With the current economic crisis and energy prices increasing, many people are considering the idea of installing solar photovoltaic power to supplement their home’s energy needs, and reduce their electricity bills.

But how much **solar panel watt** power do you need to say reduce your electricity bill by 50%? And how much will that system cost you?

To answer these questions, here is the process our friend, Mat, went through to calculate his solar watt requirements, costs, and potential power savings.

### 1) Calculate Daily Electricity Usage:

To do this you need your monthly electricity bill, and calculate your total kilowatt hour (kWh) usage for the month.

Mat had a monthly energy usage of 650 kW.

Then divide this number by the number of days in the month, to get your daily kWh usage.

In his case it was 31 days, so 650kWh / 31 = 21 kWh per day. Mat wanted to halve his electricity bill, so he only needed 21 / 2 = 10.5 kWh of solar power per day.

### 2) Calculate Total Solar Panel Watt Requirements:

To do this, first you need to find out how many usable hours of sunlight you receive per day. For your convenience, we’ve included a world solar insolation map – where you can roughly work out your area’s sunlight hours per day. Click here to see the large version in a new window.

Mat is from California, which receives approximately 5.5 hours of usable sunlight per day over the year.

Once you have your daily sunlight hours, get your daily kWh usage and divide it by your daily sunlight hours, then multiply it by 1.25*.

In Mat’s case it was 10.5kWh / 5.5hrs x 1.25 = 2.39 kW or 2390 watts.

* The 1.25 represents a 20% loss of power in the solar panel system due wiring, batteries, inverters, etc. This is an average estimate.

### 3) Calculate Total Solar Panel Cost:

The final step is to work out the cost of the solar panels. In Dec 2008, the highest retail cost for solar panels is $4.85 per watt.

So for Mat to reduce his electricity bill by 50% he would have to buy solar panels at a total retail price of $4.85 x 2390 = $11,592 ! And that’s not including the batteries, wiring, charge controller and inverter!

### 4) Take Tax Incentives Off Cost:

By now you are thinking that Mat would be crazy to buy solar panels to simply reduce his power bill by 50%. Well, thanks to government realizing the severity of global warming, they now offer state and federal tax breaks for households that reduce their carbon footprint.

In Mat’s case he could get a federal renewable energy tax credit of 40% (limited to $4,000) and a California tax credit of 10%. So, the retail solar panels will only end up costing him $11,592 x 50% = $5,796.

So you can see, there are many factors that go into calculating your **solar panel watt** requirements and costs. For Mat it would have cost him about $6,000 to reduce his power bill by 50%. But this is if he bought the solar panels. We, on the other hand sourced our solar cells at cost and made our own panels, which saved us a significant amount of money. The good news is anyone can learn to make their own solar panels and complete solar power system.

While Trudy and I have contemplated making our own step-by-step guide showing you how we did it, we feel there are already enough guides out there showing you the exact same thing. You can find out more about the top-rated guide by reading our Earth4Energy Review.

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Originally posted 2008-12-10 13:07:22.

## 13 Comments

Very useful. I stumbled and dugg this too.

@ Rana

Thanks for the encouraging words. While this might not be the most accurate way to calculate solar power costs, it does give you a rough estimate of what a commercial solar power system could cost you.

Nice…simple and easy to understand. Thanks!

Pleasure Gary.

Glad you got some value out of it. We were going to add a full-on solar power calculator for our readers to try out, but for now we think people can get a general idea of how much professionally installed solar power can cost them.

Perhaps we’ll add the calculator in the future…

No solar anything down here in the deep south. Send a group this way! I/we need a source/contact. I am getting my HERS Rater Cert soon and am the co-chair of our local Home Builders Committee on Green Building.

Really! No solar? How deep south are you, Gary?

Anyway, we’re glad to hear you’re doing your part for the environment by being part of your Green Building Committee. Keep it up!

Thanks very much for the explanation. I had a doubt though that I was hoping you could clarify. Where does the efficiency of the panel come into this? I don’t think it’s included in the 1.25 multiplication factor since those are the wiring/inverter/battery etc losses.

So if 2390 is the output power after the losses, the input power must be 2390/0.15 (for a 15% efficient panel). This equals 15933W. Let’s say 16kW. Am I correct so far? Now how do I tie this 16kW to the panel rating?

Sorry if the question doesn’t make much sense.

Hi Sam,

You seem to be confused.

Okay – When we said you need solar panels that can produce 2390 watts in total, their inefficiencies are already taken into account when calculating the cost.

Put another way, if a solar panel is rated at producing 60watts, then that is its peak output, with inefficiencies of the panel already taken into account.

Hope that answers your question ðŸ™‚

This shows how costly Solar PV can be without tax benefits. Thats why I think more emphasis should be given to solar thermal technology.

Yes,

We agree that solar thermal technology is also very important.

You can read an article on passive solar design here.

So has anyone REALLY made their own homemade solar panels? Honestly, I am scared of getting scammed with all the DIY manuals and DVDs and such. I just want to know if some one really can . I really want to try it if it is possible.

”

In Matâ€™s case it was 10.5kWh / 5.5hrs x 1.25 = 2.39 kW or 2390 watts.

* The 1.25 represents a 20% loss of power in the solar panel system due wiring, batteries, inverters, etc. This is an average estimate.

”

Did you mean to say the 1.25 represents a 25% loss of power… ? I’m a bit of a math idiot but it seems that 1.20 would add 20% to the required wattage yet 1.25 would add 25% to the required wattage (which I would interpret as a loss or inefficiency). ?

I’m asking for clarification as the more I look into this stuff the more I notice calculations done by others include inefficiency losses as well. Last I read was something like 8.5% (I don’t remember where – there is so much info out here I’m a bit bleary eyed – trying to focus on more complete formulas such as you’ve presented).

Love the useable hours of sunlight map – nice touch. Thanks!

Hi Frank,

Thanks for stopping by. Our calculation is correct. For simplicity, let’s say you have a solar panel that produces 100W. After the 20% loss due to inefficiencies, you’re left with 100W x (1 – 0.2) = 80W of power. 100W/80W = 1.25.

This article was written back in 2008, so with the advance in technology sine then, the efficiency numbers may have improved. Just plug the latest figures into the calculation. So using your figure of 8.5%, 100W would produce 100 x (1 – 0.085) = 91.5W. 100/95.1 = 1.0515. We hope that makes sense.

All the best for 2015.

Tim

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