This is probably a loaded question, but Im in the market for a new Pure Sine Wave inverter. My budget is $300-500, around that area. I was running a MSW inverter previously and it could not keep up. The biggest draw by far is the water heater. I have 4 160 AH deep cycle batteries that are probably on their way out. I expect that 3000 watts is about right for what I need (completely guessing). 2000 watts seems too low, and I do not want to get one with wattage that is too high in fear that it would draw too much from the batteries. Im a little unsure of how the wattage works with these inverters though. I understand the sine waves but I don't understand the watts. Is 3000 Watts the outputs? If so what would be the input? Anyway, the reviews on these are all over the place. Here are the ones I found:

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Power sources:

1. Alternator/Isolater

2. 300 Watt Solar on roof.

Power Draws:

1. Bosch 2.5 Gallon Electric water heater (Huge draw)

2. Dometic 12v fridge

3. Chinese Diesel heater

4. plugs/outlets/the occasional TV

5. Sureflow water pump (small draw)

6. LED lights (Small draw)

The way you have specified your needs indicates you may not fully understand what the inverter is doing in your system.

Like sparkplug says, before you starting trying to specify the inverter capacity (in watts) you need to separate out the power draws that go through the inverter. Many of the things on your list (above) are probably or definitely 12 volt Direct Current (DC) loads:

- Dometic 12v fridge (definitely)
- Chinese diesel heater (definitely)
- Sureflow water pump (? RV versions are 12 volts, but 120 volt AC units are available as well)
- LED lights (probably 12 volt if they are RV models; could be 120 volt AC if they are household ones installed in RV)

Anything that runs off of 12 volts DC is not going through your inverter. The purpose of the inverter is to take a Direct Current (DC) power source and give it an Alternating Current (AC) waveform at certain voltage. Typical DC power sources are 12 volts or 24 volts, and in North America the expected AC voltage is usually 120 volts. Less expensive inverters don't achieve the actual waveform (a sine wave), but for some devices a "Modified Sine Wave" is good enough. (You don't have a microwave oven on your list; those are one of the devices that really wants to see a "Pure Sine Wave". No harm in going with a Pure Sine Wave inverter except the extra cost; benefit is making sure as many devices as possible are compatible.)

Your "plugs/outlets/the occasional TV" are domestic 120 volt AC loads. Some of these can be substantial draws (i.e., you plug in a 1,500 watt electric heater or an 1,800 watt four-slice toaster). You need to create your AC power budget based on what all of these are and how many you expect to run simultaneously in order to size your inverter. They should each have a data plate on them that specifies their power draw in either watts or amps (at 120 volts).

The Bosch 2.5 gallon electric water heater I found on-line was in the 3000 T series; these have a 1,440 watt power draw specification. I believe that a simple resistance heating element doesn't have a substantial variation in peak versus steady-state draw. If your Bosch 2.5 gallon electric heater is a 120 volt AC unit then you need to add its power needs to your AC power budget.

If your biggest load is the Bosch, and you don't plan to run the Bosch simultaneously with something like a four-slice toaster, then a 2,000 watt inverter might be the right size for you. Only you can know whether this is true, however, by creating your AC power budget based on your actual loads, the peak usages, and which might be run simultaneously.

Most of the inverter specifications that I have seen rate the inverter by the constant load they can handle in watts. A 2,000 watt inverter should be able to meet the steady state load of a 2,000 watt device or combination of devices; this is just under 17 amps at 120 volts AC. Depending upon the inverter, it may be able to handle higher temporary loads; if so these are usually specified as a "peak" or "maximum" load in watts, along with the length of time it can handle that peak load. The power (watts) drawn from the DC power source is greater than the power supplied (in watts) to the AC devices due to inefficiencies in the inversion and voltage conversion process. If the inverter is 90% efficient, then the power drawn from the DC source to supply 2,000 watts to the AC devices would be 2,000 watts/0.90 = 2,222 watts. If your battery system is at 12 volts (you didn't specify it) then the inverter would be drawing 2,222/12 = 185 amps to supply that 2,000 watt AC load.

That kind of amp draw is pretty sizable and it's possible that what wasn't "keeping up" was the DC power supply and not your inverter. You didn't specify whether the 160 amp-hour batteries were 6 volt or 12 volts; either is possible. If they are pairs of 6 volt batteries wired in series/parallel, then you have 320 amp-hours of 12 volt battery storage, and trying to suck 185 amps out of them would be a big hit. Before you buy a new inverter thinking it will solve all of your problems, calculating the actual draw you have through your inverter and what this inverter load draws from your batteries is in order. You might just be exceeding what your batteries can reasonably supply.

Different battery technologies have different rates at which power can be drawn from them. "Deep-cycle" battery capacity is typically specified at a 20-hour discharge rate; this means you only get the rated "amp-hour" of capacity if you discharge the batteries over 20 hours. If you only have 320 amp-hours (4 6-volt 160 amp-hour batteries) then the 20-hour discharge rate is 320/20 = 16 amps. Sucking 185 amps out of them is a 320/185 = 1.7 hour discharge rate. This substantially reduces the available energy because the amount of energy the batteries can supply is reduced when you discharge them more quickly (see

https://en.wikipedia.org/wiki/Peukert's_law.) Even if your batteries are all 12-volt and you have 640 amp-hours of storage, a 185 amp draw is a 640/185 = 3.5 hour discharge rate, substantially higher than 20 hours. I would guess that you would have only about an hour of draw at this rate before your batteries would be down to 50% state of charge.