Should I use a pi pad or a tee pad?

Either works. Attenuation, match and dissipation are identical, so only the resistor values decide. Take the topology whose three values land closest to parts you can actually buy. The calculator prints the nearest E24 (5%) and E96 (1%) value under every result. A 3 dB tee asks for 8.55 Ω in series, a 3 dB pi for 292 Ω to ground, and at 20 dB it is the pi that carries the big value, 248 Ω in series. On a thin-film chip the criterion is geometry rather than stock. There the resistor is a printed strip, and its value is the sheet resistance times how many squares long the strip is, so 292 Ω on 10 Ω-per-square film needs a 29-square meander that starts to behave like a small inductor.

Why does an attenuator improve VSWR?

Because the reflected wave crosses the pad twice. A load at 3.0 : 1 sends a quarter of the power back. Behind a 3 dB pad that reflection is attenuated on the way in and again on the way out, so the return loss measures 2 × 3 = 6 dB better and the meter reads about 1.67 : 1. Nothing about the load changed. The pad only keeps the mismatch away from the stage in front of it, and it costs its own attenuation to do so.

What power rating do the resistors need?

Size for the hottest one, which above about 6 dB is the resistor closest to the source. With 1 W in, it takes 0.52 W in a 10 dB pad, 0.82 W at 20 dB and 0.94 W at 30 dB. Ratings also fall with temperature. Mini-Circuits specifies its 2 W coaxial 10 dB pad at 25 °C ambient and derates it linearly to 0.5 W at 100 °C. Use non-inductive thin-film or chip resistors, never wirewound.

Can I use a 50 Ω pad in a 75 Ω system?

Not as a matched pad. The resistor values scale with Z₀, so a 50 Ω 10 dB pad in a 75 Ω line attenuates 10.3 dB and shows about 1.44 : 1 at each port. Switch the calculator to 75 Ω for the right values. Matching 75 Ω to 50 Ω is a different job. Purely resistive, it costs at least 5.72 dB (43.3 Ω in series, 86.6 Ω across the 50 Ω side), which is why video gear reaches for a transformer when the loss matters.

How much error do standard resistor values add?

Less than you would fear, because a series value that runs high partly cancels a shunt value that runs low. Rounded to E24 parts (100 Ω and 68 Ω), the 10 dB pi pad gives 9.63 dB with a 1.01 : 1 input match. With the E96 pair (95.3 Ω and 71.5 Ω) the attenuation lands within 0.07 dB of nominal. The calculator prints the nearest value from both series under each result.

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Attenuator pad calculator

An attenuator pad is a resistor network that lowers signal level by a fixed amount while keeping both ports matched to the system impedance. This calculator gives the three resistor values for a pi or tee pad at any attenuation and impedance, and how much power each resistor has to burn.

Attenuation

Improves the return loss of whatever sits behind it by 20.0 dB.

System impedance

Input power

Topology

A pi puts a shunt resistor at each port, a tee puts a series resistor there. The attenuation is the same.

Shunt resistors R1 = R3

96.25 Ω

E24 100 Ω · E96 95.3 Ω

Series resistor R2

71.15 Ω

E24 68 Ω · E96 71.5 Ω

Power to the load

0.1 W

= 20.0 dBm

Dissipated as heat

0.9 W

90% of the input power

Hottest resistor

0.519 W

R1 · 52% of the input power

Same pad as a tee

25.97 / 35.14 Ω

series / shunt

The circuit runs from the source on the left, through the pad, into the matched load on the right, with the ground rail underneath. Green is the resistor value, amber the power it burns, and the bar beside it that power as a share of the input. Raise the attenuation and the heat visibly moves to the resistor nearest the source; switch topology and the same pad is redrawn with the resistors swapped between the signal path and ground.

Ideal resistors, both ports terminated in the system impedance. Real pads drift above a few hundred megahertz, where lead inductance and stray capacitance count as much as the resistance, and the power split assumes a matched source and load.

How it works

01
Turn the dB into a voltage ratio

Everything follows from K = 10^{A / 20}, the ratio of input to output voltage. A 10 dB pad has K = 3.162, so it divides the voltage by 3.162 and the power by 10. Only K depends on the attenuation you asked for, the rest is Ohm’s law.
02
Pick pi or tee

Both topologies give the same attenuation, the same match and the same heat. Only the resistor values differ. A 3 dB tee needs 8.55 Ω in series and a 3 dB pi needs 292 Ω to ground, and at 20 dB it is the pi that needs the large series value. Take the version whose three values sit closest to the standard values you can buy.
03
Solve the three resistors

The pi has a shunt resistor at each port and a series resistor between them. The tee is the mirror image, with a series resistor at each port and one shunt to ground. The two outer resistors are always equal, and that symmetry is what puts both ports on Z₀.
04
Size the resistors for the heat

Only 10^{−A / 10} of the power reaches the load, the rest turns into heat, and the three resistors do not share it equally. Above about 6 dB the resistor closest to the source runs hottest. It takes 52% of the input power at 10 dB and 82% at 20 dB. Rate that one for the full input power and allow for derating with temperature.

Formulas

Voltage ratio

K = 10^{A / 20}

A — attenuation, dB
K — voltage ratio (power ratio K²)
Z₀ — system impedance, Ω

Pi pad resistors

R₁ = R₃ = Z₀ · K + 1K − 1 , R₂ = Z₀ · K² − 12 · K

R₁, R₃ — shunt resistors, one at each port, Ω
R₂ — series resistor between them, Ω

Tee pad resistors

R₁ = R₃ = Z₀ · K − 1K + 1 , R₂ = 2 · Z₀ · KK² − 1

R₁, R₃ — series resistors, one at each port, Ω
R₂ — shunt resistor to ground, Ω

Worked example

10 dB pad in a 50 Ω system, 1 W in

K = 10^{10 / 20} = 3.1623
pi: R₁ = R₃ = 50 · 4.16232.1623 = 96.25 Ω, R₂ = 50 · 96.3246 = 71.15 Ω
tee: R₁ = R₃ = 50 · 2.16234.1623 = 25.97 Ω, R₂ = 2 · 50 · 3.16239 = 35.14 Ω
at the load: 1 W · 10^{−10 / 10} = 0.1 W, so 0.9 W becomes heat
split (either topology): R₁ 0.519 W, R₂ 0.329 W, R₃ 0.052 W

FAQ

Should I use a pi pad or a tee pad?: Either works. Attenuation, match and dissipation are identical, so only the resistor values decide. Take the topology whose three values land closest to parts you can actually buy. The calculator prints the nearest E24 (5%) and E96 (1%) value under every result. A 3 dB tee asks for 8.55 Ω in series, a 3 dB pi for 292 Ω to ground, and at 20 dB it is the pi that carries the big value, 248 Ω in series. On a thin-film chip the criterion is geometry rather than stock. There the resistor is a printed strip, and its value is the sheet resistance times how many squares long the strip is, so 292 Ω on 10 Ω-per-square film needs a 29-square meander that starts to behave like a small inductor.
Why does an attenuator improve VSWR?: Because the reflected wave crosses the pad twice. A load at 3.0 : 1 sends a quarter of the power back. Behind a 3 dB pad that reflection is attenuated on the way in and again on the way out, so the return loss measures 2 × 3 = 6 dB better and the meter reads about 1.67 : 1. Nothing about the load changed. The pad only keeps the mismatch away from the stage in front of it, and it costs its own attenuation to do so.
What power rating do the resistors need?: Size for the hottest one, which above about 6 dB is the resistor closest to the source. With 1 W in, it takes 0.52 W in a 10 dB pad, 0.82 W at 20 dB and 0.94 W at 30 dB. Ratings also fall with temperature. Mini-Circuits specifies its 2 W coaxial 10 dB pad at 25 °C ambient and derates it linearly to 0.5 W at 100 °C. Use non-inductive thin-film or chip resistors, never wirewound.
Can I use a 50 Ω pad in a 75 Ω system?: Not as a matched pad. The resistor values scale with Z₀, so a 50 Ω 10 dB pad in a 75 Ω line attenuates 10.3 dB and shows about 1.44 : 1 at each port. Switch the calculator to 75 Ω for the right values. Matching 75 Ω to 50 Ω is a different job. Purely resistive, it costs at least 5.72 dB (43.3 Ω in series, 86.6 Ω across the 50 Ω side), which is why video gear reaches for a transformer when the loss matters.
How much error do standard resistor values add?: Less than you would fear, because a series value that runs high partly cancels a shunt value that runs low. Rounded to E24 parts (100 Ω and 68 Ω), the 10 dB pi pad gives 9.63 dB with a 1.01 : 1 input match. With the E96 pair (95.3 Ω and 71.5 Ω) the attenuation lands within 0.07 dB of nominal. The calculator prints the nearest value from both series under each result.

A pad sets the level at the connector. What the signal does after the antenna is a terrain question. Waveshed draws line-of-sight and RF coverage over real elevation data, free in your browser.

See the path on real terrain →

Related guides

Plan radio coverage (RF / ITM) — coverage starts from the power that actually leaves the antenna, after pads, connectors and cable
LOS vs RF (FSPL & ITM) — when a few dB of level decide the link, and when geometry alone does

Sources & further reading

Mini-Circuits BW-S10W2+ — 10 dB fixed coaxial attenuator datasheet ↗ — a real 50 Ω 10 dB pad: 2 W at 25 °C, derated linearly to 0.5 W at 100 °C
Microwaves101 — Attenuators ↗ — the topologies and the closed-form resistor equations used here
Microwaves101 — Attenuator Dissipation ↗ — how the three resistors share the heat, and why the first one runs hottest
Attenuator (electronics) — Wikipedia ↗ — symmetric T-pad and pi-pad derivations, plus the minimum-loss matching pad

Attenuator pad calculator

How it works

Turn the dB into a voltage ratio

Pick pi or tee

Solve the three resistors

Size the resistors for the heat

Formulas

Worked example

FAQ

Related guides

Sources & further reading

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