Does utilizing a xover reduce watts needed?

[pewternhrata]

Does a speaker take less power to drive if a xover is set in the receiver? I'd assume it would since lower frequencies need more?

 

[lovinthehd]

To an extent as the avr's amps aren't being asked to cover as much....precisely how much or how to estimate I'd like to see more about, some say it's significant, others say not.

 

[pewternhrata]

To an extent as the avr's amps aren't being asked to cover as much....precisely how much or how to estimate I'd like to see more about, some say it's significant, others say not.

Would be interesting. when utilizing a SPL calculator, wouldn't you be getting more spl with less power? But by how much.
I'd like to see specs at full range vs 80 vs 120

 

[jinjuku]

Bottom line is a passive crossover will reduce the energy delivered directly to the transducer.

 

[Steve81]

Does a speaker take less power to drive if a xover is set in the receiver? I'd assume it would since lower frequencies need more?

Yes...and no. Obviously with some part of the audio band being filtered out, demand on the receiver's amps will be overall reduced. On the other hand, it's possible for a segment of audio to be quite high in level without having any appreciable low end energy to filter out.

 

[highfigh]

Does a speaker take less power to drive if a xover is set in the receiver? I'd assume it would since lower frequencies need more?

No. A speaker requires a certain amount of energy to make it do something but it will handle more power if the crossover removes energy in a frequency range where the speaker doesn't work well- it's the whole reason for crossovers of all types, anyway. There's no reason to try to force a driver to do something it can't and it will usually fail if this is attempted.

Lower frequencies require more power partially because humans don't hear as well at low (or very high) frequencies and because bass requires compression of the air, mids and highs don't move the air as forcefully, or as much physical volume (not audio volume level).

 

[pewternhrata]

No. A speaker requires a certain amount of energy to make it do something but it will handle more power if the crossover removes energy in a frequency range where the speaker doesn't work well- it's the whole reason for crossovers of all types, anyway. There's no reason to try to force a driver to do something it can't and it will usually fail if this is attempted.

Lower frequencies require more power partially because humans don't hear as well at low (or very high) frequencies and because bass requires compression of the air, mids and highs don't move the air as forcefully, or as much physical volume (not audio volume level).

So if a receiver is outputting 20 watts, the speakers will play at the same spl regardless if the receiver is set to full range or have a xover set?

 

[Johnny2Bad]

Most passive crossovers actually eat some power, but as that is at the higher frequencies (mids, if present, tweeter, very commonly) the net result is no change; almost all of you power will be used driving the Low Frequency driver so padding the higher frequency drivers has no effective increase or reduction.

A sub configured in certain ways can reduce the power needed by the mains, but the mains themselves still are using up significant power at LF (even if that is only at, say, 80 Hz and above) versus higher frequencies, so the net effect is as above.

If you are using an electronic crossover then the power requirements will be based on the frequencies that are sent to each driver. But somewhere along the line you still need power to drive the LF of the system. So at least one (two for stereo, and so on) amp is going to be using the same power as will a full range system. Those that are tasked with higher frequencies do not need to be as highly rated, one rule of thumb in Pro Audio is 10% to the highs vs the lows (eg 100W and 10W).* If the tweeter is crossed over high enough you could get by with 1% in a home HiFi but that is rarely the case with Pro Audio as they cover more of the mids with the HF driver.

* In practice, it's rarely that low, simply because a 100W Pro Audio amp is considered "small" and they use what they have. But some permanent installations in smaller rooms, say studio monitors, might follow the rule of thumb.

 

[Swerd]

Does a speaker take less power to drive if a xover is set in the receiver? I'd assume it would since lower frequencies need more?

Yes, but for a different reason. And like others have already said, the difference in power consumption isn't usually large.

Passive crossovers in loudspeakers have a large inductor coil for the woofer. It acts as a low-pass filter. The number of winds in the coil determine at what frequency the filter acts. The more winds there are in the coil, the lower the low-pass frequency, and the higher the inherent resistance. This resistance, called DC resistance (DCR), can consume some power. The thickness of the coiled wire also has an effect on DCR, the thinner the wire the greater the DCR. You can find inductor coils with wire gauge ranging from as thin as 20g to as thick as 12g.

As the wire gets thicker the price goes up, and the coil gets physically larger. As you can imagine a 15 or 16g inductor coil can take up more space inside a speaker cabinet than a 19 or 20g coil, while they have the same inductance value.

If a crossover is placed upstream from the amp, it can be much smaller because the voltage and current of the music signal before amplification is much smaller. As a result, the losses due to DCR are much smaller. These crossovers are often called active crossovers because they are built to require an external source of electric power.

So, in theory, a passive crossover in a speaker can eat up more power than an active crossover, however the difference isn’t so great that turning up the volume can’t correct for it. In some cases, a more powerful amp may be needed, but this is less common.

This a generalization, and like all generalizations, you can find exceptions. In a 3-way speaker with a low crossover frequency between the woofer and mid-range, less than about 400 Hz, the woofer high-pass inductor coil will be very large, and depending on the wire gauge, expensive. This is why the crossovers for subwoofers are placed upstream from their amps. It also allows you to choose different crossover frequencies.

 

[Verdinut]

No. A speaker requires a certain amount of energy to make it do something but it will handle more power if the crossover removes energy in a frequency range where the speaker doesn't work well- it's the whole reason for crossovers of all types, anyway. There's no reason to try to force a driver to do something it can't and it will usually fail if this is attempted.

Lower frequencies require more power partially because humans don't hear as well at low (or very high) frequencies and because bass requires compression of the air, mids and highs don't move the air as forcefully, or as much physical volume (not audio volume level).

Even if you don't EQ for the reduced hearing acuity at low frequencies, more power from an amplifier is required. As an example, if the frequency to be reproduced is halved, the cone excursion is increased by a factor of four.
That explains why a bigger cone area will displace more air and be more efficient in the reproduction of very low frequencies.

 

[Verdinut]

Yes, but for a different reason. And like others have already said, the difference in power consumption isn't usually large.

Passive crossovers in loudspeakers have a large inductor coil for the woofer. It acts as a low-pass filter. The number of winds in the coil determine at what frequency the filter acts. The more winds there are in the coil, the lower the low-pass frequency, and the higher the inherent resistance. This resistance, called DC resistance (DCR),can consume some power. The thickness of the coiled wire also has an effect on DCR, the thinner the wire the greater the DCR. You can find inductor coils with wire gauge ranging from as thin as 20g to as thick as 12g.

As the wire gets thicker the price goes up, and the coil gets physically larger. As you can imagine a 15 or 16g inductor coil can take up more space inside a speaker cabinet than a 19 or 20g coil, while they have the same inductance value.

If a crossover is placed upstream from the amp, it can be much smaller because the voltage and current of the music signal before amplification is much smaller. As a result, the losses due to DCR are much smaller. These crossovers are often called active crossovers because they are built to require an external source of electric power.

So, in theory, a passive crossover in a speaker can eat up more power than an active crossover, however the difference isn’t so great that turning up the volume can’t correct for it. In some cases, a more powerful amp may be needed, but this is less common.

This a generalization, and like all generalizations, you can find exceptions. In a 3-way speaker with a low crossover frequency between the woofer and mid-range, less than about 400 Hz, the woofer high-pass inductor coil will be very large, and depending on the wire gauge, expensive. This is why the crossovers for subwoofers are placed upstream from their amps. It also allows you to choose different crossover frequencies.

Also, if the DCR of the inductor in series with the woofer is high, added to the resistance of the cable between the amplifier and the crossover, the resulting reduced damping factor may be problematic if the woofer has a high Qts. With some amplifiers with a high output resistance, there wouldn't be a good transient response as the amp would be unable to control adequately the cone movement of the woofer.
When you have to use a heavy gauge wired air core inductor, the cost of each of those coils can be over US $200, and that also adds to the weight of the cabinet. Then, the money saved by avoiding those expensive coils, covers most of the cost of a pro audio active crossover. OTOH, you need one more amplifier to drive the high frequency drivers.

 

[Verdinut]

Solen, a Canadian crossover parts manufacturer, loudspeaker and other audio products distributor, makes air core inductors with wire gauges as big as 10AWG and an 18mH one cost over US $200. By the way, it was the first company to manufacture air core inductors.
Dennis Murphy uses their capacitors in some of his speaker cabinets.

www.solen.ca

 

[highfigh]

Even if you don't EQ for the reduced hearing acuity at low frequencies, more power from an amplifier is required. As an example, if the frequency to be reproduced is halved, the cone excursion is increased by a factor of four.
That explains why a bigger cone area will displace more air and be more efficient in the reproduction of very low frequencies.

I was referring to the range where a driver doesn't operate well, if at all.