My friend Irene Dubois enjoys camping (tent camping, not RV) but recently began using a CPAP machine while sleeping to treat obstructive sleep apnea. Failure to use this machine results in severe snoring, which would drive fellow campers crazy. The CPAP machine’s power adaptor runs off of AC power, which is usually unavailable at tent campsites.
Irene therefore needs a way to power the CPAP machine from an array of batteries. Here we design such an array and then test it through simulation. However, the simulation results we produced do not resemble the results expected from our design calculations, so a physical test of the design is called for. That physical test will be featured in a future post, that is, when we cobble together the funds to buy the batteries.
We would like to power the CPAP machine for two nights without charging the battery array. Furthermore, we want to ensure that these two nights of discharge do not drop the batteries below 40% of capacity (below 11.9 Volts ), since that may damage the batteries. We wish to use RV/marine lead-acid batteries due to their wide availability.
The CPAP machine requires a power supply of 12 Volts (DC) and 5 Amps. From  we conclude that 5 A is a slow discharge rate for a lead-acid battery, and therefore choose to assume that the batteries discharge linearly under this load (at least until they reach 40% of capacity). We assume that the batteries are fully charged at the beginning of the camping trip, and that each night Irene will use the machine for nine hours. If possible, we’d like to include factors of safety in the calculations (e.g. 10 hours of use per night, and a 50% battery capacity cutoff). In this design work we are not considering the use of inverters with the CPAP machine’s AC power supply—which would be lossy—instead we are using the direct DC input on the CPAP machine.
It should be noted that calculating with a 5 Amp current draw is conservative, since that is the maximum power requirement of the CPAP machine (at maximum pressure and temperature) and not necessary the in situ current demand.
1st night: 10 hours * 5 Amps = 50 Amp-hours
2nd night: 10 hours * 5 Amps = 50 Amp-hours
So we have 0.5 * battery_capacity = 100 Amp-hours, which implies battery_capacity = 200 Amp-hours.
We need a 200 Amp-hour battery. However, I do not see these for sale at Sears.com, so we will have to wire two 100 Amp-hour batteries in parallel to achieve the design goals.
Using LTspice, I attempted to confirm the design with a simulation of a 5 A current draw on two 100 Amp-hour lead-acid batteries wired in parallel:
For this simulation I used the SPICE model of a lead-acid battery provided by Helmut Sennewald at , which was the only such model I could find. However, the simulated battery array voltage reached 11.9 V (40% charge remaining) after only 1.5 hours. My guess is that the lead-acid battery model provided at  reflects the more standard automobile starter battery than a RV/marine battery, which would behave differently, but I’m not certain.
The simulation results are so different from the design calculation results that a test on a physical battery array consisting of two 100 Amp-hour batteries wired in parallel is called for, especially since it is unclear whether the model used for a lead-acid battery in the LTspice simulation is the correct one for this application.
- A critical review of using the Peukert equation for determining the remaining capacity of lead-acid and lithium-ion batteries. http://dx.doi.org/10.1016/j.jpowsour.2005.04.030