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Solar-powered wireless sensor

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lickmyeyeball

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I am starting to design my very first hobby-project, a solar-powered wireless sensor.

Being entirely new to this, I was looking for some rules-of-thumb or heuristics from you guys, as you are much wiser and capable than me.

Specifically, I was wondering what rules you follow when sizing solar arrays with respect to expected load. There are few clear sunny days, especially where I live. I expect to have to underestimate what power I will actually get.

Have you had best results when building arrays with 125% capacity relative to load? 150%? Even more?

As an illustration of what I mean, say I have a 60W load. Neglecting the power electronics likely needed between the solar array and load, I would design my solar array with a capacity of 90W if I overcompensated with 150%.

As well, do you think it is worth putting in the effort to simulate the system? I have a copy of matlab, but I feel that simulation would require a good deal of effort to learn; both about the program itself and the elements I am trying to model.

Also, I would really appreciate any general tips you might have regarding design in general. I am sort of winging it and don't know where things will take me.

I have learned some stuff in preparation for this project of mine: a bit of electronics, a bit on computer communications, a lot of control systems (not nearly as helpful as I was hoping), power electronics, and what little I could find regarding solar PV systems. I will still have to learn everything about microcontrollers, sensors, and wireless. Those things are still complete mysteries to me.

Thanks so much for your help and input. :-o First post.
 

Yes, you must factor in the proportion of cloudy days versus sunny ones.

The PV panel produces power during a few hours of daylight. Do you plan to store that power in batteries, for use later?

If so then a 90W panel could theoretically push, say, 90 / 12 = 7.5A, into a 12V battery.

If your battery has a capacity of 50 amp-hours then it would charge the battery in just over 7 hours. A bright sunny day might give enough charging time, when you take into account the sun being dimmer near sunrise and sunset.

On cloudy days it might take a few days to charge.

Solar energy is an occasional topic at this board. The bottom of this page has a list of threads with related titles. Clicking on one may lead you to others with more information.

Home-Power is an excellent resource for alternative power:

www.homepower.com
 

Here in England the average output of a outdoor solar panel is typically 5% of the rated power of the panel. This is not sunny California.

If so then a 90W panel could theoretically push, say, 90 / 12 = 7.5A, into a 12V battery.

If your battery has a capacity of 50 amp-hours then it would charge the battery in just over 7 hours.

Battery charging is not 100% efficient. Lead acid battery charging is typically about 66% efficient.
 

The PV panel produces power during a few hours of daylight. Do you plan to store that power in batteries, for use later?

Solar energy is an occasional topic at this board. The bottom of this page has a list of threads with related titles. Clicking on one may lead you to others with more information.

Home-Power is an excellent resource for alternative power:

www.homepower.com

Thanks very much for your input Brad. I especially appreciate your pointing out the relevant links the site posts at the bottom of each thread, I had never noticed that.

I hope to store the energy in a relatively small battery. The system should cycle between a sleep mode and periodically awakening to check stored energy levels and if sufficient energy is present, make a reading and relay the data. I hope that sleep mode should draw something around 3nW while readings might average somewhere around 5mW.

- - - Updated - - -

Here in England the average output of a outdoor solar panel is typically 5% of the rated power of the panel.

Wow, that yield is much smaller than I actually expected! It's really useful to know however, thank you.
I would probably expect some results similar to yours, as I am at a relatively high latitude.
 

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