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Comment by imtringued | Hacker News Reader

Guides like this explain why there are so many broken USB-C devices. The guide mentions that you do not need a PD chip for 5Vs, but then tells you that USB C is a cold connector meaning 0V is on VBUS when nothing is connected and jumps straight into the complexities of the PD protocol running over the CC pins instead of explaining how to get the 5V without the PD chip first.

Then in the section where it tells you how to do that, it fails to properly explain how to connect a load switch (10 cent component at 100 units) to get around the 10uF limit. The vast majority of applications will require less than 15 W and a good chunk of them can't get away with 10uF between VBUS and GND so a schematic how to do it in the lowest cost way would have helped here.

Edit: After reading until the very end I got the impression that this is just an ad for Texas Instruments PD controllers.

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Basic 5V devices just need a 5.1k resistor on each CC pin to GND. It is pretty obvious when reading the actual USB-C Connector Specification.

My phone's USB-C port is worn out, and so now it's a brick if I don't have a wireless charging dock around.

Is it physically impossible to get bandwidth and power out of something as durable as a Magsafe connector, even a larger-scale version?

I generally stick magnetic USB C connectors into things like keyboards, phones, and iPads.

The slight bulge is worth the reduction in insertions. Also is very convenient for snapping when I need to connect.

How do you "wear out" a usbc port? Is this phone like 15 years old? My imagination is boggled.

My old Sony xz2c's USB C port wore out after about three years use. About the same length of time they bothered to update Android. :/

It's a narrow connector and sideways force acts like a knife on the shell. blowing out the sides. and then the connector is unreliable. It is better than microusb but worse than usb-a.

So it depends on how you baby it, but a few good accidental tugs can ruin a phone.

I like to run my game controllers wired and ruined one(xbox-one controller) this way when I got a bit agitated during a game, it's replacement(8-bitdo wired) is much better designed where the plug housing goes in a snug inset in the controller. Providing strain relief. I dread having to find a new plug that fits if it goes bad. but a narrow plug should work, granted without the strain relief.

> It's a narrow connector and sideways force acts like a knife on the shell. blowing out the sides.

But the port is contained with any the body of the device and the cord is just dangling so wouldn't that kinda thing break the cord first?

I mean, everything breaks with enough use/bad luck/whatever, I've just never seen a usbc port break like that.

The other day I was using a magnetic usbc connector to a headphone and it broke by ripping the part of the connect that inserts into the port away from the body of the plug, but the port itself was fine.

A year of plugging it in once or twice a day to recharge it, and it blew.

Old rectangular USB-A was only rated for 1500 ideal insertions. USB-C is a big improvement at 10,000 ideal insertions. Most insertions are not ideal, and it is relatively common to have some leveraged pressure trying to wedge the port apart; When my phone bridges the gap between the arm of the sofa and the coffee table while plugged in, for example, it's got a bending moment. USB-C is much thinner in the up-down dimension than USB-A, and it is not designed to safely disengage when side-loaded like Magsafe. If you wanted to break it for some reason, I'm pretty sure you could do it in a few seconds while resting your elbows on the table, on grip strength alone.

Ok I'm not going to try breaking it but my intuition says it would break the cable first, not the port?

I wonder if you could bond one of those magnetic USB-C cables onto your port.

like:

https://www.amazon.com/dp/B09P9PNFS6

That’s quite unusual as the connection was specifically designed to sacrifice the cable rather than the port (unfortunately mini-A was the opposite which, when coupled with a small insertion lifetime, was pretty terrible).

I’m sure you’ve tried changing cables so you’ve had bad luck. I’m sorry for you, but you’re unlikely to have this happen again with another device

USB-C uses resistors to signal the power level for 5V. There is 5.1k for USB legacy, and I forget for 7.5W and 15W. Lots of cheap devices leave out the 5.1k resistors and depend on the same resistors in the USB-C to USB-A cable.

It is perfectly fine for USB-C device to signal USB Legacy since every charger needs to support it for microUSB to USB-C cables. The other way is for device to dynamically negotiate power level with the charger, but this is different than USB PD for higher voltage.

The device only signals that it's a power sink with its 5.1k resistors, there's only one way to put them there without implementing the full-blown PD protocol. It can't signal power levels (USB legacy / 7.5W / 15W) and doesn't need to, it's the power source that signals its capability.

You're probably confusing it with the spec requiring that the device does not attempt to consume more power than USB legacy from a source that doesn't advertise such capability - but even then there's no "negotiation" happening, it's just a matter of measuring the voltage on CC lines.

Now I want to know, how to add larger than 10uF cap?

Why? 10uF is already pretty beastly, and the point is to dampen signal verses intermittent drops and drains, not power the backing device for any amount of time.

You need a soft-start circuit between the USB connector and your big cap. When you first connect the USB device, the cap has no charge (0V) so it appears as a short to ground to the host 5V supply. The result is a large current spike which can draw down the host supply below some functional threshold or even damage it. We could oversize the host supply to deal with a 25 A transient and add cost and introduce safety concerns or push the solution to the connecting device where the problem exists. The solution is to ramp the current up "slowly" over several ms using a transistor on the device side. While the transistor is turned on slowly, it isn't fully conducting, acting as a variable resistor. The transistor is dissipating power during this transition so it must be sized correctly and timing controlled deliberately. Hence why we want the same design to contain the cap, transistor, and ramp control. A typical circuit might be: connector->small cap->soft start->big cap->main circuit. The soft start feature is often just built into a voltage regulator since there's already a transistor and control circuitry there but you can also get a dedicated load switch if you only want ~5V. The 10uF requirement is a contractual agreement between host and device that ensures compatibility between devices. If you don't follow it, a sophisticated host may simply cut power when it sees the current spike and that would be the device's problem.

Realistically, most USB devices hiccup when they exceed their current limit, especially in 5V mode. And since caps don’t loose charge that quickly typically, you can usually charge the cap up enough to deal with this transient no matter how big your cap is (unless it’s something absolutely enormous… which wouldn’t be implemented for cost reasons)

I have seen this play out with host devices with robust current limiting outputs. Sometimes it works for the connected device and sometimes the device sabotages itself when its own regulator turns on at say 3.6 V, drawing down the partially charged input cap below the regulator's UVLO and turning itself off again. Then the cap starts charging again and the cycle starts over. Depending on the host implementation, the burp mode may not reset without a full disconnection, at best leaving the connected device to sip a low average voltage and low current or persist in the above loop forever. Making sure this doesn't happen is more difficult than just implementing inrush limiting.

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