Geek Notes: It's the Law!
From time to time I will be writing entries that begin with the words Geek Notes. These are intended to let be write about some technical aspect of electronics, or pretty much any other geeky topic, to let me explain what I think I know, thus helping me grok the subject matter.
After spending many hours in class studying analog and digital electronics I've found a small handful of mathematical relations that popup again and again. I wish that early on I'd known how absolutely critical these would be in circuit analysis. So for your edification (I love that word) and my personal review let's discuss these relationships.
Before continuing there are a few prerequisites:
- Knowledge of basic algebra.
- Understanding of Scientific Notation (ie. 6.241×1018) and the related Engineering Notation.
- The use of symbols such as: pico(p), nano(n), micro(µ), milli(m), kilo(k), mega(M), and giga(G).
- Access to a scientific calculator (or equivalent utility).
- At least a vague concept of Voltage, Current and Resistance.
If you have these this will be easy. I can promise whatever math involved will be simple, with me it has to be simple.
The most important formula in electronics is Ohm's Law: V = I * R Voltage(V measured in Volts) is equal to Current(I measured in Amps) multiplied by Resistance(R measured in Ohms).
Using simple algebra you can derive the other two equations: I = V / R and R = V / I. These are rather simple, but with my poor memory I can't always remember. So I've found the following image to be helpful. It's easy to remember and let's you "see" all three equations.

What's great is if you know any two of the variables you can determine the third using one of the three equations.

An example is shown above. If you have a 6V battery connected to a 120K Ohm resistor you can calculate the current being drawn. In this case 50µA or 0.00005A. If you increase the Voltage of the battery, and keep the Resistance the same, the amount of Current will increase. Assuming 12V then it would be 12V / 120K Ohm = 100µA.

You may ask how is this useful? A simple case, show above, assumes you're going to use a 5V power supply, a commonly used voltage, to power an LED (Light Emitting Diode). If you don't include a Resistor the Current flowing through the LED will be as great as the power supply can provide. It will be a race between which will fry first; the power supply fuse or the LED. Of course it will be the LED, a power supply that cannot burn out an LED is rather pathetic. [This is not a problem in a standard handheld flashlight because the small incandecent bulb behaves as a resistor and glows as a result of current flow. But if you increase the Voltage enough the Current will increase and you will eventually burn out the bulb.]
A typical LED can handle a maximum about ~30mA, or 0.03A, before it will either be destroyed or the overheating will limit it's operating life. If you want to keep it safe you need to calculate what size Resistor you need to limit the Current to no more than ~30mA. Using R = V / I you have 5V / 30mA = 167 Ohms. The nearest commonly available resistor value is 180 Ohms.
If you change the Voltage, the Current will also change. So to protect the LED you need to recalculate the Resistance required to keep the Current under 30mA. If you drop the Voltage to 3V, from two AA batteries for example, you now need 3V / 30mA = 100 Ohms at a minimum to keep the current at 30mA or less. In this case it would be safe to leave the 180 Ohm Resistor in place, the LED will be fine, yet it will be slightly dimmer.
It was pointed out to me that because the LED is a diode it is a semiconductor device and has certain properties that are beyond the scope of this entry. However the use of a Resistor to protect components from excessive Current is all but universal. A Resistor that prevents too high a current flow is referred to as a Current Limiting Resistor.
That's all for now. Next time: Part II - Power Mad!
(I'd like to give special thanks to William E. Den Beste of Bill's Electronics Reference Library for his assistance in creating the schematic illustrations above and helping me better understand what I'm talking about. All errors are mine and mine alone.)



