Find bridging and shunt resistor values for a bridged-tee attenuator using the Bridged-Tee Attenuator Calculator, built from line impedance and target decibel attenuation values reported.
How a Bridged-Tee Network Drops Signal Level Without Changing Impedance
A bridged-tee attenuator adds two resistors to a transmission line that’s already terminated correctly, and drops the signal by a set number of decibels while leaving the impedance the source and load see unchanged.
That’s the entire appeal over building a T-pad or Pi-pad from scratch: no new termination resistors, just a shunt leg and a bridging leg layered onto an existing 50 Ω or 75 Ω line. Both resistor values fall out of one number — the voltage ratio implied by the decibel figure you’re targeting.
$$K = 10^{A_{dB}/20}$$
$K$ is the voltage ratio, $A_{dB}$ the attenuation in decibels you want the pad to produce. From $K$ and the line impedance $Z_0$, the two resistors are:
$$R_{bridge} = Z_0(K-1) \qquad R_{shunt} = \frac{Z_0}{K-1}$$
$R_{bridge}$ connects straight across from input to output, in parallel with the two $Z_0$ resistors already terminating the line. $R_{shunt}$ ties the midpoint down to ground. The two values are reciprocal images of each other scaled by $Z_0^2$ — multiply them together and you get $Z_0^2$ every time, which is a quick way to sanity-check a hand calculation.
A 10 dB Pad on 50 Ω Coax
Take a 50 Ω line — the standard for most RF test equipment, coax, and antenna feedlines — and a 10 dB pad, a common single-step attenuation value. $K = 10^{10/20} = \sqrt{10} \approx 3.1623$. That one ratio drives both resistors and everything else the calculator reports.
$R_{bridge} = 50 \times 2.1623 \approx 108.11\ \Omega$, and $R_{shunt} = 50 / 2.1623 \approx 23.12\ \Omega$. The shunt leg also converts to a conductance of 43.24 mS, and the attenuation expressed in nepers instead of decibels comes to $\ln(3.1623) \approx 1.15$ Np.
| Metric | Value |
|---|---|
| Bridging resistor | 108.11 Ω |
| Shunt resistor | 23.12 Ω |
| Shunt conductance | 43.24 mS |
| Attenuation (nepers) | 1.15 Np |
| Equivalent T-pad shunt | 35.14 Ω |
| Equivalent T-pad series (each arm) | 25.98 Ω |
| Equivalent Pi-pad shunt (each end) | 96.25 Ω |
| Equivalent Pi-pad series | 71.16 Ω |
| Power ratio | 10.00 W/W |
| S21 (voltage) | 0.32 V/V |
The T-pad and Pi-pad figures aren’t a different attenuator — they’re the equivalent resistor values you’d need if you built the same 10 dB pad as a standalone T or Pi network instead of a bridged-tee added to an existing line. Same electrical result, different parts count and different assumptions about what’s already terminated.
Reading the Power Ratio Against What Decibels Actually Mean
There’s no compliance threshold built into this calculation the way there is with an ampacity table — any positive dB value produces a valid resistor pair. What’s worth checking is whether the power ratio matches what you’d expect: decibels for power follow $10\log_{10}(P_{in}/P_{out})$, so a 10 dB pad always means a 10x power drop, a 20 dB pad always means 100x, regardless of what impedance you entered. If your power ratio doesn’t land on that pattern, the dB input is probably wrong, not the formula.
The S21 figure — 0.32 V/V in the example above — is the same information in the format RF engineers actually plug into scattering-parameter tables and simulation software. It’s just $1/K$, the inverse of the voltage ratio.
Why Line Impedance Scales Linearly and dB Doesn’t
$Z_0$ has a simple relationship to every output: double it, and both resistor values double with it, no exceptions. Attenuation behaves differently, because $K$ comes from an exponential and the resistor formulas divide by $(K-1)$.
Near low dB values, small changes swing the resistor values sharply; at higher attenuation, the same size change barely moves them. A jump from 1 dB to 2 dB reshapes the network far more than a jump from 20 dB to 21 dB does.
That’s also why the calculator won’t compute below roughly 0.001 dB. As attenuation approaches zero, $K$ approaches 1, and $R_{shunt} = Z_0/(K-1)$ heads toward infinity — which is physically correct, since a network attenuating almost nothing needs an almost-infinite shunt resistor, not a real, buildable component. Rather than return a meaningless number, the code stops and flags it.
Which equivalent network is worth building depends entirely on what you’re starting from. The bridged-tee itself only makes sense when the line is already terminated in $Z_0$ on both sides — it adds two resistors to something that already works. The T-pad and Pi-pad equivalents don’t assume any existing termination and are what you’d reach for building a standalone attenuator from bare resistors.
Frequently Asked Questions
What’s the difference between a bridged-tee attenuator and a regular T-pad?
A T-pad is a standalone three-resistor network with no assumptions about existing termination. A bridged-tee only needs two added resistors, but only works correctly when the line on both sides is already terminated in $Z_0$ — it’s piggybacking on termination that’s already there rather than replacing it.
Why do the bridging and shunt resistor values multiply to Z0 squared?
Because $R_{bridge} = Z_0(K-1)$ and $R_{shunt} = Z_0/(K-1)$ are algebraic reciprocals scaled by the same $Z_0$ term, so their product always cancels the $(K-1)$ factor and leaves $Z_0^2$. It’s a fast way to check a manual calculation without recomputing $K$.
Does a bridged-tee attenuator work on any impedance line?
The math holds for any positive $Z_0$, but 50 Ω and 75 Ω are what almost all real coax, RF test gear, and antenna feedlines use. Entering an unusual impedance produces valid numbers that won’t match any physical line you’re likely to be working with.
Why does the calculator refuse very small dB values?
As attenuation approaches 0 dB, the voltage ratio $K$ approaches 1, and the shunt resistor formula divides by a number approaching zero, sending the required resistance toward infinity. That’s a real physical limit, not a computing restriction — a near-zero attenuator genuinely needs an impractically large resistor.
How do I convert the bridged-tee values into an equivalent Pi-pad?
Pi-pad shunt resistance is $Z_0(K+1)/(K-1)$ at both ends, and series resistance is $Z_0(K^2-1)/2K$ between them. Both come from the same $K$ ratio as the bridged-tee, just arranged as a self-contained three-resistor network instead of two resistors added to an existing line.
Is higher dB attenuation always better for isolating a signal?
Not automatically — more attenuation means more signal loss, which can push a weak signal below a receiver’s noise floor. The right dB value depends on how much headroom the source has and how sensitive the downstream equipment is, not on maximizing isolation for its own sake.
These values assume ideal, purely resistive components on a line that’s actually resistive at your operating frequency — they don’t account for resistor tolerance, parasitic inductance or capacitance at high frequencies, or the power dissipation rating the resistors need to handle your signal level.