Deriving resistor values for a taper pad attenuator

Here is another take at that problem.

First let's define some quantities that will be useful later:

\[ \begin{align} A &= \frac {V_2} {V_1} \qquad\qquad B = \frac {R_L} {R_S} \qquad\qquad r_1 = \frac {R_1} {R_L} \qquad\qquad r_2 = \frac {R_2} {R_L} \qquad\qquad r_3 = \frac {R_3} {R_L} \tag{1} \end{align} \]

Since the attenuator port 1 has an input impedance that matches the source, then the source voltage is split in two at the input. Moreover, we also apply the definition of attenuation $A$.

\[ \begin{align} V_1 &= \frac{V_S}{2} \qquad\qquad V_2 = A V_1 = A \frac{V_S}{2} \tag{2} \end{align} \]

We can therefore compute the currents into the ports as:

\[ \begin{align} I_1 &= \frac{V_S}{2R_S} \tag{3} \\[2 em] I_2 &= -\frac{V_2}{R_L} = -\frac{A}{R_L} \frac{V_S}{2} \tag{4} \end{align} \]

Now let's focus on the internal node, the junction between $R_1,R_2, R_3$ (let's call it node 3). You can use KCL to compute the current in $R_3$ and then use Ohm's law to obtain $V_3$:

\[ \begin{align} I_3 &= I_1 + I_2 = \frac{V_S}{2R_S} - \frac{A}{R_L} \frac{V_S}{2} = \biggl(\frac{1}{R_S} - \frac{A}{R_L} \biggr) \frac{V_S}{2} \tag{5} \\[2 em] V_3 &= R_3 I_3 = R_3 \biggl(\frac{1}{R_S} - \frac{A}{R_L} \biggr) \frac{V_S}{2} = \frac{R_3}{R_L} \biggl(\frac{R_L}{R_S} - A \biggr) \frac{V_S}{2} = r_3 \biggl(B - A \biggr) \frac{V_S}{2} \tag{6} \end{align} \]

Now we can compute $V_3$ also using KVL applied to the input and output loops, obtaining two more independent equations for $V_3$:

\[ \begin{align} V_3 &= V_1 - R_1 I_1 = \frac{V_S}{2} - R_1 \frac{V_S}{2 R_S} = \biggl(1 - \frac{R_1}{R_S} \biggr) \frac{V_S}{2} = \biggl(1 - \frac{R_L}{R_S} \cdot \frac{R_1}{R_L} \biggr) \frac{V_S}{2} = \biggl(1 - B \cdot r_1 \biggr) \frac{V_S}{2} \tag{7} \\[2 em] V_3 &= V_2 - R_2 I_2 = \frac{A V_S}{2} - R_2 \biggl(-\frac{A}{R_L} \frac{V_S}{2} \biggr) = A \biggl(1 + \frac{R_2}{R_L} \biggr) \frac{V_S}{2} = A \biggl(1 + r_2 \biggr) \frac{V_S}{2} \tag{8} \end{align} \]

Now comparing equation 7 with equations 8 and 9, respectively, we obtain two independent equations in which the only unknown are $r_1,r_2,r_3$:

\[ \begin{align} \left\{ \begin{aligned} r_3 \biggl(B - A \biggr) \frac{V_S}{2} = \biggl(1 - B \cdot r_1 \biggr) \frac{V_S}{2} \\[2 em] r_3 \biggl(B - A \biggr) \frac{V_S}{2} = A \biggl(1 + r_2 \biggr) \frac{V_S}{2} \end{aligned} \right. \qquad\qquad &\Leftrightarrow \qquad\qquad \left\{ \begin{aligned} r_3 \biggl(B - A \biggr) = \biggl(1 - B \cdot r_1 \biggr) \\[2 em] r_3 \biggl(B - A \biggr) = A \biggl(1 + r_2 \biggr) \end{aligned} \right. \qquad\qquad \Leftrightarrow \qquad\qquad \nonumber\\[2 em] \qquad\qquad &\Leftrightarrow \qquad\qquad \left\{ \begin{aligned} r_1 = \frac{1 - r_3 \biggl(B - A \biggr)}{B} \\[2 em] r_2 = \frac{r_3 \biggl(B - A \biggr) - A}{A} \end{aligned} \right. \tag{9} \end{align} \]

Now let's focus on port 2. Looking into it we see an impedance $R_L$ which can be calculated using the usual methods of Thevenin's theorem:

\[ \begin{gather} R_L = R_2 + R_3 \parallel (R_1 + R_S) = R_2 + \frac{1}{ \frac{1}{R_3} + \frac{1}{R_1 + R_S} } \qquad \Leftrightarrow \tag{10} \\[2 em] \quad \Leftrightarrow \quad R_L - R_2 = \frac{1}{ \frac{1}{R_3} + \frac{1}{R_1 + R_S} } \quad \Leftrightarrow \quad \frac{1}{ R_L - R_2 } = \frac{1}{R_3} + \frac{1}{R_1 + R_S} \quad \Leftrightarrow \quad \frac{1}{ 1 - \frac{R_2}{R_L} } = \frac{R_L}{R_3} + \frac{1}{\frac{R_1}{R_L} + \frac{R_S}{R_L}} \quad \Leftrightarrow \nonumber \\[2 em] \qquad \Leftrightarrow \qquad \frac{1}{ 1 - r_2 } = \frac{1}{r_3} + \frac{1}{r_1 + \frac{1}{B}} \tag{11} \end{gather} \]

Now let's rewrite equations (9) like this:

\[ \begin{gather} \left\{ \begin{aligned} r_1 = \frac{1 - r_3 \biggl(B - A \biggr)}{B} \\[2 em] r_2 = \frac{r_3 \biggl(B - A \biggr) - A}{A} \end{aligned} \right. \qquad \Leftrightarrow \qquad \left\{ \begin{aligned} r_1 + \frac{1}{B} = \frac{2 + r_3 \biggl(A - B \biggr)}{B} \\[2 em] 1 - r_2 = \frac{2A - r_3 \biggl(B - A \biggr)}{A} \end{aligned} \right. \qquad \Leftrightarrow \qquad \left\{ \begin{aligned} \frac{1}{ r_1 + \frac{1}{B} } = \frac{B}{2 + r_3 \biggl(A - B \biggr)} \\[2 em] \frac{1}{ 1 - r_2 } = \frac{A}{2A + r_3 \biggl(A - B \biggr)} \end{aligned} \right. \tag{12} \end{gather} \]

substituting (12) into (11) we get:

\[ \begin{gather} \frac{1}{ 1 - r_2 } = \frac{1}{r_3} + \frac{1}{r_1 + \frac{1}{B}} \qquad \Leftrightarrow \qquad \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{1}{r_3} + \frac{B}{2 + r_3 \biggl(A - B \biggr)} \tag{13} \end{gather} \]

Now we can solve (13) for $r_3$.

\[ \begin{gather} \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{1}{r_3} + \frac{B}{2 + r_3 \biggl(A - B \biggr)} \qquad \Leftrightarrow \qquad \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{2 + r_3 \biggl(A - B \biggr) + B r_3}{r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]} \nonumber \\[2 em] \Leftrightarrow \qquad \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{2 + A r_3 }{r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]} \qquad \Leftrightarrow \nonumber \\[2 em] \qquad A r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr] = (2 + A r_3 ) \biggl[ 2A + r_3 \biggl(A - B \biggr) \biggr] \qquad \Leftrightarrow \nonumber \\[2 em] \qquad 2A r_3 + A r_3^2 \biggl(A - B \biggr) = 4A + 2r_3 \biggl(A - B \biggr) + 2A^2 r_3 + A r_3^2 \biggl(A - B \biggr) \nonumber \\[2 em] \qquad 0 = 4A - 2Br_3 + 2A^2 r_3 = 4A + 2 r_3 (A^2 - B ) \qquad \Leftrightarrow \qquad r_3 (B - A^2 ) = 2A \nonumber \\[2 em] r_3 = \frac {2A}{ B - A^2 } \tag{14} \end{gather} \]

And then substitute $r_3$ back into equations (12) to get $r_1,r_2$:

\[ \begin{gather} \left\{ \begin{aligned} r_1 &= \frac{1 - r_3 \biggl(B - A \biggr)}{B} = \frac{1 - \frac {2A}{ B - A^2 } \biggl(B - A \biggr)}{B} = \frac{B - A^2 - 2A \biggl(B - A \biggr)}{B (B - A^2)} \\[2 em] r_2 &= \frac{r_3 \biggl(B - A \biggr) - A}{A} = \frac{ \frac {2A}{ B - A^2 } \biggl(B - A \biggr) - A}{A} = \frac{ 2A \biggl(B - A \biggr) - A (B - A^2)}{A (B - A^2)} \end{aligned} \right. \qquad \Leftrightarrow \nonumber \\[2 em] \Leftrightarrow \qquad \left\{ \begin{aligned} r_1 &= \frac{B - A^2 - 2A \biggl(B - A \biggr)}{B (B - A^2)} = \frac{B + A^2 - 2AB}{B (B - A^2)} = \frac{B ( 1 + A^2/B - 2A)}{B (B - A^2)} \\[2 em] r_2 &= \frac{ 2A \biggl(B - A \biggr) - A (B - A^2)}{A (B - A^2)} = \frac{ AB + A^3 - 2A^2}{A (B - A^2)} = \frac{ A (B + A^2 - 2A)}{A (B - A^2)} \end{aligned} \right. \qquad \Leftrightarrow \nonumber \\[2 em] \Leftrightarrow \qquad \left\{ \begin{aligned} r_1 &= \frac{B ( 1 + A^2/B - 2A)}{B (B - A^2)} = \frac{ 1 + A^2/B - 2A}{B - A^2} \\[2 em] r_2 &= \frac{ A (B + A^2 - 2A)}{A (B - A^2)} = \frac{ B + A^2 - 2A}{B - A^2} \end{aligned} \right. \tag{15} \end{gather} \]

Therefore we end up with our solution for $r_1,r_2,r_3$, from which you can get $R_1,R_2,R_3$:

\[ \begin{gather} \left\{ \begin{aligned} r_1 &= \frac{ 1 + A^2/B - 2A}{B - A^2} \\[2 em] r_2 &= \frac{ B + A^2 - 2A}{B - A^2} \\[2 em] r_3 &= \frac {2A}{ B - A^2 } \end{aligned} \right. \qquad \Leftrightarrow \qquad \left\{ \begin{aligned} R_1 &= R_L \cdot \frac{ 1 + A^2/B - 2A}{B - A^2} \\[2 em] R_2 &= R_L \cdot \frac{ B + A^2 - 2A}{B - A^2} \\[2 em] R_3 &= R_L \cdot \frac {2A}{ B - A^2 } \end{aligned} \right. \tag{16} \end{gather} \]

which matches your solutions.

Yeah, that looks like it's going to be messy.

I agree with Olin: the problem starts with three variables and three conditions so, no matter how you look at it, you will end up with a system of equations.

However, you can take certain shortcuts (using your 1st picture):

$$\begin{align} R_{23}&=R_3||(R_2+R_o) \tag{1} \\ R_{13}&=R_3||(R_1+R_i) \tag{2} \\ R_i&=R_1+R_{23} \tag{3} \\ R_o&=R_2+R_{13} \tag{4} \end{align}$$

[edit]
[edit 2]
$R_{23}$ is the resistance at point X with the input disconnected, and $R_{13}$ is the rsistance at point X with the output disconnected.

Then, the input source always sees its $R_i$ in series with the equivalent $R_i$, while the output voltage is always $\frac{A}2$ over $R_o||R_o^{eq}$ ($A$ is the attenuation relative to a unity input). Thus:

$$\begin{align} I_1&=\dfrac{1}{2R_i} \tag{5} \\ I_2&=\dfrac{A}{2R_o} \tag{6} \\ I_3&=I_1-I_2=\dfrac12\left(\dfrac{1}{R_i}-\dfrac{A}{R_o}\right) \tag{7} \end{align}$$

Since the biggest unknown is $V_x$, calculate it with (3) in mind:

$$V_x=\dfrac12\dfrac{R_{23}}{R_{23}+R_1}=\dfrac12\dfrac{R_i-R_1}{R_i-R_1+R_1}=\dfrac12\dfrac{R_i-R_1}{R_i}=\dfrac12\left(1-\dfrac{R_1}{R_i}\right) \tag{8}$$

Now calculate $R_2$ and $R_3$ based on (6), (7), and (8), while considering that $V_1=\frac12$ and $V_2=\frac{A}{2}$:

$$\begin{align} R_2&=\dfrac{V_x-V_2}{I_2}=\left(\dfrac{2V_x}{A}-1\right)R_o=\left(\dfrac{1-\dfrac{R_1}{R_i}}{A}-1\right)R_o=\dfrac{\big((1-A)R_i-R_1\big)R_o}{AR_i} \tag{9} \\ R_3&=\dfrac{V_1-V_x}{I_1}=\dfrac{2V_xR_iR_o}{R_o-AR_i}=\dfrac{\left(1-\dfrac{R_1}{R_i}\right)R_iR_o}{R_o-AR_i}=\dfrac{(R_i-R_1)R_o}{R_o-AR_i} \tag{10} \end{align}$$

Since $R_i$, $A$, and $R_o$ are given, both (9) and (10) are dependent on $R_1$, only. Now, use (2) and (4) to derive the expression for $R_1$:

$$R_o\stackrel{(2,4)}{=}R_2+\dfrac{(R_1+R_i)R_3}{R_1+R_3+R_i} \tag{11}$$

After expanding and collecting the terms:

$$R_1=\dfrac{(R_o-R_2)R_i-(R_2+R_i-R_o)R_3}{R_2+R_3-R_o} \tag{12}$$

Substituting (9) and (10) in (12) takes a few lines of simplifications to give:

$$R_1=\dfrac{R_i^2\big(A^2-(2A-1)R_o\big)}{R_o-A^2R_i} \tag{I}$$

At this point, if you want to keep things simple(-ish) then use $(\text{I})$ for $R_1$ and then, sequentially, calculate (9) and (10) through substitution. Otherwise, you'll need a few more lines of simplifications to give:

$$\begin{align} R_2&=\dfrac{R_o^2+(A^2-2A)R_iR_o}{R_o-A^2R_i} \tag{II} \\ R_3&=\dfrac{2AR_iR_o}{R_o-A^2R_i} \tag{III} \
\end{align}$$

I find it interesting that all of them are divided by $R_o-A^2R_i$. Maybe there is something to it, hopefully a simplification but, now, my eyes are getting crossed by the amount of MathJax so, I'll take a break. I'll double check later on but, for now, it looks like I haven't made any mistakes.

A SPICE test with your values confirms both approaches (blue text is active, upper sources output the resistor values, k means $A$):

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[edit 3]

This is mostly cosmetic, it certainly doesn't reduce the derivation (it actually adds to it) but, inspired by the formulas from this site (which do not account for correct attenuation), $R_3$ can be calculated as in ($\text{III}$) and then $R_1$ and $R_2$ calculated from (3) and (4), based on it, resulting in slightly more palatable equations:

$$\begin{align} R_1&=\sqrt{\dfrac{\bbox[5px,border:black solid 1px]{R_i}}{R_o}}\sqrt{R_iR_o+R_3^2}-R_3 \tag{IV} \\ R_2&=\sqrt{\dfrac{R_o}{\bbox[5px,border:black solid 1px]{R_i}}}\sqrt{R_iR_o+R_3^2}-R_3 \tag{V} \end{align}$$

I'm thinking out loud here and haven't solved this yet. This answer is logging my process as I try to solve the problem. It may very well result in the same long-winded process you went thru. Nothing shrewd or insightful is promised.

There are three unknowns (R1, R2, R3), and there are fortunately three constraints:

1. The impedance looking into the input must be Rin:

       R1 + R3//(R2 + RL) = Rin

2. The impedance looking into the output must be RL:

       R2 + R3//(R1 + Rin) = RL

3. The attenuation must be A. I think it will be nicer to use gain rather than attenuation, so I'll use G = 1/A. This constraint is more tricky to write down in a single equation, so I'll use two and the intermediate value Gx. Gx is the gain from the input to the mid point:

       Gx = R3//(R2 + RL) / [Rin + R1 + R3//(R2 + RL)]

   Then the gain at the output is:

       G = Gx * RL / (R2 + RL)

Yeah, that looks like it's going to be messy.

The first step is to combine the two equations of constraint three into one, which gets rid of the intermediate value Gx I used for convenience. Actually that part is easy since its just a straight multiply. Like I said, this is thought stream dump. Breaking the third constraint into two wasn't necessary, although it does document the separate thoughts. Anyway, simplified constraint 3 is:

--- Work in progress ---
I'll get back to this. I also need to look up how to use MathJax, since these equations are getting too complicated for plain HTML.