4 ms·

Except that: (a) you don't choose how long you get because the uC manufacturer already chose. (An external brownout warning is different, and useful, but still

Except that: (a) you don't choose how long you get because the uC manufacturer already chose. (An external brownout warning is different, and useful, but still not a good solution to the problem of FS integrity.) (b) those thresholds have nothing to do with the operating thresholds of the flash (which usually require a lot more energy to do work than the uC) (c) writing to flash is a relatively high-power operation, and if you need to erase, you're doing the highest-power operation at a time when you have no power buffer (d) warmup time to do an erase is measured in milliseconds, and you probably don't have that much power buffer The best you can do at brownout is minimize the damage, because your uC is about to switch off and/or silently inject errors.

craftinator 7y ago
But you can just... Decide how many clock cycles you'll need to finish shutdown, calculate that power usage, then solder a corresponding power source in place. There really isn't guesswork.
- ianhowson 7y ago
  Flash over SPI isn't very predictable. There's a boatload of internal state -- they silently do R/M/W, there's internal stepup converters with unknown voltage levels, temperature variation greatly impacts write time, and, depending on your software stack, you might need to do an erase. Assuming you only need to execute one Flash transaction (doubtful!), covering all of the variation might take upwards of 50ms. So, yeah, you could just throw a massive cap on the power rail. Except that that conflicts with the other power, cost and size constraints implicit in the context of microcontroller systems.
- xondono 7y ago
  In some situations this might be an unsolvable equation. If your power supply is already 1.8V, most uC will not have much room for procedures when a POR/BOR is detected.
neltnerb 7y ago
I don't personally usually bother, but an EE approach would be to put a diode between the 3.3V supply and the MCU/flash power supply pins with a large bulk capacitor, and use the voltage just before the diode as your brownout detection. It can't discharge back into the power supply or other parts on the board because of the diode so the capacitor just needs to store enough charge to run for a specified time after power loss. Then you can configure how long the MCU runs after power shuts down. Put 470uF there and I imagine you'd have your milliseconds easily.
- nine_k 7y ago
  A compact and reliable 470 uF capacitor could cost comparable to to a low-end uC.
  - londons_explore 7y ago
    The low end uC's use so little power that a 1uF capacitor would be plenty for a few milliseconds.
  - ClumsyPilot 7y ago
    I mean, you get what you pay for. If your are cost-optimising that hard, you probably aren't controlling something with critical data.
    - ianhowson 7y ago
      If you aren't cost-optimizing that hard, you aren't shipping meaningful volumes yet. 470uF isn't cheap at 10k+ units. My last project couldn't use a giant cap because of size, not cost constraints.
- neltnerb 7y ago
  Another reply mentioned that many MCUs don't like it when power is applied too slowly. Here is a version I just made as an example that uses two MOSFETs and a resistor divider to delay application of power to the MCU and flash memory. It relies on the N-channel MOSFET turn on voltage along with the resistor divider to set the turn on voltage, and when it reaches the threshold the P-channel MOSFET turns on. Note the resistor R4 provides hysteresis -- after the P-channel MOSFET activates it acts to keep the N-channel MOSFET active unless the voltage drops below some level set by R2, R3, R4, and the turn on voltage of Q1. You just need to tune them appropriate for your components, and granted of course it adds some amount a bit to your BOM cost. But if you need it you need it. So basically, R2/R3/Q1 set your turn on voltage (experimentally determined or maybe with Spice) and R4 sets your turn off voltage with R2/R3/Q1. The MCU probably uses under 500mA so these parts are pretty cheap... the capacitor is about 5 cents (https://www.digikey.com/product-detail/en/kemet/ESK477M6R3AE3AA/399-6611-ND/3083026 https://www.digikey.com/product-detail/en/kemet/ESK477M6R3AE...) and the MOSFETs that looks appropriate are in the 6 cent each range. https://i.postimg.cc/Df9N8fM5/delayed-turn-on-circuit.png https://i.postimg.cc/Df9N8fM5/delayed-turn-on-circuit.png

craftinator 7y ago

But you can just... Decide how many clock cycles you'll need to finish shutdown, calculate that power usage, then solder a corresponding power source in place. There really isn't guesswork.

ianhowson 7y ago
Flash over SPI isn't very predictable. There's a boatload of internal state -- they silently do R/M/W, there's internal stepup converters with unknown voltage levels, temperature variation greatly impacts write time, and, depending on your software stack, you might need to do an erase. Assuming you only need to execute one Flash transaction (doubtful!), covering all of the variation might take upwards of 50ms. So, yeah, you could just throw a massive cap on the power rail. Except that that conflicts with the other power, cost and size constraints implicit in the context of microcontroller systems.
xondono 7y ago
In some situations this might be an unsolvable equation. If your power supply is already 1.8V, most uC will not have much room for procedures when a POR/BOR is detected.

neltnerb 7y ago

I don't personally usually bother, but an EE approach would be to put a diode between the 3.3V supply and the MCU/flash power supply pins with a large bulk capacitor, and use the voltage just before the diode as your brownout detection. It can't discharge back into the power supply or other parts on the board because of the diode so the capacitor just needs to store enough charge to run for a specified time after power loss. Then you can configure how long the MCU runs after power shuts down. Put 470uF there and I imagine you'd have your milliseconds easily.

nine_k 7y ago
A compact and reliable 470 uF capacitor could cost comparable to to a low-end uC.
- londons_explore 7y ago
  The low end uC's use so little power that a 1uF capacitor would be plenty for a few milliseconds.
- ClumsyPilot 7y ago
  I mean, you get what you pay for. If your are cost-optimising that hard, you probably aren't controlling something with critical data.
  - ianhowson 7y ago
    If you aren't cost-optimizing that hard, you aren't shipping meaningful volumes yet. 470uF isn't cheap at 10k+ units. My last project couldn't use a giant cap because of size, not cost constraints.
neltnerb 7y ago
Another reply mentioned that many MCUs don't like it when power is applied too slowly. Here is a version I just made as an example that uses two MOSFETs and a resistor divider to delay application of power to the MCU and flash memory. It relies on the N-channel MOSFET turn on voltage along with the resistor divider to set the turn on voltage, and when it reaches the threshold the P-channel MOSFET turns on. Note the resistor R4 provides hysteresis -- after the P-channel MOSFET activates it acts to keep the N-channel MOSFET active unless the voltage drops below some level set by R2, R3, R4, and the turn on voltage of Q1. You just need to tune them appropriate for your components, and granted of course it adds some amount a bit to your BOM cost. But if you need it you need it. So basically, R2/R3/Q1 set your turn on voltage (experimentally determined or maybe with Spice) and R4 sets your turn off voltage with R2/R3/Q1. The MCU probably uses under 500mA so these parts are pretty cheap... the capacitor is about 5 cents (https://www.digikey.com/product-detail/en/kemet/ESK477M6R3AE3AA/399-6611-ND/3083026 https://www.digikey.com/product-detail/en/kemet/ESK477M6R3AE...) and the MOSFETs that looks appropriate are in the 6 cent each range. https://i.postimg.cc/Df9N8fM5/delayed-turn-on-circuit.png https://i.postimg.cc/Df9N8fM5/delayed-turn-on-circuit.png