Tweeter Padding
- Mr. Peabody
Attenuation of tweeter voltage is accomplished using a voltage divider, which is to say, by placing a resistor in series with the tweeter. The tweeter’s proportional share of the voltage will be the same as its proportional share of the combined impedance. The problem that then arises is that, since the tweeter’s impedance varies with frequency, its share of the voltage will likewise vary with frequency. To mitigate this unwanted effect, a second resistor may be placed directly across the tweeter. Since this will lower the tweeter’s effective impedance, it then is necessary to apply the same proportional reduction to the series resistor, such that the nominal balance of impedance is the same as before. This begs the question of what is accomplished by placing the second resistor in parallel with the tweeter. The answer is that the tweeter’s effective impedance is better behaved, vs. the impedance of the raw tweeter. This curtails the spoiling of the frequency response by the series resistor, however the full benefit is greater than this, because the behavior of the crossover moves nearer to its ideal, theoretical behavior.
Since there are two undetermined resistors, two independent equations are needed for finding a unique solution. One of these two equations corresponds to the net impedance requirement. When the crossover hasn’t yet been designed and assembled, this value may be chosen more freely, e.g., 4 Ohms. When lowering the tweeter level for an existing speaker and wanting to avoid wholesale modifications to the crossover, it is essential for the net impedance to match the tweeter’s impedance, at the crossover point. Let’s use meaningful variable names:
Rtwtr - Impedance of tweeter (in Ohms) at crossover frequency
Rprl - Resistance added in parallel with the tweeter
Rtwtr_prl - Combined impedance of the tweeter and the parallel resistor
Rsrs - Resistance added in series with the parallel combination
Rnet - Net impedance (possibly but not necessarily equal to Rtwtr)
The first equation is easy:
(I) Rnet = Rsrs + Rtwtr_prl where Rtwtr_prl = Rprl x Rtwtr / (Rprl + Rtwtr)
The other equation corresponds to the requirement for tweeter voltage to be reduced according to some given correction factor. Let’s suppose that the voltage correction factor is .631 (corresponding to 4 dB of excess tweeter sensitivity). Since .631 is equal to the ratio of the new tweeter voltage to the original tweeter voltage, this is also true:
(II) .631 = Rtwtr_prl / Rnet
If it is not apparent why this is true, note that if you have a resistance R1 and you place a second resistance R2 in series with R1, the new voltage across R1 will equal the original voltage multiplied by the ratio R1/(R1 + R2); this obviously means that the voltage ratio is equal to R1/(R1 + R2). Note the assumption that the voltage across R1 and R2 will be the same as it was previously across R1. This won’t hold true if there is other impedance in series with R1, unless the net resistance is kept the same as the resistance of R1. This is why, when padding a tweeter with no further modifications to the existing crossover, the net resistance needs to match the resistance of the tweeter itself. The only way this is possible, when adding a resistor in series with the tweeter, is by adding a second resistor in parallel with the tweeter.
We may thus substitute ‘.631 Rnet’ into (I), in place of Rtwtr_prl:
Rnet = Rsrs + .631 Rnet => Rsrs = Rnet(1 - .631)
Rsrs is now solved, because Rnet is a given value, possibly 4 Ohms, or possibly equal to Rtwtr, but it is a given value regardless. Interestingly, Rsrs is determined by Rnet and by the voltage correction factor. If this seems suspect, note that when you place R2 in series with R1, that if you want the voltage across R1 to decrease to 1/4 its original value, that R2 will need to be 3x greater than R1, and since R1 will be 1/4 as great as the sum of R1 and R2, it follows that R2 will be 3/4 as great as the net series resistance, i.e., no matter the value of R1.
To find the other resistor, Rprl, we substitute Rnet(1 - .631) into (I) and solve for Rprl:
Rnet = Rnet(1 - .631) + Rtwtr_prl =>
.631 Rnet = Rtwtr_prl =>
.631 Rnet = Rprl Rtwtr / (Rprl + Rtwtr) =>
.631 Rnet Rprl + .631 Rnet Rtwtr = Rprl Rtwtr =>
Rprl (.631 Rnet - Rtwtr) = -.631 Rnet Rtwtr =>
Rprl = .631 Rnet Rtwtr / (Rtwtr - .631 Rnet)
We may summarize and generalize the solution, using VCF (Voltage Correction Factor) in place of the specific value .631:
Rsrs = Rnet(1 - VCF) where VCF = 10^(excess_tweeter_sensitivity_in_dB / 20)
Rprl = 1 / (1/(Rnet VCF) - 1/Rtwtr )
For the particular case where Rnet = Rtwtr:
Rprl = Rtwtr / (1/VCF - 1)
To continue with the specific example, where VCF = .631, let’s suppose that Rtwtr = 5 Ohm, and that we want for Rnet to be the same as Rtwtr:
Rsrs = Rnet(1 - VCF) => Rsrs = 5 (1 - .631) => Rsrs = 1.85 Ohm
Rprl = Rtwtr / (1/VCF - 1) => Rprl = 5 / (1/.631 -1) => Rprl = 8.55 Ohm
The nominal value for Rtwtr_prl will be 8.55 x 5 / ( 8.55 + 5) = 3.15 Ohms (which we could also have calculated by subtracting 1.85 from 5). If we guesstimate that tweeter impedance at the resonance point is 30 Ohms (6x the nominal value), Rtwtr_prl at the impedance peak will be 8.55 x 30 / ( 8.55 + 30) = 6.65 Ohms, slightly more than 2x the nominal value. To accurately know how strong the effect will be, at lowering the response peak, we would need to carry out the math that applies to the actual crossover. Short of doing this, we can get a rough sense of how strong this effect typically will be, by guesstimating the amount of equivalent series impedance introduced by the high-pass filter at the impedance peak (typically more than one octave below the crossover point). This will allow us to estimate the response peak (relative to the ideal response at that frequency), for both the raw tweeter and the padded tweeter.
Using 50 Ohms for the equivalent series impedance of the high-pass filter at the resonance point, the ideal response (if the tweeter impedance were the nominal 5 Ohms) is about -21 dB. But if the tweeter impedance is 30 Ohms at this particular frequency, the response will be -8.5 dB, or about +12 dB vs. the ideal response at this frequency. To fairly compare this with what will happen with the padded tweeter, we need to adjust the 50 Ohm value so that the ideal response will be -21 dB as before (using 3.15 Ohms for the tweeter and not forgetting about the other 1.85 Ohms). If we use 30 Ohms instead of 50, then the ideal response at the peak will be -21 dB, as before. Using 6.65 Ohms and 36.65 Ohms (in place of 3.15 and 35), the response at the peak will be about -15 dB, or about +6 dB relative to the ideal response. (I also did this using a more obvious approach, and when I did, I got the same result.)
+6dB is much better than +12 dB, and we mustn’t lose sight of the fact that at this frequency the woofer will supply the bulk of the sound pressure. Regardless there can be little doubt as to the advantage of choosing a tweeter with sensitivity at least several dB greater than the woofer, so that it will be possible to place a resistor in parallel with the tweeter. The same advantage applies to the midrange driver in a 3-way speaker. It isn’t necessary for the tweeter to be more sensitive than the midrange, only for both of them to be more sensitive than the woofer. With both the tweeter and midrange, the effect includes suppression of the rise in impedance caused by the inductance of the coil.
This simple technique can’t generally be applied to the woofer because you don’t generally want to put a resistor in series with a woofer. This would obviously attenuate the response of the woofer overall, but would also introduce a response peak at the woofer’s resonance, which otherwise does not occur notwithstanding the peak in impedance. (The reason this doesn’t occur is that at the resonance frequency, the woofer isn’t subjected to the series impedance of a high-pass filter. An exception, where there will nevertheless be a response peak at the woofer’s resonance, is when the amplifier’s output impedance is not adequately low, i.e., not less than approximately 1/10 the driver impedance.)
In cases where the woofer’s impedance is unusually great, you could theoretically place a high-wattage resistor in parallel with the woofer, so long as the effective impedance remains adequately high such that there won’t be an issue with the amplifier overheating due to the additional current. However there is no substantive reason for doing this. The only ostensible reason would be to mitigate the rise in impedance at higher frequency (caused by the inductance of the coil), at frequency where the woofer is subjected to the series impedance of the low-pass filter. The right way to deal with this is with a Zobel network. The Zobel is similar, i.e., a resistor is placed in parallel with the woofer, except that there is also a capacitor in series with this resistor. In a Zobel, the resistor value is the same as the DC resistance of the voice coil. The capacitance value is obtained by dividing the coil inductance by the square of the DC resistance. At very high frequency, the Zobel has the same impedance that the voice coil has at very low frequency. As frequency moves higher from DC, the increase in the impedance of the coil is balanced by the decrease in the impedance of the Zobel, such that the effective impedance of the coil remains constant with increasing frequency.
(Character Map ftw, unless your keyboard layout happens to include Greek letters. If your system knows Emoji, you bet it'll support these old hats, which have been in Unicode since version 1.0.)
