0
\$\begingroup\$

I was about to put a battery charging circuit into my schematic, when I read up on my MCUs VBAT pin. Quoting from the Datasheet:

3.9.6 Battery Operation: "An internal VBAT battery charging circuit is embedded and can be activated when VDD is present"

How can a miniature MCU handle battery charging? Have I missunderstood what they're trying to say?
Also, there is a table (page 26) of which peripherals are available during VBAT operation, and almost all of them are unavailable, does this mean I should implement my own charging circuit and just disregard the VBAT pin altogether (connect it to VDD)?

There is just one feature that is worth connecting the battery to VBAT pin, but I don't know how it will work when powered from VDD...:

3.15.3 VBAT battery voltage monitoring*" This embedded hardware feature allows the application to measure the VBAT battery voltage"*

\$\endgroup\$
  • \$\begingroup\$ Also, by "implementing my own charging circuit" I mean using this one :P Schematic \$\endgroup\$ – Areuz May 1 '17 at 17:26
1
\$\begingroup\$

Page 157 gives a clear hint: Charging resistor 5k or 1.5k.

VBat in a microcontroller isn't meant to let you operate the entire thing from a 50Ah Li-Ion. It's a back-up battery pin, that uses extremely little power when the main power goes away, to keep a clock or calendar running and possibly keep some or all of the RAM to its value.

It can be connected to a non-rechargeable Lithium coin cell, or to a small rechargeable battery, or to a large value capacitor.

If you connect a high-value capacitor to Vbat, such as a 1F one, that can keep the memory and/or functions in tact a while on the few-dozen-μA it needs to stay "prepared". The internal resistor can then safely recharge that capacitor once power comes back. Many of these large capacitors don't fare very well for very long when they charge with high currents, so ST just decided to include the power-path and protection resistor in the chip for your convenience.

If your entire device works on a battery, obviously it'll be big-ish, and charging it with 1.5k would take weeks if not more. Not to mention that batteries of Lithium types and several other chemistries don't do very well with a fixed resistor from a fixed voltage as a "charge system". Of course with 1.5k or 5k, if you had the patience, it might work well enough, since that's hardly any current.

Note:
Do be aware that (I didn't read more than page 157) many controllers need the Vbat pin to be powered, so if you don't use a back-up battery, check the datasheet if you then need to connect the pin to Vdd, or what else they advise. (Maybe just a 1μF cap and turn on the 1.5k?)

\$\endgroup\$
  • \$\begingroup\$ Okay, that makes sense, I'll use a external charging circuit then. \$\endgroup\$ – Areuz May 1 '17 at 17:37
  • \$\begingroup\$ Oh, and about your note, indeed the datasheet says to connect VBAT to VDD if it is not in use. I also noticed, that the maximum rating for VBAT is either 3.6 or 4V, so I can't use it to monitor the Li-Po cell I'll be using either. \$\endgroup\$ – Areuz May 1 '17 at 17:38
  • \$\begingroup\$ @Areuz Nope, That monitor is only meant to keep an eye on CR-type Litium prime or old fashion NiMH or NiCd backup packs. Or possibly LiFePO4. \$\endgroup\$ – Asmyldof May 1 '17 at 19:37

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.