Updated May 22, 2010

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On this page we will show you the equations for calculating the fractional dissipation of the individual resistors in an attenuator.

Here's an index to this page:

Tee attenuator

Pi attenuator

Reflection attenuator

Bridged tee attenuator

The overall dissipation calculation for a lumped-element attenuator seems almost trivial, what ever you put in, minus what comes out is what is dissipated by the resistor network. However, the resistors in the pi or tee network do not share the burden of heat equally, you need to consider "R_1A" could get hotter than "R_1B" (see figure below), even though they have the same ohm value. The resistor that is closest to the power generator gets hit with the most power; if this is a surprise to you, perhaps you should change careers, maybe become a microwave vendor rep or become a "product assurance engineer" at a big and wasteful company. If you are designing an attenuator for high power, unless these resistors are located incredibly close to each other, you'll need to think about the dissipation in each resistor. It's a nasty algebra problem that we've solved for you!

Tee attenuator

We solved the fractional dissipation in terms of R₁ and Z_{0. Note that R1A and R1B are different resistors, but of identical value R1}. Here's the equations:

Below are plots of the fractional dissipation of the three resistors in a tee attenuator, versus attenuation value, normalized to an input power of unity (one watt). In order to solve for actual dissipation as a function of input power, you need to multiply the curves by input power. Let's look at the curves... up to about 6 dB, the dissipation is approximately equal in R_1A and R₂. At infinite attenuation, the first resistor R_1A gets all the dissipation. R_1B never dissipates much of the power load at any attenuation value, which makes sense because it's at the attenuator's output.

Pi attenuator

Funny thing about the pi attenuator... although the resistor values are very different than those of the tee attenuator, the fractional dissipation is exactly the same! The figure below explains why R_1A dissipates the same in a tee and a pi, the remaining two resistors can be shown to have equivalent dissipation as well.

So if you were lazy, you could stop here and never calculate the pi resistor fractional dissipation equations and just substitute the results of the tee-type attenuator versus attenuation. Not us! We've solved the equations for fractional dissipation for the pi attenuator, below, but we won't repeat the plot because it would be the same.

It's time for another Microwaves101 rule of thumb! This applies to both tee and pi attenuator:

For a three dB pad (50% power dissipation), the input resistor must be sized to handle 20% of the maximum intended input power (thanks for the correction, Dave!) For a ten dB pad (90% power dissipation), the input resistor must be sized to handle 1/2 of the maximum intended input power. For a 20 dB pad (99% power dissipation), size it for 80% of the maximum input power. For higher attenuation values, size the input resistor for the full RF input power.

Reflection attenuator

For completeness, we present a plot of the fractional dissipation of resistors in a reflection attenuator. The math is trivial, theoretically the two resistors spit the dissipated power 50/50.

Bridged tee attenuator

Coming someday, but don't hold your breath!