Attenuator Calculator: T-Pad and Pi-Pad Resistor Values

A 10 dB attenuator in a 50 Ω system is a T-pad of two 25.97 Ω series resistors with a 35.14 Ω shunt resistor, or a Pi-pad of two 96.25 Ω shunt resistors with a 71.15 Ω series resistor.

For practitioners

dB
Ω
W

T-pad (two series arms, one shunt leg)

Series resistors R1 (each)
25.97 Ω
Shunt resistor R2
35.14 Ω
Voltage ratio K
3.162
Heat in the input series arm
0.5195 W
Heat in the shunt leg
0.3286 W
Heat in the output series arm
0.05195 W

Pi-pad (two shunt legs, one series arm)

Shunt resistors R1 (each)
96.25 Ω
Series resistor R2
71.15 Ω

Equal source and load impedance Z0, ideal resistors. Pick the nearest standard values and verify the match by measurement or simulation. Resistors must also survive the heat shown, and above a few hundred MHz parasitics matter.

How to use the calculator

Enter the attenuation in dB, the system impedance (50 Ω for most RF work) and the input power you expect. The calculator gives the resistor values of both pad types and shows the heat dissipated in the three resistors of the T-pad.

What a pad does

A resistive attenuator (a "pad") reduces a signal level by a fixed amount while keeping the same impedance at both ends. Typical uses are setting a signal level, protecting a sensitive input, improving the match of a source or a load, and isolating two stages so they do not interact.

The pad is built from three resistors in a T or a Pi shape. Let K be the voltage ratio for the attenuation A in dB:

K=10A/20K = 10^{A/20}

For a T-pad between equal impedances Z0:

Rseries=Z0 K−1K+1Rshunt=2 Z0 KK2−1R_{\text{series}} = Z_0\,\frac{K-1}{K+1} \qquad R_{\text{shunt}} = \frac{2\,Z_0\,K}{K^2-1}

For a Pi-pad:

Rshunt=Z0 K+1K−1Rseries=Z0 K2−12KR_{\text{shunt}} = Z_0\,\frac{K+1}{K-1} \qquad R_{\text{series}} = Z_0\,\frac{K^2-1}{2K}
T-pad: 10 dB attenuatorIN50 ohmOUT50 ohm25.9725.9735.14Pi-pad: 10 dB attenuatorIN50 ohmOUT50 ohm71.1596.2596.25The signal enters at IN and leaves at OUT with 10 dB less power;the resistors turn the rest into heat. Values in ohm, 50 ohm system.
T-pad and Pi-pad for 10 dB in a 50 ohm system.

Reference table at 50 Ω

Attenuation (dB)T-pad series (Ω)T-pad shunt (Ω)Pi-pad shunt (Ω)Pi-pad series (Ω)K
12.875433.3869.55.7691.122
38.55141.9292.417.611.413
616.6166.93150.537.351.995
1025.9735.1496.2571.153.162
2040.9110.161.11247.510

For other impedances scale every resistor value by Z0 / 50. For example, a 10 dB T-pad at 75 Ω uses 38.96 Ω series and 52.71 Ω shunt resistors.

Worked example: a 10 dB pad at 1 W

At 50 Ω, 10 dB means K = 3.162. A T-pad needs two 25.97 Ω series resistors and one 35.14 Ω shunt resistor. With 1 W at the input, the voltage is √(1 × 50) = 7.07 V and the current is 141 mA. The input series resistor dissipates I²R = 0.52 W, the shunt resistor about 0.33 W and the output series resistor 0.052 W. Together that is 0.9 W: a 10 dB pad turns 90% of the input power into heat and passes 0.1 W to the load. The calculator shows the same figures for any input power.

This is why pad resistors have to be rated well above the signal power, and why the first resistor gets the hottest.

1 W in10 dBattenuator0.1 W out0.9 W becomes heat in the resistorsThe thickness of each band is its share of the power.
Where the power goes in a 10 dB pad: 0.1 W passes, 0.9 W becomes heat.

Using an attenuator to improve a match

An attenuator placed in front of a mismatched device (the mismatch loss it hides is still there) makes the combination look better to the source. The wave that reflects from the device crosses the pad twice, so its level at the input drops by twice the attenuation. A 6 dB attenuator in front of a load with 6 dB return loss gives roughly 18 dB at the input. A 10 dB pad improves the return loss by up to 20 dB. The improvement is limited by the pad's own accuracy and by its parasitics, which become important at higher frequencies.

The same effect works against you when you measure through a pad or a cable: the return loss you read at the input is better than the load's real return loss. Calibrate at the load or de-embed the pad.

Choosing real resistors

  • Standard values. The exact values in the table do not exist as standard parts. Use the nearest E96 (1%) values, or combine two resistors, then verify the result with a measurement or a simulation. Small errors reduce both the match and the attenuation slightly.
  • Power rating. Choose resistors rated at least two to three times the dissipation shown, because heat also raises the temperature and shifts the resistance.
  • Frequency limits. Chip resistors have parasitic capacitance and inductance. Beyond a few hundred MHz the pad's layout and part size start to matter, and above a few GHz purpose-built RF attenuators are a better choice.
  • Layout. Keep the three resistors close, use a solid ground for the shunt leg and avoid long traces that add inductance.
  • Accuracy at large attenuation. For high attenuation values, a single pad is hard to build accurately because of leakage around the pad. Use cascaded pads or a commercial attenuator.

Limits of this calculator

The pad calculated here is the simple matched pad between two equal impedances. It assumes ideal resistors and real, equal source and load impedances. It does not design a minimum-loss pad between different impedances, a variable attenuator, or a pad for a reactive load.

In plain words · for students and beginners

Attenuators in plain words

What a pad is

An attenuator, or "pad", is a volume knob made of three resistors, fixed at one setting. It lowers a signal by a set number of decibels. The part of the signal that does not get through is not lost in a mysterious way: the resistors turn it into heat.

Why the resistor values are special

A random set of resistors would also reflect part of the signal. The values from the calculator are chosen so that the pad looks like 50 ohm (or whatever your system uses) from both ends. The circuits on either side then behave as if the pad were not there, except that the signal is weaker.

Why you would use one

  • To lower a level so that a sensitive input is not overloaded.
  • To set a known level for a measurement.
  • To make a poor load look better. An echo from the load has to pass the pad twice, so a 10 dB pad makes the echo 20 dB quieter.

A worked example

A 10 dB pad at 50 ohm takes 1 W in and passes 0.1 W. The other 0.9 W becomes heat, mostly in the first resistor, so the resistors must be rated for it with a good margin.

T-pad or Pi-pad

They do the same job. The letters describe the way the three resistors are arranged: a T has two resistors in the line and one to ground, a Pi has one in the line and two to ground. Choose the one whose resistor values are easier to get and place.

Mistakes beginners make

  • Forgetting the heat. A pad that is fine at 1 mW can burn at 1 W.
  • Expecting a pad built from ordinary chip resistors to be perfect at several GHz. At high frequency the layout of the parts changes the result.
  • Measuring through a pad and believing that the device behind it has a good match. The pad hides the real reflection.

Remember

  • The pad turns the unwanted power into heat.
  • The match comes from the exact resistor values, so use the nearest standard values and check.
  • An attenuator in front of a poor load improves the input match by about twice its attenuation.

Guides

Glossary

Calculators

The dBm to watts converter turns the levels before and after a pad into power and voltage, and the VSWR and return loss calculator shows what the improved match means in VSWR.

Frequently asked questions

What are the resistor values for a 10 dB attenuator at 50 ohms?

A T-pad uses two series resistors of 25.97 Ω and one shunt resistor of 35.14 Ω. A Pi-pad uses two shunt resistors of 96.25 Ω and one series resistor of 71.15 Ω. Use the nearest standard values, then check the result.

T-pad or Pi-pad, which is better?

Both give the same attenuation and the same match. In practice the choice comes down to which set of resistor values is easier to source and lay out, and to how much heat each resistor has to handle.

Does an attenuator improve the match of a poor load?

Yes. A pad of A dB in front of a mismatched load improves the input return loss by about 2·A dB, because the reflected wave crosses the pad twice. A 10 dB pad in front of a 6 dB return loss load gives roughly 26 dB at its input, limited by the pad's own accuracy.

How much power can the resistors handle?

It depends on the pad and the attenuation. In a T-pad the input series resistor and the shunt resistor carry most of the heat. The calculator shows the dissipation for the input power you enter. Choose resistors rated with a good margin above these values.

References and further reading