5 ms·

How can 100W be delivered through those tiny USB-C connector landings, especially since only a few of them are used for power?

jeffbee 3y ago
The cable must report itself as being capable of 5A, which requires it to use 20-gauge wire for the VBUS. ETA: I guess your question was specifically about the connector interface, which is less of a problem because there are 4 power and 4 ground pins. 1.25A per pin is no issue.
- dist-epoch 3y ago
  20 gauge wire is 0.9 mm diameter. A USB-C contact is much smaller (seems to be around 0.5 mm). This is the part I don't understand. If you require a 0.9 mm diameter wire, how can you have it connect to a 0.5 mm landing strip?
  - deleted 3y ago
    [deleted]
  - iancmceachern 3y ago
    Resistance is the function of conductivity of the material, cross section area of the conductor, and length. In the case of wires the length is long, so they need more cross sectional area to keep the resistance down, in the contacts the length is very short, so they need less cross sectional area to achieve the same Resistance. Often, they make the contacts from an even more conductive material than copper like gold or silver, so that helps even more.
    - ilyt 3y ago
      Also there are 4 connectors for ground and 4 connectors for power in USB-C connector
  - Dylan16807 3y ago
    The electricity is going through the plug for only a few millimeters, and is going through the rest of the cable for 500-2000 millimeters. It's okay for the plug to have more ohms per meter. It's going to have a very tiny voltage drop and emit a very tiny amount of heat.
    - ilyt 3y ago
      there are 4 connectors for ground and 4 connectors for power in USB-C connector hacks like that are not required.
      - Dylan16807 3y ago
        Hacks like what? If you're saying the plugs don't have more resistance than the equivalent length of cable, that's wrong. From what I can find it's very typical for a USB C plug to have 40 milliohms of contact resistance. Even if you divide that by 4, that's 10 milliohms in a single centimeter. 20 gauge wire is 0.3 milliohms per centimeter.
  - ilyt 3y ago
    >This is the part I don't understand. If you require a 0.9 mm diameter wire, how can you have it connect to a 0.5 mm landing strip? there are 4 connectors for ground and 4 connectors for power in USB-C connector
  - sgtnoodle 3y ago
    It's often called "necking down". The connector is higher resistance, and will indeed have more resistive loss than the wire. It's a very small length, though. The extra heat is quickly dissipated into the the nearby wire through thermal conduction, so it will be only marginally warmer than the rest of the cable. This is a very common practice on PCBs. You can use a very wide trace where there's room, and narrow it to get around tight spots like between pins on an IC. When I built solar race cars in college, we optimized the powertrain for resistive losses over weight. We used 00 AWG wire from the battery to the motor, knowing it would max out at 40 amps. We would then "neck it down" to 8 AWG to go into 40A rated connectors. To a casual observer it definitely appeared ridiculous and wrong.
- formerly_proven 3y ago
  The bigger issue with this is not the steady state but what happens when the cable is janked out carrying that current. Spark erosion of the contacts etc., the spec goes into quite some detail there.
  - Nextgrid 3y ago
    Or just makes a poor connection to begin with due to out-of-spec cables or connector wear/damage, which is likely considering the much tighter tolerance than old-school, reliable, battle-tested connectors such as barrel jacks.
dheera 3y ago
It negotiates and then switches to 20V to deliver 100W, so it only needs to run 5A through the cables. Wire gauge requirements are related to current, not power; that's why high-speed trains usually operate at 25000V or 50000V to allow their megawatts of power to be delivered through a ~1cm diameter overhead line. That said, I still hate USB-C as a power standard. I would have much preferred 2 connectors just supplying 20VDC, no questions asked, with a standard connector, just like 120VAC power works. Pick another connector for the USA low-voltage standard, none of this negotiation fuss, none of this complicated power circuitry, and much harder to get cables wrong. With the complexity of USB-C there are just way too many substandard cables flooding the market, and average Joes don't understand the difference between a 100W-capable cable and a 60W-capable cable and one that is charging-only, one that is USB 3.0, one that is USB 3.2, one that is USB 3.2 + 100W charging and all of that. People just buy "5-star" rated cables online that often don't meet the standard and it's a shitshow.
- userbinator 3y ago
  There's also the added failure mode of a PSU deciding it needs to supply 20V to a device that can only withstand 5V.
  - ilyt 3y ago
    You need to have device side protection for that anyway because broken PSU can happen in many ways you can't really predict, including "someone implemented that monster of a standard wrong"
  - userbinator 3y ago
    Care to add any counterarguments?
    - booberbinator 3y ago
      Sure. Since you asked so nicely, here's a datasheet that illustrates a device which integrates overvoltage protection. Plug a malfunctioning 20V USB-C power supply into your phone, and the phone will internally disconnect the USB-C connection to protect itself. https://www.ti.com/lit/ds/symlink/tps65987d.pdf https://www.ti.com/lit/ds/symlink/tps65987d.pdf I incinerated 250 seconds of my time, or about a billionth of my life for this answer for you - for free! Hope the little dopamine hit was enough for you today. I won't be back.
NotYourLawyer 3y ago
(Relatively) high voltage, so lowish current. I think 20 volts at 5 amps or so.

jeffbee 3y ago

The cable must report itself as being capable of 5A, which requires it to use 20-gauge wire for the VBUS. ETA: I guess your question was specifically about the connector interface, which is less of a problem because there are 4 power and 4 ground pins. 1.25A per pin is no issue.

dist-epoch 3y ago
20 gauge wire is 0.9 mm diameter. A USB-C contact is much smaller (seems to be around 0.5 mm). This is the part I don't understand. If you require a 0.9 mm diameter wire, how can you have it connect to a 0.5 mm landing strip?
- deleted 3y ago
  [deleted]
- iancmceachern 3y ago
  Resistance is the function of conductivity of the material, cross section area of the conductor, and length. In the case of wires the length is long, so they need more cross sectional area to keep the resistance down, in the contacts the length is very short, so they need less cross sectional area to achieve the same Resistance. Often, they make the contacts from an even more conductive material than copper like gold or silver, so that helps even more.
  - ilyt 3y ago
    Also there are 4 connectors for ground and 4 connectors for power in USB-C connector
- Dylan16807 3y ago
  The electricity is going through the plug for only a few millimeters, and is going through the rest of the cable for 500-2000 millimeters. It's okay for the plug to have more ohms per meter. It's going to have a very tiny voltage drop and emit a very tiny amount of heat.
  - ilyt 3y ago
    there are 4 connectors for ground and 4 connectors for power in USB-C connector hacks like that are not required.
    - Dylan16807 3y ago
      Hacks like what? If you're saying the plugs don't have more resistance than the equivalent length of cable, that's wrong. From what I can find it's very typical for a USB C plug to have 40 milliohms of contact resistance. Even if you divide that by 4, that's 10 milliohms in a single centimeter. 20 gauge wire is 0.3 milliohms per centimeter.
- ilyt 3y ago
  >This is the part I don't understand. If you require a 0.9 mm diameter wire, how can you have it connect to a 0.5 mm landing strip? there are 4 connectors for ground and 4 connectors for power in USB-C connector
- sgtnoodle 3y ago
  It's often called "necking down". The connector is higher resistance, and will indeed have more resistive loss than the wire. It's a very small length, though. The extra heat is quickly dissipated into the the nearby wire through thermal conduction, so it will be only marginally warmer than the rest of the cable. This is a very common practice on PCBs. You can use a very wide trace where there's room, and narrow it to get around tight spots like between pins on an IC. When I built solar race cars in college, we optimized the powertrain for resistive losses over weight. We used 00 AWG wire from the battery to the motor, knowing it would max out at 40 amps. We would then "neck it down" to 8 AWG to go into 40A rated connectors. To a casual observer it definitely appeared ridiculous and wrong.
formerly_proven 3y ago
The bigger issue with this is not the steady state but what happens when the cable is janked out carrying that current. Spark erosion of the contacts etc., the spec goes into quite some detail there.
- Nextgrid 3y ago
  Or just makes a poor connection to begin with due to out-of-spec cables or connector wear/damage, which is likely considering the much tighter tolerance than old-school, reliable, battle-tested connectors such as barrel jacks.

dheera 3y ago

It negotiates and then switches to 20V to deliver 100W, so it only needs to run 5A through the cables. Wire gauge requirements are related to current, not power; that's why high-speed trains usually operate at 25000V or 50000V to allow their megawatts of power to be delivered through a ~1cm diameter overhead line. That said, I still hate USB-C as a power standard. I would have much preferred 2 connectors just supplying 20VDC, no questions asked, with a standard connector, just like 120VAC power works. Pick another connector for the USA low-voltage standard, none of this negotiation fuss, none of this complicated power circuitry, and much harder to get cables wrong. With the complexity of USB-C there are just way too many substandard cables flooding the market, and average Joes don't understand the difference between a 100W-capable cable and a 60W-capable cable and one that is charging-only, one that is USB 3.0, one that is USB 3.2, one that is USB 3.2 + 100W charging and all of that. People just buy "5-star" rated cables online that often don't meet the standard and it's a shitshow.

userbinator 3y ago
There's also the added failure mode of a PSU deciding it needs to supply 20V to a device that can only withstand 5V.
- ilyt 3y ago
  You need to have device side protection for that anyway because broken PSU can happen in many ways you can't really predict, including "someone implemented that monster of a standard wrong"
- userbinator 3y ago
  Care to add any counterarguments?
  - booberbinator 3y ago
    Sure. Since you asked so nicely, here's a datasheet that illustrates a device which integrates overvoltage protection. Plug a malfunctioning 20V USB-C power supply into your phone, and the phone will internally disconnect the USB-C connection to protect itself. https://www.ti.com/lit/ds/symlink/tps65987d.pdf https://www.ti.com/lit/ds/symlink/tps65987d.pdf I incinerated 250 seconds of my time, or about a billionth of my life for this answer for you - for free! Hope the little dopamine hit was enough for you today. I won't be back.

NotYourLawyer 3y ago

(Relatively) high voltage, so lowish current. I think 20 volts at 5 amps or so.