As a heads up, I edited this question to make more sense based on what I have learned. I think this is okay since no one has answered the previous question yet.
So I'm designing a solar system and was looking at the MPPT linked below:
https://www.amazon.com/TRACER-3215RN-Solar-Charge-Controller/dp/B008KWPGAE
I read one of the points of this MPPT was that it uses
- "4- Stage charge with PWM output."
From that, I know this MPPT uses PWM to adjust the output voltage. I know this is necessary because you need to change your output voltage to fit the rest of your system (as an example system, think of a dc-to-dc step down from 36 input volts to 12 output volts or dc-to-dc step up from 36 input volts to 48 output volts). My concern is this: will this PWM significantly reduce the energy I can get from my solar panels? Let me explain why below:
I'm basically asking if I have PWM technology or a buck-boost converter that deems I need a duty cycle of 0.4 to reach the voltage I want (this means the switch is open on the buck-boost converter or PWM circuit 60% of the time), does that mean I don't generate power with my solar panels 60% of the time? For example, think of a natural gas generator that runs 8 hours a day but you only have a load plugged into it for 40% of those eight hours. That's a lot of natural gas wasted. Similarly, I'd lose 60% of my insolation on my solar panels, right? Since they only actually generate power 40% of the time since the switch is open the other 60% of the time on my PWM circuit or buck-boost converter.
Best Answer
Not if the converter is properly designed. You are probably thinking of a buck circuit that looks like this:-
simulate this circuit – Schematic created using CircuitLab
If it was this crude then you would be correct to assert that the solar energy would not be fully utilized, because current would only be drawn from the panel during the 40% PWM on-time.
I simulated this circuit in LTspice and graphed the current through M1 (which is essentially the same as the solar panel's output current):-
When the FET is turned on it draws up to 1A from the panel, but when off it draws nothing! 60% of the time the panel absorbs the solar current internally, wasting it. As a result the power output is only 1.7W, nowhere near the expected 5W.
But there is something missing from this circuit that all properly designed converters have - a reservoir capacitor across the input. After adding a 220uF capacitor across the panel, and lowering RL to 0.85Ω to maintain 2.0V output, the current through M1 looked like this:-
The reservoir capacitor eliminates the 1A current limit, so the converter is now able to draw up to 2.6A (1A from the panel plus 1.6A from the capacitor) producing 4.8W at the output - 96% of the panel's capacity. During the 60% PWM off-time the solar panel recharges the reservoir capacitor at 1A, so it is always working at full capacity.