[Looneybomber]

Simplistically speaking, wouldn't a voice coil with a lower Re generate less heat and thus be able to handle more power thermally?


I realize decreasing Re involves different wire (Al or Cu), different length of wire, and also different guages, but negating all that, if we could somehow magically keep the wire the same, but just increase/decrease the resistance, the one with lower resistance should be able to handle more power correct?


If it is correct, why would companies still create woofers with 8ohm VCs? Money?

 

[digital desire]

The closer you get to 0, the closer to a direct short. More heat.

8 ohm is easier on lower grade amps. It is also what has been used for decades, back when amps were not nearly as stout.

 

[TheEAR]

With lesser amps the classic 8 Ohm load will put no stress on the output stage,with better quality,properly designed(with no corners rounded) amps the 4 Ohm load will get you double(around) the wattage you got into a 8 Ohm load.


Most high end speakers are designed to present a nominal 4 Ohm load as quality amps can drive 4 Ohm loads with ease.


For subwoofers 2 Ohm loads are common(car audio,may I add),and high quality amps(pro amps too) can drive 2 Ohm loads as long as they are not bridged. Most pro amps will be capable of driving a bridged 4 Ohm load. They will go thermal and shut down (or worse)if atempting to drive a 2 Ohm load bridged(delivering the rated or close power).


Most my subs present a nominal 4 Ohm,easy on the amp and I get the desired minimum wattage to reach levels I will not even dare listen at.


Also the driver's VC presenting say 1 Ohm will not be a great candidate for cool running,as the VC and the amp will work hard.The closer to a short,the more heat you get,only logical.

 

[michael hurd]

It's easier to think in terms of voltages. If an amp can deliver say 40 volts into an 8, 4 and 2 ohm load respectively: (not usually the case as the lower impedance sags the rail voltage down some)


The 8 ohm speaker with 40 volts applied results in applied power of 200 watts, the 4 ohm speaker with 40 volts would result in 400 watts, and the 2 ohm speaker would get 800 watts.


Voltage x current = watts... so


200 watts / 40 volts = 5 amps

400 watts / 40 volts = 10 amps

800 watts / 40 volts = 20 amps


It's hard to actually model what happens, but as you know speakers are never a constant impedance, they are changing with frequency.


As a speaker voice coil and leads heat up, they add more resistance, and start to choke off current. This is called thermal compression.

 

[jpmst3]

When using pro amps for HT use they can also take more than the usual loads because the demands during use are often minimal compared to touring duty. I have tried loads from 8 ohm bridged to 1 ohm stereo with no issues whatsoever even with a fan mod. I am not sure an EP2500 would take a 1 ohm stereo load but my Crown certainly does.

 

[Looneybomber]



Quote:
Originally Posted by michael hurd /forum/post/15518776


It's easier to think in terms of voltages. If an amp can deliver say 40 volts into an 8, 4 and 2 ohm load respectively: (not usually the case as the lower impedance sags the rail voltage down some)


The 8 ohm speaker with 40 volts applied results in applied power of 200 watts, the 4 ohm speaker with 40 volts would result in 400 watts, and the 2 ohm speaker would get 800 watts.


Voltage x current = watts... so


200 watts / 40 volts = 5 amps

400 watts / 40 volts = 10 amps

800 watts / 40 volts = 20 amps


It's hard to actually model what happens, but as you know speakers are never a constant impedance, they are changing with frequency.


As a speaker voice coil and leads heat up, they add more resistance, and start to choke off current. This is called thermal compression.

I understand all that, VC is an inductor, ohm's law, power compression ect...


What I'm curious about is given the same amount of power (watts), a resistor of 2ohms would generate less heat than that of an 8ohm resistor, right? Less resistance = less heat?


If so, then given equal amounts of heat generated in a 2ohm resistor vs 8ohm, a load of lower resistance would require more power, right?

 

[HT Nut]

The heat generated in a circuit will always be the product of current squared times resistance. So in the example given at 40 volts heat would be


25 times 8 for the 8 0hm load or 200 watts (almost all the power into a voice coil goes to heat. The actual work of moving the cone is miniscule)


100 times 4 for the 4 ohm load or 400 watts


and


400 times 2 for the 2 ohm load or 800 watts.


For a constant number of turns in the voice coil the increase in current will provide more driving force. (Amp turns in the magnetic field). Another reason that we seem to prefer those stout amps and low resistance coils.


No matter how you compute power with ohms law, it turns out to be the same number in the end.


So the lower resistance loads indeed allow more power from the amp at constant voltage, with a penalty of a lot of heat as well.


Perhaps that rotary subwoofer is not such a bad thing after all when one considers that the rotating fan is probably a much more efficient mover of air than a magnet, coil, cone system which always has to reverse direction.

 

[Looneybomber]



Quote:
Originally Posted by HT Nut /forum/post/15522372


(almost all the power into a voice coil goes to heat. The actual work of moving the cone is miniscule)

I've said nearly that exact same phrase in the past, yet forgot about that simple fact before making this post. It's been a while since I've had a physics class or DC/AC circuits. Thanks.