Resistor network calculator — ladder & combos
Solve the resistor networks that come up on the bench: a loaded voltage divider, an R-2R ladder output, and matching T or pi attenuators.
Network type
Loaded divider output
Unloaded V_out
V_in × R2 ÷ (R1 + R2)
Loaded V_out
V_in × (R2∥RL) ÷ (R1 + R2∥RL)
Load current
V_out ÷ RL
Sag from loading
(unloaded − loaded) ÷ unloaded
Rule of thumb: keep RL at least 10× R2 so the divider stays within a few percent of its unloaded value.
About the model
The loaded divider treats the load as a pure resistance in parallel with R2; the R-2R ladder assumes the standard alternating-R network driven from a low-impedance reference; and the attenuator pads are designed to match Z at both ports. All three assume ideal resistors — real parts add tolerance, temperature drift and parasitic capacitance that shift the numbers, especially above a few hundred MHz.
For the parts themselves: resistors, resistor networks and RF components cover the divider, ladder and pad builds above.
Browse related parts:
Frequently asked questions
How do you calculate the output of a loaded voltage divider?
The loaded output treats the load as a resistance in parallel with R2: Vout = Vin × (R2∥RL) ÷ (R1 + R2∥RL), where R2∥RL = (R2 × RL) ÷ (R2 + RL). With the defaults of Vin = 5 V, R1 = R2 = 10 kΩ and RL = 100 kΩ the unloaded output is 2.50 V but the loaded output sags to about 2.38 V, and the tool reports that sag as a percentage.
What is the output of an R-2R ladder at all ones?
An n-bit ladder driven by V_ref gives Vout(all-ones) = V_ref × (1 − 1 ÷ 2^n) and an LSB step of V_ref ÷ 2^n. With the defaults of 4 bits and V_ref = 5 V, the all-ones output is 5 × (1 − 1/16) = 4.6875 V and the LSB is 312.5 mV.
How are T-pad and pi-pad attenuator resistors calculated?
For a matched attenuator with impedance Z and loss A dB the calculator computes the voltage ratio k = (10^(A/20) − 1) ÷ (10^(A/20) + 1). The T-pad uses series resistors Z×k and a shunt of Z×(1−k²)÷(2k); the pi-pad uses series resistors Z×2k÷(1−k²) and shunt resistors Z÷k. Defaults of Z = 50 Ω and 6 dB give T series resistors of about 16.6 Ω and a T shunt of about 66.9 Ω.
How much power does a 6 dB attenuator pass?
A 6 dB pad passes output power ÷ input power = A² where A = 10^(6/20) ≈ 1.995, so A² ≈ 3.98 — meaning about one quarter of the input power reaches the load. The tool reports this power ratio directly so you can check the pad's power handling. (It is the square of the voltage ratio, not of the pad factor k = (A − 1)/(A + 1) ≈ 0.33.)
Ideal resistive networks — real parts add tolerance, temperature drift and parasitic capacitance. ICBOMS provides this tool for reference only.