cancel
Showing results for 
Search instead for 
Did you mean: 

Is this decoupling configuration for the STM32F407ZET6 correct

Noah_Stoessel
Associate II

Hello im creating one of my first stm32 boards and need some help ive watched many tutorials about decoupling capacitors and have referenced what i could from the datasheet could anyone confirm if this decoupling setup would work or if modifications are needed:grinning_face:. I am using a usb-c connector for programming and power with an ams1117 3.3 volt regulator with the appropriate solid tantalum capacitors. For a little more info these are all smd 0603 capacitors and the component marked fb1 is a ferrite bead with a resitance of 120ohms at 100mhz.

 

New Screen Shot Name 2023-07-25 à 01.31.19.png

1 ACCEPTED SOLUTION

Accepted Solutions

> Wouldnt wiring the d+ and minus pins to the chip then using a 5,1k pulldown resistor to ground for the cc pins work?

Debugging requires a standalone debugger. They're cheap. I would recommend buying one, implementing the header on your board, and using that for debugging and programming. Here is one:

https://www.st.com/en/development-tools/stlink-v3minie.html

ST dev boards generally have an ST-LINK debugger built onto the dev board itself. It's a separate STM32 chip that sits between the target MCU and the USB.

If you feel a post has answered your question, please click "Accept as Solution".

View solution in original post

17 REPLIES 17
TDK
Guru

Should work just fine. The ferrite bead isn't necessary if you aren't very concerned with analog performance.

> I am using a usb-c connector for programming

Through an STLINK-V3MINIE? Note that connecting USB directly to the chip itself won't allow for debugging.

 

For general reference, in case you haven't seen it:

https://www.st.com/resource/en/application_note/an4488-getting-started-with-stm32f4xxxx-mcu-hardware-development-stmicroelectronics.pdf

If you feel a post has answered your question, please click "Accept as Solution".

Wouldnt wiring the d+ and minus pins to the chip then using a 5,1k pulldown resistor to ground for the cc pins work? then leaving the sbus1 and 2 pins unconnected should work? or not because i was looking of using usb c for programming without and extra board if there any ics that would work if my solution doesnt work and you knew of them could you give a link or a reference. thank you for your help anyways.

> Wouldnt wiring the d+ and minus pins to the chip then using a 5,1k pulldown resistor to ground for the cc pins work?

Debugging requires a standalone debugger. They're cheap. I would recommend buying one, implementing the header on your board, and using that for debugging and programming. Here is one:

https://www.st.com/en/development-tools/stlink-v3minie.html

ST dev boards generally have an ST-LINK debugger built onto the dev board itself. It's a separate STM32 chip that sits between the target MCU and the USB.

If you feel a post has answered your question, please click "Accept as Solution".
MSing.1713
Associate II

Unless my eyes are playing tricks, it looks like you have 12 VDD pins and 11 0.1u bypass caps.  Probably best to make sure there is a cap on each VDD pin.

...and FWIW, tantalum is considered a 'conflict' material.  I don't think I've see tantalum caps in a design for years.  I cannot speak to the necessity of using tantalum, though.  I can say that the LDOs I've used have generally stated that they can provide stable regulation with ceramics.  Often, though, I'm only looking for a few 100mAs. Once we get to the range of 1A, generally we place a buck regulator (e.g. ACOT) for better efficiency and these, too, operate fine with ceramic caps.

LCE
Principal

Tantalum - wasn't that something last seen in the 90s? :D

Depending on what else is powered from these 3.3V, a switching regulator might be more appropriate than the LDO. Check power dissipation.

And I would also recommend putting the SW DEBUG pins on a header.

And take care of the boot pins, at least that you can access them in case you need to.

I think folks still use tantalum because datasheets, like this one, recommend it.  I was curious about the part since I hadn't seen it placed.  It's quite inexpensive.  It also expects a large output capacitor to stabilize the output.  Modern LDOs operate fine with 1u.  This device recommends 22u.

At 1A, you're looking at (5-3.3)*1W dissipation, so roughly 2W.  Shrug.  It'll get warm.  The DS makes it look like it can handle this configuration.  If it is mounted over a large copper, it should survive.

I guess he will not need 1A, because for 1.7W you need a lot of thick copper or some other cooling.

Anyway, even for space reasons I would check what the max. current requirement is, then decide if a switching regulator makes sense - or even if a SOT23-5 LDO might be enough.
Interestingly, there are almost pin-compatible LDO / buck regulators in SOT23-5/6 out there, so for some designs I basically put the buck circuit on the board, but if an LDO was good enough, I just bridged the inductor:

V3V3_LDO-buck.png

Yep i saw it too just forgot to place one i guess:grinning_face: