Voltage Divider (Potential Divider)

A voltage divider (also called a potential divider) is a simple circuit that produces an output voltage that is a fraction of its input voltage. It is one of the most fundamental circuits in electronics.

Definition

Voltage dividers are circuits that produce output voltage(s) as a fraction of their input voltage. They consist of two or more resistors (or impedances) connected in series across a voltage source.

Circuit Configuration

Basic Unloaded Voltage Divider

flowchart LR
    subgraph "Voltage Divider"
    A[V_in] --- B[R1] --- C[V_out] --- D[R2] --- E[GND]
    end
    
    style A fill:#e1f5e1
    style C fill:#fff9c4
    style E fill:#ffccbc

Where:

  • $V_{in}$ = Input voltage
  • $R_1$ = Upper resistor
  • $R_2$ = Lower resistor
  • $V_{out}$ = Output voltage (taken across $R_2$)

Derivation

Step 1: Apply Kirchhoff's Voltage Law

For the series circuit:

$$V_{in} = I(R_1 + R_2)$$

Step 2: Express Current

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

Step 3: Output Voltage

The output voltage is the voltage drop across $R_2$:

$$V_{out} = IR_2 = \frac{V_{in}}{R_1 + R_2} \times R_2$$

Voltage Divider Formula

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

This can also be written as:

$$V_{out} = V_{in} \times \frac{R_2}{R_{total}}$$

Where $R_{total} = R_1 + R_2$ for series resistors.

Key Properties

Property Description
Output < Input $V_{out}$ is always less than $V_{in}$ (for positive resistances)
Proportional $V_{out} \propto R_2$ — output scales with the lower resistor
Ratio-dependent The output depends only on the ratio of resistances, not their absolute values
No load The formula assumes no load is connected to the output

Applications

1. Potentiometer

A variable voltage divider:

flowchart LR
    A[V_in] --- B[Variable Resistor] --- C[GND]
    B --- D[V_out<br/>Adjustable]
    
    style D fill:#fff9c4
  • Linear potentiometer — sliding contact moves along a straight resistive element
  • Rotary potentiometer — rotating shaft adjusts the tap position

2. Level Shifters

Interfaces circuits with different operating voltages:

flowchart LR
    subgraph "5V Domain"
    A[5V Logic] 
    end
    
    subgraph "Voltage Divider"
    B[5V] --- R1[1kΩ] --- C[3.3V] --- R2[2kΩ] --- D[GND]
    end
    
    subgraph "3.3V Domain"
    E[3.3V Logic]
    end
    
    A --> B
    C --> E
    
    style C fill:#fff9c4

3. Sensor Interfaces

Many sensors (like thermistors, photoresistors) are used as one leg of a voltage divider to convert resistance changes into voltage signals.

4. Biasing Circuits

Used in transistor amplifiers to set the operating point (Q-point).

Loaded Voltage Divider

When a load resistance $R_L$ is connected across the output:

flowchart LR
    A[V_in] --- B[R1] --- C[V_out]
    C --- D[R2] --- E[GND]
    C --- F[R_L<br/>Load] --- E
    
    style C fill:#fff9c4
    style F fill:#ffccbc

Analysis

$R_2$ and $R_L$ are in parallel:

$$R_p = \frac{R_2 R_L}{R_2 + R_L}$$

The equivalent resistance of the lower branch becomes $R_p < R_2$.

Loaded Output Formula

$$\boxed{V_{out} = V_{in} \times \frac{R_p}{R_1 + R_p}}$$

Loading Effect

Condition Effect
Unloaded $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$
Loaded $V_{out}$ decreases because $R_p < R_2$
Heavy load $R_L \ll R_2$ causes significant voltage drop
Light load $R_L \gg R_2$ has minimal effect

Practical Tip: For minimal loading effect, ensure $R_L \gg R_2$ (typically $R_L > 10R_2$).

Worked Examples

Example 1: Basic Calculation

Given: $V_{in} = 12$ V, $R_1 = 4$ kΩ, $R_2 = 8$ kΩ

Find: $V_{out}$

Solution:

$$V_{out} = 12 \times \frac{8000}{4000 + 8000} = 12 \times \frac{8}{12} = 12 \times \frac{2}{3}$$

$$V_{out} = 8 \text{ V}$$

Example 2: Finding Resistor Values

Given: $V_{in} = 9$ V, desired $V_{out} = 3$ V, $R_1 + R_2 = 30$ kΩ

Find: $R_1$ and $R_2$

Solution:

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

$$\frac{3}{9} = \frac{R_2}{30000}$$

$$R_2 = \frac{30000}{3} = 10000 \text{ Ω} = 10 \text{ kΩ}$$

$$R_1 = 30000 - 10000 = 20 \text{ kΩ}$$

Example 3: Loaded Divider

Given: $V_{in} = 100$ V, $R_1 = 25$ kΩ, $R_2 = 47$ kΩ, $R_L = 100$ kΩ

Find: $V_{out}$ (loaded)

Solution:

First, find $R_p$:

$$R_p = \frac{47000 \times 100000}{47000 + 100000} = \frac{4.7 \times 10^9}{147000} = 31.97 \text{ kΩ}$$

Then:

$$V_{out} = 100 \times \frac{31970}{25000 + 31970} = 100 \times \frac{31970}{56970}$$

$$V_{out} = 56.1 \text{ V}$$

Compare to unloaded: $V_{out(unloaded)} = 65.28$ V — the load causes a significant drop!

Multi-Tap Voltage Divider

For multiple output voltages:

flowchart LR
    A[V_in] --- B[R1] --- C[V1] --- D[R2] --- E[V2] --- F[R3] --- G[GND]
    
    style C fill:#fff9c4
    style E fill:#fff9c4
  • $V_1 = V_{in} \times \frac{R_2 + R_3}{R_1 + R_2 + R_3}$
  • $V_2 = V_{in} \times \frac{R_3}{R_1 + R_2 + R_3}$

Related Concepts

  • Wheatstone Bridge — Resistance measurement using balanced dividers
  • Electrical Measurements — Overview of measurement techniques
  • Ohm's Law — Foundation of voltage divider analysis
  • Resistor Networks — Series and parallel combinations

Sources

Key Takeaways

  1. A voltage divider produces an output voltage proportional to the ratio of resistances
  2. The output is always taken across the "lower" resistor (connected to ground)
  3. Loading effects must be considered when the divider drives a circuit
  4. Voltage dividers are widely used for level shifting, sensor interfacing, and biasing