Attenuator pad calculator: pi, T, bridged-T and minimum-loss pads

The resistor values for a resistive attenuator matched to 50 Ω, 75 Ω or any impedance: pi, T or bridged-T, for any attenuation in decibels. Or run it backwards, from the resistors of a pad you have to the impedance it matches and the attenuation it gives; or match two different impedances, such as a 50 to 75 ohm pad, with the minimum-loss pad. Every pad is solved as a resistor network, so the page also shows what E24 or E96 values really give and how much power each resistor dissipates.

Fig 1 — 10 dB pi pad in 50 Ω: R_sh (input) 96.25 Ω, R_se 71.15 Ω, R_sh (output) 96.25 Ω, between 50 Ω and 50 Ω. Attenuation 10.00 dB; the source sees 50 Ω, a match. With 10 dBm available, the pad dissipates 9.00 mW and the load receives 1.00 mW. Bars: the power in each resistor, from a nodal solution of the network.

Solve for

Pad type

Attenuation A (dB)

Impedance Z0 (Ω)

Round to stocked values

Available input power P_in (dBm)

Voltage ratio K = V_in/V_out = 10^(A/20) · power ratio K²: 3.162 · 10.00
Shunt resistors R_sh, one across each port: 96.25 Ω
Series resistor R_se, between them: 71.15 Ω
Power in each resistor at 10 dBm available: R_sh (input) · R_se · R_sh (output): 5.19 mW · 3.29 mW · 519 µW
Dissipated in the pad · delivered to the load: 9.00 mW · 1.00 mW (0.0 dBm)

How this is calculated

Standard: Skyworks 200312E, Design With PIN Diodes (2021), p.15, Equations 37–39; Yeh et al., J. Appl. Phys. 121, 224501 (2017), p.2, Equations 1–2; Maxim AN972, CATV Minimum Loss Pad for 75 Ω Measurements (2002), Equations 1, 2, 6, 18 and 22

$K = \frac{V_{in}}{V_{out}} = 10^{A/20}, \qquad K^2 = 10^{A/10}$: Yeh et al.: "K = Vin/Vout is the ratio of the input voltage to the output voltage", and K² the "desired amount of power attenuation". A 10 dB pad has K² = 10 and K = √10.
$A = 20\log\!\left(\frac{R_{S1} + Z_0}{R_{S1} - Z_0}\right), \qquad R_{S3} = \frac{2R_{S1}Z_0^{2}}{R_{S1}^{2} - Z_0^{2}}, \qquad R_{S1} = R_{S2}$: Skyworks Equations 38 and 39, the pi pad as printed: R_S1 and R_S2 are the shunts, R_S3 the series resistor. "Note that the minimum value for RS1 and RS2 is 50 Ω" in a 50 Ω pad. 10 dB in 50 Ω: shunt 96.25 Ω, series 71.15 Ω.
$R_{sh} = Z_0\,\frac{K + 1}{K - 1}, \qquad R_{se} = Z_0\,\frac{K^2 - 1}{2K}$: The pi pad for a wanted attenuation: Equation 38 solved for the shunt, and Equation 39 with it substituted. Derived here, and tested against the equations as printed.
$A = 20\log\!\left(1 + \frac{Z_0}{R_{S1}}\right), \qquad Z_0^{2} = R_{S1} \times R_{S2}$: Skyworks Equation 37, the bridged-T as printed: series arms of Z0, R_S1 the shunt from their junction to ground, R_S2 the bridge. "The relationship between the forward resistance of the two diodes ensures maintenance of a matched circuit at all attenuation values." Solved for a wanted attenuation (derived): R_sh = Z0/(K − 1), R_br = Z0(K − 1).
$R_1 = R_2 = Z_0\,\frac{K - 1}{K + 1}, \qquad R_3 = R_4 = Z_0\,\frac{4K}{K^2 - 1}, \qquad R_{sh} = R_3 \parallel R_4 = Z_0\,\frac{2K}{K^2 - 1}$: Yeh et al. Equations 1 and 2, the T pad. Their Fig. 1(a) draws the shunt as two equal resistors, R3 and R4, in parallel; Equation 2 is each of them, and a single shunt is their parallel value (derived). Their example: "For a 10 dB attenuator sub-stage (K² = 10) connected to a characteristic impedance of Z0 = 50 Ω, one finds R1 = R2 = 26.0 Ω and R3 = R4 = 70.3 Ω." The calculator gives 25.97 Ω, 70.27 Ω each, and 35.14 Ω as one resistor.
$R_1 = R_S - R_2 \parallel R_L, \qquad R_2 = \sqrt{\frac{R_S R_L^{2}}{R_S - R_L}}, \qquad R_S > R_L$: Maxim AN972 Equations 1 and 2, the minimum-loss pad: R1 in series on the higher-impedance side, R2 in shunt across the lower. With R_S below R_L the calculator mirrors the pad. The 75 Ω to 50 Ω example: R2 = 86.6, R1 = 43.3; the calculator gives 86.6 Ω and 43.3 Ω.
$\mathrm{Loss_{power}} = 10\log\!\left(\frac{R_S}{R_L}\right) + 20\log\!\left(\frac{R_2 \parallel R_L}{R_2 \parallel R_L + R_1}\right), \qquad \mathrm{Loss_{voltage}} = 20\log\!\left(\frac{R_2 \parallel R_L}{R_1 + R_2 \parallel R_L}\right) \;\text{or}\; 20\log\!\left(\frac{R_S}{R_1 + R_S}\right)$: Maxim Equation 18, the power loss, the same in both directions; Equation 6, the voltage loss from the higher impedance to the lower; and Equation 22, from the lower to the higher. Maxim prints them as negative decibels: −5.72 dB, −7.48 dB and −3.96 dB for 75 Ω and 50 Ω. The calculator shows them as positive losses.
$A_{min} = 20\log\!\left(\sqrt{r} + \sqrt{r - 1}\right), \qquad r = \frac{R_{high}}{R_{low}}$: The minimum-loss pad's attenuation in closed form: Equation 18 with Equations 1 and 2 substituted. Derived here, not in Maxim's note; the tests check that it equals Equation 18.
$A = 10\log\frac{P_{avail}}{P_{load}}, \qquad \Gamma = \frac{Z_{in} - R_S}{Z_{in} + R_S}, \qquad RL = -20\log\lvert\Gamma\rvert, \qquad P_R = \frac{(V_a - V_b)^2}{R}$: Derived: every pad the page shows is solved as a resistor network between R_S and R_L by nodal analysis. The attenuation is the power available from the source over the power in the load, which for a matched pad is 20 log K and for the minimum-loss pad is Maxim's power loss ("the ratio of power delivered to power available"). The power in each resistor comes from its node voltages, for the input power entered.
$Z_0 = \sqrt{Z_{oc}\,Z_{sc}}, \qquad K^2 = \frac{1 + t}{1 - t}, \qquad t = \sqrt{\frac{Z_{sc}}{Z_{oc}}}$: Derived: resistors to attenuation. For a symmetric pad, Z_oc and Z_sc are the input resistance with the far port open and shorted; Z0 is the impedance the pad matches at both ports and A = 10 log K² the attenuation between them.

Assumptions

Ideal resistors: pure resistance, no parasitic inductance or capacitance, no connection length. The results are the DC and low-frequency answer; at RF the resistor size and layout set how far the real pad departs from them (Yeh et al., p.2).
Real, resistive source and load impedances. The pi, T and bridged-T are symmetric, designed for the same impedance at both ports; the minimum-loss pad matches two different ones.
The input power is the power available from the source: what it would deliver into a matched load, the figure a signal generator is set to. The power in each resistor is computed for that.
Resistors in resistors-to-attenuation mode are symmetric: the two shunts of a pi, the two series resistors of a T and the two arms of a bridged-T are equal.
Rounding to E24 or E96 takes the nearest stocked value to each resistor on its own; a different combination can sometimes give a better match.

What a matched attenuator pad does

A resistive attenuator, or pad, is a small network of resistors that reduces a signal by a fixed number of decibels while presenting the right impedance at both ends. Put a 10 dB, 50 Ω pad between a 50 Ω source and a 50 Ω load and three things are true at once: the load receives a tenth of the power the source had available, the source still sees 50 Ω, and the load, looking back, still sees 50 Ω. That last pair is what separates a pad from a plain voltage divider. A divider drops the voltage but changes the impedance each side sees; a matched pad attenuates without reflecting anything back.

That is why pads are used as much for their match as for their loss. A reflection from a mismatched load has to pass through the pad twice, so the source sees it attenuated by twice the pad's loss: a 75 Ω load on a 50 Ω source has a return loss of 14.0 dB, and with a 10 dB, 50 Ω pad in front of it the source sees 34.0 dB. A pad also lowers a signal that is too strong for the input after it, and sets a known, broadband loss in a measurement path. Three topologies give the same result between equal impedances: the pi (a shunt resistor at each port and a series resistor between them), the T (a series resistor at each port and a shunt from the junction), and the bridged-T (two series arms equal to Z0, a bridge across them and a shunt from their junction). A fourth, the minimum-loss pad, matches two different impedances, such as 75 Ω to 50 Ω, with the least attenuation a matched resistive pad can have.

Every design starts from the voltage ratio K = V_in/V_out = 10^A/20, the definition Yeh et al. use, and K² is the power ratio. The pi and bridged-T equations are Skyworks' (Design With PIN Diodes, p.15), and the T equations Yeh et al.'s (J. Appl. Phys. 2017, p.2); the reference note below typesets both, with the inversions the calculator uses. The calculator does not trust the equations alone: it solves each pad as a resistor network between its terminations, and reports the attenuation, input impedance and return loss that network actually gives. That is what lets it tell you what a pad built from E24 values, a pad placed in the wrong system, or a pad whose resistors have drifted really does.

Pi and T attenuator resistor values, 1 to 20 dB in 50 Ω

Each row is a matched pad for 50 Ω at both ports. The pi's two shunt resistors are equal, and so are the T's two series resistors; the T shunt is a single resistor. The bridged-T's series arms are 50 Ω at every attenuation, so only its bridge and shunt are listed.

A	Pi shunt	Pi series	T series	T shunt	Bridged-T bridge	Bridged-T shunt
1 dB	869.5 Ω	5.77 Ω	2.88 Ω	433.3 Ω	6.1 Ω	409.8 Ω
2 dB	436.2 Ω	11.61 Ω	5.73 Ω	215.2 Ω	12.95 Ω	193.1 Ω
3 dB	292.4 Ω	17.61 Ω	8.55 Ω	141.9 Ω	20.63 Ω	121.2 Ω
4 dB	221 Ω	23.85 Ω	11.31 Ω	104.8 Ω	29.24 Ω	85.49 Ω
5 dB	178.5 Ω	30.4 Ω	14.01 Ω	82.24 Ω	38.91 Ω	64.24 Ω
6 dB	150.5 Ω	37.35 Ω	16.61 Ω	66.93 Ω	49.76 Ω	50.24 Ω
7 dB	130.7 Ω	44.8 Ω	19.12 Ω	55.8 Ω	61.94 Ω	40.36 Ω
8 dB	116.1 Ω	52.84 Ω	21.53 Ω	47.31 Ω	75.59 Ω	33.07 Ω
9 dB	105 Ω	61.59 Ω	23.81 Ω	40.59 Ω	90.92 Ω	27.5 Ω
10 dB	96.25 Ω	71.15 Ω	25.97 Ω	35.14 Ω	108.1 Ω	23.12 Ω
11 dB	89.24 Ω	81.66 Ω	28.01 Ω	30.62 Ω	127.4 Ω	19.62 Ω
12 dB	83.54 Ω	93.25 Ω	29.92 Ω	26.81 Ω	149.1 Ω	16.77 Ω
13 dB	78.84 Ω	106.1 Ω	31.71 Ω	23.57 Ω	173.3 Ω	14.42 Ω
14 dB	74.93 Ω	120.3 Ω	33.37 Ω	20.78 Ω	200.6 Ω	12.46 Ω
15 dB	71.63 Ω	136.1 Ω	34.9 Ω	18.36 Ω	231.2 Ω	10.81 Ω
16 dB	68.83 Ω	153.8 Ω	36.32 Ω	16.26 Ω	265.5 Ω	9.42 Ω
17 dB	66.45 Ω	173.5 Ω	37.62 Ω	14.41 Ω	304 Ω	8.22 Ω
18 dB	64.4 Ω	195.4 Ω	38.82 Ω	12.79 Ω	347.2 Ω	7.2 Ω
19 dB	62.64 Ω	220 Ω	39.91 Ω	11.36 Ω	395.6 Ω	6.32 Ω
20 dB	61.11 Ω	247.5 Ω	40.91 Ω	10.1 Ω	450 Ω	5.56 Ω

Two patterns are worth reading off it. The pi shunt never falls below 50 Ω, and Skyworks says so of its own curve: "the minimum value for RS1 and RS2 is 50 Ω". At high attenuation it approaches 50 Ω from above while the pi series resistor grows without limit, and the T mirrors that: its series resistors approach 50 Ω from below while its shunt heads for zero. And in the bridged-T the bridge times the shunt is always 50² = 2500 Ω², the matching condition Skyworks prints as Z₀² = R_S1 × R_S2.

Attenuator resistor values in 75 Ω

The same pads for 75 Ω. Every value is 1.5 times the 50 Ω one, because every design equation is Z0 times a function of K alone: a pad scales with its impedance.

A	Pi shunt	Pi series	T series	T shunt	Bridged-T bridge	Bridged-T shunt
1 dB	1.304 kΩ	8.65 Ω	4.31 Ω	650 Ω	9.15 Ω	614.7 Ω
2 dB	654.3 Ω	17.42 Ω	8.6 Ω	322.9 Ω	19.42 Ω	289.7 Ω
3 dB	438.6 Ω	26.42 Ω	12.82 Ω	212.9 Ω	30.94 Ω	181.8 Ω
4 dB	331.5 Ω	35.77 Ω	16.97 Ω	157.2 Ω	43.87 Ω	128.2 Ω
5 dB	267.7 Ω	45.6 Ω	21.01 Ω	123.4 Ω	58.37 Ω	96.37 Ω
6 dB	225.7 Ω	56.03 Ω	24.92 Ω	100.4 Ω	74.64 Ω	75.36 Ω
7 dB	196.1 Ω	67.2 Ω	28.69 Ω	83.7 Ω	92.9 Ω	60.55 Ω
8 dB	174.2 Ω	79.27 Ω	32.29 Ω	70.96 Ω	113.4 Ω	49.61 Ω
9 dB	157.5 Ω	92.38 Ω	35.72 Ω	60.89 Ω	136.4 Ω	41.25 Ω
10 dB	144.4 Ω	106.7 Ω	38.96 Ω	52.7 Ω	162.2 Ω	34.69 Ω
11 dB	133.9 Ω	122.5 Ω	42.02 Ω	45.92 Ω	191.1 Ω	29.43 Ω
12 dB	125.3 Ω	139.9 Ω	44.89 Ω	40.22 Ω	223.6 Ω	25.16 Ω
13 dB	118.3 Ω	159.1 Ω	47.56 Ω	35.35 Ω	260 Ω	21.63 Ω
14 dB	112.4 Ω	180.5 Ω	50.05 Ω	31.17 Ω	300.9 Ω	18.69 Ω
15 dB	107.4 Ω	204.2 Ω	52.35 Ω	27.55 Ω	346.8 Ω	16.22 Ω
16 dB	103.3 Ω	230.7 Ω	54.48 Ω	24.39 Ω	398.2 Ω	14.13 Ω
17 dB	99.67 Ω	260.2 Ω	56.43 Ω	21.62 Ω	456 Ω	12.34 Ω
18 dB	96.6 Ω	293.2 Ω	58.23 Ω	19.19 Ω	520.7 Ω	10.8 Ω
19 dB	93.96 Ω	330 Ω	59.87 Ω	17.04 Ω	593.4 Ω	9.48 Ω
20 dB	91.67 Ω	371.2 Ω	61.36 Ω	15.15 Ω	675 Ω	8.33 Ω

Worked example: Yeh et al.'s 10 dB T pad in 50 Ω

Yeh, LeFebvre, Premaratne, Wellstood and Palmer built thin-film attenuators to reduce the thermal noise reaching superconducting qubits, and chose a T pad because, in their thermal simulations, it had "a larger cooling power than the Π-pad designs". Their Fig. 1(a) draws the T with its shunt as two equal resistors side by side, R3 and R4, and their equations give each of them. For the 10 dB stage: "For a 10 dB attenuator sub-stage (K² = 10) connected to a characteristic impedance of Z0 = 50 Ω, one finds R1 = R2 = 26.0 Ω and R3 = R4 = 70.3 Ω." The left column is the calculator's arithmetic; the right is what Yeh et al. print.

ratio     K = 10^(10/20) = √10                     = 3.1623   
Eq 1      R1 = R2 = 50 × (K − 1)/(K + 1)           = 25.97 Ω   Yeh: 26.0 Ω
Eq 2      R3 = R4 = 50 × 4K/(K² − 1)               = 70.27 Ω   Yeh: 70.3 Ω
shunt     R3 ∥ R4 = 2K × 50/(K² − 1)               = 35.14 Ω  
check     26.0 Ω, 70.3 Ω ∥ 70.3 Ω, solved into 50 Ω = 10.00 dB 
          input resistance of that network         = 50.03 Ω

The calculator agrees with both printed values to the digit shown. The single shunt resistor of an ordinary T is the pair in parallel, 35.14 Ω, which is the calculator's T shunt; the two-resistor form is listed beside it. Built from Yeh's rounded values and solved as a network, the pad gives 10.003 dB and presents 50.03 Ω to the source, so the rounding to one decimal place costs nothing measurable.

The paper also says where the heat goes: "the resistor R1 within each 10 dB cell dissipates the most power and has the largest temperature". The nodal solution agrees. Of the power entering the pad, R1 takes 52 %, the shunt 33 % and R2 5.2 %, and the 10 % left reaches the load. That split is the same for a pi at the same attenuation, with the input shunt in R1's place, and it is what the power row of the calculator and the bars in the figure show for any pad and input power.

Worked example: Maxim's 75 Ω to 50 Ω minimum-loss pad

"CATV systems are based on 75Ω characteristic impedance, but typical laboratory equipment has 50Ω characteristic impedance", as Maxim's AN972 opens, and the note matches them with an L-section of two resistors: R1 in series on the 75 Ω side and R2 in shunt across the 50 Ω side. Its Equations 1 and 2 give the values for any R_S > R_L, and its worked example runs them for 75 Ω and 50 Ω. The calculator's defaults in minimum-loss mode are this example.

Eq 2      R2 = √(75 × 50² / (75 − 50))             = 86.6 Ω    Maxim: 86.6
Eq 1      R1 = 75 − R2 ∥ 50                        = 43.3 Ω    Maxim: 43.3
          R2 ∥ 50                                  = 31.7 Ω    Maxim: 31.69
Eq 6      20 log(R2∥50 / (R1 + R2∥50))             = −7.48 dB  Maxim: −7.48
Eq 18     10 log(75/50) + Eq 6                     = −5.72 dB  Maxim: −5.72
Eq 22     20 log(75 / (R1 + 75))                   = −3.96 dB  Maxim: −3.96

Every figure matches Table 1 of the note. One intermediate does not quite: Maxim prints 86.6 Ω ∥ 50 Ω as 31.69 in its Equation 29, where the parallel is 31.698 Ω; the print is truncated rather than rounded, and it does not move the −5.72 dB result. Solved as a network, the pad presents 75 Ω to the 75 Ω side and 50 Ω to the 50 Ω side, which is Maxim's "The 75Ω termination sees a 75Ω equivalent resistance network. Similarly, the 50Ω termination sees a 50Ω equivalent resistance network."

The point of the note is its Table 1. The power loss is the same both ways, 5.72 dB from 75 Ω to 50 Ω and 5.72 dB back, but the voltage loss is not: 7.48 dB from 75 Ω to 50 Ω and 3.96 dB the other way. Maxim's rule: "Use voltage loss when measuring in dBmV, dBµV, or dBV, or for any voltage-related measurement. Use power loss when measuring in dBm or for any power measurement."

The minimum loss depends only on the ratio of the two impedances. With r = R_high/R_low it is 20 log(√r + √(r − 1)), which is Maxim's Equation 18 with Equations 1 and 2 substituted; the algebra is done here, not in the note, and the tests check the two agree. For other impedances against 50 Ω:

R_high	Series R1	Shunt R2	Power loss	Voltage loss, high → 50 Ω	Voltage loss, 50 Ω → high
60 Ω	24.49 Ω	122.5 Ω	3.77 dB	4.56 dB	2.97 dB
75 Ω	43.3 Ω	86.6 Ω	5.72 dB	7.48 dB	3.96 dB
100 Ω	70.71 Ω	70.71 Ω	7.66 dB	10.67 dB	4.65 dB
150 Ω	122.5 Ω	61.24 Ω	9.96 dB	14.73 dB	5.18 dB
200 Ω	173.2 Ω	57.74 Ω	11.44 dB	17.46 dB	5.42 dB
300 Ω	273.9 Ω	54.77 Ω	13.42 dB	21.20 dB	5.63 dB
600 Ω	574.5 Ω	52.22 Ω	16.63 dB	27.42 dB	5.83 dB

A matched pad between two different impedances cannot lose less than this, which is why it is a minimum-loss pad. A higher attenuation between unequal impedances needs a pad of three resistors; this calculator does not design that case.

Pi vs T vs bridged-T attenuator

Between equal impedances all three give exactly the same attenuation and match, so the choice comes down to the resistor values, where the power goes, and whether the pad has to be adjustable.

Resistor values. At low attenuation the pi needs large shunts and a small series resistor: 869.5 Ω and 5.77 Ω for 1 dB in 50 Ω. The T needs the reverse, 2.88 Ω in series and 433.3 Ω in shunt. At high attenuation the extremes move to the other element: a 40 dB pi has a 2.5 kΩ series resistor, a 40 dB T a 1 Ω shunt. A very small or very large resistor is where the connections and parasitics of a real part matter most against its resistance, so the usual choice is the topology whose values stay near Z0 at the attenuation needed, or two pads in cascade.

Where the power goes. The resistor at the input, the pi's input shunt or the T's input series resistor, takes the largest share, and at high attenuation it takes nearly all of it. The split is identical for a pi and a T at the same attenuation, as shares of the available input power:

A	Input resistor	Middle resistor	Output resistor	Whole pad
3 dB	17.1 %	24.2 %	8.57 %	49.9 %
6 dB	33.2 %	33.3 %	8.35 %	74.9 %
10 dB	51.9 %	32.9 %	5.19 %	90.0 %
20 dB	81.8 %	16.4 %	0.82 %	99.0 %
30 dB	93.9 %	5.9 %	0.09 %	99.9 %
40 dB	98.0 %	2.0 %	0.01 %	100.0 %

So a 30 dB pad for a 1 W input is not three resistors sharing a watt: one of them dissipates nearly all of it, and it has to be rated for that alone. The table is derived from the network, and the calculator gives the same breakdown in watts for the power you enter.

Adjustability. The bridged-T keeps its two series arms at Z0 whatever the attenuation, and only its bridge and shunt change, always with their product equal to Z0². Skyworks uses exactly that with two PIN diodes as the variable resistors: "The relationship between the forward resistance of the two diodes ensures maintenance of a matched circuit at all attenuation values." Its range is wide, a bridge of 6.1 Ω and shunt of 409.8 Ω at 1 dB, 4.95 kΩ and 0.505 Ω at 40 dB. The bridged-T also has a property that falls out of the network rather than out of any source: when it is matched, no current flows in its output arm. At 10 dB the input arm takes 47 % of the power, the bridge and the shunt 22 % each, and the output arm nothing; the figure shows it as 0 W. Change the load and the output arm starts to conduct.

Building the pad from E24 and E96 values

Exact values are rarely stocked. The calculator rounds each resistor to the nearest E24 or E96 value and solves the rounded pad as it is, which is more useful than the rounding errors themselves: a pad is judged by its attenuation and its match, and those are what the rounding moves. For 10 dB in 50 Ω:

Pad	E24 values	Attenuation	Return loss	E96 values	Attenuation	Return loss
Pi	100 Ω, 68 Ω, 100 Ω	9.63 dB	49.6 dB	95.3 Ω, 71.5 Ω, 95.3 Ω	10.07 dB	53.8 dB
T	27 Ω, 36 Ω, 27 Ω	10.07 dB	36.4 dB	26.1 Ω, 34.8 Ω, 26.1 Ω	10.07 dB	73.9 dB
Bridged-T	51 Ω, 110 Ω, 24 Ω, 51 Ω	9.94 dB	39.6 dB	49.9 Ω, 107 Ω, 23.2 Ω, 49.9 Ω	9.96 dB	58.2 dB

The return loss is the one to look at. Rounding moves the attenuation by a few tenths of a decibel at most here, but it moves the match, and a pad fitted to improve a match should not be the thing that spoils it. The E-series calculator finds two-resistor combinations when a single stocked value is too far off.

Where the resistor model stops being valid

Parasitics. Every equation here treats the resistors as pure resistances with zero-length connections. Yeh et al. hit the limit of that directly: "For a large resistor, its parasitic capacitance and inductance become important, and a lumped element model is not valid. Microwave simulations using ANSYS's high frequency software simulator (HFSS) confirmed that the response of the attenuator degraded at high frequencies when the physical size of the resistors got too large." Their target was "a flat microwave response (less than 3 dB change) up to 12 GHz in the simulations", and the resistor geometry was chosen for it. The calculator's numbers are the DC and low-frequency answer; above that, the layout decides.

Temperature and power. Resistors drift, and a pad drifts with them. Spectrum Control's note on Weinschel attenuators quotes the military standard: "Military Standard, MIL-A-3933 for fixed attenuators calls for a TCA of 0.0004dB/dB/°C. Over a 100° C ambient temperature change, a 30 dB attenuator would change by a maximum of 1.2 dB". The arithmetic, 30 × 100 × 0.0004, is 1.2 dB. The note adds that when series and shunt share the same temperature coefficient, "the attenuation will always increase at DC, independent of the temperature and the magnitude of the TCR", and that poor coefficients "significantly degrade the SWR, with little effect on the DC attenuation." The network solution bears both out. Scale every resistor of a 10 dB, 50 Ω pi by +1 % and the attenuation rises by 0.00019 dB; by −1 % it also rises, by 0.00020 dB. The attenuation sits at a minimum against a common drift, so it barely moves. The impedance moves with the resistors: the pad now matches 50.5 Ω, and in a 50 Ω system its return loss has fallen to 47.0 dB. At ±10 % the attenuation changes by only 0.018 dB and 0.022 dB, but the return loss is down to 27.4 dB and 26.5 dB. A pad whose resistors track each other keeps its loss; it is the match that suffers.

Real chip ratings. Vishay's CZA thick-film chip attenuator, an "Unbalanced π Type" pad in one package, shows what a small part is rated for. The CZA06S's "Rated dissipation at 70 °C" is 0.075 W (0.040 W for the smaller CZA04S), the frequency range is "DC to 3 GHz", and the VSWR is "1.2 max.", a return loss of 20.8 dB. The attenuation tolerance is ±0.3 dB or ±0.5 dB, tolerance codes L and H. A 10 dB pad dissipates nine tenths of its input, so 0.075 W is reached at 83.3 mW available, 19.2 dBm, before any derating above 70 °C. The datasheet is marked "End of Life June-2021", so the part itself is a reference point rather than a recommendation; a discrete pad built from chip resistors should be checked resistor by resistor against the power table above.

Common attenuator pad mistakes

Copying Yeh's Equation 2 as the T's single shunt. It gives each of two resistors in parallel. Built with one 70.27 Ω shunt, the 10 dB, 50 Ω T attenuates only 7.39 dB and presents 62.48 Ω to the source, a return loss of 19.1 dB; the pad it has become matches 65.77 Ω, not 50 Ω.
Using the power ratio as K. K is the voltage ratio, 10^A/20: 3.162 for 10 dB, not 10. Feeding 10 into the equations designs a 20 dB pad, a pi with 61.11 Ω shunts and a 247.5 Ω series resistor instead of 96.25 Ω and 71.15 Ω. The decibel calculator converts between the two.
Quoting the minimum-loss pad's power loss for a voltage measurement. A 75 Ω to 50 Ω pad loses 5.72 dB of power but 7.48 dB of voltage in that direction. A level in dBmV or dBµV needs the voltage loss for the direction the signal travels.
Using a 50 Ω pad in a 75 Ω system. A 10 dB, 50 Ω pi between 75 Ω ports gives 10.32 dB and presents 52.04 Ω to the source, a return loss of 14.9 dB. Pads scale with Z0: design it for the system it sits in.
Matching 75 Ω to 50 Ω with an ordinary 50 Ω pad. A 6 dB, 50 Ω pi between a 75 Ω source and a 50 Ω load loses 6.18 dB and still shows the source 50 Ω, a return loss of 14.0 dB. The minimum-loss pad loses 5.72 dB and matches both sides.
Rating the resistors for the pad's total dissipation divided by three. The input resistor takes the largest share, 51.9 % of the input at 10 dB and 98.0 % at 40 dB.