0
\$\begingroup\$

I'm not sure how the VBAT pin of the GD32VF103 RISC-V MCU should be connected, exactly.

The GD32VF103 User Manual states:

The Backup domain is powered by the VDD or the battery power source (VBAT) selected by the internal power switch, and the VBAK pin which drives Backup Domain, supplies power for RTC unit, LXTAL oscillator, BPOR and BREG,and three pads, including PC13 to PC15. In order to ensure the content of the Backup domain registers and the RTC supply, when VDD supply is shut down, VBAT pin can be connected to an optional standby voltage supplied by a battery or by another source. The power switch is controlled by the Power Down Reset circuit in the VDD/VDDA domain. If no external battery is used in the application, it is recommended to connect VBAT pin externally to VDD pin with a 100nF external ceramic decoupling capacitor.

(Section 3.3.1 Battery backup domain, page 47)

Without an extra battery, I gather that I should put a decoupling capacitor between VBAT and GND and connect the main 3.3 V source also to VBAT.

What's unclear to me:

  1. What happens if VBAT is simply unconnected?
  2. What voltage range does VBAT actually support?

I've looked at the Longan nano 3302 schematic for some guidance, but I don't quite understand what they are doing with VBAT:

enter image description here

They are exposing the VBAT pin, but they also connect it to 3.3 V through resistor R8 of unspecified value. Why is it marked NC? Not-connected?


Sidenote: FWIW, the GigaDevices GD32VF103 RISC-V MCU arguably has some similarities (in its peripherals and how they are configured, register names etc.) to the GD32F103 ARM MCU which looks similar to the ST STM32F103. Hence, perhaps the battery backup domain works similar in all these MCUs, and knowledge can be transferred to some extend.

\$\endgroup\$
1

1 Answer 1

1
\$\begingroup\$

To get detailed information about the electrical characteristics of the processor, you need to look at the data sheet for the GD32VF103CBT6 processor. Perhaps that isn't available in the development board manual.

A Google search turns up the processor data sheet here.

It states in the "Operating Conditions" section that Vbat should be between 1.8V and 3.6V. To me, this implies that it should not be less than 1.8V for normal operation.

The data sheet doesn't say, but we could guess that the RTC, 32kHz oscillator, and backup registers would not function without a voltage supply connected to Vbat.

If you don't have a separate battery for backup operation, just connect it to Vdd and add the bypass cap like the data sheet recommends.

\$\endgroup\$
3
  • \$\begingroup\$ Ok, wasn't aware that there is also a separate datasheet manual. Figure 3-1 (page 47) in the user manual shows an internal power switch which selects between V_BAT and V_DD for the backup domain power. Thus, this reads like connecting V_DD to V_BAT (plus bypass capacitor) isn't strictly necessary for normal operation (when V_DD has power). First I wanted to follow the recommendation - but with the Longan Nano board I'm confused because (according to its schematic) V_BAT is already connected to 3.3V through some resistor R8. And I don't understand how this changes things. \$\endgroup\$ Apr 24 at 11:04
  • 1
    \$\begingroup\$ It appears to me that R8 is not populated by "default" on the board, which is what "NC" designates. Very often a design will place "optional" components so that the PCB design has pads for putting down these parts. You would stuff R8 if you don't have a battery to connect to +BAT. \$\endgroup\$
    – Ralph
    Apr 24 at 15:57
  • \$\begingroup\$ Makes sense. This R8 is then similar to the SJ1 power isolation jumper in the Sparkfun Arduino-Pro-Mini schematic. So one could add a 0 Ohm resistor (or short cable) as R8 if there is no battery. The Longan Nano schematic contains 4 isolated drawn 100 nF decoupling capacitors for 3.3 V. Perhaps one of them is even connected to the +3V3 end of R8 (since the MCU otherwise has 'just' 3 VDD pins marked with +3V3 where a bypass capacitor would make sense). \$\endgroup\$ Apr 24 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.