Updated May 22,
here to go to our main page on attenuators
here to go to our page on attenuator math
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:
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 "R1A"
could get hotter than "R1B" (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!
We solved the fractional dissipation
in terms of R1 and Z0. Note that R1A and R1B are
different resistors, but of identical value R1. Here's the
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 R1A
and R2. At infinite attenuation, the first resistor R1A
gets all the dissipation. R1B never dissipates much of
the power load at any attenuation value, which makes sense because
it's at the attenuator's output.
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 R1A 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.
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.
Coming someday, but don't hold