When designing solar panel systems for homes how do you convert the watts generated by the 12V DC solar panels to watts available in a 110/120V AC system? When I do the math it seems to defy logic.

Watts = amps * volts (which also means W/V=A) is the basic formula. So what is wrong with the following: The solar system generates 100 watts at 12V DC = 8.33 amps. Now I want to use that power that has been generated so I feed the power into a 12V DC to 110V AC inverter. When I multiply the amps produced of 8.33A by 110V I get 916.3 watts. I always thought that a watt was a watt (minus power loss due to resistance and inefficiencies of your circuit design of course) so do I really turn 100 watts DC into 900 watts AC? That seems to defy logic. Or do I still have only have 100W 110V AC and why? Am I overlooking a formula that should be included when Inverting DC to AC?

If my math is right then why do they say that I would need a 5kW 12V DC system for my home? That would convert to 416 amps and a 45.8kW 110V service in my home when the solar is at peak performance. (I know that I would get no power at night). My average hourly usage is 2kWh so it seems that 46kW is WAY overkill even if I save some to batteries for night time use.

  • \$\begingroup\$ See answer to a similar question \$\endgroup\$ – RedGrittyBrick Jun 13 '13 at 15:38

Your maths is wrong.

100 Watts generated by solar power at 12V.

Your 8.33 amps is the correct calculation.

The mistake you have made is assuming that this current stays the same value when added to the 110V source. You cannot just connect the systems. To feed the 12V (DC) into the 110V (AC) you need to convert them to the same thing - usually the 110V AC. This means using some form of converter (basically a transformer with the DC turned into AC by transistor switches). The POWER OUT will always be less than the POWER in so that at 110V the current will be less than 0.909 Amps NOT 8.33 Amps. (Vin * Iin >= Vout * Iout)

  • \$\begingroup\$ Your statement -- The POWER OUT will always be less than the POWER in so that at 110V the current will be less than 0.909 Amps NOT 8.33 Amps. (Vin * Iin >= Vout * Iout) -- is the Math I am missing. How does 8.33A DC Input to the inverter become 0.909 Amps Output on the AC Side?? \$\endgroup\$ – Bryant Haggen Jun 14 '13 at 7:59
  • \$\begingroup\$ Power in = 100W I = 100/12 = 8.33 A \$\endgroup\$ – JIm Dearden Jun 14 '13 at 9:45
  • \$\begingroup\$ @BryantHaggen Power in = 100W I = 100/12 = 8.33 A Assuming you could make the converter 100% efficient the Power out will be 110 x I(out) = power in = 100W. Calculating this current give 100/110 = 0.909 A . This is for 100% conversion of the power. Generally you should work on an 80% figure (there is always power loss) so the output power will be 80W (80% of 100W) and this give an output current of 80/110 = 0.73 amps (see Ian's answer below) \$\endgroup\$ – JIm Dearden Jun 14 '13 at 9:51
  • \$\begingroup\$ Thank you for your response. As a software engineer, I think in terms of algorithms that go from point A to point Z so I could not sleep knowing that a piece of the "Software Program" was missing. I had the front end and the back end of the program but not the piece in the middle. Thank you again for your assistance as it helped me finish the algorithm. Now I can sleep restfully knowing that a Watt is still a Watt (minus inefficiencies in the circuit of course as everyone was careful to point out) \$\endgroup\$ – Bryant Haggen Jun 15 '13 at 8:56

Your converter can't do magic. It converts energy from one form (current at 12V DC) to another form (current at 110V AC). It can't produce energy, to the contrary: like any physical device it is imperfect, so it will loose energy. A typical figure might be 20% loss, or 80% efficiency.

Let's assume you draw 1A of current at the 110V AC side, which amounts to 110W. To produce this, the converter needs 110W / 0.8 = 137.5W. (The other 27.5 W are the losses in the converter, which will be in the form of heat. In some circumstances that might be considered useful, but we will regard it as loss) This is delivered by your 12V input to the converter, hence it will draw 137.5 / 12 = 14.5 A (rounded).

No magic, just math.

  • \$\begingroup\$ I am still not seeing the "Middle Math" that goes on inside the converter that keeps a WATT constant (minus circuit loss of course) on the AC side of the inverter. If you were to write the full algorithm sequence (computer program if you like) that goes from the Source to the destination I might understand it better. Solar: 100W/12VDC=8.33A --> INVERTER: (what math here? How does 8.3 Amps DC input become .9 Amps AC output) --> AC_Side: .9A*110VAC=100W available. //Ignore loss for the explanation. I can factor that in later. All respondents did a good job of explaining loss. \$\endgroup\$ – Bryant Haggen Jun 14 '13 at 7:56
  • \$\begingroup\$ Ignoring the losses, energy in equals energy out. Calculate backwards from the output to the input. 9A*110V=990W. Hence at the input 990W/12A=82.5A will be drawn. If you want to calculate forward: with 12V*8.33A=100W available at the input, the output can deliver 100W/110V=0.91A (rounded). \$\endgroup\$ – Wouter van Ooijen Jun 14 '13 at 8:05
  • \$\begingroup\$ And thank you for your quick response to my initial question -- it was greatly appreciated. \$\endgroup\$ – Bryant Haggen Jun 14 '13 at 8:16
  • \$\begingroup\$ I am still not seeing the formula I hoped for. I am looking for a formula that takes 3 parameters as input 1: DC Volts. 2: DC Amps. 3: AC Volts. It would then use those values and compute the AC amps as the Output. In other words I need to convert the 8.33 Amps 12VDC = X Amps 110VAC. What is the math to solve for X. \$\endgroup\$ – Bryant Haggen Jun 14 '13 at 8:32
  • \$\begingroup\$ Ok all, I think I have it. I tried several values and this seems to work as the Math for an inverter. VDC*A_DC/VAC=A_AC In words versus algorithm -- VoltsDC times DC_Amps divided by VoltsAC equals AC_Amps output. \$\endgroup\$ – Bryant Haggen Jun 14 '13 at 8:56

You are assuming that if you feed 8 Amps into the inverter you will get 8 Amps out of the high voltage Ac side. You won't, or if you do then you will have managed to create energy out of nothing (patent this real quick and never work again!).

In your example you are feeding in 100 Watts. You will get out 100 Watts less the losses in the inverter. Assuming 80% efficiency you will get out 100/110 * 0.8 = 0.73 Amp

  • \$\begingroup\$ I understand the loss due to inefficiency. I am looking for the algorithm (formula) that happens inside the inverter that changes 8.33A DC Input to the inverter to 0.909 Amps Output on the AC Side. Yes, we both agree that the above example of 8.33 to 0.909 does not take loss into account but I can incorporate that into the formula later. I would have to multiply the inverter output amps by .8 to factor in the loss (resulting in .73A as you mentioned) before computing the max available amps I could draw, which after accounting for loss would be 80 Watts max available for use. \$\endgroup\$ – Bryant Haggen Jun 14 '13 at 8:12

protected by clabacchio Jun 14 '13 at 11:50

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