Introduction to EMC: Attenuators

Attenuators are used in RF circuits to reduce the peak-to-peak voltage at the input stage before it enters the filter and amplifier stage. It is important since, under some conditions or in noisy environments, the amplifier output could be saturated by too high noise. Attenuators are typically simple passive devices but in some cases, more complex circuits can be found in literature or market products; they should introduce as low mismatch as possible, to avoid reflection; insertion loss is instead accurately calibrated and it must be considered when performing the measure.

Since attenuators are placed between two matched transmission lines, input and output impedances must be the same and both of them must match the transmission line impedance Z0; the output resistance must be calculated to get the desired attenuation.

Attenuators are symmetric, 2-ports, linear, passive devices. Attenuation factor K is defined as (IL is Insertion Loss in dB): $$K = 10^{\frac{IL[dB]}{20}}$$ Normalized Insertion Loss is $$IL_{norm} = K^{-1} = \frac{1}{10^{\frac{IL[dB]}{20}}} = 10^{-\frac{IL[dB]}{20}}$$ Detailed calculations are shown below for the various types of attenuators.

T attenuator is made of two series resistors and a centre parallel resistor. To work properly it must satisfy the following equalities $$\begin{cases} Z_{in} = Z_{out} = Z_{0} = (Z_{0}+Z_{1b})//Z_{2} + Z_{1a} = (Z_{0}+Z_{1a})//Z_{2} + Z_{1b} \\[2ex] IL_{norm} = 10^{-\frac{IL[dB]}{20}} = \frac{(Z_{0}+Z_{1b})//Z_{2}}{(Z_{0}+Z_{1b})//Z_{2} + Z_{1a}} \frac{Z_{0}}{Z_{0} + Z_{1b}} \end{cases}$$ Since the T attenuator is symmetric, linear and purely passive, the system can be rewritten as $$\begin{cases} R_{in} = (R_{0}+R_{1})//R_{2} + R_{1} = \frac{(R_0 + R_1)R_2}{R_0 + R_1 + R_2} + R_1 \\[2ex] 10^{-\frac{IL[dB]}{20}} = \frac{(R_0 + R_1)//R_2}{(R_0 + R_1)//R_2 + R_1} \frac{R_0}{R_0 + R_1} \end{cases}$$ Expanding parallel expression we get $$\begin{cases} R_{in} = \frac{R_0 R_2 + R_1 R_2 + R_0 R_1 + R_1^2 + R_2 R_1}{R_0 + R_1 + R_2} = \frac{2 R_1 R_2 + R_1^2 + R_0(R_1 + R_2)}{R_0 + R_1 + R_2} = R_0 \\[2ex] 10^{-\frac{IL[dB]}{20}} = \frac{\frac{(R_0 + R_1)R_2}{R_0 + R_1 + R_2}}{\frac{(R_0 + R_1)R_2}{R_0 + R_1 + R_2} + R_1} \frac{R_0}{R_0 + R_1} \end{cases}$$ Simplifying the first equation we get $$\begin{cases} 2 R_1 R_2 + R_1^2 = R_0^2 \\[2ex] K^{-1} = 10^{-\frac{IL[dB]}{20}} = \frac{(R_0 + R_1)R_2}{(R_0 + R_1)R_2 + R_1(R_0 + R_1 + R_2)} \frac{R_0}{R_0 + R_1} \end{cases}$$ $$\begin{cases} 2 R_1 R_2 + R_1^2 = R_0^2 \\[2ex] K = 10^{\frac{IL[dB]}{20}} = \frac{R_1^2 + 2 R_1 R_2 + R_0 R_1 + R_0 R_2}{R_0 R_2} = \frac{R_1^2}{R_0 R_2} + \frac{2 R_1}{R_0} + \frac{R_1}{R_2} + 1 \end{cases}$$ $$\begin{cases} R_2 = \frac{R_0^2 - R_1^2}{2 R_1} \\[2ex] K = \frac{2 R_1^3}{R_0 (R_0^2 - R_1^2)} + \frac{2 R_1(R_0^2 - R_1^2)}{R_0(R_0^2 - R_1^2)} + \frac{2 R_1^2 R_0}{R_0 (R_0^2 - R_1^2)} + 1 \end{cases}$$ $$\begin{cases} R_2 = \frac{R_0^2 - R_1^2}{2 R_1} \\[2ex] K = \frac{2 R_1}{R_0 - R_1} + 1 \end{cases}$$ Final result is $$\begin{cases} R_2 = R_0 \frac{2K}{K^2-1} \\[2ex] R_1 = R_0 \frac{K-1}{K+1} \end{cases}$$

A PI attenuator is made of two parallel resistors and a central series resistor. To work properly it must satisfy the following equalities $$\begin{cases} Z_{in} = Z_{out} = Z_{0} = ((Z_{0}//Z_{2b})+Z_{1} ) // Z_{2a} = ((Z_{0}//Z_{2a})+Z_{1} ) // Z_{2b} \\[2ex] IL_{norm} = 10^{-\frac{IL[dB]}{20}} = \frac{Z_{0}//Z_{2b}}{Z_{0}//Z_{2b}+Z_{1}} \end{cases}$$ Since the T attenuator is symmetric, linear and purely passive, the system can be rewritten as $$\begin{cases} R_{in} = R_{out} = R_{0} = ((R_{0}//R_{2})+R_{1} ) // R_{2} = \frac{(\frac{R_{0}R_{2}}{R_{0}+R_{2}} + R_{1})R_{2}}{(\frac{R_{0}R_{2}}{R_{0}+R_{2}} + R_{1})+R_{2}} \\[2ex] IL_{norm} = 10^{-\frac{IL[dB]}{20}} = \frac{R_{0}//R_{2}}{R_{0}//R_{2}+R_{1}} \end{cases}$$ Expanding parallel expression we get $$\begin{cases} R_{in} = \frac{(R_0R_2+R_0R_1+R_1R_2)R_2}{R_0R_2+R_0R_1+R_1R_2+R_0R_2+R_2^2} = R_{0} \\[2ex] 10^{-\frac{IL[dB]}{20}} = \frac{R_0R_2}{R_0R_2+R_0R_1+R_2R_1} \end{cases}$$ Simplifying the first equation we get $$\begin{cases} R_0R_2^2+R_0R_1R_2+R_1R_2^2 = R_0^2R_2+R_0^2R_1+R_0R_1R_2+R_0^2R_2+R_0R_2^2 \\[2ex] 10^{-\frac{IL[dB]}{20}} = \frac{R_0R_2}{R_0R_2+R_0R_1+R_2R_1} \end{cases}$$ $$\begin{cases} R_1R_2^2 = R_0^2R_2+R_0^2R_1+R_0^2R_2 \\[2ex] KR_0R_2 = R_0R_2+R_0R_1+R_2R_1 \end{cases}$$ $$\begin{cases} R_1 = \frac{2R_0^2R_2}{R_2^2-R_0^2} \\[2ex] KR_0R_2 = R_0R_2+(R_0+R_2)\frac{2R_0^2R_2}{R_2^2-R_0^2} \end{cases}$$ $$\begin{cases} R_1 = \frac{2R_0^2R_2}{R_2^2-R_0^2} \\[2ex] K = 1+\frac{2R_0}{R_2-R_0} \end{cases}$$ Final result is $$\begin{cases} R_1 = R_0\frac{K^2-1}{2K} \\[2ex] R_2 = R_0\frac{K+1}{K-1} \end{cases}$$