Conrad Electronic, cheaper.)
Following this sterling find, Jason sent another email:
If I were to run this porchlight off a deep cycle battery with solar panel, what battery size and solar panel size would you use?
Assuming the bulb consumes 4W and the PIR 3.5mA in low use mode and 40mA when the relay energises.
The PIR has a photocell so the light will only switch on during the hours of darkness.
The LED bulb would come on sporadically in the night and turn off after 5 minutes- not very often but not sure how much it would trigger.
What battery would you use combined with which solar panel and why?
Unfortunately we would need some more numbers that only Jason can provide, so I said:
Compute/estimate the total Watt-hours that the light + everything else should consume each day.
Your battery, if the thing it to be entirely off-grid, should be several times that since you don't want to discharge lead-acid more than about 50% ever and you'll need to cope with runs of dark days.
You'll need about 1Wp of south facing panels for every 1Wh per day of energy that you will consume.
You'll have to estimate on-time or else you'll have to treat it as if on all the time in the dark as I did.
Jason was feeling a little overwhelmed by some of the terminology, and he's certainly not alone. I suggest borrowing or buying a very basic book on electronics or renewable energy, or looking at an online glossary.
In particular Jason wanted clarification on:
The Wp (Watt-peak) rating, usually on the rating plate on the back, is the most power you can draw from a given solar panel in optimal conditions, eg a sunny cloudless summer's day with the panel directly facing the sun ('normal' to the incident radiation), and (and this is often overlooked) into a perfectly-matched 'ideal' load.
(In the days when I used to edit a supercomputing trade rag, this would have been the "will not exceed" rating in the headlines!)
About 1kW of sunlight is striking every square metre (m^2) in those conditions (the same power as one bar of a typical electric fire) and the very best solar panels as of 2009 will, including all loses, return you about 180W per m^2, ie are about 18% efficient.
The rule of thumb as I mentioned above for the UK is that a well-sited and unshaded 1Wp panel will generate on average 1Wh (1 Watt-hour) per day in winter (and ~5Wh per day in summer).
That's the same as saying that the UK averages the equivalent of one full hour of bright sun per day in winter and five in summer.
To make sure that the light never fails when you need it most then you must select panels and batteries for the worst (winter) case.
1Wh (Watt hour) is enough energy to run a 1W LED/appliance/etc for one hour, or a 2W LED/appliance/etc for half an hour, so if you wanted to run a 1W LED 24 hours per day even in the depths of winter, ignoring battery and other loses, a perfectly-set-up 24Wp solar panel could provide enough energy.
1000Wh is also written as 1kWh (one kiloWatt-hour) and is one 'unit' of electricity as measured by your supply meter, and is the amount consumed by a one-bar (1kW, one-kiloWatt) electric fire each hour, and is about the solar energy hitting each m^2 of ground each hour at summer noon.
Battery capacity is commonly quoted in Amp-hours rather than Watt-hours. If you multiply the nominal battery voltage by its Ah capacity you get an idea of its maximum Wh capacity. For example, a 12V 2Ah battery holds about 24Wh of energy. (Be aware that the voltage is not exactly 12V, nor completely steady.)
However, when specifying a battery for a particular application a few wrinkles have to be kept in mind.
Firstly, the process of putting energy into a battery and getting it out again is not 100% efficient. For example, for good lead-acid batteries the "cycle efficiency" is about 80% meaning that about 20% of the energy is lost. Some battery types have better cycle efficiency (such as Li-ion I believe) and some much less good (such as NiMH, for all its other good qualities).
Secondly, some battery types do not like to be fully discharged. In particular, even with a "deep cycle" lead-acid battery you will severely shorten its life if you discharge it below 50%.
Thirdly, if you are running a completely "off-grid" application you have to allow for several days of dull and cloudy weather when the sunshine will be a tiny fraction, maybe less than 1%, of average. (This behaviour, known as 'intermittency' is one of the chief challenges in using renewable energy such as solar and wind in the grid.)
So, if we wanted to keep that 1W LED running 24x7 through the winter it would consume 24Wh of energy per day, which implies 24Wp or more of solar PV.
To allow for battery losses of ~20% we need to bump that up to at least 29Wp (24Wp * 1.2).
And to accept that energy we need 2.4Ah of lead-acid 12V battery.
To allow for runs of (say) 5 days of cloud, rain and general misery in winter we are now up to 12Ah of battery (2.4Ah * 5).
To avoid ever going beyond 50% DoD (Depth-of-Discharge) we need to double up again to 24Ah of battery.
My 4W SheevaPlug computer should by extension have about 96Ah of battery (and ~115Wp of solar PV), and as of the end of 2009 my 40Ah battery is not enough to carry me through the dark times; 200Ah would give me a decent safety margin.
As a final point, you should generally be using a solar controller to protect your expensive new battery from overcharging (or over discharging), but unless it is an MPPT (maximum power-point tracking) type then you will never be able to extract all the energy from your PV panel(s), and so you should round up your panel capacity a little more, maybe 20% again.