Find exact resistor values for Bridged-T, O-pad, and U-pad networks using the Balanced Attenuator Calculator, built from line impedance and target decibel attenuation, plus voltage.
The Resistor Math Behind a Bridged-T Attenuator
A balanced attenuator drops signal level by a set number of decibels while keeping the impedance seen by the source and the load unchanged in both directions. That second part is what separates it from a plain voltage divider — a divider changes level but also changes impedance, which causes reflections on any line long enough to care about matching. Everything below starts from one number: the voltage ratio implied by the decibel figure you enter.
$$K = 10^{A_{dB}/20}$$
$K$ is the voltage ratio, $A_{dB}$ is the attenuation you want in decibels. Squaring $K$ gives the power ratio, and its natural log gives the same attenuation expressed in nepers — three different units for the same physical loss. For the Bridged-T topology, the two resistor values come straight from $K$ and the line impedance $Z_0$:
$$R_1 = \frac{Z_0}{K – 1} \qquad R_2 = Z_0 (K – 1)$$
$R_1$ shunts the midpoint of the line to ground. $R_2$ bridges straight across from input to output, in parallel with the two existing $Z_0$ terminating resistors already on the line. That’s the appeal of a Bridged-T pad specifically: it adds only two components to a line that’s already terminated correctly, instead of building an attenuator from scratch out of three or four precision resistors.
Working a 600 Ω Line Down by 10 dB
Take a 600 Ω balanced line — the long-standing standard impedance for telephone circuits and much of professional balanced audio gear — and a 10 dB pad, a common step value sold in commercial attenuator kits. $K = 10^{10/20} = \sqrt{10} \approx 3.1623$. Every resistor value below falls out of that one number.
| Topology | Resistor | Value |
|---|---|---|
| Bridged-T | R1 (shunt) | 277.49 Ω |
| Bridged-T | R2 (bridging) | 1297.37 Ω |
| O-pad | Series (each leg) | 155.85 Ω |
| O-pad | Shunt | 421.64 Ω |
| U-pad | Series | 853.82 Ω |
| U-pad | Shunt (each end) | 1154.97 Ω |
Power ratio comes out to 10.00 W/W, voltage ratio to 3.16 V/V, and the same loss expressed in nepers is 1.15 Np — three labels on one physical fact, included because RF and telecom documentation don’t always agree on which unit to quote.
Why 600 Ω and Multiples of 10 dB Aren’t Arbitrary Choices
Judging Your Impedance Input
There’s no built-in threshold here the way there is with a code-mandated ampacity limit — pick the wrong impedance and the math still runs, it just won’t match anything real.
What’s useful instead is knowing what real systems actually use: 600 Ω is the classic telephone and pro-audio balanced-line figure, 150 Ω and 300 Ω show up in older telecom and instrumentation standards, and 50 Ω or 75 Ω dominate unbalanced RF work but rarely appear on balanced lines. If your line impedance doesn’t fall near one of those, double-check the source spec before trusting the resistor values that come out.
Judging Your dB Value
Commercial fixed pads are almost always sold in 3, 6, 10, and 20 dB steps, because those cover the common gain-staging jumps in audio and RF chains without needing a continuously variable attenuator. Entering something odd like 7.3 dB isn’t wrong — the formula doesn’t care — but it usually means you’re solving for a specific gain-matching problem rather than picking a standard part off a shelf.
Why the Attenuation Value Drives Everything, and Why Near-Zero dB Breaks It
Line impedance scales every resistor value directly — double $Z_0$ and every output doubles with it, no exceptions. Attenuation doesn’t behave that way. Because $K$ comes from an exponential and most formulas divide by $(K-1)$ or $(K^2-1)$, small changes in dB near the low end move resistor values sharply, while the same size change at high attenuation barely registers.
That’s also why the calculator refuses to compute below roughly 0.001 dB: as attenuation approaches zero, $K$ approaches 1, and $R_1 = Z_0/(K-1)$ heads toward infinity. Physically that’s correct — an attenuator built for zero loss needs an infinite shunt resistor, which isn’t a real component, so the code stops rather than returning a nonsense value.
Topology choice matters more than most of the input fields. Bridged-T needs only two added resistors on a line already terminated in $Z_0$, which keeps parts count and cost low.
O-pad and U-pad build the full network from calculated resistors with no dependency on existing termination, and split their values symmetrically across both conductors of the balanced line — O-pad splits the series arms in half, U-pad splits the shunt arms — which is the actual definition of “balanced” here: neither conductor carries a different resistor value than its twin.
Frequently Asked Questions
What makes an attenuator “balanced” instead of a regular pad?
A balanced attenuator applies matched resistor networks to both conductors of a differential pair symmetrically, so neither wire sees a different impedance path than the other. An unbalanced pad, built for coax or single-ended lines, only needs to manage one signal conductor against a common ground.
Why do O-pad and U-pad split resistor values in half?
Because the total resistance in the signal path has to be distributed evenly across both conductors to keep the line balanced. The O-pad halves its series resistors between the two legs; the U-pad halves its shunt resistors instead, which is the structural difference between the two topologies.
Can I substitute a T-pad or Pi-pad formula for a balanced line?
Not directly. T-pad and Pi-pad formulas assume a single-ended path with one series or shunt resistor per stage, while O-pad and U-pad split those same values across two symmetric conductors. Using unbalanced values on a balanced line throws off the symmetry the design depends on.
Why does the calculator reject very low dB values?
As attenuation approaches 0 dB, the voltage ratio approaches 1, and resistor values that depend on $(K-1)$ head toward infinity. That’s not a computing limit — it reflects a real circuit that would need an infinitely large resistor to attenuate almost nothing.
Which topology should I actually build?
Bridged-T is the cheapest option when the line is already correctly terminated in $Z_0$, since it only adds two resistors. O-pad and U-pad don’t rely on existing termination and are the more common choice when building a standalone attenuator module from scratch.
Does higher attenuation in dB always mean a bigger power loss?
Yes — decibels for power follow $10\log_{10}(P_{in}/P_{out})$, so every additional 10 dB corresponds to a tenfold increase in power ratio. A 20 dB pad drops 100 times the power of a signal passed straight through, versus 10 times for a 10 dB pad.
What impedance should I use for a standard telephone or audio line?
600 Ω is the long-standing default for telephone circuits and much professional balanced audio equipment, and it’s the right starting point unless your equipment’s datasheet specifies something else. Some instrumentation and older telecom gear use 150 Ω or 300 Ω instead.
These values assume ideal, purely resistive components and a line impedance that’s actually resistive at the frequencies you’re working with — they don’t account for resistor tolerance, parasitic reactance at high frequencies, or power dissipation limits on the resistors you end up using.