Sizing of and Type of Power-loss / Brown-Out Protection Capacitor

Question

I found a few resources and this question here: power-loss protection

I'm using an MCP16301 step down regulator to turn 12v to 3.3v

I'm having trouble determining I in this equation: C=I×t/ΔV

with a minimum of 6v input I get this

• ΔV = 12v-6v = 6v
• t = 5seconds
• I = ?

Question 1: If my output current is 1A max, and assuming 90% efficiency, does that mean the input current is 1A/.9 = 1.11A?

Question 2: What's the correct type of capacitor for this application?

Question 3: How do you actually detect the power loss? I'm aware than I can use an interrupt to handle the event, but what's the correct circuit?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Q1 No input current for a dc/dc convertor = 1A/.9 x Vout/Vin = 0.30525A this will increase as input voltage drops to 1A/.9 x 3.3/6 =0.6105A for the next bit use an average of 0.5A this gives a bit of margin as well.

Q2 Your cap C=I*t/dV

 dV is your brownout detect point say 11V-6V=5V

 t =5s (do you really need this long)

 I= 0.5A

 C=0.5F

This is in the supercap class go to farnell or one of the other suppliers and find the cap there.

Q3 You use a comparator one input is a divided down version of the cap voltage say 1/10 = 1.2V nominally. The other input is a stable reference (zener+resistor?) set at 1/10 your brownout detect voltage=1.1V. Make sure the comparator output is compatible with your mcu in. You may be able to find a 3.3 V supply voltage comparator.

Answer 2

First of all, capacitors are a bad choice for the sort of thing you're talking about. At 1 farad (and that's a farad, not a microfarad), your equation suggests that you will lose 1 volt/second for a 1 amp current.

And yes, a 90% efficient regulator will need an input of 1/0.9, or 1.11 tines the output current.

You're getting confused about your diagram because it's not quite correct. It should look something like

^{simulate this circuit – Schematic created using CircuitLab}

And you'll note that the power loss detector will not work when the energy storage device (shown as a capacitor) is dead. That's true. But it has to be powered by something, and obviously can't be powered by the external power source.