Attenuators are, by definition, any device or system that can reduce the
power or voltage level of a signal while introducing little or no distortion.
There are two general types:
passive and
active.
The
passive type uses only resistors, while the
active type uses electronic devices such as
transistors and integrated circuits.
Since electronics is a subject for the courses to follow, our attention here will be only on the resistive type.
Attenuators are commonly used in audio equipment (such as the
graphic and parametric equalizers introduced in the previous chapter),
antenna systems, AM or FM systems where attenuation may be
required before the signals are mixed, and any other application where
a reduction in signal strength is required.
Fig. 1: Passive coax attenuator.
The unit of Fig. 1 has coaxial input and output terminals and
switches to set the level of dB reduction. It has a flat response from dc
to about 6 GHz, which essentially means that its introduction into the
network will not affect the frequency response for this band of frequencies.
The design is rather simple with resistors connected in either a tee (T) or a wye (Y) configuration as shown in Figs. 2 and 3,
respectively, for a 50-Ω coaxial system.
Fig. 2: Tee (T) configuration.
Fig. 3: Wye (Y) configuration.
In each case the resistors are chosen to ensure that the input impedance and output impedance match
the line. That is, the input and output impedances of each configuration
will be 50 Ω. For a number of dB attenuations, the resistor values for
the T and Y are provided in Figs. 2 and 3. Note in each design
that two of the resistors are the same, while the third is a much smaller
or larger value.
For the 1-dB attenuation, the resistor values were inserted for the T
configuration in Fig. 4(a). Terminating the configuration with a
50-Ω load, we find through the following calculations that the input
impedance is, in fact, 50 Ω:
$$\begin{aligned}
R_i &= R_1 + R_2 \parallel (R_1 + R_L)\\
&= 2.9 Ω + 433.3 Ω \parallel (2.9 Ω+ 50 Ω)\\
&= 2.9 Ω + 47.14 Ω= 50.04 Ω
\end{aligned}
$$
Fig. 4: 1-dB attenuator: (a) loaded; (b) finding Ro.
Looking back from the load as shown in Fig. 4(b) with the
source set to zero volts, we find through the following calculations that
the output impedance is also 50 Ω:
$$\begin{aligned}
R_o &= R_1 + R_2 \parallel (R_1 + R_s)\\
&= 2.9 Ω+ 433.3 Ω \parallel (2.9Ω + 50 Ω)\\
&= 2.9 Ω \parallel 47.14 Ω= 50.04 Ω
\end{aligned}
$$
Fig. 5: Determining the voltage levels for the 1-dB attenuator of Fig. 4(a).
In Fig. 5, a 50- $ \Omega $ load has been applied, and the output voltage is determined as follows:
$$R^{\prime}=R_{2} \|\left(R_{1}+R_{L}\right)=47.14 \Omega $$
from above and
$$V_{R_{2}}=\frac{R^{\prime} V_{s}}{R^{\prime}+R_{1}}=\frac{47.14 \Omega V_{s}}{47.14 \Omega+2.9 \Omega}=0.942 V_{s}$$
with
$$V_{L}=\frac{R_{L} V_{R_{2}}}{R_{L}+R_{1}}=\frac{50 \Omega\left(0.942 V_{s}\right)}{50 \Omega+2.9 \Omega}=0.890 V_{s}$$
Determining the voltage levels for the 1-dB attenuator of Fig. 4(a).
Calculating the drop in $ \mathrm{dB} $ will result in the following:
$$\begin{aligned}
A_{v_{\mathrm{dB}}} &=20 \log _{10} \frac{V_{L}}{V_{s}}=20 \log _{10} \frac{0.890 V_{s}}{V_{s}} \\
&=20 \log _{10} 0.890=\mathbf{- 1 . 0 1} \mathbf{d B}
\end{aligned}$$
substantiating the fact that there is a 1-$\mathrm{dB} $ attenuation.
As mentioned earlier, there are other methods for attenuation that are more sophisticated in design and beyond the scope of the coverage of this text. However, the above designs are quite effective and relatively inexpensive, and they perform the task at hand quite well.
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