# What would happen if you connect a solar panel to a buck-boost converter?

What would happen if you connect a solar panel with the following characteristics:

Max Power:3.5W

Voltage: 6V

Open Voltage: 7.2V

To a buck-boost converter that convert an input voltage in the range 1V-6V to a voltage of 5V? Who would "win" the fight to set the voltage? The buck-boost converter can work with any input voltage and the solar panel can work at different output voltage. I can't figure a way to calculate the input impedance of the buck-boost converter.

Bellow is the representative circuit diagram for a solar panel on the left and the buck-boost converter on the right with the converter switch represented with the letter S.

• I think I am not fully understanding this. Do you connect the solar panel to the input of the buck/boost?
– jwsc
Dec 8, 2015 at 7:28
• yes, the solar panel is connected directly to the converter
– Pat
Dec 8, 2015 at 9:17
• It's a good question but incomplete. A BB converter is usually controlled and NOT operated open-loop. Usually you set Vout and Vin looks after itself. In this case the nature of the load is crucial. If the 5v is a battery or includes a large capacitor this will affect the ability of the BB to alter Vout substantially in one cycle so will affect how it behaves. In this scenario if the output is a battery a common solution is to maximise energy transfer and the BB is run in MPPT (maximum power point tracking) mode. A simple pseudo MPPT method that works quite well is to choose Vpanel at some ... Dec 8, 2015 at 10:47
• @RussellMcMahon My goal is effectively to put a battery as the load (R on the diagram). I know a MPPT would be the best solution but I'm wondering what would happen if I made it cheap and simple.
– Pat
Dec 8, 2015 at 11:35
• ... semi optimum value and control the BB so Vin is maintained at this voltage. Typically Voptimum ~= 80% x Voc so here = 7.2 x 80% = 5.76V = 5.8V. So BB is set to monitor Vin so that it remains at about 5.8V . This is close enough to optimum load for the panel in such cases. Dec 8, 2015 at 11:37

So I have actually tried is with a cheap normal buck boost with constant voltage and constant current fuction. I have a 12v volt system so I set the the output to have 14.7 CV and 3 amps CC with 50 watts solar panel. In bright sunlight with the baterry charging, it pulled the solar panel down to 8 volts (minimum operating voltage of the BB converter) while outputting at 13.2 volts and 1.5 amps. My understanding is that the panels can't provide the converter with enough power so it keeps drawing more current so the voltage drops. This is probably not optimal when the sun is shinning since it doesn't the voltage very low but I'm hoping it when keep charging the battery when it's cloudy/low in the sky with the panel voltage is lower than the baterry voltage.

I think i am missing something, but here are my two cents:

The buck/boost will operate on the input voltage given by the solar panel. The internal switch control will determine if it works as buck or as boost (obviously, if the solar voltage is lower than 5V it is a boost, if it is higher it is a buck).

Depending on the size of the load, the solar (input) voltage may drop. If the voltage gets too low for the boost to handle, the output voltage will also drop.

Solar panels shouldn't be excessive loaded, because their efficiency drops in that case. Because of that there are special buck/boost ICs which provide point of load control for the solar panel (BQ24650 for example). In essence they lower their output power to not overly load the solar panel.

The answer is a lot wider than I would like to write.

I have a Ebay cheap PWN, this is a dc-dc converter. I also have a a buck. Now the difference between the two should be known. a PWN (dc-dc) just turns on and off to change the voltage and the current remains the same. a Buck does the same, however the difference is that there is a inductor (L) and the drop in voltage increases the current (i).

The Load (R) is dependent on the panel and power (W) of the two. if they are not equal then there is a sag on current or voltage. these are rather the same over apposing differences as the sag is inclusive of W.

There is NO point in a boost converter. Any set up using this is pointless unless its for SOS needs. a low power panel Boosting its voltage, with lower the current, typically the lower voltage panel with have a low current in the first instance and therefore you will brown out.

If you use a higher power panel, for example; 255w, Voc38 Vmp33.6 Isc8.8 Imp 8.3

I used a Buck with CC and CV To charge a battery (14.7v). my current read 7.47, my solar input voltage was 34.4.

Therefore as the voltage was divided the current went up. (38/14.7=2.585) If my current read 7.47 I divide this by 2.585 and this would give me the current (2.889). Therefore my total power on the solar was... (42.478) as was the power into the battery (-2%).

Solar has voltage but not current, the more sun you get the more current you get. if you were using a boost converter, then the difference would be taken from the current (ohms law) say we have a 12v panel, and wanted to charge a 24v battery (28.6) the Voc is 17.85 and the panel is 80w. (I is W/V) 2.1917

we will say that its a bit cloudy and and your getting 40% (80-60%=32w) (0.8767 I/2) and we need to double the voltage. we then get 25v (28v) with a 0.438 current.

Now depending on the type of battery you have, if you have less than 5 amps you can damage a battery trying to charge it. however this could be used for a non use standby battery (if they accept the low (trickle) charge).

You could use Super caps

• What does PWN stand for? Jan 6, 2022 at 19:07
• Could be a typo for PWM? Dec 30, 2022 at 19:48