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I'm trying to work out how much current my solar circuit could sustain under normal conditions.

I have an monocrystalline solar panel claiming 5W @ 12V (17.8V Vmp, 5Wp) based on their data sheet.

The data gathered by CEDA from MIDAS near to where I am suggests that the average (incl. night) for the whole year was 470KJ/m2 per hour or ~130.55W/m2 given \$P_{(W)} = \frac{1000\times470_{KJ}}{3600_{s}}\$

The data provided by MIDAS gives "solar irradiation amount" measured in "KJ/m2" in hourly increments for a given recording station.

Based on this information and the power curve in the data sheet, I would expect on average I would be able to draw around 150-200mA @ 12V.

The Buck/Boost module I use contains an LM2596 which advertises ~80% efficiency when the input is 12V with a 5V 3A load.

Does this mean I could expect ~80% of the current from the solar panel to be available at the output side of the Buck?

To this end, do I potentially have at least 120mA @ 5V given input of 150mA @ 12v available on average?

Adam Leyshon
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  • That's not how buck convertors work : 150mA at 12V is 1.8W or 360mA at 5V : your buck will give 80% of that. (On a dull day, that's optimistic of course...) –  Mar 18 '21 at 12:02
  • 80% of the power, not 80% of the current, at best. – jwh20 Mar 18 '21 at 12:05
  • Thanks for the advice, I will make a note that it's power not current that the efficiency represents. – Adam Leyshon Mar 18 '21 at 12:37
  • One thing to be aware of, if your buck converter doesn't have MPPT(Maximum Power Point Tracking), in dim light or when the load is high, the converter will draw too much power and decrease the output voltage of the solar panel. This can greatly decrease your total output and also cause the circuits to operate in a continuous brownout state, potentially destroying them. – K H Mar 19 '21 at 03:26
  • The panel will have inadequate output at minimum for two periods of time during each light cycle, so whatever you attach to it should be protected from brownout if necessary. You could put a low voltage cutout between the solar panel and buck converter, but you would want to make sure it didn't oscillate too quickly. The best solution would be to add an MPPT circuit and or battery. – K H Mar 19 '21 at 03:32
  • [This question](https://electronics.stackexchange.com/questions/65824/can-electronics-be-damaged-by-undervolting-it) is about damage that can be caused by undervoltage. – K H Mar 19 '21 at 03:34

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The first part of this answer is to show that the OP misrepresented the numbers from his data by suggesting that the daily energy average was 470 kJ/m2: -

...suggests that the average (incl. night) for the whole year was 470KJ/m2

  • 470 kJ is energy for the whole year per square metre.

  • So divide that by 365 to get energy for 1 day = 1288 joules.

  • Then divide that by 24 to get energy per hour = 53.7 joules.

  • Then divide that by 3600 to get the equivalent average energy per second = 0.015 joules.

This is a power of 15 mW per square metre.

I think you might be somewhat overestimating what power you will receive. Maybe your numbers are wrong or you are misinterpreting them: -

You're right, I've clarified that the 470KJ/m2 was the average per hour. – Adam Leyshon

Here are some solar energy numbers from NASA for the Midlands in the UK (about the same for Wales I would imagine): -

enter image description here

We can see that a reasonable average daily energy per square metre is around 2.5 kWh for the UK. That's an average daily power of 104 watts per square metre.

enter image description here

Image taken from this site and 104 watts per square metre ties in nicely as an average.

If we factor in the solar panel's energy efficiency (maybe 20%), it becomes about 21 watts per square metre.

Given that the solar panel linked to in the question has dimensions of 185 mm x 250 mm, that's an area of 0.04625 square metres so, the viable average power that could be liberated is 0.963 watts.

Given also the efficiency of the converter is 80% then you can only really liberate about 0.77 watts

If you are wanting 12 volts, then the average current available will be about 64 mA.

Andy aka
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  • This doesn't sound right. It says "UK hourly solar radiation data". So that would be the energy for an hour. 15 mW per square meter would be incredibly low. – Lars Hankeln Mar 18 '21 at 11:55
  • @LarsHankeln please explain that to the OP - I'm using his figures. – Andy aka Mar 18 '21 at 12:00
  • @Andy you appear to be confusing "average" with "total". As a sanity , 1288J is about 2 seconds worth of full sun, and even in Argyll we get more than that! It's a plausible figure for a daily average but I haven't read the linked report to find out. –  Mar 18 '21 at 12:03
  • You're right, I've clarified that the 470KJ/m2 was the average per hour. – Adam Leyshon Mar 18 '21 at 12:04
  • OK interesting, at 5V 0.77w would give ~150mA, that would be fine with me. Still higher than I expected, being optimistic of course. – Adam Leyshon Mar 18 '21 at 12:24
  • @BrianDrummond I was showing the OP that his interpretation of the the figures he found was flawed. – Andy aka Mar 18 '21 at 12:29
  • @Andyaka given that his figure was 135W/m^2 for Glamorgan and your newer figure is 104W/m^2 for the Midlands, they aren't so very far apart. –  Mar 18 '21 at 12:50