Authors: Charles Platt
Make: Electronics (11 page)
Let’s look at one other data sheet, as not all of them are written the same way. I’ll choose a different LED, the Kingbright part WP7113SGC. Click on the link to the manufacturer’s site, and I find on the second page of the data sheet a typical forward voltage of 2.2, maximum 2.5, and a maximum forward current of 25 mA. I also find some additional information: a maximum reverse voltage of 5 and maximum reverse current of 10 uA (that’s microamps, which are 1,000 times smaller than milliamps). This tells us that you should avoid applying excessive voltage to the LED the wrong way around. If you exceed the reverse voltage, you risk burning out the LED. Always observe polarity!
Kingbright also warns us how much heat the LED can stand: 260° C (500° F) for a few seconds. This is useful information, as we’ll be putting aside our alligator clips and using hot molten solder to connect electrical parts in the near future. Because we have already destroyed a battery, a fuse, and an LED in just four experiments, maybe you won’t be surprised when I tell you that we will destroy at least a couple more components as we test their limits with a soldering iron.
Anyway, now we know what an LED wants, we can figure out how to supply it. If you have any difficulties dealing with decimals, check the Fundamentals section “Decimals,” on the next page, before continuing.
Background
The origins of wattage
James Watt (Figure 1-70) is known as the inventor of the steam engine. Born in 1736 in Scotland, he set up a small workshop in the University of Glasgow, where he struggled to perfect an efficient design for using steam to move a piston in a cylinder. Financial problems and the primitive state of the art of metal working delayed practical applications until 1776.
Despite difficulties in obtaining patents (which could only be granted by an act of parliament in those times), Watt and his business partner eventually made a lot of money from his innovations. Although he predated the pioneers in electricity, in 1889 (70 years after his death), his name was assigned to the basic unit of electric power that can be defined by multiplying amperes by volts. See the Fundamentals section, “Watt Basics,” on
page 31
.
Figure 1-70.
James Watt’s development of steam power enabled the industrial revolution. After his death, he was honored by having his name applied to the basic unit of power in electricity.
How Big a Resistor Does an LED Need?
Suppose that we use the Vishay LED. Remember its requirements from the data sheet? Maximum of 3 volts, and a safe current of 20mA.
I’m going to limit it to 2.5 volts, to be on the safe side. We have 6 volts of battery power. Subtract 2.5 from 6 and we get 3.5. So we need a resistor that will take 3.5 volts from the circuit, leaving 2.5 for the LED.
The current flow is the same at all places in a simple circuit. If we want a maximum of 20mA to flow through the LED, the same amount of current will be flowing through the resistor.
Now we can write down what we know about the resistor in the circuit. Note that we have to convert all units to volts, amps, and ohms, so that 20mA should be written as 0.02 amps:
V = 3.5 (the potential drop across the resistor)
I = 0.02 (the current flowing through the resistor)
We want to know R, the resistance. So, we use the version of Ohm’s Law that puts R on the left side:
R= V/I
Now plug in the values:
R = 3.5/0.02
Run this through your pocket calculator if you find decimals confusing. The answer is:
R = 175Ω
It so happens that 175Ω isn’t a standard value. You may have to settle for 180 or 220Ω, but that’s close enough.
Evidently the 470Ω resistor that you used in
Experiment 3
was a very conservative choice. I suggested it because I said originally that you could use any LED at all. I figured that no matter which one you picked, it should be safe with 470Ω to protect it.
Cleanup and Recycling
The dead LED can be thrown away. Everything else is reusable.
Fundamentals
Decimals
Legendary British politician Sir Winston Churchill is famous for complaining about “those damned dots.” He was referring to decimal points. Because Churchill was Chancellor of the Exchequer at the time, and thus in charge of all government expenditures, his difficulty with decimals was a bit of a problem. Still, he muddled through in time-honored British fashion, and so can you.
You can also use a pocket calculator—or follow two basic rules.
Doing multiplication: move the decimal points
Suppose you want to multiply 0.03 by 0.002:
1.
Move the decimal points to the ends of both the numbers. In this case, you have to move the decimal points by a total of 5 places to get 3 and 2.
2.
Do the multiplication of the whole numbers you have created and note the result. In this case, 3 x 2 = 6.
3.
Move the decimal point back again by the same number of places you counted in step 1. In this case, you get 0.00006.
Doing division: cancel the zeros
Suppose you need to divide 0.006 by 0.0002:
1.
Shift the decimal points to the right, in both the numbers, by the same number of steps, until both the numbers are greater than 1. In this case, shift the point four steps in each number, so you get 60 divided by 2.
2.
Do the division. The result in this case is 30.
Theory
Doing the math on your tongue
I’m going to go back to the question I asked in the previous experiment: why didn’t your tongue get hot?
Now that you know Ohm’s Law, you can figure out the answer in numbers. Let’s suppose the battery delivered its rated 9 volts, and your tongue had a resistance of 50K, which is 50,000 ohms. Write down what you know:
V = 9
R = 50,000
We want to know the current, I, so we use the version of Ohm’s Law that puts this on the left:
I = V/R
Plug in the numbers:
I = 9/50,000 = 0.00018 amps
Move the decimal point three places to convert to milliamps:
I = 0.18 mA
That’s a tiny current that will not produce much heat at 9 volts.
What about when you shorted out the battery? How much current made the wires get hot? Well, suppose the wires had a resistance of 0.1 ohms (probably it’s less, but I’ll start with 0.1 as a guess). Write down what we know:
V = 1.5
R = 0.1
Once again we’re trying to find I, the current, so we use:
I = V/R
Plug in the numbers:
I = 1.5/0.1 = 15 amps
That’s 100,000 times the current that may have passed through your tongue, which would have generated much more heat, even though the voltage was lower.
Could that tiny little battery really pump out 15 amps? Remember that the battery got hot, as well as the wire. This tells us that the electrons may have met some resistance inside the battery, as well as in the wire. (Otherwise, where else did the heat come from?) Normally we can forget about the internal resistance of a battery, because it’s so low. But at high currents, it becomes a factor.
I was reluctant to short-circuit the battery through a meter, to try to measure the current. My meter will fry if the current is greater than 10A. However I did try putting other fuses into the circuit, to see whether they would blow. When I tried a 10A fuse, it did not melt. Therefore, for the brand of battery I used, I’m fairly sure that the current in the short circuit was under 10A, but I know it was over 3A, because the 3A fuse blew right away.
The internal resistance of the 1.5-volt battery prevented the current in the short circuit from getting too high. This is why I cautioned against using a larger battery (especially a car battery). Larger batteries have a much lower internal resistance, allowing dangerously high currents which generate explosive amounts of heat. A car battery is designed to deliver literally hundreds of amps when it turns a starter motor. That’s quite enough current to melt wires and cause nasty burns. In fact, you can weld metal using a car battery.
Lithium batteries also have low internal resistance, making them very dangerous when they’re shorted out. High current can be just as dangerous as high voltage.
Fundamentals
Watt basics
So far I haven’t mentioned a unit that everyone is familiar with: watts.
A watt is a unit of work. Engineers have their own definition of work—they say that work is done when a person, an animal, or a machine pushes something to overcome mechanical resistance. Examples would be a steam engine pulling a train on a level track (overcoming friction and air resistance) or a person walking upstairs (overcoming the force of gravity).
When electrons push their way through a circuit, they are overcoming a kind of resistance, and so they are doing work, which can be measured in watts. The definition is easy:
watts = volts × amps
Or, using the symbols customarily assigned, these three formulas all mean the same thing:
W = V × I
V = W/I
I = W/V
Watts can be preceded with an “m,” for “milli,” just like volts:
Number of watts | Usually expressed as | Abbreviated as |
0.001 watts | 1 milliwatt | 1mW |
0.01 watts | 10 milliwatts | 10 mW |
0.1 watts | 100 milliwatts | 100 mW |
1 watt | 1,000 milliwatts | 1W |
Because power stations, solar installations, and wind farms deal with much larger numbers, you may also see references to kilowatts (using letter K) and megawatts (with a capital M, not to be confused with the lowercase m used to define milliwatts):
Number of watts | Usually expressed as | Abbreviated as |
1,000 watts | 1 kilowatt | 1 KW |
1,000,000 watts | 1 megawatt | 1 MW |
Lightbulbs are calibrated in watts. So are stereo systems. The watt is named after James Watt, inventor of the steam engine. Incidentally, watts can be converted to horsepower, and vice versa.
Theory
Power assessments
I mentioned earlier that resistors are commonly rated as being capable of dealing with 1/4 watt, 1/2 watt, 1 watt, and so on. I suggested that you should buy resistors of 1/4 watt or higher. How did I know this?
Go back to the LED circuit. Remember we wanted the resistor to drop the voltage by 3.5 volts, at a current of 20 mA. How many watts of power would this impose on the resistor?
Write down what you know:
V = 3.5 (the voltage drop
imposed by the resistor)
I = 20mA = 0.02 amps
(the current flowing through the resistor)
We want to know W, so we use this version of the formula:
W = V × I
Plug in the values:
W = 3.5 × 0.02 = 0.07 watts (the power being dissipated by the resistor)
Because 1/4 watt is 0.25 watts, obviously a 1/4 watt resistor will have about four times the necessary capacity. In fact you could have used a 1/8 watt resistor, but in future experiments we may need resistors that can handle 1/4 watt, and there’s no penalty for using a resistor that is rated for more watts than will actually pass through it.
Experiment 5: Let’s Make a Battery
Long ago, before web surfing, file sharing, or cell phones, kids were so horribly deprived that they tried to amuse themselves with kitchen-table experiments such as making a primitive battery by pushing a nail and a penny into a lemon. Hard to believe, perhaps, but true!