TAPER-PAD ATTENUATOR

The taper-pad uses \(R_1\), \(R_2\) and \(R_3\) to match \(R_{IN}\) to \(R_L\). Circuit gain (\(A_{12}\)) is from input (1) to output (2). For example, if \(A_{12}\) is set to 0.25 then this is a 4:1 attenuation (-12.0412 dB).

\(\hspace{9cm}\)

Taper-pad attenuator Circuit

Relevant Formulas

$$A_{12(MAX)} = \frac{R_L}{R_{IN}}\left[1-\sqrt{1-\frac{R_{IN}}{R_L}}\text{ }\right]\hspace{0.5cm}R_L>R_{IN}$$ $$A_{12(MAX)} = 1-\sqrt{1-\frac{R_L}{R_{IN}}}\text{ }\hspace{2.0cm}R_{IN}>R_L$$
$$R_1 = R_L\left[\dfrac{1 + \frac{R_{IN}}{R_L}A_{12}^2 - 2\cdot A_{12}}{\frac{R_L}{R_{IN}}- A_{12}^2}\right]$$
$$R_2 = R_L\left[\dfrac{\frac{R_L}{R_{IN}}+ A_{12}^2 - 2\cdot A_{12}}{\frac{R_L}{R_{IN}}- A_{12}^2}\right]$$
$$R_3 = R_L\left[\dfrac{2\cdot A_{12}}{\frac{R_L}{R_{IN}}- A_{12}^2}\right]$$
   
\(R_{IN}\) Ω \(\boxed{R_1=}\) Ω
\(R_L\) Ω \(\boxed{R_2=}\) Ω
\(A_{12}\) V/V \(\boxed{R_3=}\) Ω
\(A_{12}\text{ (dB)}\) dB \(\boxed{A_{12(MAX)}=}\) V/V

Transfer values (applicable when analysing as two back-to-back L-Pads)
\(\hspace{1cm} R_X\hspace{1cm}=\)Ω \(\hspace{1cm} R_3\hspace{1cm}=\)Ω
\(\hspace{1cm} R_{3A}\hspace{0.9cm}=\)Ω \(\hspace{1cm} R_{3B}\hspace{0.8cm}=\)Ω
\(\hspace{1cm} A_{1X}\hspace{0.9cm}=\)V/V \(\hspace{1cm} A_{X2}\hspace{0.8cm}=\)V/V

Basic gain equations
At point X, \(R_3\) is in parallel with \(R_2+R_L\) and forms a potential divider with \(R_1\). Importantly, the sum of \(R_1\) and \(R_3||(R_2+R_L)\) equals \(R_{IN}\) hence: $$\hspace{3cm}A_{1X} = \dfrac{R_{IN}-R_1}{R_{IN}}\hspace{3cm}$$
The attenuation between point X and the output is simply this: $$A_{X2} = \dfrac{R_L}{R_L+R_2}$$
Therefore, the overall gain is: \(A_{1X}\times A_{X2}\) $$A_{12}=\dfrac{R_L}{R_{IN}}\cdot\dfrac{R_{IN}-R_1}{R_L+R_2}\tag{1}$$
From (1) we can solve for \(R_1\) and \(R_2\):




$$R_1 = R_{IN}\cdot\left[1-\dfrac{R_L+R_2}{R_L}\cdot A_{12}\right]$$ $$R_2 = R_{L}\cdot\left[\dfrac{R_{IN}-R_1}{A_{12}\cdot R_{IN}}-1\right]\tag{2}$$

Solve \(A_{1X}\) by using \(R_3\):


Noting that \(R_{IN} = R_3||(R_2+R_L) + R_1\):





Multiply by \(A_{X2} = \dfrac{R_L}{R_2+R_L}\) to obtain \(A_{12}\):



Rearrange:
$$A_{1X}=\dfrac{R_3||(R_2+R_L)}{R_3||(R_2+R_L) + R_1}$$ $$A_{1X}=\dfrac{R_3||(R_2+R_L)}{R_{IN}}$$ $$A_{1X}=\dfrac{R_3\cdot(R_2+R_L)}{R_{IN}\cdot(R_2+R_3+R_L)}$$ $$A_{12}=\dfrac{R_L}{R_{IN}}\cdot\dfrac{R_3}{R_2+R_3+R_L}$$ $$$$ $$R_3=\dfrac{R_2+R_L}{\frac{R_L}{R_{IN}\cdot A_{12}}-1}\tag{3}$$

Transfer Impedance analysis
At this point, the formulas above are all interdependent so, as a means of distilling them, we formulate a second analysis. This means considering the transfer impedance (\(R_X\)): -

(1) Resistor \(R_3\) is split into two parts (a and b)

(2) Use formulas from the resistive matching L-pad

(3) Modify and apply to left and right sections.
For the left hand side
$$R_1=R_{IN}\sqrt{1-\dfrac{R_X}{R_{IN}}}\tag{4}$$ $$R_{3A} = \dfrac{R_{IN}\cdot R_X}{R_1}$$
For the right hand side
$$R_2=R_{L}\sqrt{1-\dfrac{R_X}{R_{L}}}\tag{5}$$ $$R_{3B} = \dfrac{R_{L}\cdot R_X}{R_2}$$
Using equation (4) $$R_X=\dfrac{R_{IN}^2-R_1^2}{R_{IN}}=\dfrac{(R_{IN} - R_1)(R_{IN} + R_1)}{R_{IN}}$$ Using equation (5) $$R_X=\dfrac{R_{L}^2-R_2^2}{R_{L}}=\dfrac{(R_{L} - R_2)(R_{L} + R_2)}{R_{L}}$$
Noting that \(A_{1X} = \frac{R_{IN}-R_1}{R_{IN}}\),

$$R_X = A_{1X}\cdot (R_{IN} +R_1)$$

Noting that \(A_{X2} = \frac{R_L}{R_L+R_2}\), $$R_X = \dfrac{R_L-R_2}{A_{X2}}$$
Equating the two above formulas together like this: -
$$A_{1X}\cdot (R_{IN} +R_1)= \dfrac{R_L-R_2}{A_{X2}}$$ $$A_{12}=\dfrac{R_L-R_2}{R_{IN}+R_1}\tag{6}$$ Then, we substitute in equation (6) for \(R_2\) from equation (2): -
$$A_{12}=\dfrac{R_L-R_L\left[\dfrac{R_{IN}-R_1}{A_{12}\cdot R_{IN}}-1\right]}{R_{IN}+R_1}$$ $$A_{12}^2\cdot\dfrac{R_{IN}}{R_L}\cdot (R_{IN}+R_1)=R_1+2\cdot A_{12}\cdot R_{IN}-R_{IN}$$ $$A_{12}^2\cdot\dfrac{R_{IN}^2}{R_L}=R_1\left[1-A_{12}^2\cdot\dfrac{R_{IN}}{R_L}\right]+R_{IN}\left[2\cdot A_{12}-1\right]$$ $$R_1 = \dfrac{A_{12}^2\cdot\frac{R_{IN}^2}{R_L}-R_{IN}\left[2\cdot A_{12}-1\right]}{1 - A_{12}^2\cdot\frac{R_{IN}}{R_L}} \hspace{1cm}=\hspace{1cm}R_L\left[\dfrac{1+A_{12}^2\cdot\frac{R_{IN}}{R_L}-2\cdot A_{12}}{\frac{R_L}{R_{IN}}-A_{12}^2}\right]\tag{7}$$ QED