Inverter efficiency
firerescue712
Solar Expert Posts: 95 ✭✭
I have two MSW inverters. One is a Harbor Freight 1500-watt. The other is a WalMart cheapo 300-watt. I was curious to see how efficient they are. Using a Kill-A-Watt meter, I was pulling 3.34 amps for a ceiling light circuit. I used a clamp meter to check the amps coming from the batteries. The HF used 11.73 amps. The 300- watter used 9.67 amps. When the lights were turned off, the clamp meter dropped back to .4-.5 amps. I am running many small 12v circuits off the batteries continually. This is the residual amp reading. Are the MSW inverters this poor at efficiency? Or is it just mine? I am planning on upgrading to TSW soon. Maybe sooner than I planned. Cariboocoot, i am all ears. Thank you.
Comments
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Re: Inverter efficiency
You may find the equivalent TSW inverter has upwards of 2x the idle current of a MSW inverter.
Nature of the beast.
Running small continuous loads, look at dedicating a small inverter to run them and only turn on the larger inverter when large loads need the power.
Look at the specs. for the MorningStar 300 Watt TSW 12 volt input inverter... It has a "search mode" which pulses the AC line about once a second looking for loads greater than ~6 watts--And turns on 100% when loads are found.
Losses are a huge issue for solar and when you add them all up--Derating the solar panel marketing number to AC at the outlet by 1/2 is about right.
If you have specialized needs (like 24x7 AC power), you do need to take those extra inverter losses into account (larger solar array, larger battery bank).
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: Inverter efficiency
What Bill said.
In simple terms, PSW inverters have more components that consume more power (the engineers can stop cringing now). A difference between a few square wave steps for output and a whole lot of square wave steps smoothed into a sine wave.
You've already noticed one thing: the bigger inverter inevitably consumes more power than the smaller one even for the same output. If they were two inverters of the same size and type from different companies there could still be a significant difference. It's like 'inverter roulette'.
Curiously, if the load were induction instead of resistance the over-all efficiency could be better on pure sine than MSW. It's that power factor thing again. A motor might draw 20% more on MSW than PSW, so there goes the advantage of the 'simpler' inverter.
So look for both the stand-by or idle current use and the run current use and consider them when weighing up what you need. No sense powering 20 Watts of inverter to supply 200 Watts of power when you only need to use 6 Watts. -
Re: Inverter efficiency
Gentlemen, thank you for your inputs. I will put this knowledge to good use when i upgrade. We are all born unknowledged. It is a relief some jave learned more and share that with others. -
Re: Inverter efficiency
Thought I'd put this in should anyone wonder how this stand-by vs. run consumption plays out in the real world (from my own system):
Outback VFX3524. Stand-by consumption 6 Watts, running consumption 20 Watts. Difference: 14 Watts.
Minimum power need of running the refrigerator requires 20 minutes every hour. In one day that's 8 hours in running mode and 16 hours in stand-by. Savings: 16 hours * 14 Watts = 224 Watt hours or approximately 9.3 Amp hours on the 24 Volt system. About 3% of the total battery capacity.
However, that discounts all the other things that will be on and demanding power for approximately half the day, divide those numbers by two. Total savings in running just the refrigerator with the inverter in stand-by mode overnight: 4.65 Amp hours or 1.5% total battery capacity.
Everyone's system and usage is different. I'm just trying to show how the two modes play out. -
Re: Inverter efficiency
Parasitic load on a small system is a killer. I've measured all my inverters, MSW & PSW. None are great, but as already mentioned, the MSW inverters use less.
When my solar array was smaller, and inverter idle current mattered more, I cheated a little.
Rather than leave a 400W inverter on all the time, I connected it to a remote controlled relay. Left the living room lights plugged into it all the time. When I wanted the lights on, I'd push a button on the relay. Push the other button to turn them off. Zero parasitic load when the lights were off, as the inverter was off too. The amount of power the remote receiver used was negligible.
The MSW worked great for this purpose. I measure greater than a 90% efficiency, and the CFL bulbs did not hum of flicker despite the MSW waveform.
Pics, details, and a short video of this setup here: http://2manytoyz.com/gridless.html -
Re: Inverter efficiencyfirerescue712 wrote: »I was curious to see how efficient they are. Using a Kill-A-Watt meter, I was pulling 3.34 amps for a ceiling light circuit. I used a clamp meter to check the amps coming from the batteries. The HF used 11.73 amps. The 300- watter used 9.67 amps.
I may be missing something here - everyone else focused on the idle current draw, but my impression is you were asking about the current draw differences here. If I'm off, just ignore me!
These numbers don't make sense, if the 3.34A load was on each of the inverters. Keep in mind the battery amps will be higher than AC amps since you're stepping the voltage up. But it should be a factor of 10 for a 12V battery bank!
With your numbers, you had an AC load of 120V x 3.34A according to the KaW - 400W. But the inverters were only drawing 12V x 11.73A = 140W and 12V x 9.67A = 116W. Pretty good efficiency, if you ask me! :cool:
I have heard Kill-a-Watt meters don't like MSW - perhaps it's reading incorrectly due to that? I'd be guessing you had about 100W load on the inverters at the time? -
Re: Inverter efficiency
Yes, MSW can confuse a Kill-a-Watt...
But also remember that Volts * Amps does not equal Watts.
For a Kill-a-Watt meter, the actual equation should be:- Volts * Amps * Power Factor = Watts
- Volts * Amps = VA
The AC side of the inverter (more or less) responds to the VA of the load. The DC side, (again more or less) responds to the Watts (times 1/efficiency of the inverter).
And Inverters are constant power devices... I.e., both battery voltage and current are important... If you have a 120 watt AC load and two different battery voltages, you will get two different amp readings on the DC side of the inverter:- 120 watts * 1/0.85 inv eff * 1/14.5 volts = 9.7 amps (inverter+battery charging)
- 120 watts * 1/0.85 inv eff * 1/10.5 volts = 13.4 amps (inverter at bat cutoff voltage)
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: Inverter efficiency
He's looking at two different inverters of two different sizes powering the same load. The idle current may be similar on both, but the larger (less efficient perhaps) unit consumes more current to produce the same amount of output power (or similar, for as Bill pointed out power isn't always what it seems to be).
The draw of a larger inverter when running may be much larger than that of a smaller inverter, but with output at peak capacity it becomes a smaller percentage of the total.
Example:
Morningstar 300 using 6 Watts equals 2% of its total output power.
Outback 3500 using 20 Watts equals 0.57% of its total output power.
The Outback being asked to produce 300 Watts will still consume 20 Watts, a whopping 6.6% of the power at that level.
(Not precise calculations; just demonstrating the relationships.)
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