Level, filters & alignment

CD-horn equaliser

Bypass capacitor across the L-pad to restore the top octaves of a constant-directivity horn.

Sheet: 23 / 25
Units: Metric
Air: 20 °C

Fig. 1Driver roll-off, network and result

EqualisedDriver on CD hornNetwork alonedrag on plot to read

Bypass capacitor $C$: 8.31 µF
Nearest E12 $C$: 8.20 µF
Series $R_1$: 5.47Ω
Parallel $R_2$: 3.70Ω
Flat up to $f_p$: 11.1kHz
Extension: 1.66octaves

Fig. 2L-pad with bypass capacitor

How it works

A compression driver puts out roughly constant acoustic power up to its mass break point, typically 3 to 4 kHz, and 6 dB per octave less above it. An exponential horn hides this by beaming: it squeezes the falling power into a narrowing angle, so the on-axis response stays flat. A constant-directivity horn does not, and its on-axis response follows the power response downwards.

The usual passive fix costs one extra part. The driver has to be padded down to match the woofer anyway; bypass the series resistor of the L-pad with a capacitor and the attenuation fades away at high frequencies, giving a rising shelf that mirrors the roll-off.

Design

Start with the L-pad for the mid-band attenuation $A$ :

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

With the capacitor $C$ across $R_1$ , the network's response rises at 6 dB per octave between a zero at $f_z$ and a pole at $f_p$ :

H(s) = K\,\frac{1 + s R_1 C}{1 + s K R_1 C}, \qquad f_z = \frac{1}{2\pi R_1 C}, \qquad f_p = \frac{f_z}{K}

Placing the zero on the driver's break frequency $f_b$ cancels the roll-off:

C = \frac{1}{2\pi f_b R_1}

The correction runs out at $f_p$ , where the pad is fully bypassed. More attenuation therefore buys more extension: 10 dB gives 1.7 octaves, enough to carry a 3.5 kHz break to 11 kHz.

Practical notes

Find $f_b$ from a measurement of the raw driver on the horn; it is where the response is 3 dB below the mid-band plateau.
At the top of the band the pad disappears and the amplifier sees the driver directly, so impedance and power handling there are those of the bare driver.
Use a film capacitor. Its value is small and tolerance matters more than voltage rating.
Active crossovers achieve the same with a first-order shelving filter set to the same two frequencies.

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How it works

Design

Start with the L-pad for the mid-band attenuation $A$ :

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

With the capacitor $C$ across $R_1$ , the network's response rises at 6 dB per octave between a zero at $f_z$ and a pole at $f_p$ :

H(s) = K\,\frac{1 + s R_1 C}{1 + s K R_1 C}, \qquad f_z = \frac{1}{2\pi R_1 C}, \qquad f_p = \frac{f_z}{K}

Placing the zero on the driver's break frequency $f_b$ cancels the roll-off:

C = \frac{1}{2\pi f_b R_1}

The correction runs out at $f_p$ , where the pad is fully bypassed. More attenuation therefore buys more extension: 10 dB gives 1.7 octaves, enough to carry a 3.5 kHz break to 11 kHz.

Practical notes

Find $f_b$ from a measurement of the raw driver on the horn; it is where the response is 3 dB below the mid-band plateau.
At the top of the band the pad disappears and the amplifier sees the driver directly, so impedance and power handling there are those of the bare driver.
Use a film capacitor. Its value is small and tolerance matters more than voltage rating.
Active crossovers achieve the same with a first-order shelving filter set to the same two frequencies.

CD-horn equaliser

Inputs

How it works

Design

Practical notes

Inputs

How it works

Design

Practical notes

§ How it works

Design

Practical notes

§ How it works

Design

Practical notes

How it works

How it works