Sorry in advance for what may be an elementary question. I'm designing a circuit to provide a few hundred milliseconds of backup power to an eMMC module in case of power loss. This is to provide the module a chance to flush any pending writes to prevent corruption. My thought was to simply lay a sizable capacitor in parallel with VCC and GND. But I'm assuming I'll need to do something to ensure that power supplied to the eMMC while the cap is discharging is cut as soon as the voltage dips below 3.3v, so the eMMC module doesn't flicker on and off during that time.

So first, is this in fact an issue I need to worry about, or does it not matter if the eMMC flickers for a short duration or is supplied too low of a voltage? Second, assuming it does matter can someone suggest an MCU or something to handle this?


2 Answers 2



\$ \frac{dv}{dt} = \frac{i}{C} \$

You'll have to measure i while it is writing.

Then you'll need to know for how long you want it to last.

The problem with capacitors as backup power is that the voltage will decrease quickly as the capacitor discharges. So if the nominal voltage is 3.3V, and the minimum allowed voltage is 3.1V, only a small fraction of the charge inside the cap can actually be used.

However, you want to protect against sudden power loss, which means your circuit is not running from batteries, and thus there isa good chance your 3.3V is derived from a higher voltage (say, 5V or more).

In this case you can place the capacitor on this higher voltage, then use a regulator to power the eMMC. For example, using 10mA current, and a 100µF cap:

di/dt = 100V/s = 0.1V/ms

Time for cap to go from 3.3V to 3.1V: 2ms

Time for cap to go from 5V to 3.3V: 17ms

In the latter case, having more voltage margin means we can use more of the charge stored inside the cap (but an extra LDO is required, and also a schottky diode to make sure the cap doesn't discharge back into the power supply). Something like that:


simulate this circuit – Schematic created using CircuitLab

U2 could also be a switching converter, but at this point the added complexity would be a bother. One could also use a coin cell battery.

  • \$\begingroup\$ Thanks for the excellent answer. Looking at the datasheet it turns out the supported range for the component is 2.7-3.3v. So assuming a worst case of 100mA the time to drop from 3.3v to 2.7v should be 600 ms with a 100µF cap. So in this case I should be in good shape even without charging at a higher voltage and using a regulator. But great info, and good point regarding the diode. \$\endgroup\$
    – stdout
    May 19, 2018 at 18:38
  • 2
    \$\begingroup\$ @stdout, I am afraid you are mistaken by a factor of 1000. Happens all the time. As peufeu said, dV/dt = i/C. Plug numbers for dV=0.5 V, i=0.1 A and C=0.0001 F. You will get 5*e-4, or 500 us. \$\endgroup\$ May 19, 2018 at 19:46
  • \$\begingroup\$ Michael: "I must've put a decimal point in the wrong place or something. ----, I always do that. I always mess up some mundane detail." Peter: "Oh! Well, this is not a mundane detail, Michael!" LOL, indeed I did. Seemed a bit too good to be true. Thanks Ali. \$\endgroup\$
    – stdout
    May 19, 2018 at 20:15

Capacitor is fine. Use CV=It. I is of all the load on the rail, V is 10% or your voltage. If you feel 10% drop is too much, just change it. Be sure current will not flow to additional circuits which are not normally powered from same rail.


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