Why two resistors
A USB-C cable connects exactly one of the two CC lines, and which one depends on which way up the plug went in. Fit a resistor to only one and the board works half the time, which is worse than not working at all.
The plug goes in either way up
That reversibility is the best thing about USB-C, and it is bought by putting two of most contacts in the connector — one set for each orientation. The socket bridges most of those pairs on the board, so D+ is D+ whichever way the plug went in.
The CC pins are the exception. They are not bridged, because they are what tells the source which way round the cable is. A USB-C cable carries a single CC wire, and it lands on CC1 or CC2 depending on the orientation.
So both need a resistor
If only CC1 carried a 5.1 kΩ to ground, the board would work when the plug went in one way and be completely dead the other. Half the time, with no pattern the user can see, and both states looking identical from the outside.
That is why there are two. One does the handshake; the other sits with its resistor doing nothing at all, waiting for you to turn the plug over and swap their jobs.
Why 5.1 kΩ exactly
The source has its own resistor pulling CC up towards 5 V. Yours pulls down. Together they make a divider, and the source decides whether a device is present by looking at where the voltage lands.
For that to work, everybody has to agree on the pull-down. The specification names 5.1 kΩ, and a source checks the voltage falls in the window that value produces. Something close but wrong — 4.7 kΩ, or two 10 kΩ in parallel — lands near the edge of that window. Lenient chargers accept it; strict ones do not.
Both resistors on this board are 5.1 kΩ, fitted at the factory, and that is the part of the product you are actually paying for.
What they cannot do
These two resistors say one thing: a device is here. They cannot ask for 9 V or 12 V, and they cannot report back what the charger said. A source that sees them switches on 5 V and nothing else.
Everything past that — the higher voltages, the negotiation, the fast charging — needs a chip that talks the Power Delivery protocol. A passive resistor cannot hold a conversation.
When it does not work
Correct, every time. The cable wires through one CC line only; the other is left open at the plug end, so its 5.1 kΩ has nothing to divide against and sits at ground. Turn the plug over and the two readings swap. A board where both read the same voltage is the odd one, not this.
Because the source is looking for a value inside a window, and 5.1 kΩ is the number the USB-C specification names. A 4.7 kΩ will usually work with a lenient charger and fail on a strict one, which is the worst kind of substitution — it passes on your bench and comes back as a support email. The resistors are already fitted here, so the question only arises if you are building your own.
Do not. Joining them puts the two 5.1 kΩ resistors in parallel, which is 2.55 kΩ, and that is outside the window a source is looking for. Worse, on a cable that carries VCONN the source may be driving one of those pins to power an active cable, and you have just shorted that to its partner.
The board does not, and does not care — that is the point of having two identical resistors. Something reading the CC pins can work it out, which is exactly what the sketch in this book does, and is the only reason the pins are brought out at all.
The same divider, read the other way: what the charger's own resistor is telling you about its limits.
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USB-C Breakout Board, 12-Pack with 5.1 kΩ CC Resistors
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