Potential Divider

A potential divider (PD), also known as a voltage divider, is an electrical circuit, which usually divides the voltage between the ground rail and the power rail. The division of voltage is proportional to the values of the series resistors across it. The application of such a circuit involves calculating the resistor values to correspond to the required midpoint voltage. A simple passive voltage divider consists of a pair of resistors in series; however, active voltage dividers may include Zener diodes or integrated circuits (IC) in the network. The most widely used applications are in the field of electronic engineering where simple and low-cost resistors provide biasing voltages for transistors. This method of biasing also applies to operational amplifiers (op-amps). In analogue electronics, a simple midpoint voltage divider circuit consists of two resistors of equal values in series to produce a voltage level that is midpoint between the ground and the power rail, and such a voltage is useful for biasing transistors as well.

Light-dependant resistors (LDR), and thermistors are components that also require installation within a potential divider network for their proper biasing and operation. There is more information about their use in the following sections of the article.

This article is for beginners, and focuses on simple resistive networks. I decided to write it because my nephews came running to me one day asking, "How do Potential Dividers Work? And how are they used?" It might be useful for GCSE Design and Technology student, Physics GCSE, KS4, and just about anyone new to potential divider circuits and theory, as I have tried to make it as simple as possible.

Circuit Equation

The principle, or how it works, is very simple to understand, and the key is to understand that for an unloaded potential divider the same current "I" flows through both resistors R1 and R2 as shown in the animation above.

Step 1: Calculate the total resistance of the series network. This is simple as you just add the values of R1 and R2 to give the total resistance Rtot.

Step 2: To calculate the current flowing through the circuit we use the Ohm's Law in the following form.

I = Vs / Rtot

Where, Vs is the supply voltage, and Rtot is the value you calculated in Step 1.

Step 3: Using Ohm's Law again, the voltage across each resistor is the product of the current flowing through it and its resistance value. Hence, the following equation gives the voltage across R2.

V2 = I × R2

Similarly, the following equation gives the voltage across R1.

V1 = I × R1

Since the total voltage is given by the following expression.

Vs = V1 + V2

All you have to do is to find either V1 or V2, and then substitute into the above equation. Hence, this provides us with the following two expressions.

V2 = Vs - V1

V1 = Vs - V2

Circuit Formula

Another way to calculate the signal value, also known as output voltage (Vout), is by using the standard formula shown above. In essence, the output voltage is the voltage across R2, and as you can see, a simple circuit consists of two resistors in series between ground and supply voltage. A standard formula is OK to use and is just manageable for a two-resistor network. It is important to understand where each value comes from and what it is being calculated.

Potential Divider Calculator

With 3 Resistors

If you understand the basic principle, then it does not matter how many resistors there are in the series. Multi-resistor networks are usually found in electronic meters to provide a range function.

Step 1: Just as before, calculate the total resistance Rtot by adding the values of R1, R2, and R3.

Step 2: Use Ohm's Law, calculate the Current "I" by using the following formul.

I = Vs / Rtot

Step 3: Use Ohm's Law again, calculate the voltage across each resistor as following.

V1 = I × R1

V2 = I × R2

V3 = I × R3

The laws of Physics dictate that Vs must be the sum of the individual voltages across the resistors, given by the following expression.

Vs = V1 + V2 + V3

Parallel Resistor Circuit

Sometimes a divider circuit might have a parallel resistor network of the type shown above. In this case, your first step should be to add the values of the parallel resistors to form a single effective resistance. Adding resistors in parallel is very easy. Just use the formula shown above, or use the Resistors in Parallel article, which has a calculator.

In the example above a parallel resistor network consisting of Ra and Rb combine to form a total effective resistance R1, thus forming an equivalent circuit consisting of R1 and R2. You then follow the same principle as shown in the top part of this article to find the voltage output.

Experiment Circuit

Connect the circuit diagram sown above to carry out a potential divider experiment. You will need the following components and apparatus.

• Breadboard
• Digital Meter
• 9 V PP3 Battery
• 100 Ω Resistor
• 200 Ω Resistor

Experiment Breadboard Layout

Investigation

First, set the meter function to measure resistance, and measure the values of R1 and R2 and record them in the experiment results table below.

Set the meter function to measure DC Voltage, and measure the supply voltage Vs across the battery terminals and record that information in the table below as well.

Click on the icon above to see a breadboard layout for this experiment. Connect the circuit shown above and then measure the voltage V1 across the 100 Ω resistor. Then measure the voltage V2 across the 200 Ω resistor. Record these values in the table below. In addition, investigate to see what happens when R1 and R2 are the same values, and measure the voltages across the resistors.

Experiment Results

Essay

The resistor values, measured by a digital meter, were 102 Ω and 198 Ω respectively. This was within their 5 % tolerance rating. Taking into account the measuring errors, and component tolerance, the voltage V1 is rounded to 3 V and Voltage V2 rounded to 6 V.

In this calculation I am using the ideal values of the resistors and the voltage source Vs. The following expression calculates the total resistance Rtot to be 300 Ω.

Rtot = 100 Ω + 200 Ω

The Current "I" flowing through the resistors is 0.03 A, which is found using the following calculation.

9 / 300

Using Ohm's Law, the following calculations give the voltages across the resistors.

0.03 × 100 = 3 V

0.03 × 200 = 6 V

Therefore, in conclusion, you can see that the voltage divides in the same proportion as the resistor values.

Disadvantage

In transistor circuits, there may be many potential divider networks, using this simple method of biasing. With low resistor values, the current "I" is much greater, and the battery depletes faster. To prevent this from happening engineers use higher resistor values such as 100 kΩ and 200 kΩ in biasing circuits. Resistive networks work well for DC and low frequency applications but are not ideally suited for AC or high frequency circuits.

When using resistors it is important to note that they have tolerance values, and this will introduce a slight error and therefore may not be a suitable method where a precise reference voltage is required.

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