Circuit Theory II · 120 h · Topic 4 of 8

Attenuators and compensators

A 10 dB attenuator for 50 Ω needs 25.97 and 35.14 Ω. With the nearest commercial resistors it gives 10.07 dB: the rounding hardly matters, but you have to calculate it.

T attenuator π attenuator E24 and E96 series Lead network Equalizer

01Lowering the level without causing a mismatch

An attenuator reduces the amplitude of a signal by a fixed number of decibels, and also presents the correct impedance on both sides. A simple resistive divider attenuates, but it changes the impedance seen by the generator and the load: in a 50 Ω system that produces reflections, as seen in transmission lines, and the real attenuation is no longer the calculated one. They are used to protect the input of an instrument, to lower the level of a generator below its minimum, to improve the match between two pieces of equipment and to measure gains by comparison.

02T and π attenuators

A symmetric attenuator designed for an impedance \( Z_0 \) has an image impedance of \( Z_0 \) on both sides: terminated in \( Z_0 \), it presents \( Z_0 \) at the input. With \( K = 10^{A/20} \), the voltage ratio corresponding to an attenuation of \( A \) decibels:

\[ R_{serie} = Z_0\,\frac{K - 1}{K + 1} \] \[ R_{der} = Z_0\,\frac{2K}{K^2 - 1} \] For the T attenuator: two equal series resistors and one shunt resistor in the middle.

\[ R_{der} = Z_0\,\frac{K + 1}{K - 1} \] \[ R_{serie} = Z_0\,\frac{K^2 - 1}{2K} \] For the π attenuator: two equal shunt resistors at the ends and one series resistor between them.

The T is convenient for small attenuations, where the shunt resistor is large; the π, for large ones. Both are symmetric two-ports and, as seen in two-port networks, in cascade they add their decibels only if each one is properly terminated. The exact values almost never exist as commercial resistors, and rounding them changes the attenuation and the match slightly.

Lab · designing an attenuator with commercial resistors

You choose the topology, the impedance and the attenuation. The exact values are calculated, rounded to the nearest E24 (5 %) or E96 (1 %) series value and the real attenuator is recalculated between a generator and a load of \( Z_0 \). The schematic shows the rounded values.

Attenuation =

03Other attenuators

Type	What it is for
Bridged T	Variable attenuation by changing only two resistors
Asymmetric minimum-loss	Matching two different impedances, for example 50 and 75 Ω, with the least possible attenuation (5.7 dB)
Stepped	Switched attenuators in steps of 1, 2, 4, 8 dB, in generators and analyzers
Voltage-controlled	PIN diodes or integrated circuits, in automatic gain control

04Compensators

A compensator is a network that corrects the frequency response or the phase of another block. The simplest case was already seen in the oscilloscope: the compensated attenuator, with a capacitor in parallel with each resistor so that the divider is the same at all frequencies. In control, phase lead and lag networks are used:

\[ H(s) = \alpha\,\frac{1 + s R_1 C_1}{1 + s\,\alpha R_1 C_1} \] \[ \sin\varphi_m = \frac{1 - \alpha}{1 + \alpha} \] The lead network: \( R_1 \) in parallel with \( C_1 \), in series, and \( R_2 \) to ground, with \( \alpha = R_2/(R_1 + R_2) \). Between the zero and the pole it advances the phase, with a maximum \( \varphi_m \) at the geometric mean of the two frequencies. With \( \alpha = 0.1 \), the maximum phase lead is 55°.

Cable equalizers are compensators of the same kind: a long cable attenuates high frequencies more, and a network with the inverse response restores them. Every high-speed data receiver has one, fixed or adaptive.

05In the lab

Activity 1 · A 20 dB attenuator

Build a 20 dB π for 50 Ω with E24 resistors, measure the attenuation with a generator and an oscilloscope terminated in 50 Ω, and compare with the lab.

Activity 2 · Matching 50 to 75 Ω

Design and build a minimum-loss attenuator between 50 and 75 Ω and measure the reflection with and without it.

Activity 3 · Lead network

Build a lead network with α = 0.1 and measure the phase as a function of frequency. Locate the maximum.

06Common mistakes

Using a simple divider where an attenuator is needed. It changes the impedance and reflections appear.
Forgetting the power rating of the resistors. The first resistor of an attenuator connected to a transmitter dissipates almost all the power.
Terminating the attenuator in a different impedance. The attenuation and the match are no longer the design values.
Ignoring parasitic inductances. At hundreds of megahertz, surface-mount resistors and short traces are essential.

07Self-assessment

What resistors does a 6 dB T for 50 Ω have?

With \( K = 2 \): series \( 50 \cdot 1/3 \approx 16.7\ \Omega \) and shunt \( 50 \cdot 4/3 \approx 66.7\ \Omega \).

How much power does a 10 dB attenuator receiving 10 W dissipate?

1 W comes out: it dissipates 9 W.

What maximum phase lead does a network with α = 0.25 give?

\( \arcsin(0.75/1.25) = \arcsin 0.6 \approx 37° \).

Why do two badly terminated 3 dB attenuators not give 6 dB?

Because the impedance each one sees is no longer the design one, and their attenuation and match change.

08Further reading

Mac E. Van Valkenburg. Network Analysis. 3rd ed., Prentice Hall, 1974 (Spanish edition, “Análisis de redes”, Limusa). Image impedances and symmetric networks, the basis of attenuator design.
Katsuhiko Ogata. Modern Control Engineering. 5th ed., Prentice Hall, 2010 (Spanish edition, “Ingeniería de control moderna”, Pearson). Lead and lag networks applied to compensator design.
H. Ward Silver (ed.). The ARRL Handbook for Radio Communications. ARRL, annual editions. Tables and practical construction of attenuators for radio frequency.

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