“Ultimate” Surge Protector with Solar Cells

You want a solar powered power supply that provides 40 W for 10 hours every day, for 400 W-h per day. Obviously all this power originally comes into the system via the solar panel, so it must be sized accordingly. Let's say the switching power supplies in the system are a total of 70% efficient. Then there is power lost in storing and later retrieving it from the battery. Let's say that's another 70%. Combining those two, you have about 50% efficiency from solar panel output to ultimate load.

Now you know the solar panel has to produce about 800 W-h per day. With a very large battery, it only needs to produce this averaged over a long time. The smaller the battery, the smaller the averaging window where the panel still has to produce this power. How much is reasonable depends on factors you haven't told us.

Let's say you've sized the system so that you need the 800 W-h/day average over a few days. You left out a lot of information, like what latitude this is at and therefore what the minimum length of sunlight in the winter is, what probability of failure you can tolerate, what minimum percentage full sun your location expects averaged over a few days, etc. For example, if you conclude that worst case over a few days you can only count on the equivalent of 1 hour full sun per day, then the panel needs to be able to put out 800 W in full sunlight.

The next question is the battery. From the previous example, it looks like the battery should be able to run the system without any input power for at least a full day usage, which is 400 W-h into the load. Let's say half the total switching power supply loss of 70% assumed above is between the battery and the load, which means from battery to load is 84% efficient. 400 W-h / 84% = 480 W-h, which is what the battery has to be able to produce without input power and without it being exceptional and therefore significantly degrading the battery.

Let's see how the numbers work out for a 12V lead-acid battery. 48W / 12V = 4A drain when the load is powered. Since the load needs to run at this power level for 10 hours, that represents 40 A-h capacity. However, that needs to be significantly derated. A new lead-acid 40 A-h battery can do this once perhaps at the right temperature, but running it down to empty will kill it. For lead-acid you want a "deep cycle" battery but still derate significantly. Something like a 80 A-h "marine" battery might do it. Other battery technologies have different tradeoffs with how fully they can be discharged, operating temperature range, life time, life cycles, cost, availability, etc, etc.

Transmogrifying your

• "should there be any voltage?"

into

• "will there be any voltage?"

the answer is "There might be, it all depends on circumstances and equipment involved".

I don't know if "a surge protector" tends to be a reasonably specific device in the UK but here in the antipodes at the dawn of time it could mean a wide range of things.

BUT

• IF a surge protector consists of a device which conducts when line to line voltage exceeds normal max by a significant margin - such as back to back zeners, a MOV, a transzorb, a gas discharge tube, a neon or similar.

• but which is of very high resistance when no voltage is applied

• and IF the surge protector contained X and/or Y capacitors (line to line or line to ground)

then YES voltage would be very likely to be present, because the capacitor(s) would probably retain some charge if the mains was disconnected with the load switched off.

If mains is disconnected by opening a switch or pulling out a plug, then line to line voltage and this capacitor voltage could be anything from about +330 to about -330 V as the disconnection timing is not synchronised with mains zero crossing. A capacitor connected across the line or leg to ground capacitors will be left with the mains value at the time of disconnection. If there is no load present this voltage could remain "for some while".