Deriving resistor values for a taper pad attenuator
A taper pad is a resistive attenuator that maintains impedances on both ports and provides a specific amount of gain-loss ($A_{12}$): -
I have derived formulas for each resistor (that I know to be correct) and have checked with micro-cap using DC analysis: -
"So what" you might think. Well the problem is really that it took me ages to derive the formulas and, I am convinced that there must be a simpler approach than the method I took so, what I'm looking for is a shrewd and insightful way of finding (say) the value of R1 given: -
- The required gain-loss i.e. $\frac{V_2}{V_1}$ or $\frac{V_{OUT}}{V_{IN}}$
- The input impedance, $R_{IN}$
- The output load impedance, $R_{L}$
I don't want answers that say if you "do this" you can "find that" then it's easy to drill-down to what you want. I want to see the actual math. I've done it (and my algebra is correct) but, it was very long-winded and I'm sure I missed a trick along the way.
I will also add one more important thing that I've come to realize: none of the existing calculators out there (apart from mine) get the formulas correct. They show correct formulas for equal input and output impedance but, they screw up when the input and output impedances are different.
3 answers
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.
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Andy aka | (no comment) | Oct 13, 2022 at 17:23 |
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 $A$):
[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:
- The impedance looking into the input must be Rin:
R1 + R3//(R2 + RL) = Rin
- The impedance looking into the output must be RL:
R2 + R3//(R1 + Rin) = RL
- 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.
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