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.
Wye (Y) 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:
$$\begin{aligned} R_i &= R_1 + R_2 \parallel (R_1 + R_L)\\ &= 2.9 Ω + 433.3 Ω \parallel (2.9 Ω+ 50 Ω)\\ &= 2.9 Ω + 47.14 Ω= 50.04 Ω \end{aligned} $$
Wye (Y) configuration.
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
$$V_{R_{2}}=\frac{R^{\prime} V_{s}}{R^{\prime}+R_{1}}=\frac{47.14 \Omega V_{s}}{47.14 \Omega+2.9 \Omega}=0.942 V_{s}$$
$$V_{L}=\frac{R_{L} V_{R_{2}}}{R_{L}+R_{1}}=\frac{50 \Omega\left(0.942 V_{s}\right)}{50 \Omega+2.9 \Omega}=0.890 V_{s}$$
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.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024
Privacy Policy