Skip to main content
ELECTRICAL

Voltage Divider Calculator

Calculate output voltage and resistor ratios in unloaded voltage divider circuits with Vout = Vin·R2/(R1+R2). Step-by-step electronic DC circuit solutions.

DOCUMENTED FORMULA intermediate
Interactive Workbench
intermediate
Quick PresetsClick to autofill values
Parameters3 variables
Source voltage applied across the resistor network.
Upper resistor connected to source voltage in Ohms.
Lower resistor connected to ground in Ohms.
Calculation ResultAwaiting Input
Output Voltage (V_out)
--.--

Calculated output voltage across resistor R2.


On this page Jump to section ▾

Practical Worked Example

Step-by-step practical problem and numerical calculation scenario.

Problem Statement

A DC input voltage of Vin=24.0 VV_{\text{in}} = 24.0\text{ V} is applied across a series resistor network consisting of R1=30.0 kΩR_1 = 30.0\text{ k}\Omega (30,000 Ω30{,}000\text{ }\Omega) and R2=10.0 kΩR_2 = 10.0\text{ k}\Omega (10,000 Ω10{,}000\text{ }\Omega). Calculate:

  1. The unloaded output voltage (VoutV_{\text{out}}) taken across resistor R2R_2.
  2. The voltage division ratio.
  3. The total series resistance and steady-state loop current.

Step-by-Step Solution

Step 1: Identify Input Parameters

  • Input Voltage (VinV_{\text{in}}): 24.0 V24.0\text{ V}
  • Upper Resistor (R1R_1): 30,000 Ω30{,}000\text{ }\Omega (30 kΩ30\text{ k}\Omega)
  • Lower Resistor (R2R_2): 10,000 Ω10{,}000\text{ }\Omega (10 kΩ10\text{ k}\Omega)

Step 2: Calculate Total Series Resistance (RtotalR_{\text{total}})

Because R1R_1 and R2R_2 are connected in series, their resistances add directly: Rtotal=R1+R2=30,000 Ω+10,000 Ω=40,000 Ω=40.0 kΩR_{\text{total}} = R_1 + R_2 = 30{,}000\text{ }\Omega + 10{,}000\text{ }\Omega = 40{,}000\text{ }\Omega = 40.0\text{ k}\Omega


Step 3: Determine the Voltage Division Ratio

The attenuation ratio depends solely on the relative values of R1R_1 and R2R_2: Ratio=R2R1+R2=10,000 Ω40,000 Ω=0.250(25.0%)\text{Ratio} = \frac{R_2}{R_1 + R_2} = \frac{10{,}000\text{ }\Omega}{40{,}000\text{ }\Omega} = 0.250 \quad (25.0\%)


Step 4: Calculate the Output Voltage (VoutV_{\text{out}})

Applying the voltage divider equation: Vout=Vin×(R2R1+R2)=24.0 V×0.250=6.00 VV_{\text{out}} = V_{\text{in}} \times \left(\frac{R_2}{R_1 + R_2}\right) = 24.0\text{ V} \times 0.250 = 6.00\text{ V}

The unloaded output voltage across R2R_2 is 6.00 V6.00\text{ V}.


Step 5: Compute Loop Current & Power Dissipation

  • Series Current (II): I=VinRtotal=24.0 V40,000 Ω=0.00060 A=0.60 mAI = \frac{V_{\text{in}}}{R_{\text{total}}} = \frac{24.0\text{ V}}{40{,}000\text{ }\Omega} = 0.00060\text{ A} = 0.60\text{ mA}
  • Total Power Dissipation (PP): P=Vin×I=24.0 V×0.00060 A=0.0144 W=14.4 mWP = V_{\text{in}} \times I = 24.0\text{ V} \times 0.00060\text{ A} = 0.0144\text{ W} = 14.4\text{ mW}

Step 6: Critical Model Assumption (Loading Effect)

  • Unloaded Condition: This calculation assumes an open circuit at the output terminal (Iout=0I_{\text{out}} = 0).
  • Loading Effect: If an external load resistor (RloadR_{\text{load}}) is connected in parallel with R2R_2, the effective lower resistance drops to R2,eff=R2⋅RloadR2+RloadR_{2,\text{eff}} = \frac{R_2 \cdot R_{\text{load}}}{R_2 + R_{\text{load}}}, pulling VoutV_{\text{out}} below 6.0 V6.0\text{ V}. To keep loading error under 1%1\%, ensure Rload≥100×R2R_{\text{load}} \ge 100 \times R_2 or buffer the output with an operational amplifier.

Variables & Units Reference

Input parameters, calculated outputs, and supported SI/US customary units.

Mathematical Variables and Units Summary Table
Variable Symbol Parameter Name & Description Role Supported Units
vIn Input Voltage (V_in) Source voltage applied across the resistor network. Input mV, V, kV
r1 Resistor R1 Upper resistor connected to source voltage in Ohms. Input mΩ, Ω, kΩ, MΩ
r2 Resistor R2 Lower resistor connected to ground in Ohms. Input mΩ, Ω, kΩ, MΩ
result Output Voltage (V_out) Calculated output voltage across resistor R2. Output V
ℹ

Result Interpretation & Practical Meaning

Physical significance, evaluation guidelines, and domain context for the calculated output.

The output indicates the unloaded open-circuit potential difference (Vout) established across the lower resistor in a series two-resistor network. This voltage represents the ideal ratio Vout = Vin · R2 / (R1 + R2). When a load circuit or measuring instrument with finite resistance (Rload) is connected across R2, the effective parallel impedance drops to R2 || Rload, pulling Vout below the theoretical unloaded value. To maintain voltage accuracy within 1%, Rload should exceed R2 by at least a factor of 100.

What is a Voltage Divider?

Definition:
A voltage divider is a fundamental linear passive circuit that produces an output voltage proportional to an input voltage, governed by Vout=Vin⋅R2R1+R2V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}, where VinV_{\text{in}} is input potential, and R1R_1 and R2R_2 are series resistor values in Ohms.

A resistive voltage divider (or potential divider) is a fundamental linear circuit composed of two resistors connected in series across an input voltage source. It produces an output voltage that is a fixed fraction of the input voltage, determined strictly by the ratio of the two resistances.

Voltage dividers are widely used in analog and digital electronics to create reference voltages, scale high potentials down to safe microcontroller Analog-to-Digital Converter (ADC) ranges, bias transistors, and interface resistive sensors (such as photoresistors, thermistors, and strain gauges).


Circuit Topology & Governing Equations

             Input Voltage (V_in)
                    |
                    +
                    |
                   [ ]  Resistor R1 (Upper Leg)
                   [ ]
                    |
          Output ---+---> Output Voltage (V_out)
                    |
                   [ ]  Resistor R2 (Lower Leg)
                   [ ]
                    |
                   === Ground (0V)

1. Kirchhoff’s Voltage Law & Series Current

Applying Ohm’s law and Kirchhoff’s Voltage Law (KVL) to the unloaded series loop, the total series resistance is Rtotal=R1+R2R_{\text{total}} = R_1 + R_2, and the continuous quiescent current flowing through the divider is:

I=VinR1+R2I = \frac{V_{\text{in}}}{R_1 + R_2}

2. Canonical Voltage Divider Equation

Because the output voltage VoutV_{\text{out}} is measured across the lower resistor R2R_2 to ground:

Vout=I⋅R2=Vin⋅(R2R1+R2)V_{\text{out}} = I \cdot R_2 = V_{\text{in}} \cdot \left(\frac{R_2}{R_1 + R_2}\right)

Where:

  • VinV_{\text{in}}: Source input voltage applied across the network (Volts, V\text{V}).
  • R1R_1: Upper series resistor connected between VinV_{\text{in}} and the output node (Ohms, Ω\Omega).
  • R2R_2: Lower resistor connected between the output node and ground (Ω\Omega).
  • VoutV_{\text{out}}: Resulting output voltage measured at the junction node (V\text{V}).

Downstream Load Impedance & The 10x Rule

The standard voltage divider equation assumes an unloaded ideal condition (i.e., the output is connected to an infinite input impedance device, such as a high-impedance CMOS op-amp or digital multimeter).

When an external load resistance RLR_L is connected in parallel with R2R_2, the effective lower resistance becomes:

Req=R2∥RL=R2⋅RLR2+RLR_{\text{eq}} = R_2 \parallel R_L = \frac{R_2 \cdot R_L}{R_2 + R_L}

The actual loaded output voltage drops to:

Vout,loaded=Vin⋅(ReqR1+Req)<Vout,idealV_{\text{out,loaded}} = V_{\text{in}} \cdot \left(\frac{R_{\text{eq}}}{R_1 + R_{\text{eq}}}\right) < V_{\text{out,ideal}}

The 10x Engineering Rule of Thumb

To prevent excessive voltage sag without using an active buffer amplifier, the load resistance must be significantly larger than R2R_2:

RL≥10⋅R2R_L \ge 10 \cdot R_2

(When RL=10R2R_L = 10 R_2, the loaded voltage drops by less than 10%10\%. For precision measurements, RL≥100R2R_L \ge 100 R_2 is recommended).


Variable Definitions & Standard Units

VariablePhysical QuantityStandard SI UnitCommon Engineering UnitsDescription
VinV_{\text{in}}Input VoltageV\text{V}mV\text{mV}, kV\text{kV}Supply rail or signal potential applied across the divider.
R1R_1Upper ResistorΩ\OmegakΩ\text{k}\Omega, MΩ\text{M}\OmegaTop resistor in series string (pull-up leg).
R2R_2Lower ResistorΩ\OmegakΩ\text{k}\Omega, MΩ\text{M}\OmegaGround-connected resistor across which VoutV_{\text{out}} is tapped.
VoutV_{\text{out}}Output VoltageV\text{V}mV\text{mV}Stepped-down potential at the intermediate node.
IIQuiescent CurrentA\text{A}mA\text{mA}, μA\mu\text{A}Continuous current drain through the divider (Vin/(R1+R2)V_{\text{in}} / (R_1+R_2)).

Step-by-Step Worked Example

Problem:

A 5.0 V5.0\text{ V} UART digital logic signal needs to be stepped down to 3.3 V3.3\text{ V} to interface safely with a 3.3 V3.3\text{ V} ARM Cortex microcontroller GPIO pin. Using R2=3.3 kΩR_2 = 3.3\text{ k}\Omega (3,300 Ω3,300\ \Omega), calculate the required resistance for R1R_1, the resulting output voltage, and the continuous power dissipated in R1R_1.

Step 1: Calculate Required Upper Resistor R1R_1

Rearranging the divider equation to solve for R1R_1:

VoutVin=R2R1+R2  ⟹  R1=R2⋅(Vin−VoutVout)\frac{V_{\text{out}}}{V_{\text{in}}} = \frac{R_2}{R_1 + R_2} \implies R_1 = R_2 \cdot \left(\frac{V_{\text{in}} - V_{\text{out}}}{V_{\text{out}}}\right)

R1=3,300 Ω×(5.0 V−3.3 V3.3 V)=3,300×(1.73.3)=1,700 Ω=1.70 kΩR_1 = 3,300\ \Omega \times \left(\frac{5.0\text{ V} - 3.3\text{ V}}{3.3\text{ V}}\right) = 3,300 \times \left(\frac{1.7}{3.3}\right) = 1,700\ \Omega = 1.70\text{ k}\Omega

Step 2: Verify Output Voltage (VoutV_{\text{out}})

Vout=5.0 V×(3,300 Ω1,700 Ω+3,300 Ω)=5.0×(3,3005,000)=5.0×0.66=3.30 VoltsV_{\text{out}} = 5.0\text{ V} \times \left(\frac{3,300\ \Omega}{1,700\ \Omega + 3,300\ \Omega}\right) = 5.0 \times \left(\frac{3,300}{5,000}\right) = 5.0 \times 0.66 = 3.30\text{ Volts}

Step 3: Compute Quiescent Current & Resistor Power Dissipation

  • Current: I=5.0 V5,000 Ω=0.001 A=1.0 mAI = \frac{5.0\text{ V}}{5,000\ \Omega} = 0.001\text{ A} = 1.0\text{ mA}
  • Power in R1R_1: PR1=I2⋅R1=(0.001 A)2×1,700 Ω=1.70 mWP_{R1} = I^2 \cdot R_1 = (0.001\text{ A})^2 \times 1,700\ \Omega = 1.70\text{ mW}

Interpretation: The 1.7 mW1.7\text{ mW} power dissipation is well within standard 1/8 W1/8\text{ W} (125 mW125\text{ mW}) metal film surface-mount resistor ratings.


Common Calculation Mistakes

  • Swapping Resistor Positions in Formula: Putting R1R_1 in the numerator (R1R1+R2\frac{R_1}{R_1 + R_2}) instead of R2R_2. The output voltage is measured across R2R_2; therefore, R2R_2 must be in the numerator.
  • Neglecting Downstream ADC Input Impedance: Connecting a 100 kΩ100\text{ k}\Omega divider to an ADC with a 10 kΩ10\text{ k}\Omega sample-and-hold input impedance, causing severe voltage measurement errors. Use low impedance dividers (≤10 kΩ\le 10\text{ k}\Omega total) or buffer with an op-amp voltage follower.
  • Selecting Overly Small Resistors (High Power Loss): Using 10 Ω10\ \Omega and 20 Ω20\ \Omega resistors on a 12V12\text{V} rail. While the voltage ratio (1/31/3) is identical, total current is 400 mA400\text{ mA} and power dissipation is 4.8 Watts4.8\text{ Watts}, causing rapid resistor thermal destruction.

Practical Engineering Applications

  • Microcontroller ADC Signal Conditioning: Attenuating 0–12V0\text{--}12\text{V} automotive battery potentials down to 0–3.3V0\text{--}3.3\text{V} ADC input ranges for battery state-of-charge monitoring.
  • Resistive Sensor Interfacing: Constructing half-bridge circuits with Light Dependent Resistors (LDRs), NTC thermistors, and flex sensors to convert variable resistance into readable voltage swings.
  • Audio Volume Controls: Employing logarithmic potentiometers as variable voltage dividers to attenuate line-level audio signals smoothly across human hearing perception curves.


References & Technical Sources

  1. Horowitz, P., & Hill, W. (2015). The Art of Electronics (3rd ed.). Cambridge University Press. (Chapter 1: Foundations — Voltage Dividers & Thevenin Equivalent Circuits).
  2. Sedra, A. S., & Smith, K. C. (2019). Microelectronic Circuits (8th ed.). Oxford University Press. (Chapter 1: Signals and Amplifiers — Voltage Division & Input/Output Impedances).
  3. Alexander, C. K., & Sadiku, M. N. (2020). Fundamentals of Electric Circuits (7th ed.). McGraw-Hill Education. (Chapter 2: Elementary Circuit Laws — Voltage and Current Division).

Mathematical Derivation & Formula

Theoretical formulation derived from standard physics and engineering textbooks and models.

Step-by-Step Mathematical Proof

Step 1

Series Circuit Loop Current

I=VinR1+R2I = \frac{V_{\text{in}}}{R_1 + R_2}

Kirchhoff's voltage law determines the common current circulating through the series resistance chain.

Step 2

Voltage Drop across Lower Resistor

Vout=I×R2=Vin(R2R1+R2)V_{\text{out}} = I \times R_2 = V_{\text{in}} \left( \frac{R_2}{R_1 + R_2} \right)

Applying Ohm's law across R2 yields the division transfer function.

Common Engineering & Calculation Pitfalls

Frequent calculation errors, unit confusion, and model boundary conditions.

Ignoring Load Impedance Loading Effect

Why it happens: Connecting a low-impedance sensor or load directly to a high-resistance divider without buffering.

Correct approach: Ensure the load impedance is at least 100× larger than R2, or insert an op-amp unity-gain buffer.

Using Voltage Dividers as High-Current Power Supplies

Why it happens: Trying to step down voltage to power motors or microcontrollers directly via passive resistors.

Correct approach: Use switching buck regulators or linear LDO regulators; resistive dividers dissipate excessive heat under load.

Component Tolerance Stacking

Why it happens: Using 5% tolerance resistors when setting reference voltages for precision analog-to-digital converters.

Correct approach: Select 0.1% or 1% thin-film resistors for precision ADC voltage division.

Real-World Engineering Applications

Industry deployment scenarios, practical design constraints, and field applications.

Microcontroller Interfaces

High-Voltage ADC Signal Attenuation

Stepping down 12V or 24V industrial bus voltages to fit within 0–3.3V microcontroller analog inputs.

Analog Sensing

Resistive Sensor Interfacing (NTC Thermistors, LDRs)

Converting variable sensor resistance into a measurable proportional voltage signal.

Frequently Asked Questions

Technical explanations grounded in governing physical laws and textbook derivations.

Frequently Asked Questions

How does downstream load resistance (RloadR_{\text{load}}) distort the expected output voltage?

A basic voltage divider assumes zero current leaves the node between R1R_1 and R2R_2. When an external load (e.g., an ADC pin, microcontroller GPIO, or sensor amplifier) with finite resistance RloadR_{\text{load}} is connected across R2R_2, it places RloadR_{\text{load}} in parallel with R2R_2. The effective lower leg resistance becomes:

R2,eff=R2×RloadR2+Rload<R2R_{2,\text{eff}} = \frac{R_2 \times R_{\text{load}}}{R_2 + R_{\text{load}}} < R_2

This parallel combination reduces the equivalent resistance, causing the actual measured VoutV_{\text{out}} to drop (“voltage sag”) below the unloaded theoretical formula. To keep loading error under 1%1\%, circuit designers enforce the thumb rule: Rload≥100×R2R_{\text{load}} \ge 100 \times R_2, or buffer the node with a unity-gain operational amplifier.

What trade-offs govern the selection between megaohm and kilohm resistor values?

Selecting resistor values involves a strict engineering compromise between static power dissipation and signal bandwidth/noise:

  • Low Resistance (e.g., 1 kΩ1\text{ k}\Omega): Stiff against load currents and immune to stray electrostatic pickup, but draws continuous quiescent power from the battery (P=V2/(R1+R2)P = V^2 / (R_1 + R_2)), causing heat and draining battery-powered electronics.
  • High Resistance (e.g., 1 MΩ1\text{ M}\Omega): Conserves microamps of battery current, but increases Johnson-Nyquist thermal noise (vn=4kBTRΔfv_n = \sqrt{4 k_B T R \Delta f}) and creates a slow low-pass RC filter with stray PCB capacitance (CparasiticC_{\text{parasitic}}), severely limiting ADC high-speed sampling rates.

Can a resistive voltage divider safely step down voltages to power an active microcontroller?

No, never. A resistive voltage divider must only be used as a signal reference or sensing attenuation network, not as a power supply rail. Because microcontrollers dynamically change their current consumption (e.g., jumping from 50 μA50\text{ }\mu\text{A} in deep sleep to 80 mA80\text{ mA} during Wi-Fi transmission), the effective load resistance fluctuates violently, causing the supply voltage to collapse out of regulation and inducing immediate brownout resets. Active power delivery requires a linear low-dropout regulator (LDO) or switched-mode DC-DC buck converter.

Under what conditions are capacitive or reactive voltage dividers preferred over resistors?

In high-voltage AC electric utility substations and RF oscilloscope probe tips, capacitive dividers (Vout=Vin⋅C1/(C1+C2)V_{\text{out}} = V_{\text{in}} \cdot C_1 / (C_1 + C_2)) are preferred because capacitors dissipate zero real average power (Wattage=0\text{Wattage} = 0), preventing severe thermal runaway under kilovolt stresses. Furthermore, 10X10\text{X} passive oscilloscope probes incorporate an adjustable shunt trimmer capacitor in parallel with the 9 MΩ9\text{ M}\Omega probe resistor to match cable capacitance, maintaining a flat attenuation ratio across gigahertz bandwidths.

Academic References & Textbooks

Curated academic sources, engineering textbooks, and relevant technical references.

  • The Art of Electronics (3rd Edition)
    Authors: Paul Horowitz, Winfield Hill • Source: Cambridge University Press • Year: 2015