Posts tagged class

Solar Power Class: Semiconductors and the P/N Junction

Holiday’s over, class, and it’s time to get back to our solar power lessons.  Today, let’s talk about semiconductors, and a slew of other things along the way!

What is a semi-conductor?  Pretty much what it sounds like.  If a substance which conducts electricity (think copper wires) is a conductor, and one that does not (rubber) is called an insulator, then a semi-conductor is right in between. Basically, atoms with only 1 or 2 electrons in their valence (outermost) orbit conduct electricity more easily than those which have a full or almost full outer valence (7 or 8).  The first are conductors, the latter are insulators.  So semiconductors are those with 3, 4, or 5 valence electrons, which means that they are neither particularly inclined nor disinclined toward conductance.

By taking silicon (which has 4 valence electrons) and doping it, as it is called, with an element which has either 3 or 5 electrons, materials with a net positive or negative charge can be created. This has to do with silicon’s natural crystal forming tendencies.  If the material used had 3 valence electrons, then the overall material has a positive charge and it is called a P-type material.  If it had 5, then the material holds a negative charge, and is called an N-type material.

All interesting enough, but things really get going when you place the two together.  The P/N Junction is the innovation that pretty much ushered in the electronic age.  The particular application of this material to solar power was actually one of the first experiments done with it, by Bell Labs back in the 50s.  When a P-type material and an N-type material are placed back to back, and a circuit is completed wiring the two pieces together, the sun excites electrons in the negatively charged layer (remember, this layer has extra electrons) and as the voltage (or pressure) builds to a point at which they overcome barrier resistance, they begin to jump to the positively charged layer where holes in the crystal structure await them.  From there, the circuit exits the P-type material and travels back to the N-type material via the wire circuit to restore the electron balance in each material.  If we hook up a battery amongst the wire circuitry, it is those bounced electrons which we store and call energy.

Okay, time for a break.  Join me back here tomorrow for the next installment of our solar power class.  In the meantime, for your reading pleasure, check out The Light Revolution, a great book about the necessary influence of the sun in our health and buildings.  Full review soon!

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Solar Panel Primer

Looking to buy a solar panel for your home, or maybe build your own? It can be confusing trying to sort out the actual capacity of a panel and what it will or won’t power. So here are a few facts about solar panels to get you started on your way to energy freedom.

Solar panels are composed of individual solar cells, wired in some combination of series and parallel to achieve the desired voltage and amperage that produces the wattage you need for operation. In order to charge a 12 volt battery, you need a panel that outputs 18 volts. Here is a fact you’ll want to know: a solar cell, no matter how large or small, puts out a total voltage of 0.55 volts. For the sake of making math easy, let’s call it 0.5 volts.

So in order to get 18 volts, you need to create a voltage multiplier circuit (which for those of you not following the photovoltaic course on this site means a SERIES circuit). Between 34 and 36 cells wired positive to negative will achieve this. (0.5 * 36 = 18 volts) Let’s say you have a bunch of cells that have 0.5 volts each, and 1.5 amps each. If you simply wire 36 in series, as is recommended, you will end up with a panel that puts out 18 volts with total amperage of… 1.5 amps. All together, this means you will get 27 watts (per hour of full sun).

If you’ve checked the back of any of your appliances recently, you may realize that this wattage is not enough for your needs. After all, even a compact fluorescent 12 volt bulb will probably use at least 9 or 10 of those watts each hour it operates, which is 1/3 of your total available power. (For the sake of this example we will ignore the fact that you probably wouldn’t need a light while the sun is shining!) Use three bulbs, and your power will run out with the sunshine at dusk. So how do you increase your available power?

Most photovoltaic system batteries are designed to operate at 12 volts (similar to a standard car battery), so you don’t want to up the voltage unless you want to start upping the number of batteries as well. If wattage (available power) is equal to voltage * amps, and you can’t change the voltage, then it stands to reason that upping the amps is the way to go. To do this, you wire groups of cells together in parallel. In order to wire things up in parallel, the things wired together should be the same, meaning you will be wiring together groups of 36 cells (wired in series). To double the amperage, you would wire two sets of 36 cells together, for a total of 72 cells, 18 volts, and 3 amps (1.5 amps * 2). Now, you have an available 54 watts (18 * 3) per hour of full sun. For every additional set of 36 panels in series that you add (in parallel) to your array, you get an additional 1.5 amps, and your total wattage is increased by 27 watts per hour of full sun.

There are a few other things to remember when choosing a panel for your needs. The first is kind of obvious: the sun only shines so many hours per day. In most places, you can expect between 5 and 8 hours of “full sun” (the conditions under which your panel produces its maximum output) per day. So if you have a 50 watt panel and 6 hours of full sun, you can expect around 300 watts of power production daily sent to your battery (or perhaps a little more from a few hours of partial sun production). On cloudy days, or if you live in an area of high particulate pollution such as a large city, you will get less.

Which brings me to fact #2: to get the full output from your panel, they must be kept clean. Over time, dust settles on the panels and reduces the amount of light hitting the surface. Or in the winter, in cold areas, you will have snow to contend with, and in the fall, you’ll have to keep an eye out for leaves falling atop the panel. Birds are another common culprit. We’ve all seen the ground under areas where birds congregate… it isn’t pretty. Basically, your panel will output at the level of the cell getting the LEAST sunlight. So if you have one cell in full shade (or fully covered with something), and others in the sun, your total output will be reduced more than you’d intuitively think. Save yourself some grief and locate the panels in a place easy to access for cleaning, and outside the regular path of avian neighbors.

Lastly, since solar panel setups are Direct Current (DC) systems, and since most household appliances run on Alternating Current (AC), you will usually need an inverter to convert DC to AC power your appliances can use. But there is a good reason to consider investing in some 12 volt appliances and lights. In the process of inverting the power to AC, about 15% of the power is lost. So if you have a total of 500 watts to use, and you invert it all to AC, you will have about 425 watts available for powering loads (appliances). There is an increasing variety of 12 volt appliances on the market, from light bulbs and radios to refrigerators. The more 12 V appliances you use, the more use you get for your available power.

This is by no means a complete list of considerations when buying a solar generation system, but it should be enough to help you sort through the eBay flotsam and find a panel that meets your particular needs. Stay tuned for next time, when we’ll be discussing batteries!

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Solar Power Class: Combined Circuits

The next installment of our electrical theory class is about combined circuits. Though studying all the simple circuits of the past few posts (see Ohm’s Law, Kirchoff’s Laws the Formulae Roundup if you’d like a review) is well worth the effort so that you understand the reasons why your power system is or is not working, most often in a power generation setting, you’ll be dealing with things hooked up in combined circuits, which means in combinations of series and parallel that produce the required output wattage to run a house circuit.

So how do you deal with the laws we just learned when you see a circuit that doesn’t seem to fit either category?

Solving Combo Circuits:

For the above example, can you show the remaining measurements for each resistor, and also calculate the total resistance for the circuit? To do this, we must reduce the circuit to a simple circuit of only one type. Before we get started, let me share one more little formula with you. Don’t worry, this one is designed to make your life easier. In a parallel circuit with only two resistors, instead of solving for total resistance in the usual way, you can find it using the following formula, also known as the “product over sum” method:
Rt = (R1 * R2) / (R1 + R2)

First, let’s work on the total resistance measurement. For a combination circuit, you use all the knowledge you’ve been storing up about simple series and parallel circuitry, and simply solve from the outside in, treating each circuit as though it is independent of the rest.

The outermost two resistors are hooked up in parallel (R5 and R6). Because there are only two resistors in this circuit, we can solve for the “equivalent” resistance by using the formula above. Therefore, R5-6 = 30*60 / 30 + 60 = 20Ω.

The next step “into” the center of the circuit is a series circuit composed of R3, R4, and the equivalent value for R5-6. Since in series circuits, we can simply add the partial resistnaces to get the total, we arrive at: 20 + 10 + 20 = 50Ω

This leaves us with a simple series circuit with three resistors – R1, R2, and the equivalent of R4-5-6. Again, we have a series circuit, so we simply add up the partial resistances to arrive at our final Total Resistance for the entire combination circuit. Therefore, 50 + 30 + 50 = 130Ω = Rt

Now that we know the total resistance, finding the total current through the circuit is as easy as using Ohm’s Law: E = I * R. Therefore, 240w = I * 130Ω, and I = 1.846 amps. Now we can use Ohm’s Law’s other formula to figure out the total power. P = I * E = 1.846 * 240 = 443.1 watts.

Now we know all of the totals for the circuit. They are as follows:
Rt = 130Ω
Et = 240v
It = 1.846 A
Pt = 443.1w.

From here you could go on to solve the current or voltage drops at any particular point along the circuits, or similarly figure out the work done at any point along the circuit’s path.

There you have it: you can now solve practically any circuit you’ll encounter.  And remember, practice makes perfect, so keep doing examples until you feel thoroughly comfortable with all these past few lessons.  If you’d like another explanation of combination circuits, check out the links below:

Physics Classroom Tutorial

Understanding and Calculating Combination Circuits

Electrical Engineering Training Series

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Photovoltaic Class

To all my fellow travelers on the road to solar living! If you’re following along with the solar power class, be sure to note that the class now has its own page on the site (it’s at the top of the page on the right-hand side), where you can find all the lessons in one easy to remember place. If you are following along, stop by and say hello to your other classmates by commenting on that page, so we can facilitate discussion and learning for all involved. Also, if you find great resources that everyone will want to know about, post them there!

For my fellow Los Angeles classmates who are checking in, though all the information now posted is publicly available to all who visit the site, there will eventually some information and resources posted which are specific to our class that will be available only via password. By commenting and saying hello, I will know your email address to send you a password when such information is posted.

See you there!

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Solar Power Class: Formulae Roundup

Okay, so back to the classroom, while it’s fresh on your mind. Yesterday we talked about Kirchoff’s Laws, as well as a few other formulas that you’ll use when trying to figure out your electrical generation setup and capacity. To review, here’s a listing of formulas:

Voltage = “Electromotive Force” = E
Current = “Amperes” = I
Resistance = “Ohms” Ω = R
Power = “Watts” = P

Ohm’s Law (for DC power): voltage = current * resistance ( E = I * R ),
power = current * voltage ( P = I * E )

Total Circuit Resistance: SERIES: Rt = R1 + R2 + … + Rn
PARALLEL: 1/Rt = 1/R1 + 1/R2 + … + 1/Rn

Voltage Divider Formula (Series): Erx = Et ( Rx / Rt ) (solve for resistor “x”)
Current Divider Formula (Parallel): Irx = It ( Rt / Rx ) (solve for resistor “x”)

Kirchoff’s Voltage Law (Series): Et – E1 – E2 – … – En = 0 (circuit w/ “n” resistors)
Kirchoff’s Current Law (Parallel): It – I1 – I2 – … – In = 0 (circuit w/ “n” resistors)

Whoa, boy, that’s a lot of formulas! Using these, you can pretty much figure out whatever you want about a simple series or parallel circuit. Of course, the circuits designed for power generation are rarely simple like the above, but we’ll get to that in a second!

So now that you’ve beat these formulas in your head, how do you use them? Here’s a sample problem:

“A 220V series circuit has three resistors. The first has a resistance of 25Ω, the second has a resistance of 50Ω, and the third has a resistance of 35Ω. What is the current and resistance of the circuit, and what is the voltage drop across each resistor? How much power will it produce?”

First, you diagram everything to see what you’re working with. You see that it’s a series circuit, which means that the current is common and the total resistance is the sum of all the partial resistances.

Rt = 25Ω + 50Ω + 35Ω = 110Ω

and since E = I * R, 220v = I * 110Ω, or I = 220/110 = 2 amps.

So now you know the voltage (220 V), the resistance (110 Ω), and the current (2 amps). To figure out the power produced, you simply use Ohm’s Law, (P = I * E), or P = 220 V * 2 amps = 440 watts. We’ll add that to the list: power (440 W).

The final part of the work is to figure out the voltage drops across the circuit at each resistor. This will require the voltage divider formula, since it’s a series circuit. ( Ex = Et ( Rx / Rt ) )

Here are the values plugged in for resistor 1: Er1 = 220V ( 25Ω / 110Ω ) = 50V.
Here are the values plugged in for resistor 2: Er2 = 220V ( 50Ω / 110Ω ) = 100V.
Here are the values plugged in for resistor 2: Er3 = 220V ( 35Ω / 110Ω ) = 70V.

You can check your work on this by adding all the voltage drops together: they should equal the voltage given in the problem: 50V + 100V + 70V = 220V. Good! Now for extra credit you can figure out the power produced at each resistor by using your power formula and the voltage and resistance of each resistor in the circuit. If you do, post your answer in the comments section!

This is basically the same process as for parallel circuits. Let’s use the same problem as above, but now the circuit is hooked up in parallel:

In this case, to get total resistance, you use the formula 1/Rt = 1/R1 + 1/R2 + 1/R3. This looks scarier than it is. to get a reciprocal (one divided by a number) on a calculator, you simply plug in the number, hit the divided by sign, then the equals sign. The calculator will do all the dirty work for you, but you should try doing them in your head sometimes, ’cause you never know when the calculator might stop working when you need to figure something out! So:

1/Rt = 1/25Ω + 1/50Ω + 1/35Ω = .04 + .02 + 0.0286 = 0.0886

This is the reciprocal of the total resistance (this is also called the “conductance”), so we take the reciprocal of this and get a total resistance: 11.287Ω. A total resistance of less than each of the parts? Yes, this example illustrates two things about parallel circuits: first, the numbers are rarely as “pretty” as in series circuits. Secondly, the total resistance of a parallel circuit is always less than the lowest partial resistance. If it’s not, then you calculated something incorrectly.

So now, as before, we know the voltage (220V) and the resistance (11.287Ω). To figure out the total current, we use Ohm’s Law ( E = I * R ): 220V = I * 11.287Ω so I = 19.49 amps. You can now easily figure out the power produced: P = 19.49A * 220V = 4287.8 watts of power (or 4.3 kW).

Using the same resistors and voltage, you’re generating more than ten times the power by hooking things up in parallel!

Rounding out the lesson, you’re now trying to figure out the current drop across the resistors (remember, in a parallel circuit, current is divided, not voltage). Use the Current Divider Formula: Irx = It ( Rt / Rx )

Here are the values plugged in for the first resistor: Ir1 = 19.49A ( 11.28Ω / 25Ω ) = 8.794 A.
Here are the values plugged in for the second resistor: Ir2 = 19.49A ( 11.28Ω / 50Ω ) = 4.397 A.
Here are the values plugged in for the third resistor: Ir3 = 19.49A ( 11.28Ω / 35Ω ) = 6.281 A.

To check your numbers, use Kirchoff’s Current Law, and add up the current drops:
8.794 A + 4.397 A + 6.281 A = 19.472 A, which is the same as above, save my rounding. This is why it’s important to keep numbers true to the highest number of decimal places you can… your work will be that much more accurate.

Again, if you want to figure out the power available at any point, use the same process as for the series circuit.

Congratulations! You’ve now mastered simple series and parallel circuits. That’s a lot to know. Next time, we’ll take a look at how to deal with more complex circuits, like the ones you might encounter in your alternative energy generating system, and also talk about why all this math is important to you when all you want to do is drop off the grid and tune out. I found that this stuff didn’t really cement in my head until I tried figuring out practical problems with them, so I’d recommend you do a little practice session, perhaps using the problems found at Open DNS: Ohm’s Law Practice 1, and Ohm’s Law Practice 2. Or try this site if you like to “play and learn”: Ohm’s Law Battleship Game, where your shots won’t land unless you can solve some problems along the way. See you next time, captain!

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Solar Power Class: Kirchoff’s Laws

Whew! It’s been a busy couple of weeks in my photo-voltaic class. I was starting to fear that I’d have to take a math class to keep up with all the formulas! When I posted the first week’s lesson, I realized later that I’d given out misinformation, which is the danger of posting about something you don’t yet understand! So, from this point out, I’ll just post the lessons after I’ve been tested in the contents, that way you’re always getting the information someone has TOLD me I understand. Since we had a test this past weekend, here’s a new dose of mathematical fun!

This week: Kirchoff’s Laws. Last time, I discussed Ohm’s Law of DC power, which interrelated voltage, current, amperage, and power, and provided several formulas you can use to figure out any of the above for a circuit. If you’d like to review, check out the original post here. Now, let me repeat a few pertinent facts: a series circuit is when you basically hook everything up in a big loop, positive end to negative end in a chain. See the diagram below:

A parallel circuit is one in which the positive and negative ends are “shunted” together (parallel circuits are sometimes called shunts) creating a ladder effect. Again, see the diagram below:

Series circuits are called voltage divider circuits, because though a common current flows across the wire, at each stop along the way, voltage is dropped. These are two important concepts: 1. current is common. 2. voltage is divided along the circuit. Parallel circuits are the opposite. Though a common voltage flows through all the wires, the current is divided between the different potential paths. Therefore, in a parallel circuit, 1. voltage is common, and 2. current is divided along the circuit. Series circuits are called “voltage dividers” and parallel circuits are called “current dividers”. VERY IMPORTANT is you want to know how to manipulate these circuits later on.

There are even two formulae which will help you to calculate a voltage or current at any particular point along a circuit. Say you have three resistors along your circuit. In a series circuit, a voltage divider, if you want to know the voltage of resistor “b”, you would use the voltage divider formula: Erb = Et (Rb/Rt), where t represents total and Er is the voltage drop. Here’s an example:

In a series circuit with a total resistance of 100 Ohms, and a voltage of 120V, resistor b has a resistance of 25 Ohms. The total voltage drop across resistor b would be:

Erb = 120 v ( 25 Ohms / 100 Ohms ) = 30 V

Now, if you aren’t sure what the resistance of a particular resistor on the circuit is, then Kirchoff’s Law of Voltage (for series circuits only!) comes into play. His law states that the total voltage minus the voltage of each resistor, etc on the circuit will always equal zero. In other words, the Total Voltage equals the sums of all the voltage drops along the path. Here’s the official equation: Et – E1 – E2 – … – En = 0, where the circuit has n resistors. So if you know that one resistor has a voltage drop of 25 v and the third has a voltage drop of 50 v, and the total voltage is 130 v, then 130 – 25 – E2 – 50 = 0, and E2 = 55 v. Got it?

Now, on to parallel circuits, ones you’ll see a lot of in battery configurations. Because parallel circuits are current dividers, they need a separate formula for figuring out current drops around the circuit. This is called the Current Divider Formula (using Current at Resistor b): Ib = It ( Rt / Rb ). As with the voltages of a series circuit, if you need to know the current drop at a particular point, Kirchoff had a law for that, too. It’s called Kirchoff’s Current Law (for parallel circuits), and it states that the total current minus the current drops along the way, equals zero. So It – I1 – I2 – … – In = 0, where the circuit has n resistors.

Now, this is a LOT of information to absorb, especially in practice, so let me stop here for now, and we’ll pick up here tomorrow with the rest of the lesson. It seems like way too formulas to ever be useful, but once you get to solving practical equations with them, it’s not too bad. But let’s save that for the next lesson, sleep on it, and I’ll see you in class tomorrow~

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