit inrush or anything, but to prevent the two regulators from causing oscillations (one turns off because the voltage looks good, which causes the other to cycle too).