Easy, huh? We used a variant of Ohm's Law, the formula that we use to determine how electricity behaves. We just plugged our numbers into the following equation: R = (Vs - Vf) ÷ I

In this case R represents the impedance of our resistor; Vs represents our source voltage; Vf represents the device voltage; and I represents the device's current draw. As a reminder, always start with equations in brackets and work your way outward.

We also need to know our resistor's power capacity, or its wattage (W) rating. To solve for wattage we still subtract the device voltage from the supply voltage (Vs - Vf). Only instead of dividing that figure by the device's current rating (I) we multiply it. Our 10V difference multiplied by 0.025 (25mA) indicates 0.25 watts.

What we did was use another variant of Ohm's Law, only this time we solved for watts, not impedance. It looks like this: W = (Vs - Vf) x I

By the numbers we need a 400-ohm resistor with a 1/4-watt rating. Unfortunately it doesn't exist so we have to choose the resistor with the next greater impedance-in this case a common 470-ohm resistor. One with the next greater wattage is a good idea, in this case 1/2-watt.

So in the end, this is what our circuit looks like. You should recognize the resistor from before. The triangular job radiating arrows is the LED. More on that symbol later.

Resistors In Series Circuits

More than possible, it's actually favorable to use a resistor in a circuit with more than one device. Of the two ways we can do that, the most accessible way is in series.

When wired in series, the output from one device feeds another one's input. That second device either completes the circuit or feeds another device that in turn completes the circuit or feeds another device. And so on.

Here's what LEDs wired in series look like in schematic:

Series circuits are convenient because the voltage of the devices cumulates. Adding a second identical 3.6V LED to our earlier example would increase the device voltage to 7.2. The difference between 13.6 V and 7.2 is 6.4, which would require a less powerful resistor.

That's a good thing because resistors are sort of the enemy. As noted earlier, resistors dissipate power in the form of heat, which is waste to a lighting circuit. So instead of dissipating excessive energy as heat with resistors we can dissipate it as more light with LEDs. The circuit may still need a resistor, but it would be far smaller and waste less power.

The formula to calculate a resistor for devices wired in a series resembles the formulas to calculate a resistor for a single device. Only this time we combine the devices' operating voltages before we subtract that figure from our supply voltage. We then use the current load from just one LED, not both, to solve for the resistor's impedance and wattage. (That the formula references the current draw of just one device indicates something: the devices must maintain similar values.)

So if the LEDs were 25mA, we'd divide 6.4 (the surplus voltage) by 0.025 to determine the resistor's impedance; we'd multiply those two figures to determine the wattage. Rather than confuse things with the symbols, here's what the real numbers would look like in the formula (remember to work from the centermost brackets outward):

256 = [13.6 - (3.6 + 3.6)] ÷ 0.025

0.166 = [13.6 - (3.6 + 3.6)] x 0.025

So we'd need at least a 256-ohm, 0.166-watt resistor. The next larger would be a 270-ohm, 1/4-watt resistor.

More devices can be added to a circuit, but only to a point: the combined voltage of the components shouldn't exceed the supply voltage. Some maintain that the unregulated supply voltage must exceed the combined component voltage by at least a couple volts simply to accommodate the resistor. Case in point, three 3.6-volt resistors would amount to 10.8 volts, or about 2.8 volts. The tiny 120-ohm, 1/8-watt resistor necessary would hardly waste any power.

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