Driving current through a resistance

If you place a voltage source across an electronic component that has measurable resistance (such as a light bulb or a resistor), the force of the voltage will push electrons through the component. The movement of gobs of electrons is what constitutes electric current. By applying a greater voltage, you exert a stronger force on the electrons, which creates a stronger flow of electrons — a larger current — through the resistance. The stronger the force (voltage V), the stronger the flow of electrons (current I).

This is analogous to water flowing through a pipe of a certain diameter. If you exert a certain water pressure on the water in the pipe, the current will flow at a certain rate. If you increase the water pressure, the current will flow faster through that same pipe, and if you decrease the water pressure, the current will flow slower through the pipe.

It’s constantly proportional!

The relationship between voltage (V) and current (I) in a component with resistance (R) was discovered in the early 1800s by Georg Ohm (does his name sound familiar?). He figured out that for components with a fixed resistance, voltage and current vary in the same way: Double the voltage, and the current is doubled; halve the voltage, and the current is halved. He summed up this relationship quite nicely in the simple mathematical equation that bears his name: Ohm’s Law.

Ohm’s Law states that voltage equals current multiplied by resistance, or

What this really means is that the voltage (V), measured across a component with a fixed resistance, is equal to the current (I) flowing through the component multiplied by the value of the resistance (R).

For example, in the simple circuit in Figure 6-1, a 9-volt battery applied across a 1 kΩ resistor produces a current of 9 mA (which is 0.009 A) through the circuit:

Ohm’s Law is so important in electronics that you’d be wise to repeat it over and over again, like a mantra, until you’ve mastered it! To help you remember, think of Ohm’s Law as a Very Important Rule.

When using Ohm’s Law, watch your units of measurement carefully. Make sure that you convert any kilos and millis before you get out your calculator. If you think of Ohm’s Law as , you’ll be okay. And if you’re brave, you can also use , which works just as well (because the millis cancel out the kilos).

But if you aren't careful and you mix units, you may be in for a shock! For instance, a lamp with a resistance of 100 Ω passes a current of 50 mA. If you forget to convert milliamps to amps, you’ll multiply 100 by 50 to get 5,000 V as the voltage across the lamp! Ouch! The correct way to perform the calculation is to convert 50 mA to 0.05 A, and then multiply by 100 Ω, to get 5 V. Much better!

Ohm’s Law is so important (did I say that already?) that I’ve created the following list to help you remember how to use it:

There’s a reason Georg Ohm has his name associated with resistance values as well as the law. The definition of an ohm, or unit of resistance, came from Georg Ohm’s work. The ohm is defined as the resistance between two points on a conductor when one volt, applied across those points, produces one amp of current through the conductor. I just thought you might like to know that. (Good that Georg’s last name wasn’t Wojciehowicz!)

One law, three equations

Remember your high-school algebra? Remember how you can rearrange the terms of an equation containing variables (such as the familiar x and y) to solve for one variable, as long as you know the values of the other variables? Well, the same rules apply to Ohm’s Law. You can rearrange its terms to create two more equations, for a total of three equations from that one law!

These three equations all say the same thing but in different ways. You can use them to calculate one quantity when you know the other two. Which one you use at any given time depends on what you’re trying to do. For example:

• To calculate an unknown voltage, multiply the current times the resistance (). For instance, if you have a 2 mA current running through a 2 kΩ resistor, the voltage across the resistor is (or ) .
• To calculate an unknown current, take the voltage and divide it by the resistance (). For example, if 9 V is applied across a 1 kΩ resistor, the current is .
• To calculate an unknown resistance, take the voltage and divide it by the current (). For instance, if you have 3.5 V across an unknown resistor with 10 mA of current running through it, the resistance is .

Calculating current through a component

In the simple circuit you saw in Figure 6-1, a 9-volt battery is applied across a 1 kΩ resistor. You calculate the current through the resistor as follows:

If you add a 220 Ω resistor in series with the 1 kΩ, as shown in Figure 6-2, you’re restricting the current even more.

To calculate the current flowing through the circuit, you need to determine the total resistance that the 9-volt battery is facing in the circuit. Because the resistors are in series, the resistances add up, for a total equivalent resistance of 1.22 kΩ. You use this equivalent resistance to calculate the new current, as follows:

By adding the extra resistor, you’ve reduced the current in your circuit from 9 mA to 7.4 mA.

That double squiggle symbol (≈) in the equation just given means “is approximately equal to,” and I used that because I rounded off the current to the nearest tenth of a milliamp. It’s usually okay to round off the tinier parts of values in electronics — unless you’re working on the electronics that control a subatomic particle smasher or other high-precision industrial device.

Calculating voltage across a component

In the circuit that was shown in Figure 6-1, the voltage across the resistor is simply the voltage supplied by the battery: 9 V. That’s because the resistor is the only circuit element other than the battery. Adding a second resistor in series (refer to Figure 6-2) changes the voltage picture. Now some of the battery voltage is dropped across the 220 Ω resistor (R1), and the rest of the battery voltage is dropped across the 1 kΩ resistor (R2). I labeled these voltages V₁ and V₂, respectively.

To figure out how much voltage is dropped across each resistor, you use Ohm’s Law for each individual resistor. You know the value of each resistor, and you now know the current flowing through each resistor. Remember that current (I) is the battery voltage (9 V) divided by the total resistance (R1 + R2, or 1.22 kΩ), or approximately 7.4 mA. Now you can apply Ohm’s Law to each resistor to calculate its voltage drop:

Note that if you add the voltage drops across the two resistors, you get 9 volts, which is the total voltage supplied by the battery. That isn't a coincidence; the battery is supplying voltage to the two resistors in the circuit, and the supply voltage is divided between the resistors proportionally, according to the values of the resistors. This type of circuit is known as a voltage divider.

There’s a quicker way to calculate either of the “divided voltages” (V₁ or V₂) in Figure 6-2. You know that the current passing through the circuit can be expressed as

You also know that:

and

To calculate V₁, for example, you can substitute the expression for I shown above, and you get

You can rearrange the terms, without changing the equation, to get

Similarly, the equation for V₂ is

By plugging in the values of R1, R2, and V_battery, you get and , just as calculated earlier.

The following general equation is commonly used for the voltage across a resistor (R1) in a voltage divider circuit:

Many electronic systems use voltage dividers to bring down a supply voltage to a lower level, after which they feed that reduced voltage into the input of another part of the overall system that requires that lower voltage.

In Chapter 5, you see an example of a voltage divider that reduces a 9-volt supply to 5 volts using 15 kΩ and 12 kΩ resistors. You can use the voltage divider equation to calculate the output voltage, V_out, of that voltage divider circuit, which is shown in Figure 6-3, as follows:

The circuit in Figure 6-3 divides a 9 V supply down to 5 V.

Calculating an unknown resistance

Say you have a large flashlight that you’re running off a 12-volt battery, and you measure a current of 1.3 A through the circuit. (I discuss how to measure current in Chapter 16.) You can calculate the resistance of the incandescent bulb by taking the voltage across the bulb (12 V) and dividing it by the current through the bulb (1.3 A). It’s a pretty quick number-crunch:

Analyzing complex circuits

Ohm’s Law comes in handy when analyzing more complex circuits than the simple light bulb circuit discussed earlier. You often need to incorporate your knowledge of equivalent resistances to apply Ohm’s Law and figure out exactly where current is flowing and how voltages are being dropped throughout your circuit.

Look at the series-parallel circuit shown in Figure 6-6. Say you need to know exactly how much current is flowing through each resistor in the circuit.

You can calculate the current running through each resistor, step by step, as follows:

1. Calculate the equivalent resistance of the circuit.

   You can find this value by applying the rules for resistors in parallel and resistors in series (refer to Chapter 5 for details), like this:

   

2. Calculate the total current supplied by the battery.

   Apply Ohm’s Law, using the battery voltage and the equivalent resistance of the circuit:

   

3. Calculate the voltage dropped across the parallel resistors.

   You can do this calculation either of two ways, and you get pretty much the same result (the slight difference is due to rounding error):

   • Apply Ohm’s Law to the parallel resistors. You calculate the equivalent resistance of the two resistors in parallel, and then multiply that by the supply current. The equivalent resistance is 688 Ω, as shown in the first step. So the voltage is
     
   • Apply Ohm’s Law to R1 (the 220 Ω resistor), and subtract its voltage from the supply voltage. The voltage, V₁, across R1 is
     

     So the voltage, V₂, across the parallel resistors is

     

4. Finally, calculate the current through each parallel resistor.

   To get that result, you apply Ohm’s Law to each resistor, using the voltage you just calculated (V₂). Here’s what it looks like:

   

   Note that the two branch currents, I₂ and I₃, add up to the total supply current, I₁: . That’s a good thing (and a good way to check that you’ve performed your calculations correctly).

Designing and altering circuits

You can use Ohm’s Law to determine what components to use in a circuit design. For instance, you may have a series circuit consisting of a 9 V power supply, a resistor, and an LED, as shown in Figure 6-7.

As you see in Chapter 9, the voltage drop across an LED remains constant for a certain range of current passing through it, but if you try to pass too much current through the LED, it will burn out. For example, suppose your LED voltage is 2 V and the maximum current it can handle is 20 mA. What resistance should you put in series with the LED to limit the current so it never exceeds 20 mA?

To figure this out, first you have to calculate the voltage drop across the resistor when the LED is on. You already know that the supply voltage is 9 V and the LED eats up 2 V. The only other component in the circuit is the resistor, so you know that it will eat up the remaining supply voltage — all 7 V of it. If you want to limit the current to be no more than 20 mA, you need a resistor that is at least . Because it’s hard to find a 350 Ω resistor, suppose you choose a commonly available resistor that is close to, but higher in value than, 350 Ω, such as a 390 Ω resistor. The current will be , or about 18 mA. The LED may burn a little less brightly, but that’s okay.

Ohm’s Law also comes in handy when tweaking an existing circuit. Say your spouse is trying to sleep but you want to read, so you get out your big flashlight. The bulb in your flashlight has a resistance of 9 Ω and is powered by a 6 V battery, so you know that the current in the flashlight circuit is . Your spouse thinks the light is too bright, so to reduce the brightness (and save your marriage), you want to restrict the current flowing through the bulb a bit. You think that bringing it down to 0.45 A will do the trick, and you know that inserting a resistor in series between the battery and the bulb will restrict the current.

But what value of resistance do you need? You can use Ohm’s Law to figure out the resistance value as follows:

1. Using the desired new current, calculate the desired voltage drop across the bulb:
   

2. Calculate the portion of the supply voltage you’d like to apply across the new resistor.

   This voltage is the supply voltage less the voltage across the bulb:

   
3. Calculate the resistor value needed to create that voltage drop given the desired new current:
   

4. Choose a resistor value that is close to the calculated value, and make sure it can handle the power dissipation:

   As you see in the next section, to calculate the power dissipated in an electronic component, you multiply the voltage dropped across the component by the current passing through it. So the power that your 4.2 Ω resistor needs to be able to handle is

   

   Result: Because you won’t find a 4.2 Ω resistor, you can use a 4.7 Ω 1 W resistor to reduce the brightness of the light. Your spouse will sleep soundly; let’s hope the snoring won’t interfere with your reading!

Using Joule’s Law to choose components

You’ve already seen how to use Joule’s Law to ensure that a resistor is big enough to resist a meltdown in a circuit, but you should know that this equation comes in handy also when you’re selecting other electronic parts.

Lamps, diodes (discussed in Chapter 9), and other components also come with maximum power ratings. If you expect them to perform at power levels higher than their ratings, you’re going to be disappointed when too much power makes them pop and fizzle. When you select the part, you should consider the maximum possible power the part will need to handle in the circuit. You do this by determining the maximum current you’ll be passing through the part and the voltage across the part, and then multiplying those quantities together. Then you choose a part with a power rating that exceeds that estimated maximum power.

Joule and Ohm: perfect together

You can get creative and combine Joule’s Law and Ohm’s Law to derive more useful equations to help you calculate power for resistive components in circuits. For instance, if you substitute I × R for V in Joule’s Law, you get:

That gives you a way to calculate power if you know the current and resistance but not the voltage. Similarly, you can substitute V/R for I in Joule’s Law to get

Using that formula, you can calculate power if you know the voltage and the resistance but not the current.

Joule’s Law and Ohm’s Law are used in combination so often that Georg Ohm sometimes gets the credit for both laws!