4 ms·

Seems like this problem could have easily been solved by putting a diode on the device side.

by bdonlan 8y ago

gsich 8y ago
This is the correct solution.
codedokode 8y ago
There is a voltage drop on a diode, around 0.2 - 0.7 volts, cannot this become a problem?
- Faaak 8y ago
  Then the supply before the diode supplies 3.3V + the voltage drop; for example 3.6V.
  - gizmo686 8y ago
    So now you need a 3.6V power regulator, in additition to the 3.3V one you likely already have for all the components built for 3.3V.
    - 0110011 8y ago
      Or another diode to put in-line with the DP circuitry. It's probably not ideal from a low-power perspective but if that's what you gotta do to make robust and compliant hardware, you do it.
    - Faaak 8y ago
      A Schottky will drop .3V; That's within the range (±10%) specified in the specs
      - kevincox 8y ago
        Yes. But now you have given up all of your tolerance so you better have a perfect (or over-voltage) power supply. This means that you now have less choice in power supply.
        Faaak 8y ago
        As a parent commenter said, you could also put a diode for your circuitry. You would then have a single 3.7V supply.
        snovv_crash 8y ago
        And 10% higher power draw just for this one feature. No, there are better ways to solve this.
      - deleted 8y ago
        [deleted]
simias 8y ago
Diodes incur a voltage drop which may not be desirable, especially if you already have a 3.3V power source that you can use. I would expect high-end equipment to have a dedicated power supply path for the external connector which would avoid these issues but of course cheaper (or size-constrained) circuits might cut corners and just connect whatever 3.3V power source they have lying around.
simcop2387 8y ago
You can likely get around the diode drop that this would impart by clever use of a P-channel mosfet. https://hackaday.com/2011/12/06/reverse-voltage-protection-with-a-p-fet/ https://hackaday.com/2011/12/06/reverse-voltage-protection-w... This ends up using the body-diode that's present in all mosfets (it exists because you can't manufacture one without it) to get the initial protection, that then lets you turn on the mosfet fully, getting a much smaller voltage drop. You'd still need to be a bit cleverer to handle both sides but it shouldn't be too hard.
- hatsunearu 8y ago
  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.
hatsunearu 8y ago
Or, you know, make sure cables are ratifying standards, or if you're a GPU manufacturer, make sure the 3V3 supply is tolerant to being externally driven and doesn't backpower anything in the GPU (not exactly hard).
- ThisIs_MyName 8y ago
  > if you're a GPU manufacturer, make sure the 3V3 supply is tolerant to being externally driven and doesn't backpower anything in the GPU That's exactly what a "diode on the device side" does.
  - hatsunearu 8y ago
    Which adds loss. There are better ways to solve this than the "Undergrad Freshman" approach.
etaioinshrdlu 8y ago
Or a nice chip that does something like this one... http://www.ti.com/lit/ds/symlink/tps2032.pdf http://www.ti.com/lit/ds/symlink/tps2032.pdf N Channel (more efficient than P channel) with integrated charge pump for the gate driver.

Seems like this problem could have easily been solved by putting a diode on the device side.

gsich 8y ago

This is the correct solution.

codedokode 8y ago

There is a voltage drop on a diode, around 0.2 - 0.7 volts, cannot this become a problem?

Faaak 8y ago
Then the supply before the diode supplies 3.3V + the voltage drop; for example 3.6V.
- gizmo686 8y ago
  So now you need a 3.6V power regulator, in additition to the 3.3V one you likely already have for all the components built for 3.3V.
  - 0110011 8y ago
    Or another diode to put in-line with the DP circuitry. It's probably not ideal from a low-power perspective but if that's what you gotta do to make robust and compliant hardware, you do it.
  - Faaak 8y ago
    A Schottky will drop .3V; That's within the range (±10%) specified in the specs
    - kevincox 8y ago
      Yes. But now you have given up all of your tolerance so you better have a perfect (or over-voltage) power supply. This means that you now have less choice in power supply.
      - Faaak 8y ago
        As a parent commenter said, you could also put a diode for your circuitry. You would then have a single 3.7V supply.
        snovv_crash 8y ago
        And 10% higher power draw just for this one feature. No, there are better ways to solve this.
    - deleted 8y ago
      [deleted]

simias 8y ago

Diodes incur a voltage drop which may not be desirable, especially if you already have a 3.3V power source that you can use. I would expect high-end equipment to have a dedicated power supply path for the external connector which would avoid these issues but of course cheaper (or size-constrained) circuits might cut corners and just connect whatever 3.3V power source they have lying around.

simcop2387 8y ago

You can likely get around the diode drop that this would impart by clever use of a P-channel mosfet. https://hackaday.com/2011/12/06/reverse-voltage-protection-with-a-p-fet/ https://hackaday.com/2011/12/06/reverse-voltage-protection-w... This ends up using the body-diode that's present in all mosfets (it exists because you can't manufacture one without it) to get the initial protection, that then lets you turn on the mosfet fully, getting a much smaller voltage drop. You'd still need to be a bit cleverer to handle both sides but it shouldn't be too hard.

hatsunearu 8y ago
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.

hatsunearu 8y ago

Or, you know, make sure cables are ratifying standards, or if you're a GPU manufacturer, make sure the 3V3 supply is tolerant to being externally driven and doesn't backpower anything in the GPU (not exactly hard).

ThisIs_MyName 8y ago
> if you're a GPU manufacturer, make sure the 3V3 supply is tolerant to being externally driven and doesn't backpower anything in the GPU That's exactly what a "diode on the device side" does.
- hatsunearu 8y ago
  Which adds loss. There are better ways to solve this than the "Undergrad Freshman" approach.

etaioinshrdlu 8y ago

Or a nice chip that does something like this one... http://www.ti.com/lit/ds/symlink/tps2032.pdf http://www.ti.com/lit/ds/symlink/tps2032.pdf N Channel (more efficient than P channel) with integrated charge pump for the gate driver.