# Resistive load

greenthumb76
Solar Expert Posts:

**40**✭
Could someone please explain (or point me to an explanation) of what problems there are with a modified sinewave inverter and a resistive load?

Thanks!

Thanks!

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## Comments

30,058adminTypically, no problems with MSW inverters and resistive loads (filament lamps, electric/resistance heaters, etc.)--Other than the large amounts of power resistive loads (typically heaters) vs what can be supplied by other fuel sources (fuel, solar thermal, wood, etc.).

Do you have some specific application you wish to support?

-Bill

814✭✭✭Green thumbs I am going to try give simple explanation as to the difficulties of certain resistive loads.. this example it not a scientific one .ok??

ok you have a electric wood bench saw ,, now when you start it its not under much load .ok? its only under load and drawing a lot of current when actually cutting the wood.

so its on load time is most likely less than its off load time . so while the motor gets hoter when under load it self cools with its built in fan when not in use.

Now lets take a stove element its cold load is high it settles as it heats up BUT it is drawing its full rated power the whole time its on, (for this example its on HIGH ) OK? Cheap inverters are not designed to provide power to such devices. Have a look at any available ones the fans are too small and the usual aluminium case is too small to disipate the heat.., that is why a 1000 w inverter is unhappy with heater elements above 500w,, If you really want to get an inverter designed for heavy duty use you need to look at industrial inverters..but at 3 to 4 times the price for same size..

30,058adminYes--what John says--Inexpensive inverters are not usually very happy running near 100% load for extended periods of time (100% power for minutes/hours is virtually impossible when running a skill saw--very easy with a heating element/resistive lighting load/etc.).

-Bill

40✭I had a different post, regarding the fuse size needed, based on having a 1500watt /3000 peak MSW inverter. I wanted to run my 1200 w toaster oven (at about 350F for 5-10 minutes) or my 1000w bucket heater for 10-15 minutes. (With about 200 amp hours of batteries, and 70 watts of solar panels, I obviously would not be doing this very often.) I was told the inverter would not like this. I was curious as to the reasons why.

30,058adminI am going to run a lot of math by you as way of explanation of the issues. It is all pretty straight forward, and uses typical derating factors that give good results.

Usually it is much less confusing to present a set of requirements (Watts*Hours per day; location, and seasonal use--summer vs winter) to give a balanced system design.

What I have done below is sort of backwards, showing what would be required (roughly) to meet your needs--And showing how your existing setup would have issues running the desired loads.

If this is too confusing--Let us know and we can figure instead design a system that would support your needs (I.e., in Northern Nevada, need solar 9 months of the year to run a 1,200 watt toaster oven for 20 minutes per day without using a generator).

Assuming that this is a 12 volt battery bank, 85% efficient inverter, 10.5 volt cutoff, and needing a 1.25x safety factor:

- 1,500 watts * 1/10.5 volt cutoff * 1/0.85 eff * 1.25 fuse/wiring safety factor = 210 amp fuse / breaker / wiring

Obviously, if you run a surge for 5-10+ minutes, then you are looking at needing ~420 amp fusing to avoid false trips.And, inexpensive inverters (and some expensive ones) actually have problems running over 80% of rated load for any length of time.

And high loads on a 12 volt battery bank is very difficult--Run the battery down to 11.5 volts (less than full charge, and voltage drop from heavy loads), gives you a 1.0 wiring drop for typical inverter with 10.5 volt cutoff.

For flooded cell batteries, some good (rough) rules of thumb:

- C/10 to C/20 are efficient loads
- C/8 is about the maximum continuous load (battery may overheat on load or charging in excess of C/8)
- C/2.5 is about the maximum surge current supported by flooded cell batteries--More current and you may collapse the battery voltage

So, worst case loading for 1,200 watts, 85% efficient, 10.5 volt cutoff inverter:- 1,200 watts * 1/10.5 volts * 1/0.85% = 134.5 amps

And the battery bank size (12 volt) based on above rules of thumb:- Capacity (20 Hour Rate) = 134.5 amps * 20 (5% discharge rate) = 2,690 AH
- Capacity (20 Hour Rate) = 134.5 amps * 10 (10% discharge rate) = 1,345 AH
- Capacity (20 Hour Rate) = 134.5 amps * 8 (12.5% discharge rate) =1,076 AH
- Capacity (20 Hour Rate) = 134.5 amps * 2.5 (40% discharge rate) =336 AH

So, a battery bank that was expected to run such a load only a couple times a day for a minimum amount of time with a flooded cell battery bank would be looking at 336-1,076 AH @ 12 volts minimum...If you have AGM batteries, some of them can support upwards of C*4 current:

- Capacity = 134.5 AH / 4 (400% discharge rate) = 33.6 AH

Would it actually work, and could you pull 135 Amps from a 33 AH battery... Pretty heavy wire for a very small battery.Note that this would fully discharge the battery in 15 minutes--This is UPS type loading--Can be done, but probably not a good choice for daily use (running battery dead is never a good idea).

Another way of sizing a battery bank is based on the available charging current... A good rule of thumb is 5-13% rate of charge. For your system, 70 watt panel and assume 0.77 system efficiency:

- 70 watts panel * 1/14.5 volts charging * 0.77 efficiency * 1/0.13 charge rate = 28.6 Amp*Hours @ 12 volts minimum battery capacity
- 70 watts panel * 1/14.5 volts charging * 0.77 efficiency * 1/0.05 charge rate = 74 Amp*Hours @ 12 volts

So, for a 70 watt panel, the 200 Amp*Hour battery bank would really like to see a solar array of:- 200 AH battery * 14.5 volt charging * 1/0.77 derating * 0.05 rate of charge = 188 watts of solar panel minimum
- 200 AH battery * 14.5 volt charging * 1/0.77 derating * 0.13 rate of charge = 490 watts of solar panel maximum

And, what would be the amount of time you could run a 1,200 watt load from a 70 watt solar panel. Assuming 0.52 system efficiency (solar panel to charger to battery to AC inverter), 4 hours of sun (reasonably sunny weather):- 70 watts * 4 hours of sun per day * 0.52 derating = 164 Watt*Hours per day
- 164 Watt*Hours per day / 1,200 Watt Load = 0.137 Hours per day = 8.2 minutes per day

And assuming your bucket heater is used during winter--you may be looking at closer to 2 hours of sun (or less) per day--Or 1/2 the power listed above.There are a lot of reasons why what you asked for is difficult to do... Solar power is expensive and, typically, for simple heating projects, a solar thermal heating setup (insulated bucket with south facing plastic panel to let sunlight in to heat bucket) would probably work better.

And for running a toaster oven--a 20lb tank of propane will last you quite a while running a little catalytic oven.

-Bill

814✭✭✭Green thumb I thought I explained it in this thread post#3

put simply again the inverter is just going to overheat. it will then either shut down or do a melt down.. lack of large enough case on inverter to dissipate the heat and inadequate fans

40✭John--Thank you for your help. Please don't get all bent out of shape. I'm answering Bill's question:

Bill--thank you so much for taking the time to explain this all, especially with the numbers that pertain to my system! It gives me what I need to know in terms of expanding my system, or learning to live with what I've got.

814✭✭✭sorry greenthumbs just thought you not understood so was trying to be helpful..:cool: