This is a subset of “how do I run this load” but since it comes up frequently it will be addressed here as a separate case.

• compressor fridges are a part-time load; as a rule of thumb we can assume the compressor will be running 1/3rd of the time.
• they will consume a given number of Wh (Watt-hours) over a 24-hour period
• you must have at least enough charging capacity to replace those Wh
• and have enough battery bank capacity to make it through periods where charging is absent

The fridge specs (or reviews) will report the power required when the compressor is running. This is usually something like 45w-60w. We will use the 60w example from here on.

The compressor only runs as needed. It will run more in hot ambients and less in cool ambients. Until you observe it in actual conditions we will assume it runs 1/3rd of the time.

60w x 0.33 duty cycle x 24 hours = 480Wh required each day to run the fridge. We will use the 480Wh example from here on, and assume this is the only load. In real life the fridge will be only one part of your daily power requirements. We also assume the fridge will be run off 12vdc. If you are running it off an inverter then add ~20% to the Wh requirements to make up for inversion losses.

The battery bank capacity requirement is dependent on a few factors:

• how deeply the chemistry can be discharged (~50% for lead, ~80% for LiFePO4)
• nominal voltage (12.0v for lead, 12.8v for LiFePO4)
• days of “autonomy” (discussed separately)

Storing 480Wh of energy would require:

• ≥80Ah of lead (480Wh / 0.5 depth of discharge / 12.0v nominal)
• ≥47Ah of LiFePO4 (480Wh / 0.8 depth of discharge / 12.8v nominal)

In practical terms 100Ah of LFP is cheaper by the Ah than 50Ah, so we would probably round up to 100Ah. This would allow more aggressive charging from the alternator¹⁾ and automatically give us another day of autonomy (see below).

The capacities above assume the bank is fully charged each day (“one day of autonomy”). If you can only count on getting a full charge every 2 days (“two days of autonomy”) then multiply the Ah required by 2, every 3 days by 3, etc.

Ah requirements for days of autonomy with LFP:

• fully charging every day = 47Ah, as above
• fully charging every other day = 94Ah
• fully charging every third day = 141Ah
• etc

Many folks will round up from 141Ah of LFP to 200Ah since that is a common size (and even less expensive by the Ah).

Note: the math is simple with LiFePO4, since it is not affected by partial states of charge. With lead banks larger capacities require stronger charging setups.

LiFePO4 is easier to charge than lead; it is more efficient, usually faster, and failure to reach 100% will not affect the battery negatively. LFP charges quite well by alternator alone.

Lead requires ~20% more energy to replace the same Wh/Ah, and takes many hours to charge fully. For this reason lead is best charged by a combination of solar+alternator charging.

The simplest case will be charging LFP from a DC-DC charger. Using a 30A DC-DC as an example, it would take about 1.25 hours (75 minutes) of driving each day to replace 480Wh. 480Wh / [30A x 12.8v])

Charging LFP directly from the alternator with a combiner works but the time required is not strictly predictable. Charging current will be highest at low State of Charge and lowest at high SoC; this is called the “current taper”.²⁾ Bonus: combiners usually make self-jumpstarting trivially easy.

Assuming there is enough time (usually 5-6 hours) a lead bank can be charged by DC-DC alone. The long (and mandatory) Absorption period is the tail that wags the dog.

Usually only an option for “power stations”, the ciggy port output is predictable but underpowered. Because it's usually fused at 10A, the charge rate will be ~120w. It would take 4 hours of driving to replace 480Wh through the ciggy port (480Wh / 120w).

Charging from solar is predictable on average if one knows where/when one will be camping. Because December has the lowest insolation in the Northern hemisphere we base our panel requirements on that month. If you are not full-timing then use the month closest to winter solstice.

You will need to use PVwatts or similar to calculate how much sun (hours of Full Sun Equivalent, aka “kWh/day”) is available in your actual wintering spot, but we will use Belle Fourche, South Dakota since it is the geographical center of CONUS.

December in Belle Fourche has only 1.52 hours of FSE. It would take ~375w of panel³⁾ to replace 480Wh year-around. (480Wh / 1.52 hours of FSE / 0.85 overall efficiency)

The good news is there would be excess solar harvest in the other months of the year:

  Solar wattage	375
  Month	Daily Wh Avg   
  Jan	 574    
  Feb	 806    
  Mar	1329    
  Apr	1658    
  May	1865    
  Jun	2298    
  Jul	2324    
  Aug	1983    
  Sep	1562    
  Oct	1010    
  Nov	 641    
  Dec	 485    
  Average	1378     

table based on the PVwatts link above

Often a mix of charging sources yields superior results for less money. It also offers redundancy in case one charging source is not available.

Adding 30mins of 30A DC-DC charging a day would drop the panel requirement to ~236W. This could mean easier mounting, a smaller solar charge controller, smaller-gauge wiring, etc.

And since the solar would be handling higher-voltage / longer duration charging duties one could fall back to a less expensive combiner for the alternator side. The alternator doing the heavy lifting might mean one could use a less expensive PWM controller.

Or one could use a combo DC-DC/MPPT controller that handles both charging sources in one unit. Some can also maintain the vehicle's starter battery from solar.