Level, filters & alignment

L-pad attenuator

Two resistors to bring a horn down to the woofer’s level at constant load, with E12 values and power split.

Sheet: 22 / 25
Units: Metric
Air: 20 °C

Fig. 1L-pad between crossover and horn driver

Series $R_1$: 5.47Ω
Parallel $R_2$: 3.70Ω
Nearest E12 $R_1$: 5.60Ω
Nearest E12 $R_2$: 3.90Ω
With E12 parts: 9.93dB
Load seen: 8.22Ω

Fig. 2Where 50 W goes

68 %22 %

R1 34.2 WR2 10.8 WDriver 5.00 W

Choose resistor power ratings at least twice the continuous dissipation shown, and use non-inductive types.

How it works

Compression drivers on horns are typically 10 to 15 dB more sensitive than the woofers they are paired with. The L-pad is the standard cure: two resistors that throw away the surplus level while presenting the crossover with the same load impedance as the bare driver, so the filter does not have to be redesigned.

Design equations

Let $K$ be the voltage ratio corresponding to an attenuation of $A$ decibels. Requiring the pad's input impedance to equal the driver impedance $Z$ fixes both resistors:

K = 10^{-A/20}, \qquad R_1 = Z\,(1-K), \qquad R_2 = \frac{Z\,K}{1-K}

$R_2$ in parallel with the driver makes $KZ$ ; adding $R_1$ in series restores $Z$ , and the voltage divider between the two parts gives exactly $K$ .

Where the power goes

P_{R_1} = (1-K)\,P, \qquad P_{R_2} = K(1-K)\,P, \qquad P_{driver} = K^{2}P

At 10 dB of attenuation the driver receives a tenth of the power and the series resistor burns more than two thirds of it. The resistors, not the driver, set the power handling.

Practical notes

The read-outs give the nearest E12 preferred values and the attenuation and load they actually produce.
$Z$ should be the driver's impedance in the crossover region, not the nominal figure on the label. On a horn it is often well above nominal.
The parallel resistor also damps the impedance peaks of the driver, which makes the crossover behave closer to theory. For that reason many designers prefer a pad to a series resistor alone.
To equalise a constant-directivity horn at the same time, bypass $R_1$ with a capacitor: see the CD-horn equaliser sheet.

Loading sheet…

How it works

Design equations

Let $K$ be the voltage ratio corresponding to an attenuation of $A$ decibels. Requiring the pad's input impedance to equal the driver impedance $Z$ fixes both resistors:

K = 10^{-A/20}, \qquad R_1 = Z\,(1-K), \qquad R_2 = \frac{Z\,K}{1-K}

$R_2$ in parallel with the driver makes $KZ$ ; adding $R_1$ in series restores $Z$ , and the voltage divider between the two parts gives exactly $K$ .

Where the power goes

P_{R_1} = (1-K)\,P, \qquad P_{R_2} = K(1-K)\,P, \qquad P_{driver} = K^{2}P

At 10 dB of attenuation the driver receives a tenth of the power and the series resistor burns more than two thirds of it. The resistors, not the driver, set the power handling.

Practical notes

The read-outs give the nearest E12 preferred values and the attenuation and load they actually produce.
$Z$ should be the driver's impedance in the crossover region, not the nominal figure on the label. On a horn it is often well above nominal.
The parallel resistor also damps the impedance peaks of the driver, which makes the crossover behave closer to theory. For that reason many designers prefer a pad to a series resistor alone.
To equalise a constant-directivity horn at the same time, bypass $R_1$ with a capacitor: see the CD-horn equaliser sheet.

L-pad attenuator

Inputs

How it works

Design equations

Where the power goes

Practical notes

Inputs

How it works

Design equations

Where the power goes

Practical notes

§ How it works

Design equations

Where the power goes

Practical notes

§ How it works

Design equations

Where the power goes

Practical notes

How it works

How it works