In our last post we looked at the various components for a solar setup: the solar panel that supplies the energy, the battery that stores it, the charge controller (regulator) that allows charge to the battery when needed and prevents overcharging when full, and the inverter that changes direct current (DC) to alternating current (AC). So far so good.
The next thing we need to work out when putting together a solar setup is what sizes our components should be. The basic calculations go as follows:
- Find the Wattage of your appliances: Make a list of all the electrical appliances you’ll use in a typical day, and how many Watts they each consume. You can usually find this on the appliance itself or in it’s handbook.
(list appliances Wattage)
NOTE: If you can only find a figure for Amps, simply multiply this by 12 (volts), to convert it to Watts. (Amps x volts = Watts)
- Calculate your daily total Watt-hour requirement: Estimate how many hours you use each appliance in a day. Multiply each appliance’s Wattage by the hours you’ll use it for in a day. Then add all the totals together to get the final daily total Watt-hours you require.
(Watts x hours used = daily Watt-hour requirement)
NOTE: To convert this number to Amp-hours: (Watt-hours ÷ Volts = Amp-hours)
- Next calculate your panel size: Simply divide the daily total Watt-hours you require by the hours of usable light you expect in an average day. This will give you your minimum panel wattage.
(daily Watt-hour requirement ÷ sunlight hours = minimum solar panel Watts)
NOTE: Here in the UK usable sunlight hours have been worked out at approximately 4 during longer summer days and 1 during the shortest winter days.
- Then your battery size: Multiply your daily Watt-hour requirement by the amount of days you want your battery to run on reserve, then divide this by 12 (volts) to convert to Amp-hours, which batteries are rated in. Multiply by two (because you don’t want to drain your battery more than half) to give the minimum battery size.
NOTE: If you use all your Watt-hours in a short time within a day you should take into account the effect of discharge rates (explanation in components post).
(daily Watt-hour requirement x 7 [days] ÷ 12 [Amp conversion] x 2 [battery drain less than half] = battery Amp-hour size)
- Size your charge controller according to the Amps produced by your panel. Calculate the Amps produced by dividing the panel wattage by its voltage.
(solar panel Watts ÷ solar panel Volts = controller Amps)
- Finally your inverter size: You may not need an inverter (for example if you only use 12v appliances) but if you do need one: add together the Watts for the items you may want to run at the same time. Use the total Wattage, then add 10-20%, as your minimum requirement (because inverters typically draw 10-15% more energy from your battery than the devices require).
(Watts of appliances used simultaneously + 20% = inverter Watts)
1. List appliances wattage:
|Daily Watt-Hour requirement|
|Laptop||50||2||100 + 10% for inverter = 110|
|Mobile||5||2||10 (separate solar charger)|
|264 total, rounded up = 300
2. Watts x hours used = daily Watt-hour requirement: 264
We’ve rounded that figure up to 300Wh.
3. Daily Watt-hours ÷ sunlight hours = minimum solar panel Watts: 300 ÷ 1 = 300
Assuming year round use we’ve used winter hours in this calculation. A quick look at panels reveals that one of the sizes they come in is 160 watts, so two of those would cover our needs giving us 320 watts.
4. Daily Watt-hours x 3 [days] ÷ 12 [Amp conversion] x 2 [battery drain protection] = battery Amp-hour size: (300 x 3) ÷ 12 x 2 = 150Ah
If your setup is sized right you should theoretically be able to use 1 day in this calculation, in practice there may be occasional days where light is reduced to below usable levels, so we are using 3 days to be sure we have power in worst case scenarios. So we’re looking for a battery size of a minimum 150Ah, and the closest rounded up size we could find right now is 160Ah.
5. Solar panel Watts ÷ solar panel Volts = controller Amps: 320 ÷ 17.9 = 17.8 Amps
To work out what amp controller we need, the calculation results at just under 19 amps so we’d round that up to a 20 Amp controller (this example is based on the 2 panels wired in parallel, more about this when we post about installation).
6. Watts of appliances used simultaneously + 10-20% = inverter Watts: 50 + 10 = 60 Watts
The only thing we’d be running through an inverter from our example list is the laptop, so we would round up and get an inverter with a 100 watt rating.
Here’s another example using 4 hours usable sunlight (you might want to use summer daylight hours if you only use your camper in that season), and 7 days reserve charge:
300 Watt-hours per day ÷ 4 hours of sunlight = 75Wh
(75Wh x 7 days) ÷ 12 Amp conversion x 2 battery drain level = 87.5Ah battery
320W ÷ 17.9V = 17.87 Amp controller
50W + 10 = 60 Watt inverter
Components come in round figures so here are those values rounded up. This should also help compensate for losses within the system =
100 Watt panel
100 Ah battery
20 Amp controller
100 Watt inverter
Things to take into consideration if changing from this formula:
- When you have invested in a solar panel for your motorhome, it would be a shame to see the energy generated go to waste simply due to the limited capacity of your battery. If your 12v solar panel starts charging the battery in the morning and completes the job in a few hours, the rest of the energy generated throughout the day is wasted. This is a complete waste, especially if you need the energy in the evening or at night for lighting etc. So think about whether your battery capacity is adequate for the size of the solar panel you use and how quickly it will charge the battery to 100%.
- On the other hand, if your batteries are oversized for the system, they may never get fully charge.
- Batteries often have a limit on the initial charging current. Typically gel/AGM batteries have a limit on charging current equal to 20% of their capacity. So a 100ah battery cannot be charged by a current higher than 20amps (this would be the equivalent of a 320W solar system current). This will not normally be a concern unless the battery you use is really small, however, it’s sensible to check the specifications of your battery to see whether there are any restrictions on the charging current.
- One characteristic associated with lead acid batteries is that the capacity will decrease as the rate of discharge increases. On the battery case you should find capacity figures. If you find just one Ah figure such as 100ah, it is normally rated for discharge over 20 hours. Multiple figures such as “K5 75Ah – K20 100Ah – K100 110Ah” show the Ah at different discharge rates, (so in this case if discharged over 5 hours you get 75Ah, but if discharged over 100 hours you get 110Ah). So keep in mind whether you use all your power at once or spread over a longer period.
Check everything very carefully.
How to calculate the length of time it will take your battery to charge
Let’s use a 100ah 12v battery at 50%, and 320w 18v panel for this example. The battery needs charging, but has no load drawing off of it while charging:
The quick and wrong answer would be to figure out the Watt hours of the battery that need replacing and divide by the wattage of panel.
To convert the batteries Ah to Wh, take the battery Voltage x Amp-hours = Watt-hours. Then divide by 2 to work out 50% of charge.
12 (V) x 100 (Ah) = 1200 (Wh) ÷ 2 (50%) = 600 Wh
600 (Wh) ÷320 = 1.87 hours
-Under Ideal Conditions-
The reality is different because even in ideal conditions all the power leaving the panel does not get converted into stored power. For example some is lost as heat. If you are using a MPPT controller the maximum efficiency you can achieve is 66%. So with 320 Watt solar panels the most energy you can generate in an hour is:
320 x .66 = 211 watts.
So if our panels can generate 211 Watts per hour , the calculation changes to:
600 (Wh) ÷ 211 = 2.84 hours.
If you were to take into account the losses mentioned above – cloudy conditions, your panel not directed at the sun, or if panel Voltage is not well matched to your battery – this would increase charging time significantly. Adjustments taking into account all these losses have been estimated at 2.5. So we would multiply charging hours by 2.5.
600 (Wh) ÷ 320 x 2.5 (losses) = 4.7 hours
We hope this post was useful to you. We’ll be concluding the series with a final post about installation, so keep an eye out for that.