I have a 5V super capacitor which I'm trying to extract the most energy from in the most efficient way possible. This capacitor is tied to the input of some regulator (which is the question of this post) that is regulating a (say) 2.5V output rail @ ~1mA. Something along these lines.
I've looked into parts from common Semiconductor companies (Linear, TI, etc.), and it seems that I can go down two paths with this mystery regulator:
- Use a buck-boost regulator (such as LTC3530).
- First boost the voltage to a voltage slightly higher than 2.5V (such as LTC3400-1) then buck the voltage down to 2.5V (e.g. LTC3406A). When the input capacitor voltage is greater than 2.5V, the boost regulator is bypassed and only the buck is active. Conversely, when the input cap voltage is less than 2.5V, the boost is active, outputs a voltage greater than 2.5V (say 3.3V), and the buck then steps the voltage down to 2.5V.
(Note: the parts selected above are for illustrative purposes only).
Option 1 would give me an effective input voltage range of (5V-1.8V) = 3.2V, while Option 2 would give a wider range of (5V-0.8V) = 4.2V. If I have a 1F input cap, these ranges correspond to 10.88J and 12.18J of useable energy respectively.
It would initially seem that Option 2 is the preferred method, but adding two power stages (boost, then buck) instead of using a single combined stage (buck-boost regulator) adds more losses, which is where the gain in useable energy from Option 2 becomes a wash.
I've tried crunching the numbers using the efficiency plots provided in the datasheets (for said parts and other parts from other manufacturer), but there are a lot of variables that affect these values, which can drastically change which Option is more beneficial. There are also other factors to consider such as losses in the inductor and capacitors, which I am unwisely neglecting.
Does Option 1 extract more energy from the input capacitor than Option 2? Or, how can I make Option 2 more energy efficient such that is outperforms Option 1?