Is the electrical system pressed more with lower ohm configuration?

[ultraSoundWave]

Hi! Comparing two amplifiers, one rated at 2000 watts with an impedance of 0.5 ohms and another rated at the same power but with a 2-ohm impedance, I am curious about the impact on the electrical system. When considering these options, does the load on the electrical system differ between the two amplifiers? Which amplifier would be a more suitable choice, taking into account the potential draw from the electrical system?

Example:
Mmats M2000.5 vs Mmats M2000.2
on a stock alternator and two batteries, Which would be better or are they the same as same amount of wattage is in use!
Thanks

 

[Blumlein 88]

Welcome to ASR.

Which impedance are you talking about? 2000 watts is 2000 watts. If the ratings are under comparable conditions the draw from the battery will be virtually the same.

 

[Blumlein 88]

Okay looked at the two models you are referring to. One is rated for 2 ohm loads, I assume 2 ohm woofers or several paralleled for that equivalent. The other for .5 ohm loads. The .5 rated model would put out more amperage at lower voltage for that rating if genuine.

For instance 32 volts at .5 ohm is 64 amps and 2048 watts at the amp output.

64 volts at 2 ohms is 32 amps and 2048 watts at the amp output.

On the supplying system, two batteries both would draw the same amperage. They claim 83% efficiency so both would pull something like 2500 watts from the battery. I'd assume that is some peak rating. 200 amps for a 12 volt battery or 100 amps for 24 volts if the batteries are series connected. Either way both pull equally from the car electrical supply system.

 

[ultraSoundWave]

Welcome to ASR.

Which impedance are you talking about? 2000 watts is 2000 watts. If the ratings are under comparable conditions the draw from the battery will be virtually the same.

Thanks

 

[ultraSoundWave]

Okay looked at the two models you are referring to. One is rated for 2 ohm loads, I assume 2 ohm woofers or several paralleled for that equivalent. The other for .5 ohm loads. The .5 rated model would put out more amperage at lower voltage for that rating if genuine.

For instance 32 volts at .5 ohm is 64 amps and 2048 watts at the amp output.

64 volts at 2 ohms is 32 amps and 2048 watts at the amp output.

On the supplying system, two batteries both would draw the same amperage. They claim 83% efficiency so both would pull something like 2500 watts from the battery. I'd assume that is some peak rating. 200 amps for a 12 volt battery or 100 amps for 24 volts if the batteries are series connected. Either way both pull equally from the car electrical supply system.

Ok I think I understand that draw from the electrical system would be the same. Got it, Really appreciate your time. Thanks

 

[DonH56]

Power = Voltage * Current = Voltage^2 / Impedance (ohms) = Current^2 * Impedance

For the same power at different impedances, voltage and current from the amplifier delivered to the speaker will be different, but the power drawn from the wall (if that is what you mean by the "electrical system") will be the same.

For 2000 W at 0.5 ohms voltage is 31.62 Vrms and current is 63.25 Arms. At 2 ohms, voltage is 63.25 Vrms and current is 31.62 Arms. Since impedance is a factor of 4 (2/0.5) different, and volts and amps relate to power and impedance by sqrt(), the factor of 4 becomes a factor of 2 (sqrt(4) = 2) so the voltage and current numbers simply flip. The amp is delivering 2000 W in either case, and draw from the wall (or battery) will be the same. You'd likely want to use larger diameter wire for the 0.5-ohm speaker, however, given its lower impedance and higher current flow.

To go just a bit further, most amplifiers are designed to provide a maximum voltage, limited by the power rails. The maximum output current is usually dictated by the current capacity of the power supply, output devices (transistors), and thermal (heat) management. That means the amplifier that can deliver 2 kW into 2 ohms can likely deliver much more power into 0.5 ohms, at least briefly before current limiting or protection circuits cut in. If (a big if!) the amplifier could deliver 63.25 Vrms into 0.5 ohms, that would be 8 kW (!) into 0.5 ohms. At least before the amp blew up, or the speakers did.

HTH - Don

Edit: Oops, @Blumlein 88 already answered, sorry for the repeat.

 

[DVDdoug]

With normal program material, average power is typically around 10% of the peak so at full-power you might be averaging around 200W from the alternator/battery and the battery (without the alternator) might last longer than "expected". But you still need momentary current for the peaks.


...And I don't trust amplifier power ratings unless they are independently measured.

 

[ultraSoundWave]

Power = Voltage * Current = Voltage^2 / Impedance (ohms) = Current^2 * Impedance

For the same power at different impedances, voltage and current from the amplifier delivered to the speaker will be different, but the power drawn from the wall (if that is what you mean by the "electrical system") will be the same.

For 2000 W at 0.5 ohms voltage is 31.62 Vrms and current is 63.25 Arms. At 2 ohms, voltage is 63.25 Vrms and current is 31.62 Arms. Since impedance is a factor of 4 (2/0.5) different, and volts and amps relate to power and impedance by sqrt(), the factor of 4 becomes a factor of 2 (sqrt(4) = 2) so the voltage and current numbers simply flip. The amp is delivering 2000 W in either case, and draw from the wall (or battery) will be the same. You'd likely want to use larger diameter wire for the 0.5-ohm speaker, however, given its lower impedance and higher current flow.

To go just a bit further, most amplifiers are designed to provide a maximum voltage, limited by the power rails. The maximum output current is usually dictated by the current capacity of the power supply, output devices (transistors), and thermal (heat) management. That means the amplifier that can deliver 2 kW into 2 ohms can likely deliver much more power into 0.5 ohms, at least briefly before current limiting or protection circuits cut in. If (a big if!) the amplifier could deliver 63.25 Vrms into 0.5 ohms, that would be 8 kW (!) into 0.5 ohms. At least before the amp blew up, or the speakers did.

HTH - Don

Edit: Oops, @Blumlein 88 already answered, sorry for the repeat.

Great knowledge! Thanks

 

[unpluggged]

I am curious about the impact on the electrical system

The power consumption rating of each specific amplifier (as with any electric appliance) is stated on its back panel and in the documentation included in the package.

 

[voodooless]

So the real question is; what is your load? You can only get whatever power is specified if you offer that load to the amp. The 2 ohm amp may not be stable at 1 ohm, and the .5 ohm amp will have less power with 2 ohm load.