In its most basic form, the **L-pad Attenuator** is nothing more than a very simple voltage divider circuit used in many electrical and electronic circuits to generate a lower voltage.

The difference this time is that this type of attenuator is used in frequency dependent circuits to create loss (attenuation) in a transmission line or to match the impedances of unbalanced source and load networks.

The *L-pad attenuator* consists of two purely resistive elements in series with each other connected across a voltage source with the ratio between these two resistances forming a voltage divider network as shown below.

Related Products: Fixed Attenuator | Voltage Variable Attenuators

We can see that the L-pad attenuator design is identical to the voltage divider circuit used to reduce its input voltage by some amount. The two resistors are connected in series across the whole of the input voltage, while the output signal or voltage is taken across just one resistance, with the two resistive elements forming the shape of an inverted letter “L” and hence their name, “L-pad Attenuators”. For this types of circuit, attenuation is given as Vout/Vin.

Input resistor R1 is in series with the output, while resistor R2 is in parallel with the output and therefore the load. Then the output voltage provided by this “L” shaped arrangement is divided by a factor equal to the ratio of these two resistor values as shown.

As the L-pad attenuator is made of purely resistive components, there is no phase shift in the attenuator. The insertion of the attenuator between the source and the load must not alter the source voltage and therefore the resistance seen by the source must remain the same at all times. As the two resistive elements have constant values, if the impedance of the load is not infinite, the attenuation is altered and so to is its impedance. As a result the L-pad attenuator can only supply an impedance match in one direction only.

**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.

This is because the arrangement of the resistive elements does not produce the same impedance looking into the network from both directions. In other words, the L-pad attenuator is an asymmetrical attenuator and therefore, if an attenuation network is required to match two unequal impedances in both directions, other types of attenuator such as the symmetrical “T-pad” or the “Pi-pad” attenuator should be used.

As mention previously, a passive attenuator is a resistive network designed to reduce the power or signal level of an audio or radio frequency signal without introducing any distortion to the signal. Sometimes the output from an audio amplifier maybe too high and attenuation is required to feed this signal into a loudspeaker. For example lets assume we want to reduce the power being delivered from an amplifier which has an output source impedance, ( Z_{S} ) of 8Ω feeding a loudspeaker load, ( Z_{L} ) of 8Ω by 6dB. The values of resistors R1 and R2 are as follows.

The equation for the L-pad attenuator circuit connected between two equal impedances ( Z_{S} = Z_{L} ) looking in the direction of the source impedance, Z_{S} will be.

To simplify the design of the attenuator, a “K” value can be used in the attenuator equation above to simplify the maths a little. This “K” value is the ratio of the voltage, current or power corresponding to a given value of attenuation. The general equation for “K” is given as:

Then in our example the “K” value for a voltage attenuation of 6dB will be 10^{(6/20)} = 1.9953. Substituting this value for attenuation into the two equations gives.

Then between two equal impedances looking in the direction of the source impedance Z_{S}, the value of the series resistor, R1 is 4Ω and the value of the parallel resistor, R2 is 8Ω.

The problem with this type of L-pad attenuator configuration is that the impedance match is in the direction of the series resistor R1, while the impedance “mismatch” is towards the parallel resistor R2. The problem with this is that as the level of attenuation is increased this mismatch becomes increasingly larger and at high values of attenuation the value of the parallel resistor will become fractions of an Ohm.

For example, the values of R1 and R2 at an attenuation of -32dB would be 7.8Ω and 0.2Ω, that’s 200mΩ effectively shorting out the loudspeaker which could have a serious effect on the amplifiers output circuitry.

One way to increase attenuation without overloading the source is to impedance match the circuit in the direction of the load impedance, Z_{L}. However, as we are now looking into the L-pad attenuator circuit from the parallel resistor side, the equations are slightly different. Then between equal impedances and with the impedance match looking from the load, the values or resistors R1 and R2 are calculated as follows.

If we know increase attenuation to -32dB, the value of the resistors will become, R1 = 310Ω and R2 = 8.2Ω respectively, and these values are safe enough for the source circuit to which it is connected.

Thus far we have looked at connecting the **L-pad Attenuator** between to equal impedances in order to provide attenuation of a signal. But we can also use the “L-pad attenuator” to match the impedances of two unequal circuits. This impedance match may be in the direction of the larger or the smaller impedance but not both. The configuration of the attenuator will be the same as before, but the equations used in matching the two unequal impedances are different as shown.

Between two unequal impedances, the impedance matching is towards the smaller of the two impedances from the source.

Between two unequal impedances, the impedance matching is towards the larger of the two impedances from the load.

A signal transmission line which has a source impedance of 75Ω is to be connected to a signal strength meter of impedance 50Ω which has a maximum display of -12dB. Calculate the values of resistors required in an L-pad attenuator circuit to operate the meter at maximum power.

With the impedance match towards the smaller 50Ω value, resistors R1 and R2 are calculated as follows.

Then resistor R1 is equal to 59.6Ω and R2 is equal to 22.2Ω, or the nearest preferred values.

The **L-pad attenuator** can be used to perfectly match one impedance to another providing a fixed amount of attenuation, but the resulting circuit is “lossy”. However, if a fixed amount of attenuation is of no importance and only the minimum insertion loss is required between the source and the load, the **L-pad attenuator** can be used to match two impedances of unequal values using the following equations to calculate resistors, R1 and R2.

Where: resistor R1 is on the side of the larger impedance and resistor R2 is on the side of the smaller impedance and in our example above that would be 75Ω and 50Ω respectively. The minimum insertion loss in decibels of an L-pad attenuator connected between a source and a load is therefore given as:

In this tutorial we have seen that a **L-pad attenuator** circuit is a passive and purely resistive network which can be used to reduce the strength of a signal while matching the impedances of the source and load. L-pad attenuators are commonly used in audio electronics to reduce the audio signal produced by an amplifier delivered to a speaker or headphones.

However, one of the main disadvantages of the “L-pad attenuator” is that because the L-pad attenuator is a constant impedance device, at low power settings the attenuator converts all of the energy not sent to the load into heat which can be considerable. Also, at much higher frequencies or where an attenuator circuit is required perfectly match the input and output, other improved attenuator designs are used.

In the next tutorial about **Passive Attenuators**, we will look at another type of attenuator design called the **T-pad Attenuator** that uses three resistive components to produced a balanced attenuator.

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Hi Sir,

I believe there was a mistake on the problem solving part. 10^{12/20}=3.98

Yeah, thanks Chenelyn, 🙂

Can you explain where the equations came from? I’ve tried to find an explanation in the prior articles and can’t find one.

Also can you explain “looking in”? I see the comment you made below, but you didn’t really explain what the term meant

As explained, attenuators are resistive networks designed to provide impedance matching between a source and a load to reduce a signals amplitude or for maximum power transfer. All the equations used are standard attenuation equations with Z representing the impedances and k as an attenuation constant giving a relative current, voltage, power or energy decrease in decibels.

“looking in” means exactly that, looking into the attenuator circuit from the outside, either as a source feeding the attenuator or as the load connected to it.

Hi,

Please forgive me if I am committing an error beforehand, just trying to help.

I liked very much the article, but I find it a bit confusing in some aspects, specially for the newbie. This led me to research lots of hours to understand what I was missing, and why my calculations were failing, and i believe I have found some imprecision’s that could be corrected/improved to make things clear:

1 – The author talks frequently about Attenuation, while often it’s referring to Vout/Vin, which is gain (for positive) or damping for (negative). It would be Attenuation if we were talking about the inverse, Vin/Vout. This might seem a bit of preciosity and play with words, but that actually made a pretty big confusion while doing the calculations for the Attenuation and applying the K-factors substitutions.

2- Also there’s a section that could be improved, which is the algebra approach to go from the divider formulas, work it, and get where it needs to directly apply the K’s substitutions (ex: K-1). This could be improved with an actual example.

3 – Also the author refers frequently the directions when calculating impedances: “towards”, “in the direction of Z…”, “in the direction of R1”, etc…

Sometimes this can be confusing, specially if you’re dyslexic like me. I would suggest adding some extra diagrams to explain this clearly.

I hope this helps!

Cheers,

Artur

Hello Artur, thank you for this.

1. By definition, an attenuator is a circuit of pure resistors used to reduce a signals strength without introducing distortion. Attenuation is the reduction of gain (negative gain) in decibels, that is, the output voltage is less than the input voltage. Damping relates to the response of a circuit when subjected to a change. It can be under-damped, over-damped or critically damped.

2. K is the impedance factor used for calculating attenuator loss and is the ratio of voltage, current, or power for a given value of attenuation in decibels. A simplified table of common K values is given as an example in the previous tutorial.

3. Impedances are calculated “looking” into the attenuator circuit either from the source end or the load end to ensure correct impedance matching for maximum power transfer.

If anyone is reading this, please please please HELP ME!

I salvaged some L pads from old and small speakers. I hooked them up to the tweeters on other speakers but the dial gives a really narrow response. I mean on the dial, only 1/10th of it is used to increase or reduce the volume. What can I do to increase the overall function? Thank you

my guess would be that you have a mismatch in the intended ohm load, and your actual ohm load…i.e. the l pads you are using are intended for a series crossover, where the l pad is meant for a 4 ohm driver, and you are using an 8 ohm driver, etc

Thanks for replying. I think you’ve nailed it, the Lpads came from 4 ohm speakers iirc. So if I solder a 4 ohm resistor(???) in line, will that fix the problem? Or just give up fiddling with I know nothing about, lol!

How large a hole do I need to install an l-pad????