Page 7
Individual Driver and Summed System Response
The above graph was produced with the microphone at a distance of 45", on the tweeter’s axis. It was preferred to
use the slightly farther then 1 meter (39.37") distance to be more in the far field, to get a better representation of how
the drivers will blend. The response is within the +/- 2.0 dB window to 12KHz. In fact, the response is commendably
flat from 300 Hz to 3.5KHz. It can be seen that the acoustic crossover occurs at 2.1KHz. These curves have been
smoothed 1/6th octave.
The woofer’s peak at 3,200 Hz is still present, and it should be noted that a parallel notch filter could be used in the
low pass circuit to reduce it’s effect. One was omitted to keep in the spirit of a low cost design, since this would add
three more crossover components per speaker. The peak’s audibility was found to be relatively benign however.
It’s worth discussing the parallel notch filter because it is a handy circuit that is used to control frequency response
peaks. A parallel notch filter consists of a capacitor, an inductor, and a resistor all wired in parallel with each other,
then this circuit is placed in series with the driver.
For a 8 ohm woofer with a typical “peak” (about 5 dB), use the following formula for a starting reference...
C = 1/ (33 * freq of peak)
L = .025 / (freq of peak squared * C)
With the 6-1/2" woofer, mounted on the BR-1 baffle, the peak is centered at 3,200 Hz. So …
C = 1 / (33 * 3200)
C = .0000094 farads or 9.4 uF
L = .025 / (10240000 * .0000094)
L = .000259 henries or .259 mH
To find the value of R, you first must find the Q of the peak. Do this by finding the –3dB points on both sides of the peak.
For the 6-1/2" woofer, these –3dB points are 2,500 Hz and 3,700 Hz. Now use this formula to calculate the peak’s Q...
Q = freq of peak / (Fh – Fl)
Q = 3200 / (3700- 2500)
Q = 2.67
We can now calculate the resistor value by:
R = Q / square root of C/L
R = 2.67 / square root of .0000094 / .000259
R = 14.05 ohms
