# Voltage Divider

### Key Points:

- Voltage dividers are passive circuits that output a fraction of the input voltage.
- They are used to generate reference points and new circuit branches with inputs at a particular voltage.
- Voltage dividers are really helpful for gaining intuition about circuits, because they can be used as a model to quickly calculate or estimate voltage in many situations.
- The rule/formula for the output voltage of a voltage divider is given by:

V_{out}=V_{in}\frac{R_2}{R_1+R_2}

### Why it’s important:

Voltage dividers are simple and common circuits that are used to generate a variety of voltages. They can produce voltage levels needed for other components or circuits. They are also used to provide points of measurement for reference or troubleshooting. Voltage dividers are a great real-world application that can help us to understand Ohm’s Law. The best thing about voltage dividers is that they can help you develop an intuitive understanding of the voltage at different points along a resistive branch of a circuit. You can model ALL of the resistors above AND below the points as being ONE resistor on either side. Then, with your intuition about voltage dividers, you can do a good job of estimating the voltage at any point along the branch.

## Anatomy of a Voltage Divider

A voltage divider consists of resistors in series connected to a power source with at least one connection point between the resistors. The input voltage is designated V_{in}, and the output voltage is designated V_{out}.

There are two things that make voltage dividers powerful, and why they are so great to learn about early on in your electronics journey. 1) We can simplify **any **circuit or branch with resistors in series using a two resistor voltage divider as a model. Just sum all of the resistances above the measurement point and call it R_{1}, and sum the resistances below and call it R2. 2) We can use an easy formula to calculate the output voltage knowing ONLY the input voltage and resistance values.

Let’s start with the classical approach, using Ohm’s Law.

## Calculating V_{out} Using Ohm’s Law

There are three ways to calculate Vout. Two of them use Ohm’s Law, and the third (easiest) method uses the **voltage divider rule**.

### Voltage Dividers Analysis Method # 1: The Painful Approach

In the example shown above, we have two 10 Ohm resistors in series, and a single connection point between them. This connection point provides a reference point or source voltage, known as V_{out}. Keep in mind that V_{out} does not necessarily need to have a wire or conductor leading to a test point. You’ll get the same result if the only connection is between R_{1} and R_{2}, and you just measure the voltage at the output (negative terminal) of R_{1} or the input (positive terminal) of R_{2}.

The voltage of Vout can be found using Ohm’s Law. Remember that the current depends on the total equivalent resistance of the circuit. First, we need to find the current in the circuit. The current is just equal to the battery voltage divided by the total resistance R_{T}, which is the sum of R_{1} and R_{2}.

I = \frac{V}{R_T} = \frac{V_{source}}{R_1+R_2}=\frac{20V}{10\Omega + 10\Omega} = 1A

Now that we have the current through the circuit, we can find the voltage drop across R_{1}, which we call V_{1}.

V_1 = IR_1 = (1A)(10\Omega)=10V \\ V_2 = IR_2 = (1A)(10\Omega)=10V

Since V_{out} is directly connected to the output of R_{1}, V_{out} is equal to V_{in} – V_{1}:

V_{out}=V_{in}-V_1 = 20V-10V=10V

We can check our result by adding up the V_{1} and V_{2}. The total voltage drop across both resistors should equal the source voltage of the battery, V_{in}– 20 volts.

V_T = V_1 + V_2 = 10V + 10V = 20V

We did it. In theory, you *could *use this simple method based on Ohm’s Law to calculate the voltage anywhere on a voltage divider.

### Voltage Dividers Analysis Method # 2: Less Painful

If you’re paying careful attention, you might notice that V_{out} is actually equal to V_{2}. So we could have skipped some of the calculations and calculated Vout more directly:

V_{out} = V_2 = IR_2 = (1A)(10\Omega) = 10V

Why the heck didn’t we just do that in the first place? Well, sometimes a voltage divider has a more complex circuit below V_{out}. In this case, the first method can be used so you don’t have to deal with calculating the resistance below V_{out}.

There’s a better way to calculate V_{out} if you know R_{1} and R_{2}.

Deriving the formula requires you to understand method #1. One we have the formula, we can use it easily calculate V_{out} directly.

Why do we need **better **way to calculate the output voltage? It can be a little tedious to calculate the current and voltage drops across the resistors and then subtract from the input voltage.

So let’s learn about a simple formula that allows us to quickly calculate V_{out} using only V_{in} and the total resistance above and below the measurement point. This also called the voltage divider rule.

## The Voltage Divider Rule

The formula to determine the output voltage if you know the total resistance above and below the position of V_{out} is:

V_{out}= V_{in}*\frac{R_2}{R_1+R_2}

In this rule, ‘R_{1}‘ represents the total resistance of the circuit above the point of connection for V_{out}, and ‘R2’ represents the total resistance below V_{out}.

It’s hard to emphasize how useful this rule really is. You can quickly get a sense of how a circuit is going to function if you have this rule in mind.

For instance, let’s look at three cases: one where R_{1} is much bigger than R_{2}, another where R_{2} is much bigger than R_{1}, and another where R_{1} is about equal to R_{2}.

### Voltage Dividers Case 1: R_{1} >> R_{2}

In this case, R_{1} is much bigger than R_{2}. What happens to V_{out}?

V_{out}= V_{in}*\frac{R_2}{R_1+R_2}

V_{out} is equal to V_{in} times R_{2} over R_{1} plus R_{2}. V_{in} stays the same.

Inside the fraction, the numerator (just R_{2}) is small, but the denominator is really big because R_{1} is a high value.

With a small number over a really big number, we can approximate the fraction as being **zero**. Mathematically, this is like taking the limit of R_{1} approaching infinite. To any mathematicians in the audience: try not to have a heart attack.

If the fraction goes to zero, V_{out} also goes to zero. When R_{1} is much greater than R_{2}, the voltage after R_{1} is almost zero.

How to interpret this result: Remember that the voltage is always the measurement of potential comparing one point to another. In this case we are comparing the voltage after R_{1} to ground, which we also assume is the voltage at the negative terminal of the battery. R_{1} is so big that we drop most of the voltage supplied by the battery. By the time we get to the output of R_{1}, the voltage is close to zero.

### Voltage Dividers Case 2: R_{1} << R_{2}

In this case, R_{1} is much smaller than R_{2}. Now what happens to V_{out}?

V_{out}= V_{in}*\frac{R_2}{R_1+R_2}

Let’s look at the fraction again. R_{2} gets really big, making the numerator really big. However, R_{2} is also in the denominator making the bottom term really big too. In the limit that R_{2} goes to infinite, the fraction becomes R_{2}/R_{2} which is equal to 1.

V_{out} becomes V_{in} times 1, which is just V_{in}.

When R_{2} is much bigger than R_{1}, the output voltage V_{out} can be estimated as just the input voltage V_{in}.

### Voltage Dividers Case 3: R_{1} = R_{2}

In our final case study, let’s see what happens when R_{1} is about equal to R_{2}. Let’s say you quickly scan a circuit and notice that R_{1} should about the same as R_{2}, you can use this case to approximate what the output voltage will be.

Let’s look one more time at our voltage divider rule:

V_{out}= V_{in}*\frac{R_2}{R_1+R_2}

This time, if R_{1} = R_{2}, our fraction will become R_{1}/2R_{1}, which is equal to 1/2. So when the two resistance are close to equal, V_{out} will be approximately 1/2 V_{in} (V_{in}/2).

### Developing Intuition Using Voltage Dividers

Why did we do all that (hand-wavy) math? It wasn’t to (hopefully not) confuse you. It was so that we can start to develop a deep intuition about resistors in series using voltage dividers. Here’s how I think about it:

- The closer R
_{1}and R_{2}are to each other, the closer that that V_{out}will be to 1/2 V_{in}. - The bigger R
_{1}gets (compared to R_{2}), the more V_{out}will approach zero (0). V_{out}gets closer to the voltage at the negative terminal of the source. - The bigger R
_{2}gets (compared to R_{1}), the more V_{out}will approach V_{in}. V_{out}gets closer to the voltage at the positive terminal of the source.

The more you can integrate this kind of intuition into your mental circuit analysis pattern, the easier it will become.

### Derivation of the Voltage Divider Rule

Let’s get back to basics and derive the voltage divider rule.

\textrm{Note that that Vin is equal to the current times the sum of R1 and R2:} \\ (1): V_{in}=I(R_1+R_2) \\ ~\\ \textrm{The voltage out is equal to Vin minus V1. We substitute Vin from the above equation:}\\ (2): V_{out} = V_{in}-V_1 = I(R_1+R_2)-IR_1=IR_2 \\ ~ \\ \textrm{We also rewrite the first equation in terms of I (current):} \\ (3): I = \frac{V_{in}}{R_T} = \frac{V_{in}}{R_1+R_2} \\ ~ \\ \textrm{We write equation (2) and then substitute I using eqution 3, solving for Vout.} \\ V_{out} = IR_2 = \frac{V_{in}}{R_1+R_2}R_2 = V_{in}\frac{R_2}{R_1+R_2}

*Lesson 0*: Introduction to Module 3

*Lesson 1*: Introduction to DC Circuits

*Lesson 2*: Series and Parallel Circuits

*Lesson 3*: DC Power Sources and Batteries

*Lesson 4*: Resistors, Capacitors, and Inductors

*Lesson *5: Resistors in Series

*Lesson *6: Resistors in Parallel

*Lesson *7: Voltage Dividers

*Lesson *8: Kirchoff’s Current Law

*Lesson *9: Kirchoff’s Voltage Law

*Lesson *10: Capacitors

*Lesson *11: Dielectric Materials

*Lesson *12: Capacitors in Parallel

*Lesson *13: Capacitors in Series

*Lesson *14: Capacitors in Series and Parallel