Voltage Divider Fundamentals

What is a voltage divider?

A voltage divider is a fundamental building block for electrical circuits. The elementary resistor voltage divider is defined as a two-resistor circuit used to reduce voltage. For the resistive circuit shown in Figure 1, the output voltage is always less than the input voltage. The name “voltage divider” offers a good description of the circuit as the voltage is divided or split between two resistors.

The voltage divider is among the first lessons in electronics. It follows the concepts of Ohm’s Law and the fact that current flows in a complete circuit. In fact, the voltage divider is often used as a tool to reinforce Ohm’s law.

In this post we will explore the elementary voltage divider. We will then show common application errors associated with the circuit. Finally, we will show that the voltage divider concepts are applicable to nearly every real-world circuit.

Figure 1: Circuit diagram for the voltage divider. The output voltage is taken from the junction of the two resistors.

Tech Tip: The circuit shown in Figure 1 does not include a load on the output signal. As explored in the next section, this can lead to unexpected errors. This included something as simple as measuring the voltage. Technically, we can say that a voltmeter disturbs the circuit leading to a lower than anticipated voltage measurement.

Quick solutions

If you are looking for a quick way to solve for the voltage divider, please use the DigiKey online calculator as shown in Figure 2. With this tool you can quickly calculate the output voltage for any input voltage and combination of resistors. There is also a tab (not shown) for situations when a voltage divider is operating under load.

Figure 2: The DigiKey voltage divider calculator may be used to quickly solve a voltage divider problem.

Grok the voltage divider

Chances are high that you are a student if you are reading the page. As an educator I encourage you to avoid using the online resistor calculators. In the long run, they are not particularly useful to your education. Instead, you need to grok the resistive voltage divider through study and a good deal of repetition as you work through the mathematics. This is essential as your future studies are filled with concepts that rest on the voltage divider mechanisms. You truly need the ability to work the mathematics without thinking. More importantly, you need to have an instinctual feel for the circuit, which is the true definition of grokking.

The resistor divider is a shortcut

The voltage divider is a shortcut that allows us to solve the voltage divider system using a single equation. If we dig deeper, we can discover the two underlying equations.
To begin, we recognize that the current flowing in the circuit is defined by:

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

We also recognize that the voltage across resistor R2 is defined as:

V_{R2} = I \times R_2

If we combine the equations, we see that:

V_{R2} = V_{out} = \frac{V_{in}}{R_1 + R_2} \times R_2

As a final step, we rearrange the equations to obtain the familiar (standard) form of the voltage divider formula:

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

Expanding the resistive voltage divider to calculate Vin, R1, or R2

The voltage divider is a system with 4 variables. Given any three, you must be able to calculate the fourth as demonstrated by the following four equations. As a practical algebra exercise, you should be able to derive equations each equation from the standard form of the voltage divider equation.

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

V_{in} = \frac{V_{out} \times (R_1 + R_2)}{R_2}

R_1 = \left(\frac{V_{in}}{V_{out}} - 1\right) \times R_2

R_2 = \frac{V_{out} \times R_1}{V_{in} - V_{out}}

In practice it may be simpler and less error prone to keep the voltage divider in standard form (1st equation) and then perform the algebra. However, the previous four examples show the outcome of the steps you must take to solve for the unknown. There is something to be said for consistency. Also, the person grading your solution may not be thinking about the voltage divider in terms of the three additional equations.

Voltage droop with divider loading

A classic design error is to think about the voltage divider in terms of an isolated system. Specifically, it fails to consider the loading on the voltage divider. This load could be an external circuit or even something as innocuous as a voltmeter or oscilloscope probe.

This loading effect is best demonstrated in Figure 3. Here we see a collection of independent voltage dividers consisting of the R1/R2 pair, the R3/R4 pair, and the R5/R6 pair. The output voltage of each divider is measured using a voltmeter featuring a simulated 1 MΩ input resistance. The resistor pairs are chosen to increase by a factor of 10 for every stage as we move from (left to right).

In our ideal (unloaded) calculations, each resistor pair yields a 5 VDC output. In the real world we see observe a droop in voltage for the higher resistor pairs. The 1 kΩ pair yields a 5.00 VDC reading while the 100 kΩ pair yields a 4.76 VDC reading.

Figure 3: This circuit presents three individual voltage dividers where the voltage is measured using a 1 MΩ input impedance voltmeter. Observe that the high resistance R5 / R6 pair has a significant voltage drop.

Tech Tip: The input resistance or impedance of a voltmeter or oscilloscope probe may load down the circuit. The DigiKey calculator does include an option for calculating the output voltage for a loaded voltage divider.

Next steps

We could continue this conversation to include Thevenin’s Theorem. The Figure 3 example would be particularly insightful as it provides a convenient way to visualize the circuit. We could also expand our voltage divider to encompass reactive elements such as inductors and capacitors. The challenge increases when we allow the frequency to change. If we make those allowances, the simple resistive voltage divider transforms into a filter such as this representative low-pass device. Perhaps we can explore these concepts another day.

Parting thoughts

The resistive voltage divider encapsulates the early lessons in electronics. As a student of electronics, you will be tasked to solve thousands of related problems. You will grok electronics by working through these problems enthusiastically and looking for those relationships / applications outside the immediate problem. With that intuitive understanding you will see the impact of loading on the voltage divider and no longer be surprised when the voltage is lower than expected. This same concept arrives fully formed in the future when we explore filters and the detrimental impact of circuit loading.

Be sure to read this follow up article involving voltage divider calculations involving three or more resistors.

Please leave your comments and suggestion in the space below. Also, be sure to answer the questions and critical thinking questions at the end of this note.

Best wishes,

APDahlen

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About this author

Aaron Dahlen, LCDR USCG (Ret.), serves as an application engineer at DigiKey. He has a unique electronics and automation foundation built over a 27-year military career as a technician and engineer which was further enhanced by 12 years of teaching (interwoven). With an MSEE degree from Minnesota State University, Mankato, Dahlen has taught in an ABET-accredited EE program, served as the program coordinator for an EET program, and taught component-level repair to military electronics technicians. Dahlen has returned to his Northern Minnesota home and thoroughly enjoys researching and writing articles such as this.

Questions

The following questions will help reinforce the content of the article.

  1. What is the equation for the voltage divider?

  2. Could a variable resistor be used as a voltage divider?

  3. Solve for the output voltage given:
    A) Vin = 13.8 VDC, R1 = 5 Ω and R2 = 10 Ω
    B) Recalculate the output voltage when a load of 15 Ω is applied to the voltage divider.

  4. Repeat the previous problem without using the voltage divider shortcut.

  5. Solve for R1 given and unloaded system, Vin = 100 VDC, Vout = 75 VDC, and R2 = 25 Ω.

  6. Construct and then solve five unique voltage divider problems.

  7. What two equations are used to develop the voltage divider shortcut?

  8. Describe the derivation of the voltage divider equation solving for R2. For full credit show you work.

  9. Locate the specifications for the voltmeter used in your lab. What is the input impedance? Is the 1 MΩ example in Figure 3 a reasonable estimate?

  10. With regards to the previous question, recalculate using the specifications for a classic analog meter such as the Simpson 260.

  11. Identify and then describe all of the circuit laws or fundamental concepts that are used to build the voltage divider equation.

Critical thinking questions

These critical thinking questions expand the article’s content allowing you to develop a big picture understanding the material and its relationship to adjacent topics. They are often open ended, require research, and are best answered in essay form.

  1. Return to question #4 and recalculate R1 so that the voltage divider continues to provide 75 VDC when it is loaded with a parallel connected 100 Ω and a 120 Ω resistor.

  2. What does this resistor pair article tell us about the relationship between Vin and Vout?

  3. Is it possible to have an unsolvable voltage divider system? For example, if we select an R1 along with Vout and Vin, can we encounter a situation where there is no solution for R2? Does your answer change if the voltage divider is loaded? Hint: The obvious Vout > Vin condition is not the answer to this question. Also, does your answer change if we are free to change the voltage on the lower side of R2?

  4. How is a voltage divider related to a filter?

  5. Looking forward, what is Thevenin’s Theorem and how does it apply to the loading of a voltage divider?

  6. What advice would you offer the DigiKey team with regards to their voltage divider calculator?

  7. Describe the attributes of the non-inverting op amp and how it could be used to reduce the burden on a voltage divider.