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I have an outdoor sensor (which will have to work in the winter on Poland, however, we usually have temperature much over -10 Celsius degrees) which needs 5V.

The sensor operates 24 hours per day and consumes

  • about 300 mA when working, for about 10 seconds every 5 minutes and

  • less than 1 mA the rest of the time.

I want to power it using solar panels.

What solar panels should I use (what voltage)?

Do I need any DC-DC boost converters? step-down? step-up? Could you name some examples? How to look for them?

Are there any ICs for that purpose? For example, does BQ25504 would be helpful here?

What type of rechargeable batteries should I use? Are Ni-MH a good choice?

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  • 1
    \$\begingroup\$ Does it really need 5V or will it work on a lipo range (3~4.2)? \$\endgroup\$
    – Passerby
    Sep 5, 2016 at 16:10
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    \$\begingroup\$ Unfortunately, the device consists of the sensor part which needs 5V and the electronic part where linear regulator 5V->3.3V is placed. So, yes, it needs 5V. \$\endgroup\$
    – Defozo
    Sep 5, 2016 at 16:24
  • \$\begingroup\$ Since you require 5V, there are 5V lipo modules that have a built in solar panel and usb input, lipo charging and boost to 5V in a under 1" square package. So what you really need to figure out is the battery requirements. Based on 300mA 10 seconds, 1mA 290 seconds, that's an average of 11mA draw. A 300 mAh battery would be sufficient for 24 hours, assuming your solar setup can charge it. Also you need to add in the charging modules current in that calculation. \$\endgroup\$
    – Passerby
    Sep 5, 2016 at 16:38
  • \$\begingroup\$ Easiest solution obrazki.elektroda.pl/5925982800_1473095587.jpeg \$\endgroup\$ Sep 5, 2016 at 17:18
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    \$\begingroup\$ @Defozo You say "It should work 24 hours of every day" ie 300 mA for 10s every 5 minutes + 1 mA otherwise 24 hours every day. Yes? || [ = 1 mA x 24h + 299 extra in 10/300th of time = 24 mAh + 239 mAh = 263 mAh/day approx. So say 250 to 300 mAh/day. GOOD - this is what I and others assumed below. \$\endgroup\$
    – Russell McMahon
    Sep 6, 2016 at 23:37

2 Answers 2

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I have an outdoor sensor (which will have to work in the winter on Poland, however, we usually have temperature much over -10 Celsius degrees) which needs 5V.

The sensor operates 24 hours per day and consumes about 300 mA when working, for about 10 seconds every 5 minutes and less than 1 mA the rest of the time.

Energy required:

Operate: 10s in 5 mins = 10/300 = 1/30th duty cycle.
Sleep: 1 mA when not operating.

Can be seen as 1 mA continuous + 299 mA x 10/300 = 10.97 ~= 11 mA average

mAh/day = 24 hours x 11 mA = 264 mAh.
Say 250 mAh/day for calculations.

Warsaw in December averages 0.67 sunshine hours per day - Wow! - that's low.

Due to losses from panel to sun alignment, panel to battery matching, battery store & retrieve efficiency and more, storable solar energy is about 50% of the panels nominal maximumcapacity or less.
To get 250 mAh/day you want say 500 mAh + from PV panel.

If you use a 9V panel with linear regulator that's energy of
9V x 500 mAh = 4.5 Watthour per day.

As there is as little as 0.67 hours of average sun panel size needs to be
4.5 wh/0.67 SSH ~= 6.5 Watts.
A 10 Watt PV panel may be "safe".

That is far larger than many would feel was necessary. It may not be.

I said 9V panel out above. I had in mind a 7.4V LiIon battery - 2 series cells. This will charge from 9v. It can use a linear regulator. This is wasteful of PV capacity but more easily implemented than a switching regulator type solution.

In sub zero temperature conditions (snowy) PV panel efficiency will increase slightly over nominal but all batteries will either have much degraged energy output capacity or "just not work". Manufacturer's specification sheets should be consulted and a good idea of minimum temperatures is required.
Lithium based batteries are about as good as any at sub-zero temperatures. Claims vary by manufacturer and model but as a guide LiIon may be rated to -10C and LiFePO4 to -20C.


Battery capacity for 1 day is >= 250 mAh as above. Somewhat higher is "most wise".

If you want operation over N days of no sun you need N x 250 mAh of battery.

http://www.gaisma.com/en/location/warsaw.html


You're saying about 50% storable solar energy. What does it mean?

The apparent energy you can get from a PV panel of a given rating and the amount you can use after storing and recovering it vary by a factor of say Ks. In real world situations, if the panel does not track the sun (usually it doesn't), and if it is not perfectly clean & snow free, and if the panel optimum operating voltahge Vmp is not exactly the battery optimum charge voltage AT ALL TIMES then available energy and mominally available energy will vary. I am saying that starting with Ks = 0.5 is in the order of right. eg a clean optimally pointed 10 Watt panel at 25C in full midday summer sun for one hour will deliver ABOUT 5 Watt hours of energy out of a typical battery system - and not the 10 Watt hours you might expect. Battery to panel matching can make a useful difference - and this is what a MPPT (maximum power point tracking) controller does. MPPT is effectively an "electronic gearbox" that finds the optimum V & I operating point for a panel in current situation.

BTW I can only find solar cells which have only over a dozen % of efficiency. Where should I put that efficiency in my calculations?

Panel output per solar input is what is relevant to you.
You are interested in how much power you can get at a given insolation (light power) level. While efficiency influences panel size an eg 10 Watt panel has the same power output whether it is 10% or 20% efficient - the 10% efficint panel will be about twice as large - and this matters substantially in some applications but is not very important in others.
In this case efficiency is (probably) not too important in it's own right - unless perhaps size constraints apply.

eg if you have 1000 W/m^2 of insolation and a 10% efficient 10 Watt panel you will get 10 Watts from it. The 10 W depends on the efficiency and area but you do not need to know either to calculate output as the manufacturer has included those parameters when rating the panel. A 20% efficient 10 W panel could have half the area of active material - but it's still a 10 Watt panel. So, efficiency is not needed in your calculations if you use the manufacturers core ratings. - usually Wmp, Vmp, Imp.
Usually at 25C panel temperature and AM 1.5 (an optical path to panel spec).


Durability in outdoor use:

"Crystalline Silicon" PV panels are assumed. See comments at end.

Note that if the panel will be "out in the weather" continually and if you want years of service there are some panel types that are extremely good and some that are completely unsuitable.

Epoxy resin encapsulated = VERY poor lifetime. These have a rounded edge and no frame. Lifetime outdoors is a few years for a "good" one and can be 6 months in some cases. Suitable (if then) for toys and products not constantly exposed to the sun.

"PET" encapsulated. Plastic outer layer similar to that used on softdrink bottles. Heat laminated using (usually) "EVA" adhesive/encapsulant. Will withstand "years of exposure". Quality varies with manufacturer.

Glass front sheet + EVA encapsulant. The "typical" industry standard method of producing PV panels. Usually with an aluminum frame. 20+ year lifetime typically. 30+ years with care. 40+ years with luck and decreasing output (I have one). Heavier and more fragile. The best solution for outdoor long term use.

Other: There are other "frontsheets" and panel types that are less liable to be encountered. Fluro-plastic (FEP etc) can have excellent results. Usually from specialised suppliers - not common on retail / hobby market.

Amorphous Silicon - smooth dark layer on glass with fine grooves between "cells". Older technology. Works OK but has few advantages other than having low cost per Watt, sometimes. Low efficiency, fragile, output drops with age unless 'tempered' occasionally. Unlikely to be a good idea.

CIGS, CdTe, ... Less likely on retail market. May be flexible or low cost. Best versions are competitive in efficiency with crystalline silicon.

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  • \$\begingroup\$ There are modules that do solar to lipo, lipo to 5V boost. So the switching is already handled. \$\endgroup\$
    – Passerby
    Sep 5, 2016 at 17:04
  • \$\begingroup\$ You're saying about 50% storable solar energy. What does it mean? BTW I can only find solar cells which have only over a dozen % of efficiency. Where should I put that efficiency in my calculations? \$\endgroup\$
    – Defozo
    Sep 6, 2016 at 22:35
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Start with a spec for power output and input, budget and complexity requirements so a suitable make/buy choice is easier.

  • 5V & 3V Out, (5v main load)
  • solar PV input
  • _% tolerance
  • 300mA for 10s every 300s or 10mA avg.
  • 1mA standby current or 11mA tot. avg. @5V
  • equiv 55mW avg. power consumption
  • climatic: -10'C to ?
  • low budget prototype ?
  • sustained storage for X days with low solar input for storage energy
  • if X=3 days, storage requirement= 792mAh
  • Consider charge up Time = 10 hrs

  • Consider 2 series LiPo cells (16850) with 6.6 to 7.4V out and PV Voc=9-12V (open cct) with LiPo charger (3 stage) to charge LiPo with USB reg out(5V) that can charge up

  • some options

    • Samsung INR18650-25R $2.43
    • Samsung INR18650-30Q 3000mAh $3.45
    • Sony US18650VTC5 2600mAh $3.58
    • Sony US18650VTC4 2100mAh $2.98
    • LG-HE2 18650 2500mAh $2.58
    • LG-MJ1 18650 3500mAh $3.72
    • LG-HE4 18650 2500mAh $2.68
    • LG-HG2 18650 3000mAh $3.63
    • Panasonic NCR18650B 3400mAh $3.05

If derating capacity to <50% for cold temp and want charge up in 4 hr then PV charge for 1000mAh in 4h = 250mA or a minimum of 20mAh per day average

250mA*9V= 2.25W Choose 3W min. $9 online

e.g. http://www.ebay.com/itm/9V-3W-3Watt-Mini-poly-solar-Panel-small-solar-cell-PV-module-for-DIY-solar-Kits-/251946289450

Increase W rating to reduce charge recovery time.

Buy Option $27 CND or AUD. add USB or DC Jack and 3.3V LDO

http://obrazki.elektroda.pl/5925982800_1473095587.jpeg

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  • \$\begingroup\$ Tony - most of that is very good (as I'd expect) and most is similar to what I suggested - except I suggested PV-> battery directly and battery-> load with linear regulator with no boost to maximise low power efficiency. || The main area we differ is on PV panel. He is in Poland, as I see you are aware. Using Warsaw as an example, typical December insolation is only 0.67 hours/day. With a fixed aligned panel, less than 100% PV->battery-> load efficiency, a bit of allowance for snow clearing and more the available energy ... \$\endgroup\$
    – Russell McMahon
    Sep 6, 2016 at 0:00
  • \$\begingroup\$ Thanks Russ. Then PV capacity must be derated accordingly \$\endgroup\$ Sep 6, 2016 at 0:02
  • \$\begingroup\$ I wonder if retrofit into a reflective foil DTV satellite dish, a small PV in Winter only, put in LNA position can boost efficacy of PV, not unless it is a cloudy day \$\endgroup\$ Sep 6, 2016 at 0:14
  • \$\begingroup\$ ... is liable to be equivalent to 100% insolation for say around 0.3 hours/day. You suggested 9-12V PV and I said 9V so say 11 mA average x 24 hours x 9V =~ 2.4 Watts. Multiply that by my insolation derived factor of 1/0.3 gives ~= 8 Watts of PV for 1 days average operation. Even if one regards my 0.3 hours average effective insolation / day as low. If living "day to day" you need allowance for eg high-snow days. | ie the PV panel needs to be apparently ludicrously large. Even 10W marginal? ... \$\endgroup\$
    – Russell McMahon
    Sep 6, 2016 at 0:14
  • \$\begingroup\$ 20% nominal efficacy of 1kW/m2 is 200W/m2 then reduced by directional loss on a 8hr maximum winter day and factor elevation losses in Winter then climatic diffusion loss from clouds, the insolation ratio Dec/July is ~13% for Warsaw, 7.5 bigger is needed in Dec compared to July. \$\endgroup\$ Sep 6, 2016 at 0:27

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