Hacker News Reader

Comment by bdonlan | Hacker News Reader

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

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.

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-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.

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

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.

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.

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.

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).

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).

> 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.

Which adds loss. There are better ways to solve this than the "Undergrad Freshman" approach.

Or a nice chip that does something like this one... http://www.ti.com/lit/ds/symlink/tps2032.pdf

N Channel (more efficient than P channel) with integrated charge pump for the gate driver.

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

Then the supply before the diode supplies 3.3V + the voltage drop; for example 3.6V.

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.

A Schottky will drop .3V; That's within the range (±10%) specified in the specs

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.

As a parent commenter said, you could also put a diode for your circuitry. You would then have a single 3.7V supply.

And 10% higher power draw just for this one feature. No, there are better ways to solve this.

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.

This is the correct solution.

Reply on news.ycombinator.com