Proline 5000
™
Inverter by Wagan Tech
®
9
www.wagan.com
4. To get an estimate of the maximum current (in amps) that a battery bank must be capable of
delivering to the inverter, divide the AC load watts by ten (for 12V battery). For example, a
1,500 watt AC load will need 150 amps at 12 volts DC. This relationship holds for 12V DC
inverters with 90% efficiency.
Using the 1,500 watts (or 150 amps) for 10 hours example, 150 amps are needed for 10 hours.
This provides us with the basic amp-hours (AH) of battery life that is required. Ten hours at
150 amps equals 1,500 Amp-hours (AH). This answer is just a starting point because there are
additional factors that determine actual run time. These include:
• Cable gauge and length (cable losses)
• Charge level of the batteries (between use, chargers have to be able to fully charge the
batteries)
• Temperature of the batteries (colder batteries provide fewer amps)
• Age and condition of the batteries (older batteries lose AH capacity)
• Use of DC appliances
• Compliance with turning off unnecessary AC and DC loads.
DERATING THE BATTERY BANK
Most lead-acid batteries have a rating expressed in amp-hours (AH). The most common rating of
AH is “at the 20 hour rate”.
NOTE: Despite several internet explanations, there is no relationship between cold cranking
amps (CCA) and ampere-hours (AH).
For example; if a 20 AH battery is discharged at a 1 amp rate, is will take 20 hours to discharge
that battery. The terms “charged” and “discharged” relate to actual battery voltage. This means
that the output voltage of a nominal 12 volt battery starts at 13.2 volts (fully charged) then drops
to 10.6 volts (discharged). If the load on the battery causes the battery to discharge faster than
the 20 hour rate, the capacity (AH) of the battery is measurably reduced (derated). Derating is
a major run time factor. The following curve can help to determine what the battery bank can
deliver under load. The results are used to estimate how much additional battery capacity is
needed to achieve the desired run time.
The left vertical numbers of the curve represents percentage of the battery capacity at the 20
hour rate. In this example, the user needs a one hour run time. If the example battery is 220 AH
(20 hour rate), and the load is 220 amps that is 100% (horizontal number) of the AH (20 hour
rate), starting at the 100% horizontal point and looking up to the curve the results are that only
56% of the battery capacity is available. This means that a higher battery capacity is required to
get the desired run time: one hour. The curve also shows that a load of 200% of the 20 hour rate
yields only 31% of the battery capacity. The installer must carefully plan the capacity of battery
bank or the run time may be seriously affected. To the inexperienced installer, several trial
battery capacities may be required to make sure a large enough battery capacity is available to
achieve the desired run time.
