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Worst Case Calculation and Selection of Resistors for a Voltage Divider Circuit


A voltage divider is a simple yet fundamental circuit in electronics, used to create a specific fraction of an input voltage. The basic configuration involves two resistors, 

R1R_1 and R2R_2, connected in series across a voltage source VinV_{in}. The output voltage VoutV_{out} is taken from the junction of R1R_1 and R2R_2.

Basic Theory

The output voltage VoutV_{out} of a voltage divider is given by: Vout=VinR2R1+R2V_{out} = V_{in} \cdot \frac{R_2}{R_1 + R_2}

Worst Case Calculation

Worst case analysis ensures that the circuit performs reliably under the most extreme conditions. For a voltage divider, the worst-case scenario occurs due to the tolerance of the resistors. The tolerance indicates the permissible deviation from the nominal value, typically expressed as a percentage.

Steps for Worst Case Calculation

  1. Determine Nominal Values and Tolerances:

    • Let R1nomR_{1_{nom}} and R2nomR_{2_{nom}} be the nominal values of R1R_1 and R2R_2, respectively.
    • Let T1T_1 and T2T_2 be the tolerances of R1R_1 and R2R_2, respectively.
  2. Calculate Extremes:

    • Maximum R1R_1: R1max=R1nom(1+T1)R_{1_{max}} = R_{1_{nom}} (1 + T_1)
    • Minimum R1R_1: R1min=R1nom(1T1)R_{1_{min}} = R_{1_{nom}} (1 - T_1)
    • Maximum R2R_2: R2max=R2nom(1+T2)R_{2_{max}} = R_{2_{nom}} (1 + T_2)
    • Minimum R2R_2: R2min=R2nom(1T2)R_{2_{min}} = R_{2_{nom}} (1 - T_2)
  3. Calculate Worst Case VoutV_{out}:

    • Calculate VoutV_{out} for the maximum and minimum possible values of R1R_1 and R2R_2.

    Voutmax=VinR2maxR1min+R2maxV_{out_{max}} = V_{in} \cdot \frac{R_{2_{max}}}{R_{1_{min}} + R_{2_{max}}} Voutmin=VinR2minR1max+R2minV_{out_{min}} = V_{in} \cdot \frac{R_{2_{min}}}{R_{1_{max}} + R_{2_{min}}}

Example Calculation


  • Vin=10VV_{in} = 10V
  • R1nom=1kΩR_{1_{nom}} = 1k\Omega, T1=5%T_1 = 5\%
  • R2nom=2kΩR_{2_{nom}} = 2k\Omega, T2=5%T_2 = 5\%

Calculate extremes:

  • R1max=1kΩ×1.05=1.05kΩR_{1_{max}} = 1k\Omega \times 1.05 = 1.05k\Omega
  • R1min=1kΩ×0.95=0.95kΩR_{1_{min}} = 1k\Omega \times 0.95 = 0.95k\Omega
  • R2max=2kΩ×1.05=2.1kΩR_{2_{max}} = 2k\Omega \times 1.05 = 2.1k\Omega
  • R2min=2kΩ×0.95=1.9kΩR_{2_{min}} = 2k\Omega \times 0.95 = 1.9k\Omega

Calculate worst case VoutV_{out}:

  • Voutmax=10V2.1kΩ0.95kΩ+2.1kΩ=10V2.13.056.89VV_{out_{max}} = 10V \cdot \frac{2.1k\Omega}{0.95k\Omega + 2.1k\Omega} = 10V \cdot \frac{2.1}{3.05} \approx 6.89V
  • Voutmin=10V1.9kΩ1.05kΩ+1.9kΩ=10V1.92.956.44VV_{out_{min}} = 10V \cdot \frac{1.9k\Omega}{1.05k\Omega + 1.9k\Omega} = 10V \cdot \frac{1.9}{2.95} \approx 6.44V

Selection of Resistor Combination

Selecting the best combination of resistors involves considering several factors:

  1. Tolerance:

    • Lower tolerance resistors (e.g., 1% or 0.1%) will provide a more stable and precise VoutV_{out}.
  2. Power Rating:

    • Ensure the resistors can handle the power dissipation. The power dissipated in a resistor PP is given by P=V2RP = \frac{V^2}{R}.
  3. Temperature Coefficient:

    • Choose resistors with low temperature coefficients to minimize variation in resistance due to temperature changes.
  4. Availability and Cost:

    • Balance between precision and cost. High precision resistors are typically more expensive.

Example of Selecting Resistors

For a specific application, suppose you need Vout=5VV_{out} = 5V from Vin=10VV_{in} = 10V. The ideal ratio is: R2R1+R2=0.5\frac{R_2}{R_1 + R_2} = 0.5

Choosing R1=1kΩR_1 = 1k\Omega and R2=1kΩR_2 = 1k\Omega:

  • Nominal Vout=10V1kΩ1kΩ+1kΩ=5VV_{out} = 10V \cdot \frac{1k\Omega}{1k\Omega + 1k\Omega} = 5V

If we select resistors with 1% tolerance:

  • R1max=1kΩ×1.01=1.01kΩR_{1_{max}} = 1k\Omega \times 1.01 = 1.01k\Omega
  • R1min=1kΩ×0.99=0.99kΩR_{1_{min}} = 1k\Omega \times 0.99 = 0.99k\Omega
  • R2max=1kΩ×1.01=1.01kΩR_{2_{max}} = 1k\Omega \times 1.01 = 1.01k\Omega
  • R2min=1kΩ×0.99=0.99kΩR_{2_{min}} = 1k\Omega \times 0.99 = 0.99k\Omega

Worst case VoutV_{out}:

  • Voutmax=10V1.01kΩ0.99kΩ+1.01kΩ=10V1.012=5.05VV_{out_{max}} = 10V \cdot \frac{1.01k\Omega}{0.99k\Omega + 1.01k\Omega} = 10V \cdot \frac{1.01}{2} = 5.05V
  • Voutmin=10V0.99kΩ1.01kΩ+0.99kΩ=10V0.992=4.95VV_{out_{min}} = 10V \cdot \frac{0.99k\Omega}{1.01k\Omega + 0.99k\Omega} = 10V \cdot \frac{0.99}{2} = 4.95V

This combination provides a VoutV_{out} very close to the desired value, with minimal deviation.


Performing worst-case calculations for a voltage divider involves determining the maximum and minimum possible values of the output voltage considering resistor tolerances. Selecting the best resistor combination requires balancing precision, power rating, temperature stability, availability, and cost. By carefully considering these factors, you can design a robust and reliable voltage divider circuit.

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