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Associate III
February 25, 2026
Solved

STM32F723 USART TX and RTS drive capability

  • February 25, 2026
  • 6 replies
  • 347 views

 

Couldn't see a definitive answer in the datasheet - What is the source \ sink capability of a USART TX \ RTS pin?

The nearest answer I could find was page 128 of DS13313 Rev 5 -

"GPIOs (general purpose input/outputs) can sink or source up to ±8 mA, and sink or source up to ±20 mA"

but there isn't anything specifically that mentions alternate function pin capability.

Can TX and RTS pins directly source or sink an LED via a 220 Ohm resistor (that would be ~15mA each) ?

I'm thinking I may have to add external drives like high side MOSFETs otherwise.

Thanks.

Best answer by Bags

Excellent.

Was running out of PCB space so thats good news. 6 LEDs makes 180mA, so just got under that too :o)

6 replies

TDK
Super User
February 25, 2026

AF functionality uses the same GPIO circuitry to source/sink current. The drive strength is the same as what you've found in the datasheet.

> Can TX and RTS pins directly source or sink an LED via a 220 Ohm resistor (that would be ~15mA each) ?

Yes, provided the sum of all sources/sink remains within limits. Something like 200 mA.

"If you feel a post has answered your question, please click ""Accept as Solution""."
BagsAuthorBest answer
Associate III
February 26, 2026

Excellent.

Was running out of PCB space so thats good news. 6 LEDs makes 180mA, so just got under that too :o)

Andrew Neil
Super User
February 26, 2026

Basically the same question (and answer) here: STM32L Series LPUART Voltage Level.

As @TDK said, it's the GPIO specifications - not specific to the AF in use.

A complex system that works is invariably found to have evolved from a simple system that worked.A complex system designed from scratch never works and cannot be patched up to make it work.
TDK
Super User
February 26, 2026

Why do you keep marking your own replies as the solution?

"If you feel a post has answered your question, please click ""Accept as Solution""."
BagsAuthor
Associate III
February 26, 2026

Why does your platform allow marking of own replies to solutions?

If its a platform bug you should raise an engineering change request to fix it.

TDK
Super User
February 26, 2026

It's not a bug. If your own answer was the solution, you should mark it as such. But that's not what is happening.

 

"If you feel a post has answered your question, please click ""Accept as Solution""."
Chris21
Associate II
February 26, 2026

"6 LEDs makes 180mA, so just got under that too"

Did you though?  Limit for F723 looks to be 120 mA.

Also note: "This current consumption must be correctly distributed over all I/Os and control pins. The total output current must not be sunk/sourced between two consecutive power supply pins referring to high pin count LQFP packages."

Chris21
Associate II
February 26, 2026

Although if LEDs are 15 mA each 6 * 15 = 90 mA...

 

BagsAuthor
Associate III
February 28, 2026

Sorry, 180mA was my estimated max for the entire PCB with a little bit of added margin, but I can crank down the 6 LEDs a bit (they are Optos) before edges become an issue.

I forgot to check the total current sunk for the alternative H723 MCU - luckily its 20mA higher than a F723 and not less.

I'll breadboard an Opto to a UART TX\RTS to verify that area before I finalize the PCB.

Visitor II
February 28, 2026

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gbm
Principal
February 28, 2026

Do you realize that any modern LED supplied with 15 mA current would make you room very bright? For modern LEDs used as status lights, the typical reasonable current is 0.5..3 mA. In most of my boards I use 1k5 or 2k2 series resistors for driving LEDs.

My STM32 stuff on github - compact USB device stack and more: https://github.com/gbm-ii/gbmUSBdevice
BagsAuthor
Associate III
March 2, 2026

Optos don't light up rooms unless you made a serious wiring mistake.

15mA is the upper limit but I probably can crank it down to 10mA without impacting max baud rate...

Opto Propagation Delays.png