I'm building a USB-powered demo board with the Atmel SAM E70 controller, and a single hurdle I'm facing is the battery-powered operation. I intend to use a simple 3V coin cell battery as a backup power source, and want to preserve MCU's current date/time in a low-power mode.
I have a minor STM microcontroller background and kinda expected Atmel MCU to have similar capabilities, but E70 (and the whole line of Cortex-M7-based Atmel MCUs, it seems) lacks the VBAT input. E70 datasheet has a separate section (22.4.4 – "Using Backup Batteries/Backup Supply") on the topic, but it doesn't provide any ready answers – only a sketchy schematic and some power sequencing advice.
E70 has a sister chip – the V71, and I've found that V71's demo board from Atmel ("V71 Xplained Ultra") actually does provide a detailed schematic, which is far from being easy & compact:
It seems that Atmel expects system engineers to build a backup power system of their own; it's a bit of a letdown, as STM F4/F7 line has an embedded backup powering scheme.
Thus, 2 questions:
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Is there any easier (than building a scheme with an opamp and 2 additional ICs) way or a tried-and-tested schematic to provide a backup battery power for Atmel S70/E70/V70?
Note: I was advised to use a separate RTC IC (like STM M41T62LC6F), but don't feel comfortable with adding an additional IC to the board.
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(in case any Atmel engineer happens to visit this question) What was the reasoning behind leaving out the VBAT/low-power controller out of the MCU?
Best Answer
The idea behind the Vbat input on chips such as the STM32 is that the RTC and other low-power peripherals will normally be powered from the main Vdd, automatically switching to Vbat when Vdd is absent. On-chip, this can be implemented with something as simple as a diode-OR:
simulate this circuit – Schematic created using CircuitLab
To save pins (I assume), the SAME70 doesn't provide a separate connection for the RTC, instead having it take power from the main Vdd. However, as they show in the datasheet section you referenced, you can add the power-switching logic externally:
simulate this circuit
The schematic you posted is very complicated, because it's trying to do something slightly different: C103 on that diagram is a supercapacitor (100 mF!), so much of the circuitry is responsible for charging it at a reasonable rate. You're using a coin cell, so none of that is relevant.
There are several practical considerations with the schematic I posted. First, I recommend Schottky diodes, because you lose a lot less voltage that way. Second, I assume that you're generating a 3.3 V rail from the USB Vbus with an LDO linear regulator. If so, you'll want to take the feedback for the regulator from after the diode, so that the regulator can compensate for the diode drop. It would look something like this (ignoring bypass capacitors):
simulate this circuit
Third, this will try to power the entire chip from the coin cell when USB power isn't present. You have to be very careful to enter Backup mode as soon as you notice Vbus missing, or you'll drain the coin cell quickly. You'll want to use a GPIO pin to detect VBus (see schematic 2 in section 58.2.8).