USB-C breakout/The board in your hand/05. The power LED, and what it proves
The board in your hand · 05 of 11

The power LED, and what it proves

The indicator runs on about six tenths of a milliamp, which is a thirtieth of what an LED usually gets. That is deliberate, and it tells you one narrow thing that people routinely read too much into.

What is actually fitted

One red 0603 LED between VBUS and a 5.1 kΩ resistor down to ground. That is the whole circuit, and the resistor value is the same one used on both CC lines — the board carries three 5.1 kΩ parts, two doing the handshake and one feeding the light.

What the power LED draws
VBUS 5.00 V
VBUS5.00 V
Series resistor
5.1 kΩ
LED current
0.61 mA
Costs you
3.0 mW
It is dim on purpose. The same 5.1 kΩ value that does the handshake is used to feed the LED, which puts about six tenths of a milliamp through it at 5 V — roughly a thirtieth of what an indicator LED usually gets. On a board whose whole job is to hand the source’s current to your circuit, an indicator that ate 20 mA would be taking a real bite out of a 500 mA budget. Faint is the correct amount of light here.

At 5 V, with a red LED dropping something like 1.9 V, that leaves about 3.1 V across 5.1 kΩ — roughly 0.6 mA. An indicator LED normally gets fifteen to twenty. So it glows rather than shines, and in daylight you may have to look at it directly.

Why so faint

Because the board's entire purpose is to hand the source's current to your circuit, and a plain USB-C source may only offer 500 mA. An indicator burning 20 mA of that is four per cent of your budget spent on a light nobody is looking at. Six tenths of a milliamp is a tenth of one per cent.

What it proves, and what it does not

It proves the handshake worked. If the LED is lit, the CC resistors have done their job, the source has accepted the board and switched VBUS on. That is genuinely useful — it is the difference between a dead cable and a dead project, answered from across the room.

It does not prove you have current. The LED needs half a milliamp. It will sit there glowing contentedly while a source that only offers 500 mA is being asked for two amps by your circuit, and while VBUS sags to four volts under the load. A lit LED means voltage is present, not that the supply is coping.

The test that answers the second question is a meter on VBUS with the load running. The light cannot tell you that, and no indicator on any board can.

When it does not work

The LED is very dim compared to my other modules

It is meant to be. Its series resistor is 5.1 kΩ — the same value as the CC resistors — where an ordinary indicator uses a few hundred ohms. That puts under a milliamp through it instead of fifteen or twenty. On a board whose job is handing the source's current to your circuit, an indicator that ate 20 mA would be taking a real bite out of a 500 mA budget.

The LED is on but my circuit does not work

The LED only proves VBUS has voltage on it. It says nothing about how much current the source will give you, and it will stay lit while the supply sags under a load that is too big. If the light is on and the circuit is dead, measure VBUS with the load connected rather than trusting the LED.

The LED is off but I measure 5 V on VBUS

Check you are measuring against the board's own GND pad and not some other ground. If VBUS really is at 5 V against that pad and the LED is dark, the LED or its resistor has a bad joint — but on a board where neither was touched by you, that is rare. Far more often the meter's black lead is somewhere else.

Can I make it brighter?

You could replace the 0603 resistor with something smaller, and it is not worth it. Every milliamp you give the LED is a milliamp your project does not get, and on a 500 mA source that is the difference you will feel first. If you want a visible power light, put one on your own circuit where you control the budget.

Where this goes next

A cable connects one CC line and you never get to choose which, so both need their own.

Why two resistors

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