The ebook FEEE - Fundamentals of Electrical Engineering and Electronics is based on material originally written by T.R. Kuphaldt and various co-authors. For more information please read the copyright pages.

T-Section Attenuator

The T and Π attenuators must be connected to a Z source and Z load impedance. The Z-(arrows) pointing away from the attenuator in the figure below indicate this. The Z-(arrows) pointing toward the attenuator indicates that the impedance seen looking into the attenuator with a load Z on the opposite end is Z, Z=50 Ω for our case. This impedance is a constant (50 Ω) with respect to attenuation-- impedance does not change when attenuation is changed.

The table in Figure below lists resistor values for the T and Π attenuators to match a 50 Ω source/ load, as is the usual requirement in radio frequency work.

Telephone utility and other audio work often requires matching to 600 Ω. Multiply all R values by the ratio (600/50) to correct for 600 Ω matching. Multiplying by 75/50 would convert table values to match a 75 Ω source and load.

Formulas for T-section attenuator resistors, given K, the voltage attenuation ratio, and ZI = ZO = 50 Ω.

The amount of attenuation is customarily specified in dB (decibels). Though, we need the voltage (or current) ratio K to find the resistor values from equations. See the dB/20 term in the power of 10 term for computing the voltage ratio K from dB, above.

The T (and below Π) configurations are most commonly used as they provide bidirectional matching. That is, the attenuator input and output may be swapped end for end and still match the source and load impedances while supplying the same attenuation.

Disconnecting the source and looking in to the right at VI, we need to see a series parallel combination of RO, RI, RO, and Z looking like an equivalent resistance of Z, the source/load impedance: (Z is connected to the output.)

Z=RO + (R2 ||(R1 + Z))

For example, substitute the 10 dB values from the 50 Ω attenuator table for R1 and R2 as shown in Figure below.

Z = 25.97 + (35.14 ||(25.97 + 50))
Z = 25.97 + (35.14 || 75.97 )
Z = 25.97 + 24.03 = 50

This shows us that we see 50 Ω looking right into the example attenuator (Figure below) with a 50 Ω load.

Replacing the source generator, disconnecting load Z at VO, and looking in to the left, should give us the same equation as above for the impedance at VO, due to symmetry. Moreover, the three resistors must be values which supply the required attenuation from input to output. This is accomplished by the equations for R1 and R2 above as applied to the T-attenuator below.

10 dB T-section attenuator for insertion between a 50 Ω source and load.

Last Update: 2010-11-19