Wednesday, November 29, 2023


Explaining converters, inverters and generators – Part 2

Do I need an inverter?
By Wolfe Rose
with Mike Sokol

In my last column, I explained why there are two separate AC and DC power systems in your RV, which accounts for half the RV being unpowered in transit or at sites without electric hookup. While the CONverter turns 120V AC into DC used around the RV, the ability to cross that bridge the other direction is provided by an INverter, which turns 12V DC into 120V AC. You’re probably wondering, “That sounds great! Why wouldn’t you have one in every trailer already?”

The first and lesser reason is simply price. If you paid over $100,000 for your RV, you’re more likely to get an inverter included. If you only paid $20,000, adding $1000 for a decent inverter is not something manufacturers are going to just throw in, especially when many buyers stay plugged into shore power most of the time.

The second reason is that the 3600 watts available from a 30-amp shore power connection is actually a HUGE amount of power, and (dirty secret!) most RVs would draw a lot more than their shore power plug “allowed” if you turn on your 1000-watt water heater, 1500-watt air conditioner, 1000-watt microwave, refrigerator, and then maybe add an electric griddle, coffeemaker, blow-dryer, etc. As an aside, I have an alarmed ammeter on my shore power because I’ve melted plugs from just getting too close to maximum amperage.

The third (and fatal) factor is putting 3600 watts of shore power draw up against your pitiful battery capacity. Most travel trailers come with a nominal 100 Amp-hour battery (100 Ah), and to keep from damaging your battery from over-discharge, you only discharge down to half-battery capacity. So that suggests you have maybe 50 Amp-hours at 12 Volts to work with, which works out to 600 watt-hours of usable capacity at long discharge rates. Even using unrealistically perfect math, to equal your 30 amps of shore power, you’d kill your battery every 10 minutes, if it didn’t simply melt down before then. So with a typical usage scenario at an impossible 100% efficiency, your 100-amp-hour battery can only operate your microwave for 36 minutes before the battery is dead.

But that’s still at an unrealistic 100% efficiency. Modified Sine Wave (MSW) inverters run at around 85% efficiency, so the inverter pulls around 100 amps of current to make 1000 watts. And because of lead acid battery chemistry, at a 100-amp discharge you’ll only get around 35% of your “100 Ah at 5 Amps x 20 hrs” battery’s rated capacity, or 420 watt-hours of usable power, which works out to around 21 minutes of run time or 1000 watts of inverter output. But the microwave doesn’t “like” choppy MSW power due to all the harmonics and only works at around 50% efficiency, running twice as long for the same heating effect. So your 1000-watt microwave is now acting like a 500-watt microwave, all the while overheating its own power transformer and shortening its lifespan. So in an attempt to get your morning jolt of caffeine you’ve drained and most likely damaged your house battery, just to heat a large cup of coffee?

A high-end “Pure Sine Wave” inverter running at 90% efficiency helps, but not enough. Because it’s getting a pure sine wave without harmonics, the microwave will now heat normally at its 1000 watts of rated power draw, so you might squeeze out 23 minutes of microwave run time from a 100-amp-hour battery with a 90% efficiency Sine Wave inverter.    

[Special thanks to for their inverter efficiency numbers. And here’s a link to their article on Pure Sine Wave and Modified Sine Wave Inverters.]

So the bottom line is that a 400-watt convenience inverter might be okay in bursts, but clearly, no significant inverter producing a lot of wattage is going to run from house batteries for long. In the third part of this series, I’ll discuss what the heck an “inverter generator” is and how it works.





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Wolfe (@guest_14044)
6 years ago

The focus here was to discuss how the inverter’s amperage draw is prohibitive unless you have substantial solar/wind/running alternator to replenish your battery against the inverter’s draw. Discussing the tail-chasing effect of Peukerts Law diminishing capacity against increasing draws was judged too technical for here.

TMolnar: Yes, I’m aware that MSW inverters will run MOST things acceptably, just inefficiently. Some switching power supplies heat up excessively, and a few electronics get interference. Pure sine ARE better, but 5X cost. Are you *really* surprised an inverter vendor insisted you need the most expensive for every use? LOL.

I have 150/400/4000W MSW inverters myself, and they work fine for what *I* use them for (not electronics), and the 4KW ONLY against my 600W dual alternators. Mine may be on the extra cheap side, but my inverters only claim 60% efficiency, and their surge capacities are total lies… I don’t usually exceed 1/2 the sustained wattage.

Tommy Molnar (@guest_14008)
6 years ago

I’ve been using a Cobra 2500 watt inverter for over 20 years now. It’s modified sine wave. We rarely (like almost never) use the nuker (microwave) even when hooked up so that’s not a problem. I’ve been charging cell phones, computers, camera batteries, you name it, with no apparent ill effects. We run our Dish box off it all day long, for playing the music. The TV only comes on to set up the dish when first establishing a campsite. I’ve been toying with the idea of a pure sine wave inverter but the reviews of most of them don’t sound like there’s much of an advantage. I’ve got 375 watts of solar on the roof feeding my two Trojan T-145 golf cart batteries.

Robert Henry (@guest_14002)
6 years ago

Good article! I’m aware of Peukert’s Law regarding true capacity under various loads. Is there a way to determine VOLTAGE drop, given a short, high amperage load? I.e. inverter low voltage shut-down when trying to use a single cup coffeemaker (2 mins). Thanks!

chris p hemstead (@guest_13990)
6 years ago

You’d be lucky to get even a few minutes of MW running from a single battery due to the huge voltage drop at such a high amperage draw.

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