Electronic – Design considerations for low-current SMPS application

In addtion to the concerns brought up in the comments (incorrect P-FET polarity, no catch diode/MOSFET), I have some at-a-quick-glance concerns:

• The microcontroller won't be able to drive the gate of Q1 very hard (usually GPIO pins can only source a few milliamps) so your turn-on and turn-off will be very slow. This will limit how well your high-side switch will behave.

• You don't have a gate-to-source resistor on Q1, so you're solely dependent on the GPIO keeping the MOSFET on or off. If the GPIO pin goes high-impedance, the MOSFET may turn itself on if the gate picks up a charge from the environment.

• If your 70R P-channel gate resistor is solidly on (if Q1 is saturated), it's going to burn

  \$ D \cdot \dfrac{(16V)^2}{70 \Omega} = D \cdot 3.65W\$

  which is crazy high power since D is going to be high (input is close to output). Also, the 225mA or so that will flow will also be burned in Q1, which isn't healthy since it's a relatively small device.

  (You need \$V_{GS}\$ of around 4V to draw ~400mA through Q1, and you need \$V_{GS}\$ of -7.5V for 40A in Q4).

  • Your purely resistive feedback network is a bad idea. You really need some compensation and/or filtering. Your comparator will be hyper-fast and could react to switching noise, pickup, ripple, etc. - since you don't seem to be using an error amplifier with compensation to control the gain and phase, you're going to need some cap across R5 (and some luck).

  • You don't have any current monitoring or over-current protection in your power train.

  • You don't have any over-voltage protection in your power train.

  • You don't have any over-temperature protection in your power train.

  • You don't have input reverse-polarity protection and an input fuse in your power train. Big no-no, especially when the source is battery-based (big short-circuit sourcing capability).

This is a simpler project if you use an off-the-shelf analog synchronous buck controller. I don't understand why you would want to use the ATtiny for this.

That being said, this isn't a simple project by any stretch. Your schematic is largely incomplete and lacks basic safety protection that any power supply (especially ones that run at high power levels like yours) will need.

Think about your requirements, calculate all the losses, design in some protections and come back with rev. 2.

You have wrong calculations: In the first case, the usable energy is 10.88J; In the second case - 12.18J. (The full energy of the capacitor charged to 5V is 12.5J) $$ E_{start} = U_0^2.C/2 $$ $$ E_{end} = U_1^2.C/2 $$ $$ E_{usable} = E_{start} - E_{end} = (U_0^2 - U_1^2).C/2 $$

So, the difference is not so big. One more example, if you use only buck regulator discharging the capacitor to 2.5V the energy will be 9.375J.

So, the efficiency can be very very important in this case. If you have one regulator with 85% efficiency, the final gain is 10.88J*0.85 = 9.248J

But if you have two stages by 85% you will have 12.18J*0.85*0.85 = 8.8J

So, to resume: In order to use two stage regulator you need every stage to have efficiency higher than 90%.

Even using one stage, buck regulator and discharging to 2.5V can be a good solution if you have regulator with high enough efficiency.

Notice for low power: As a rule, the efficiency of the regulators degrades on low powers. The general purpose regulators are usually not optimized for extremely low output currents. (Although 1mA is still not so small current, so standard solution still can exists).

If the needed current is very small (for example 5uA), I would suggest designing of your own switching regulator, where the regulator consumption could be made small enough in order to not affect the efficiency. Check "The Art of electronics" - there is a good example of micro power design.