voltage controlled switch

Re: voltage controlled switch

Actually, there is a possibility that the 1,500 watt inverter is not really that over-sized...

Basically, for AC power you have the power equation that we all know and love:

• Power= Volts * Amps
• V=P/A=310 watts/4.76 amps = 65 volts

Note, for the numbers you gave, the results do not make sense. I am guessing that you have a 117 volt or so AC inverter, not a 65 volt inverter...

It turns out that the above power equation is not really correct for AC (or non-steady state DC power either). The missing part is the phase angle between the voltage and current since waves (if you remember your math, Volts and Amps are vectors and you use trig and Pythagorean Theorem to do the math). This missing term is the Cosine (angle) or Power Factor:

• Power = Voltage * Current * Cosine (phase angle)
• Power = Voltage * Current * Power Factor
• Cosine (phase angle) = Power Factor

So, in real terms, a Power Factor that is less than 1.0 increases the current flow through the wire and inverter electronics. And is typically found to be less than 1.0 for motors and many electronic power supplies (battery chargers, wall transformers, etc.).

Many inverters, transformers, and generators will have their output defined in both maximum Watts and VA (Volt*Amps or VA). For smaller Inverters and generators, many will have their Watt rating = VA rating. For some units (and larger devices) they may assume PF = 0.6 or so and show

• Watts=VA*PF=VA*0.6

UPS manufacturers will typically advertise VA ratings because it makes the UPS "look bigger" than it really is. It makes the battery capacity look larger (battery energy consumed only uses Watts, not PF or VA).

Note, non-sine wave current (such as electronic power supply input current) which is more of a series of peaks than a sine wave) can also be represented by the Power Factor number.

And MSW inverters with their Modified Square Wave also make PF worse for motors/inductors/transformers/some power supplies.

If you have a Kill-a-Watt meter there is a PF and V*A (VA) setting too. Using your numbers:

• Power = Volts * Current * PF
• PF = P/(V*A) = 310 watts / (117v * 4.76a) = 0.56

And the VA rating (sometimes called VAR for utility power for Volt Amp Reactive) of your load is:

• VA = Volts * Amps = 117v * 4.76a = 557 VA

Also, an MSW inverter will make the PF worse for motors and such...

Plus, you need a larger inverter because the current into an AC motor goes way up during starting (typically 4x or more). How the inverter and loads react the starting current is different--so there is a bit of hit and miss trying to guess which units will work together.

So a 750 Watt inverter may be a bit small for your application.

If you can find an inverter that meets your needs and has a remote control on/off input would be the best way to handle the problem... Most likely though, it will be the more expensive MSW and TSW inverters that will have the remote switch option.

I don't know your situation, but if you have a lot of solar panels (say around 2-3kW or more) and your utility supports Net Metering--You are approaching the point where a Xantrex XW Hybrid (GT and off-grid capable) Inverter woudl makes sense.

Regarding the load control outputs on typical solar charge controllers--they are only rated for 8 amps or so DC--not the hundred or so you need.

Wire size for a 750 Watt inverter, you can use a Voltage Drop calculator to figure out wire gauge. You should also follow the NEC tables for minimum wire gauge ratings too.

For the Voltage Drop Calculator. Say you need:

• 750 watts * 1/10.5 min input votlage * 1/0.80 eff at max power * 1.25 NEC safety factor = 112 amp rated circuit minimum
• 750 watts * 1/10.5 min input votlage * 1/0.80 eff at max power = 89 amps maximum

Say your battery goes down to 11.5 volts under full load--and the inverter cutoff is 10.5 volts, the maximum drop allowed would be 1.0 volts in the wiring. Assume you have the inverter 4' from the battery bus/bank connection. Using the Voltage Drop Calculator we find:

• 4' run (8' round trip) of 10 awg wire will drop 0.9 volts

From one of the NEC tables for general wiring, we see that a minimum size for 112 amps would be around 2-1/0 depending on insulation type...

So, in this case, you are limited by NEC minimum current support. If you had a longer wire run, you would be limited by voltage drop.

-Bill

Re: voltage controlled switch

I did not address the contactor--because you guys have done a nice job of that already already... I was addressing the sizing of the inverter and the wiring which he also typed about.

Also, mechanical contactors can take 6-12 watts (or more?) to operate their relays--for smaller systems, those loads cannot be ignored.

The alternative was to find an Inverter that has a remote inhibit switch (typically 0v/Vbatt) which is pretty easy to turn on and off with the small voltage controlled switch. But all of that adds money to the overall costs too...

-Bill