In the last lesson we learned about resistors; what they are, what they do, and their basic characteristics such as tolerance, power rating, and value. In this lesson we’ll take that a step further and look at how resistors combine in series and parallel, learn to analyze DC circuits that contain resistors using Ohm’s and Kirchhoff’s Laws, and demonstrate how to use them in real-world applications. Finally, we’ll go into the lab (video segment) and demonstrate some of the things we learned on the bench.
Resistors on their own are of no value. For a resistor to carry out it’s function, it must be part of an electrical circuit. The circuit can use Alternating Current (AC), Direct Current (DC), or a combination of both. For these early lessons, we’ll stick with DC circuits.
A Basic Circuit
Let’s draw a simple schematic diagram of a resistor circuit. The standard schematic symbol for a resistor (the American standard anyway) is shown on the right of the circuit and labeled “R”. The symbol can be drawn vertically, as shown below, horizontally, or even at an angle (we usually avoid drawing components at an angle unless we are indicating a specific configuration such as a “Delta” or “Y”). Also, recall that a resistor has no polarity and can be drawn either way round.
The symbol on the left of the schematic diagram, is the standard symbol for a battery. The “V” indicates voltage. A battery symbol, like the battery itself, has positive and negative terminals indicated by “+” and “-” signs. The long and short horizontal lines indicate battery cells. The longer line is also used to indicate the positive terminal while the shorter line indicates the negative terminal. The battery symbol can also be drawn vertically or horizontally.
The lines between the components represent wires or electrical connections and are usually drawn horizontally or vertically with right angles as shown.
The schematic represents a completed circuit, in that there is at least one continuous path for current to flow from the positive terminal of the power source (in this case a battery) through the circuit, and back to the negative terminal of the battery.
When current, measured in Amperes, or Amps for short, and represented using the letter “I”, flows through the resistor it induces a voltage across the resistor. The end that the current flows into has a positive voltage with respect to the end that the current flows out of. This is indicated below by the + and – signs at the top and bottom of the resistor. This is called a voltage drop because the voltage on one end of the resistor is lower that the voltage on the other end of the resistor – it drops.
Ohm’s Laws
Ohm’s Laws allow us to relate voltage (V) in Volts, current (I) in Amperes or Amps, resistance (R) in Ohms, and power (P) in Watts so that given any two parameters, we can calculate the others. The basic formula that relates voltage, current, and resistance is:
Given the circuit above, where V = 10V and R = 10Ω, we can calculate I.
Let’s look at another example. In this case, we know the voltage and want to choose a resistor value that will result in a 250mA current. The lower case “m” means “milli” as in milliamps. 1mA is 1/1000th of an amp.
In this case, we simply rearrange the formula so that R=V/I and convert 250mA to Amps by dividing by 1,000 giving 0.25A.
Ohm’s Law for power, relates power, current, and voltage. Again, given any two, we can calculate the third.
Let’s take a look at a circuit and calculate the required power rating for the resistor.
This is why calculating power is so important in component selection. A 1/4W resistor would literally explode into a ball of flame in this example. Here we need to choose a 10W resistor as the absolute minimum. More likely, we would include a safety margin of 50% and use a 15W resistor or mount the resistor to a heat sink to improve it’s power handling ability.
Let’s look at a more complicated example.
In this case we know the current (I) but we don’t know the voltage (V). How can we calculate P = V x I without knowing V? Luckily, both of Ohm’s formulas are related and we can substitute values from one into the other like so:
Because we can relate the two formulas to each other, given any two values of V, I , R, or P, we can calculate the others.
Now that you know Ohm’s Laws and how to apply them, you can analyze simple circuits. Next we’ll look at how to analyze more complex circuits that contain multiple resistors.
Resistors in Series & Parallel
Very few real-world circuits contain only a single resistor and a power source. Real circuits are more complex than that so we need to understand how multiple components, in this case resistors, behave when they are combined in series, parallel, or any combination thereof.
To start, let’s look at a circuit that combines two resistors in series. In this case, rather than show a battery, we are showing a generic 10V power source. It can be a power supply, solar cell, battery, or any other type of voltage supply. For purposes of analysis, we can treat them all just like we treated the battery.
In order to apply Ohm’s Law to this circuit, we need to know the total resistance of R1 and R2 combined. Once we know that we can calculate the current by dividing the voltage (10V) by the total resistance.
Resistors in series are additive. The total resistance is the sum of all of the resistors in series given by the formula:
Now, solving for the current (I), is a simple matter of adding the values of the series resistors to get the total resistance and then dividing the voltage (V) by the resistance (R). Let’s look at and solve an example circuit.
It’s that simple. Now let’s look at what happens when we combine resistors in parallel.
Resistors in parallel behave differently than resistors in series. To use our piping analogy again, two pipes in parallel can carry water faster than two pipes in series and also faster than just one pipe. Resistors in parallel behave similarly. The total resistance for resistors in parallel will always be lower than the smallest resistor and is given by the formula:
While it looks complicated, it’s really not. Just plug in the values for all of the resistors in parallel and you can calculate the total resistance. In fact, in a case where you are dealing with only two parallel resistors, the formula reduces to:
Let’s solve an example circuit.
First we’ll calculate the total resistance and then we can solve for the current (I).
But real circuits rarely contain only series or only parallel components. Real circuits contain endless combinations and permutations of components connected in a wide variety of ways. Analyzing these more complex circuits is a simple matter of reducing the circuit to a simpler form by calculating and substituting equivalent values as we did in the above series and parallel examples.
Let’s take a look at a series-parallel circuit example and use the techniques we learned to simplify the circuit so we can analyze and solve it.
We start on the right side of the circuit and reduce the two parallel resistors to a single equivalent resistor.
This results in the following, simplified equivalent circuit.
Now we can calculate the total resistance and then calculate the current.
The next step is to expand the circuit diagram back to its original form while calculating the voltage drops and currents in each branch of the circuit. First we’ll calculate the voltage drop across the 1KΩ resistor.
And if there is 6.55V across the 1kΩ resistor, the remainder must be across the parallel resistors. And once we know the voltage across the parallel resistors, we simply use Ohm’s Law to calculate the current.
And we end up with a fully solved circuit – we know the voltage drop across, and the current through, each component.
Kirchhoff’s Laws
We were able to solve the above example because of the assumptions we made regarding the remaining voltage and how the current splits between the two resistors. We can make these assumptions thanks to Kirchhoff’s Voltage Law and Kirchhoff’s Current Law.
Kirchhoff’s Voltage Law states that, “The sum of the voltages around a closed loop is equal to zero”. Essentially, any voltage that isn’t across one component must be across the others. Looking at our above example, there are two loops, one on the left and one on the right.
Going around each loop, first the left and then the right, we can see Kirchhoff’s Voltage Law at work:
Kirchhoff’s Current Law states that, “The sum of the currents entering and leaving a node is equal to zero”. Basically, any current coming in must go out. Again looking at our example we see:
Using Ohm’s and Kirchhoff’s Laws we can solve almost any passive (resistor, capacitor, inductor) circuit – both DC and AC.
Practical Application 1
Let’s look at a few practical examples where we can apply what we’ve learned.
First, we’ll assume that we’ve designed a 12V power supply and we want to include a blue Light Emitting Diode (LED) to indicate that the 12V output is active. If we tried to apply 12V directly across the LED we’d end up frying it as soon as we turned the power on. We need to reduce the voltage across the LED and limit the current passing through it with a series dropping resistor shown in the circuit diagram below.
The data sheet for the blue LED recommends it be powered with 3.2V at 20mA. That means we need to drop the remainder across the resistor. Applying Kirchhoff’s Voltage Law:
So we need to drop 8.8V across the resistor. Now we need to select the resistor so that at 8.8V we end up with 20mA of current flowing through it. Applying Ohm’s Law to calculate both the resistance and power we find:
Since 440Ω is not a standard value, we will use the next higher value standard resistor which is 470Ω at 1/4W. Notice that we went with a higher, rather than lower, value resistor. We would rather pass less current through the diode rather than more. Slightly lower than recommended current will also extend the life of the LED. More current would have the opposite effect.
Just to be safe, we check the circuit using the new resistor value and find that it results in a current of 18.7mA through the diode, well within the acceptable range.
Practical Application 2
Let’s consider another practical example. In this case, we want to power a device that requires a 3.3V supply. Any voltage over 3.6V will ruin the device. However, the rest of the circuit uses 5V and we already have a 5V regulator . We know from the specifications that the device has an input impedance (the term used for resistance with other than resistors) of 1MΩ.
To create the 3.3V supply we’ll use a voltage divider circuit – two resistors in series with the load (the device connected to the output) attached to the output terminals. The circuit diagram looks like this:
Since the right ratio of R1 to R2 can be obtained with a great variety of different resistor pairs, we need to make an engineering tradeoff to aid us in the selection. On the one hand, we can pick small value resistors to ensure that the load has only nominal effect on the voltage. On the other hand, selecting smaller value resistors will result in higher than needed current which wastes energy (a key consideration in battery powered products) and creates unwanted heat which could require additional thermal management (although these are pretty small numbers so it’s not an issue in this case). Usually, two orders of magnitude difference between parallel resistors (R2 & RL) would mean we could ignore the load resistance, but in this case we want to be very close. Three orders of magnitude should achieve that. If we choose R2 to be 1kΩ with a 0.1% tolerance we can ignore the load resistance in our calculations and still have a maximum current draw less than 4mA. Here’s the circuit with the current calculated (using Ohm’s Law) for both R2 and RL. Note that the overall contribution to current draw from RL is insignificant (0.1% of total current) compared to the current through R2.
According to Kirchhoff’s Voltage Law that leaves 1.7V across R1. Since we are ignoring the load current, we can assume that the current through R1 is the same as the current through R2 or 3.3mA. Using Ohm’s Law yields a resistor value of:
While 515Ω is not a common value, a search of the DigiKey electronics supplier website yields a 515Ω 0.1% tolerance through hole resistor in addition to two 511Ω surface mount resistors, any of which will work. In fact, we can easily go down to 500Ω or a little lower and still be within spec.
Well, that’s it for this installment. You can find the link to the video demonstrating some of these techniques in the lab below and on YouTube.
Cheers
Dominick
