Resistive Voltage Divider | Qucs-S S-parameter Viewer Documentation

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# 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.