Attenuator Calculator (Pi and T Pads)

Resistor values for a Pi or T attenuator pad — for equal impedances or between two different ones — derived from the image-match conditions rather than copied from a table, together with the nearest E24 or E96 parts, the attenuation and return loss those real parts actually give, and the power each resistor has to dissipate.

Last updated 24 Sep 2026

attenuator pad

dB and Z → three resistors

TopologyBoth give identical performance. The Pi is usually easier on a PCB at RF because two of its three resistors go straight to ground.

Attenuation wanted (dB)

Impedances

Source impedance (Ω)

Load impedance (Ω)

Standard resistor series

Power into the pad (W)For the dissipation figures. In a matched pad the first resistor takes the biggest share — often more than half of everything you put in.

The pad you chose, with the nearest standard resistor values written on it and the source and load impedances at the two ends. In the Π pad R1 and R2 are the shunt arms and R3 the series arm between them; in the T pad R1 and R2 are the series arms and R3 the shunt to ground. The resistor carrying the most power turns amber — in a matched pad that is almost always the one nearest the source, and at high attenuation it takes most of the input on its own. No moving dots are drawn: the current here is RF, not a steady DC.

96.25ΩExample

a 10 dB Π pad in 50 Ω, built from E24 parts, with 1 W going in

Both pads, from the image-match conditions

K = 10^A/10 (power ratio) N = √K = 10^A/20 (voltage ratio)
T: R₁ = (Z₁(K+1) − 2√(KZ₁Z₂))/(K−1), R₂ likewise with Z₂, R₃ = 2√(KZ₁Z₂)/(K−1)
Π: R₃ = ½(K−1)√(Z₁Z₂/K), R₁ = Z₁(K−1)√Z₂ / ((K+1)√Z₂ − 2√(KZ₁))
equal impedances: T: R₁=R₂=Z(N−1)/(N+1), R₃=2NZ/(N²−1); Π: R₁=R₂=Z(N+1)/(N−1), R₃=Z(N²−1)/(2N)
real parts: A = 20 log₁₀(|A·Z₂ + B + C·Z₁Z₂ + D·Z₁| ÷ 2√(Z₁Z₂)) from the ABCD matrix
minimum loss: A_min = 20 log₁₀(√r + √(r−1)), r = Z_hi/Z_lo

K: the POWER ratio, 10^(A/10) — not the voltage ratio. Half the published pad formulas use one and half the other, which is the commonest way these equations get copied out wrong
Z1, Z2: source and load impedance. A pad is image-matched: it presents Z₁ to the source and Z₂ to the load at the same time
A min: the attenuation of the minimum-loss L pad. Between unequal impedances no matched resistive network can do better, and at exactly this value the T pad’s far series arm falls to zero and it becomes that L pad

Worked example

a 10 dB Π pad in 50 Ω, built from E24 parts, with 1 W going in
K = 10^(10/10) = 10 and N = √K = 3.1623
The two shunt arms are Z(N+1)/(N−1) = 50 × 4.1623 ÷ 2.1623 = 96.248 Ω, and the series arm is Z(N²−1)/(2N) = 71.151 Ω
The nearest E24 parts are 100 Ω and 68 Ω. Solving the network with those fitted gives 9.629 dB rather than 10 — 0.371 dB out — and an input impedance of 50.33 Ω, a return loss of 49.6 dB
With 1 W in, R1 takes 503.3 mW, R3 takes 333.3 mW and R2 takes 54.46 mW; 108.9 mW reaches the load. The hottest resistor is carrying 50.3% of everything you put in, which is why a pad is rated by its first resistor and not by its total
The T pad for the same job is 25.975 Ω in each series arm with 35.136 Ω to ground — identical performance, different parts

Pads people actually build, in 50 Ω

The last column is the share of the input power that lands in the Π pad’s first shunt resistor, which is the one that has to be rated. It rises towards 100% as the attenuation rises, because a big pad is essentially a load at the input with a small sample taken off the far end.
Attenuation	Π: R1 = R2	Π: R3	T: R1 = R2	T: R3	Power in the first resistor, per watt in
1 dB	869.55 Ω	5.77 Ω	2.88 Ω	433.34 Ω	5.8%
2 dB	436.21 Ω	11.61 Ω	5.73 Ω	215.24 Ω	11.5%
3 dB	292.40 Ω	17.61 Ω	8.55 Ω	141.93 Ω	17.1%
6 dB	150.48 Ω	37.35 Ω	16.61 Ω	66.93 Ω	33.2%
10 dB	96.25 Ω	71.15 Ω	25.97 Ω	35.14 Ω	51.9%
20 dB	61.11 Ω	247.50 Ω	40.91 Ω	10.10 Ω	81.8%
30 dB	53.27 Ω	789.78 Ω	46.93 Ω	3.17 Ω	93.9%

Minimum-loss pads between common impedances

An L pad — one series resistor and one shunt resistor — is the cheapest matched connection between two unequal impedances, and this is what it costs. The 75 Ω to 50 Ω case is the one you can buy: Mini-Circuits’ BMP-5075+ is specified at 5.7 ± 0.1 dB, which is this row. Nothing resistive does better; a transformer or a quarter-wave section does, over a narrower band.
From	To	Minimum loss	Series (high side)	Shunt (low side)
75 Ω	50 Ω	5.72 dB	43.3 Ω	86.6 Ω
93 Ω	50 Ω	7.20 dB	63.2 Ω	73.5 Ω
300 Ω	50 Ω	13.42 dB	273.9 Ω	54.8 Ω
600 Ω	50 Ω	16.63 dB	574.5 Ω	52.2 Ω
300 Ω	75 Ω	11.44 dB	259.8 Ω	86.6 Ω
600 Ω	75 Ω	14.77 dB	561.2 Ω	80.2 Ω

Three resistors, three conditions

An attenuator is not a volume control. Turning a signal down with a potential divider changes the impedance the source sees and the impedance the load sees, and at radio frequencies that produces reflections that matter more than the level change you wanted. A pad is the fix: a network that presents the right impedance at both ports while delivering exactly the loss asked for. Three resistors give exactly three degrees of freedom, which is why both the Π and the T have three and why the answer is unique.

Where the formulas come from. Write the three conditions — input impedance equals Z₁ with Z₂ on the far end, output impedance equals Z₂ with Z₁ on the near end, power ratio equals K = 10^A/10 — and solve. That is the image-impedance treatment, and it is equivalent to writing the network’s ABCD matrix and demanding the image impedances and the image attenuation come out right. The results are the expressions above. This page does not copy them from a table: they are re-derived in its build script and then checked by solving the actual resistor network node by node, so every pad on this page has been verified to present exactly Z₁ and exactly Z₂ and to lose exactly A decibels.

K is a power ratio. The single commonest error in published pad formulas is mixing 10^A/10 with 10^A/20. The unequal-impedance formulas above use K, the power ratio. The equal-impedance simplifications use N = √K, the voltage ratio. Both are printed so you can see which is which.

Real resistors. The ideal values do not exist. A 10 dB Π pad in 50 Ω wants 96.2 Ω and 71.2 Ω, and the nearest E24 parts are 100 Ω and 68 Ω. This page fits those real values back into the network and tells you what they actually do — the attenuation, the input impedance and the return loss at both ports. Usually the attenuation error is negligible and the return loss is the thing that suffers, because an attenuator is often bought precisely to improve the match seen by something badly behaved. A pad of A decibels improves a load’s return loss by 2A, so a 6 dB pad turns a 3:1 mismatch into something you can live with — but only if the pad’s own match is better than what you are hiding.

Power. In a matched pad the first resistor carries most of the heat. At 10 dB in a 50 Ω Π pad, the input shunt resistor alone dissipates 52% of everything you put in, and the load gets 10%. A pad sold as “10 dB, 2 W” is rated by that first resistor; it does not mean the three parts share 2 W. At higher attenuation it gets more lopsided still, and above about 20 dB the pad is essentially a matched load with a small tap.

Unequal impedances have a floor. Between 75 Ω and 50 Ω no matched resistive network can lose less than 5.72 dB. Push the attenuation below that and the design equations return a negative resistance, which is the arithmetic telling you the job is impossible. At exactly the minimum, the T pad’s far series arm falls to zero and what is left is an L pad — one series resistor and one shunt resistor — and that is the minimum-loss pad you can buy off the shelf. The reason is simple enough: a resistive network that matches a low impedance to a high one has to throw away the difference, and the wider the ratio the more it throws.

For what a mismatch costs before you decide to pad it out, the VSWR and return loss converter; for converting the levels themselves, the dBm to watts converter and the decibel calculator; and for the cable the pad usually sits in line with, the coax impedance calculator.

Frequently asked questions

What is the difference between a Pi pad and a T pad?

Nothing you can measure. Given the same impedances and the same attenuation they have identical two-port behaviour; they are duals of each other. The choice is practical: the Π has two of its three resistors going straight to ground, which suits a PCB layout and surface-mount parts, while the T needs only one ground connection. At low attenuation the T’s series arms get very small and the Π’s shunt arms get very large, so the T is usually kinder for 1–3 dB and the Π for 10 dB and up.

How do I calculate attenuator resistor values?

With K = 10^(A/10) and N = √K: a T pad in a single impedance Z has series arms Z(N−1)/(N+1) and a shunt arm 2NZ/(N²−1); a Π pad has shunt arms Z(N+1)/(N−1) and a series arm Z(N²−1)/(2N). For unequal impedances use the fuller expressions above, which reduce to these when Z₁ = Z₂.

Why can’t I make a 3 dB pad between 50 and 75 ohms?

Because the minimum-loss pad between those two impedances is 5.72 dB, and nothing resistive and matched at both ports can do better. Ask for less and the equations hand back a negative resistance. If you need the match without the loss, use a transformer or a quarter-wave transmission-line section instead — both give up bandwidth in exchange.

What power rating do the resistors need?

Rate them individually, not as a set. In a matched pad the resistor nearest the source takes the largest share, and at 10 dB in 50 Ω that is over half the input power in the Π topology. This page prints the dissipation in each resistor for the input power you state; derate generously, because thin-film chip resistors lose their value as well as their life when they run hot.

Do standard E24 resistors give the right attenuation?

Usually to well within a tenth of a decibel, which is finer than the resistors’ own tolerance. What suffers first is the match, not the level. This page fits the nearest standard values back into the network and reports the attenuation, the input and output impedance and both return losses, so you can see which of the two is limiting.

Does an attenuator improve VSWR?

Yes, by twice its attenuation in decibels, because the reflected wave passes through it twice. A 6 dB pad in front of a load with 6 dB of return loss presents 18 dB — provided the pad’s own return loss is better than that. Padding a badly matched load is the standard fix when the load cannot be improved and the loss can be afforded.

Related calculators

References

Steer MB. Microwave and RF Design III — Networks, 3rd ed. (NC State University / Engineering LibreTexts, 2020, ISBN 9781469656946), §5.5 Terminations and Attenuators: the T and Π design equations for unequal impedances in terms of K = P_in/P_out, their equal-impedance simplifications, and a worked 20 dB pad in 75 Ω giving R₁ = R₂ = 61.4 Ω and R₃ = 15.2 Ω — reproduced exactly by this page’s build script as 61.36 Ω and 15.15 Ω.
Mini-Circuits BMP-5075+ datasheet, 50 Ω to 75 Ω minimum loss pad, DC–2000 MHz: nominal attenuation 5.7 ± 0.1 dB. That is the minimum-loss figure this page computes as 5.719 dB, which is a useful confirmation that the floor is real and not a modelling artefact.
Pozar DM. Microwave Engineering, 4th ed., Wiley, 2011 — Chapter 4 for ABCD parameters of two-port networks and the conversion between ABCD and scattering parameters, which is where the transducer-loss expression used for the standard-value check comes from.
IEC 60063:2015, Preferred number series for resistors and capacitors — the E12, E24, E96 and E192 series this page rounds to.