I'm not understanding the concept of L-Pad attenuator other than seeing it as another ordinary voltage divider to attenuate a voltage. Electronics-Tutorials: L-pad Attenuator gives the following description:

L-pad attenuators are commonly used in audio applications to reduce a larger or more powerful signal while matching the impedance between the source and load in provide maximum power transfer. However, if the impedance of the source is different to the impedance of the load, the L-pad attenuator can be made to match either impedance but not both.

I understand maximum power transfer occurs when input impedance matches output impedance (or \$ Z_{in} = Z_{out} \$), and that \$ \text{dB} = 20log({V_{out} \over V_{in}}) \$. In terms of impedance matching, looking into the attenuator from the left, \$ Z_{in} = (Z_L || R_2)+R_1 \$ and from the right, \$ Z_{out} = (Z_s + R_1) || R_2 \$ in series. Hence for the two impedances to match and if I'm given a voltage attenuation ratio, I'll have exactly the following two equations to solve for two unknown values of \$ R_1 \$ and \$ R_2 \$:

  1. Matching input and output impedance: \$ (Z_L || R_2)+R_1 = (Z_s + R_1) || R_2 \$.
  2. Voltage attenuation ratio(): \$ V_{out} = V_{in} \big({{R_2 || Z_L} \over {({R_2 || Z_L})+Z_S+R1}}\big) \$

Aren't these two equations all I need to match input and output impedance while achieving a specific attenuated voltage? Why are there these logarithmic equations with a 'K' value solving for \$ R_1 \$ and \$ R_2 \$? And what does it mean when they say

L-pad attenuator can be made to match either impedance but not both

enter image description here

  • \$\begingroup\$ I didn't see the equation \$Z_{out} = Z_s + R_1 + R_2\$ on that website, and it appears to me to be incorrect. \$\endgroup\$ Jan 4, 2021 at 6:22
  • \$\begingroup\$ @MathKeepsMeBusy the equation is not from the website. I'm not understanding the tutorial and hence the question. I came up with that equation from the circuit - if I look into the attenuator from the right (output), the 8hm, R1 and R2 are in series, and I don't see anything wrong with it. \$\endgroup\$
    – KMC
    Jan 4, 2021 at 6:25
  • \$\begingroup\$ From the speaker's point of view, R2 is in parallel with R1, the amplifier impedance and the voltage source. We can ignore the voltage source for calculating impedance, so \$Z_{out} = R_2 || (R_1 + Z_s)\$. \$\endgroup\$ Jan 4, 2021 at 6:30
  • \$\begingroup\$ @MathKeepsMeBusy oh yeas you're right, I've amended \$ Z_{out} = R_1 + R _2 + Z_s \$ to \$ Z_{out} = (R_1 + Z_s) || R_2 \$ \$\endgroup\$
    – KMC
    Jan 4, 2021 at 6:41
  • \$\begingroup\$ The example is not practical. An amplifier driving an 8 ohm speaker would have output impedance of almost 0 ohms. \$\endgroup\$
    – Justme
    Jan 4, 2021 at 6:42

2 Answers 2


L-pad attenuator can be made to match either impedance but not both

What it means is that a simple L-pad made from two resistors can do one of two things: -

  • Provide an attenuation
  • Match two impedances

But you cannot simultaneously match two impedances AND have independent control of the desired attenuation level. The reason is because there are only two resistors and there are not enough variables (R1 and R2) to get both a defined attenuation AND a defined impedance matching circuit.

To get both, you need at least three resistors. This webpage gives an example and shows the two formulas that define R1 and R2: -

enter image description here

As you should be able to see, if you define \$R_{IN}\$ and \$R_L\$ from the outset, you get unique values for R1 and R2. Those unique values are fixed by \$R_{IN}\$ and \$R_L\$ and produce an attenuation that is fixed. In other words, it is the value of \$R_{IN}\$ and \$R_L\$ that produce R1 and R2 and that, in turn, means no control over what the attenuation is.


Andy Aka's answer is definitely correct and provides good insight into the problem. Here I will just try to provide an answer to

Why are there these logarithmic equations with a 'K' value solving for \$R_1\$ and \$R_2\$?

Such equations provide a direct link between the attenuation you want to obtain and the resistors in your L-pad attenuator when source and load impedances are the same (\$Z_S=Z_L\$, as you can see in the tutorial you mention) and you want to match your attenuator to the source impedance (not an optimal choice, as the tutorial itself points out).

\$K\$ is defined as an attenuation, i.e. the inverse of the "gain" \$V_{OUT}/V_{IN}\$ (if we choose to consider voltages): \$K\triangleq V_{IN}/V_{OUT}\$. In this way, what the tutorial calls \$dB\$ is of course just \$K\$ expressed in decibel: \$dB=20\log(K)\$.

Then, \$V_{OUT}=V_{IN}\frac{R_2||Z_L}{R_1+R_2||Z_L}=V_{IN}\frac{Z_S-R1}{Z_S}\$ since \$Z_S=R_1+R_2||Z_L\$ is the condition for source matching. After some algebra we obtain \$R_1=Z_S\frac{K-1}{K}=Z_S\frac{10^{dB/20}-1}{10^{dB/20}}\$ as reported in the tutorial.

The formula for \$R_2\$ can be obtained by substituting the above result for \$R_1\$ into \$Z_S\$ and expanding the parallel resistor \$R_2||Z_L\$: \$Z_S=Z_S\frac{K-1}{K}+\frac{R_2Z_L}{R_2+Z_L}\$. Then recall that \$Z_S=Z_L\$ and solve in \$R_2\$.

The above allows you to set the desired attenuation \$K\$ through a circuit that is matched to your source only, as you can easily see by taking the values obtained in the tutorial: \$Z_S=Z_L=R_2=8\,\Omega\$ and \$R_1=4\,\Omega\$.

  • \$\begingroup\$ Wish I could check both answers as yours answers the other part of the question (the derivation). Mine came out different, until I realize the \$ dB \$ is calculated with the inverse of the power/voltage ratio as you pointed out. With \$ K > 1 \$, \$ dB \$ ends up always to be a positive "magnitude" of attenuation (instead of being negative). Just out of curiosity, why do we use \$ K \$ (and needing to remember the magnitude is an attenuation) rather than going with the "standard" \$ dB \$ (where we can see gain or attenuation intuitively from the sign of the value)? It's really strange. \$\endgroup\$
    – KMC
    Jan 5, 2021 at 16:58
  • 1
    \$\begingroup\$ @KMC I'm not sure I see your point, but mathematically they're two different things: K is a voltage ratio that appears in the circuit equation, while dB is its value expressed in decibel through a specific function, 20log(K), thus can't be used as it is in the above equations. Moreover, please note that you can't see gain or attenation even from dB itself, because it still depends on how it was defined: Whether from Vout/Vin or Vin/Vout. Anyway, given any definition, positive or negative dB is mapped to K larger or smaller than 1 through the logarithm. Does that answer your question? \$\endgroup\$
    – DavideM
    Jan 6, 2021 at 13:56
  • 1
    \$\begingroup\$ yes sir: the statement "depends on how it was defined" cleared my confusion. Before, I took \$ dB = 10log(P_{out}/P_{in}) \$ as an universal definition. From all the intro circuit analysis textbooks I've read so far - gain is always positive and attenuation negative; what goes in (input) must be on the denominator side, and what comes out (output) has to be the numerator. But as you pointed out, that's not and there's no such standard. \$\endgroup\$
    – KMC
    Jan 6, 2021 at 15:38

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