Resistive Voltage Divider

Overview

The resistive voltage divider reduces an input voltage into a smaller output voltage using two resistors: one series and one shunt.

This calculator helps you design voltage dividers with power dissipation analysis, supporting parallel resistor combinations for both upper and lower branches.

Basic Circuit Configuration

  • Upper resistor(s): Connected between \(V_{DC}\) and \(V_{out}\)

  • Lower resistor(s): Connected between \(V_{out}\) and ground

The output voltage is taken from the node between the two resistors.

Voltage Divider Formula

For a basic two-resistor divider, the output voltage is calculated using:

\[V_{out} = V_{DC} \times \frac{R_{lower}}{R_{upper} + R_{lower}}\]

Where:

  • \(V_{DC}\) is the input voltage

  • \(R_{upper}\) is the resistance of the upper branch

  • \(R_{lower}\) is the resistance of the lower branch

  • \(V_{out}\) is the output voltage

Parallel Resistor Combinations

This calculator supports multiple resistors in parallel for both the upper and lower branches. When multiple resistors are connected in parallel, the equivalent resistance is calculated using:

\[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}\]

This allows to

  • Distribute the current (i.e. power dissipation) across multiple components

  • Add degrees of freedom to fine-tune the output voltage

Current and Power Calculations

Current Through Each Resistor

Upper Branch: For each resistor in the upper branch, the current is:

\[I_{upper} = \frac{V_{DC} - V_{out}}{R_{upper}}\]

Lower Branch: For each resistor in the lower branch, the current is:

\[I_{lower} = \frac{V_{out}}{R_{lower}}\]

Power Dissipation

The power dissipated by each resistor is given by:

\[P = I^2 \times R\]

Warning

Real-world resistors come with a power dissipation rating (e.g. 50 mW, 250 mW, etc.). It’s critical to keep the power dissipation with enough margin (typically 50-75% derating).

Example Calculation

Consider a voltage divider with the following parameters:

Parameter

Value

Input Voltage (\(V_{DC}\))

5.0 V

Upper Resistor

1 kΩ

Lower Resistor

1 kΩ

Max Power per Resistor

50 mW

Output Voltage:

\[V_{out} = 5.0 \times \frac{1000}{1000 + 1000} = 2.5 \text{ V}\]

Current Through Each Resistor:

\[I = \frac{V_{DC}}{R_{upper} + R_{lower}} = \frac{5.0}{2000} = 2.5 \text{ mA}\]

Power Dissipation:

\[P_{each} = I^2 \times R = (0.0025)^2 \times 1000 = 6.25 \text{ mW}\]
\[P_{total} = 2 \times 6.25 = 12.5 \text{ mW}\]

Since 6.25 mW < 50 mW, standard 1/8 W (125 mW) resistors would be more than adequate.