How it works
In a divider made of two series resistors, the output is the voltage across the bottom resistor R2. With no load, the equations are:
Vout = Vin × R2 / (R1 + R2)I = Vin / (R1 + R2)PR1 = I² × R1, PR2 = I² × R2
With a load RL, R2 is effectively in parallel with RL: R2′ = R2 × RL / (R2 + RL), and Vout = Vin × R2′ / (R1 + R2′). If the load equals R2, the output drops to two-thirds of its no-load value.
For the reverse calculation, R2 is fixed and the ideal value is R1 = R2 × (Vin / Vout − 1). The closest E24 or E96 standard value is then chosen, and the calculator shows the actual Vout and the error against the target (%). E96 has finer spacing, so the error is smaller.
Worked example
With Vin 12 V and R1 = R2 = 10 kΩ, the no-load Vout is 6.00 V, the current is 0.600 mA, and each resistor dissipates 3.60 mW. Adding RL = 10 kΩ gives R2′ = 5 kΩ, so Vout becomes 4.00 V, a drop of 33.3 %.
For a 3.3 V target with R2 = 10 kΩ, the ideal R1 is 26.4 kΩ. The E24 value 27 kΩ gives about 3.24 V (−1.7 %). The E96 value 26.1 kΩ gives about 3.32 V (+0.7 %).
Design tips
- A divider current of at least 10 times the feedback input bias current or the load current is typical. Too little current makes the bias error large; too much increases standby loss.
- Loads often connect directly to the divider output. If the load resistance is less than about 10 times R2, the voltage drop is several percent, so consider a buffer amplifier.
- For precision voltages, use E96 values and 0.1 % resistors. Matched resistors from the same lot reduce ratio error.
- With a high input voltage, most of the voltage falls across the top resistor. Check the voltage and power ratings of the chip resistor.
Limitations
- The load is assumed to be a simple resistor. A current-source or non-linear load gives different results.
- E24 and E96 are standard approximations. The real combination error must be judged together with the resistor tolerance.
- Temperature coefficients and self-heating of the resistors are not included.
Frequently asked questions
How much does the output drop when a load is connected?
The load is effectively in parallel with R2, so R2 behaves as if it were smaller. For example, with R2 = 10 kΩ and RL = 100 kΩ, the drop is about 5 %. The calculator shows the voltage drop directly.
How much current should the divider carry?
Typically at least 10 times the feedback input bias current or the load current. Too little current causes bias error; too much increases standby loss.
In the reverse calculation, should I use E24 or E96?
E24 has wide spacing, so the error against the target can be large. E96 has fine spacing and a much smaller error. Use E96 for precision voltages, and E24 if you want fewer resistor types.