One of the more useful new technologies for preppers is photovoltaic power generation. Twenty years ago this technology was in the early stages with very limited applicability due to a high cost/benefit ratio. Today, due to a number of significant advances, a solar power generation capability on some scale is entirely practical.

At the same time, entirely off-grid solar power is not yet practical unless one is prepared to make some significant sacrifices in living circumstances. This can be illustrated by simply referring to your monthly electric bill where you will find a wealth of information regarding your electrical power usage.


I have a four-bedroom home and my July electric bill shows an average daily usage of 52 kWh: this means, on the average, I use 52,000 watts in a 24 hour day.

It is worth noting that my circumstances are different than most homes in my area: I purchased the home from a mid-level manager at the local power company. While most houses in my area use a combination of electrical power and natural gas, when my home was built he specified all electric appliances: the hot water tank, stove, and “furnace” (electric heat pump) are all electric. I do not even have natural gas from the street. As a result, my electricity usage is greater than most homes in my area.

Regardless, by referring to your electric bill you can determine your typical demand for electrical power. While power usage is technically measured in watt-hours (100 watts for one hour equals 100 watt hours), the usage of a typical home (particularly in the summertime when air conditioning is in use) is too great to measure in watt hours: it is presented in kilo watt hours: 1 kWh = 1,000 watt-hours.

Even if your average daily usage during the summertime is only 24 kWh some quick math reveals the problem: under optimum conditions twenty 100 watt solar panels can produce 24,000 watts in 12 hours of sunlight (2,000 watts X 12 hours): this is enough to satisfy the requirement of 24 kWh in one day. Ah, but wait: many locations do not have 12 hours of unbroken sunlight every day; the output on a rainy day may be as little as 10% of the output in full daylight. Furthermore, no electrical system is 100% efficient: in order to obtain 24,000 watts of electric power it will be necessary to generate enough extra to account for the inevitable losses.

Very well. You have sized up the number of panels to suit your climate: instead of twenty 100 watt solar panels you have fifty. But wait, you have another issue: you need to size your charge controller to handle the maximum solar output. Can the charge controller handle 5,000 watts when the solar panels are fully producing?

Once you have overcome this hurdle you have a much more difficult obstacle: you need to have the battery capacity to store the energy that is produced by the solar panels and delivered by the charge controller. This cost is steadily decreasing as battery technology advances; however, the necessary quantity of batteries alone for this application can cost $20K or more (probably more: you may need extra batteries to tide you over the cloudy/rainy days). Furthermore, unlike the solar panels which have a useful life of 25-30 years the batteries will need to be replaced every 5-10 years.

In short, a complete solar power generation capability including installation for the modest application described above could easily cost $40,000-$50,000 or more with a very substantial additional outlay periodically to replace the batteries. This does not even begin to address the challenge of finding a suitable location for fifty 100 watt solar panels (or the large battery bank).

By comparison, fossil fuel-generated grid power is cheap and solar power generation has a very long way to go to close the gap. At present the best that can be achieved is this: solar panels may be installed on the roof of a residence with the power generated fed back to the grid and the value of the power subtracted from the homeowner’s monthly bill. This eliminates the charge controller / battery storage aspect of solar power. With government subsidies offsetting the cost of the panels and installation you might be able to recoup your investment in 7-10 years. Yippee.

(Could you utilize the government subsidies to offset the cost of a grid-feed system that could be converted to off-grid in the event the grid goes down: i.e. purchase the charge controller and batteries separately to perform the conversion if the grid goes down while utilizing the subsidized grid-feed system as long as the grid is up? That is an intriguing question. Let’s set that aside for another discussion.)


All of this is not to say that solar power generation is impractical but only to say that generation and storage equal to the economy of the power grid is impractical. However, we plan for contingencies including the loss of the electric grid for hours, days, weeks, or longer. Is it feasible to have solar generation capability that is less than full grid power but still effective at a lower usage level? The answer is an unqualified “yes”.

The key to sizing a back up solar generation system is determining the best combination of utility versus cost. Is it feasible to use the system to run your whole house air conditioning? No. Is it feasible to have a solar generation system that runs several fans? Yes. Can the system run an electric stove? No. Can it run an Instant Pot? Absolutely. Can you alternate between running a refrigerator and a chest freezer? Maybe. How badly do you need a refrigerator and chest freezer?

In order to determine your electric needs there are two methods. Method one utilizes the information included on every electronic device that specifies the voltage/current/power requirements of the device. This is of limited value: does my Instant Pot draw 8 amps of electrical current? Yes … for the first 8 minutes while it warms and pressurizes. After that, the device requires bursts of 8 amps with periods between where very little electrical current is needed. This is important information: my power source must be capable of delivering 8 amps when required. However, it will not be required continuously and this must be taken into account as well.


This brings us to the second method that utilizes an inexpensive but extremely useful device: the Kil-A-Watt meter. This device plugs into the electrical outlet. The device to be tested is then plugged into the meter. The meter may then be configured to display Voltage, Current (Amps), Power (Watts), and (most useful of all) KWH: kilowatt hours. Press this button to reset the counter and then turn on the device plugged into the Kil-A-Watt meter: the meter will measure the power required over time.


This is a good place to present a simple primer on electrical power. Voltage is the measure of the “pressure” that is exerted to push electrons through a circuit (a path from the power source through the device and back to the power source). It is measured in Volts.
Current is a measure of the “volume” of electrons through a circuit at a given moment in time: it is measured in Amps.

There are two kinds of electrical sources: direct current and alternating current. A car battery is direct: the output voltage is approximately 12 volts DC (direct current) and does not change except to slowly decrease as the battery loses charge. A receptacle in your home is alternating: with the right instrument you could see the voltage start at 0, increase to approximately 160 volts, decrease to 0, continue to decrease to a negative 160 volts, and then increase back to 0. This is repeated 60 times per second (in the US; in other countries such as China this happens 50 times each second). The shape of this wave is described as a sine wave and the average of this wave (technically the “root mean square”, a mathematical measurement of the “area” under the sine wave) is 110 – 120 volts AC (alternating current).

Limiting the flow of electrons through an electrical circuit is a load: it may be resistive, capacitive, inductive, or some combination of the three.

For those who are wise in the ways of science there is much more to this topic including resistive/reactive loads, phase shift, etc. For the rest of us there is a rule of thumb method that is sufficient as long as we remember one thing: an inductive load (think anything with an electric motor) requires a great deal of current (amps) to start and then decreases to a much lower current unless something hinders that motor from turning. When a fan is turned on there is a spike in current draw and then the current decreases once the fan reaches operating speed. If a circular saw is turned on there is a spike in current draw which decreases when the saw reaches operating speed; however, when the saw encounters a load because it is cutting material the current increases again. If the saw is stalled the motor is in a “locked rotor” condition and can draw a very large current. The solar generator must be capable of delivering the maximum current required by the load, even if this current is only required for a short time.

With this in mind, we can apply a simple formula: power (watts) is equal to voltage (volts) times current (amps). This is technically only true for a resistive load, but, with the proviso above, it will serve as a rule of thumb for our purposes.

To illustrate I will refer to the Instant Pot again. When first activated it will draw approximately 8 amps at 120 volts (AC) through a resistor element that becomes hot due to the “friction” of the electrons passing through. This will in turn heat the pot, increasing both heat and pressure in the sealed pot. This will last for approximately 8 minutes until the pot reaches the desired temperature/pressure. Applying the formula we see that the Instant Pot requires 960 watts of power during this initial phase. Since 8 minutes is 13% of an hour this translates to 128 watt hours of electrical power.

The remainder of the 10-minute cooking cycle to hard boil one dozen eggs requires another 100-watt hours for a total of 228-watt hours to hard boil a dozen eggs.

(To be concluded tomorrow, in Part 2.)