which is harder on elec. ? ohm wise....

[garychoffmann 10+ year member]

Don't forget about amplifier efficiency. General rule of thumb is amplifiers become less efficient at lower ohm loads.

 

[fasfocus00 5,000+ posts]

It's about current draw and efficiency. Power is power. So 2400 watts at half ohm is the same as 2400 watts at 2 ohms. The reason electrical setups need to be so much beefier on .5 ohm loads is bc the amps efficiency drops making it have to draw more current to produce the same power. Amp A draws 100 amps of current at 14 volts and produces 1200 watts at 2 ohms. Amp B draws 100 amps of current at 14 volts and produces 1200 watts at .5 ohm. This would only happen if the efficiency of both amps were equal with respect to impedance. Real world would have amp B having a 130+ current draw to overcome the inefficiency at a half ohm load

 

[basswiigee]

Ohms Law I=V/R Where I is your current (like what your alt can produce) V is your voltage (like what your amp is producing) and R is resistance (which your subwoofers are creating)
Most car audio amps are a constant VOLTAGE source. Looking at the formula above that means they try to maintain a constant voltage given any load they can handle (when they can't handle it they shut off or break, hence .1ohms resististance would break about anything lol) If you keep voltage the same and drop your impedence you dividing by a smaller number, hence your I value, or draw from your alt, doubles, as well as your wattage. This is because "watts" is a measurement of power (P) where P=IxV (remember voltage was constant, so since I became 2x as large, so does P).

I need to get a avatar, I post tech stuff all the time and nobody ever remembers me lmao.

laymans terms please....

 

[T3mpest 10+ year member]

laymans terms please....

You end up pulling more current with Lower ohm loads. Not double or anything but more current for sure. Voltage is a squared term so halving the impedance doesn't make up for the squared voltage so the higher you go the bigger the discrepancy. So 10000 watts at 1ohm and 10000 at 4 the 4ohm load is a good deal more efficient all things equal. This is also why amps lose efficiency as you lower the load. Same idea backwards.

 

[basswiigee]

You end up pulling more current with Lower ohm loads. Not double or anything but more current for sure. Voltage is a squared term so halving the impedance doesn't make up for the squared voltage so the higher you go the bigger the discrepancy. So 10000 watts at 1ohm and 10000 at 4 the 4ohm load is a good deal more efficient all things equal. This is also why amps lose efficiency as you lower the load. Same idea backwards.

thank you!!!!!

 

[Slo_Ride 5,000+ posts]

I prefer to run my amps at 0 ohms

 

[T3mpest 10+ year member]

thank you!!!!!

Welcome. Just as an easy way to think of it. A amp can be 90% efficient at 4ohms drop it to one a.d it may only be 82%. This is because the extra current draw in amps ramps up faster proportionally than the voltage it puts out. You could get the same wattage at a higher impedance using less current. However the amount of voltage you need goes through the roof and it makes for an expensive amp to have rails that can handle it. Hence most amps are run at less efficient low ohm loads, the don't need to handle as much voltage this way.. the simply have to dissipate more heat hence the big heatsinks.

 

[Timmy13091 10+ year member]

Not disagreeing with you here, but could you then explain why it takes such a beefy electrical system to operate an amplifier at half an ohm? I'd run my amp at .5 (it is capable with the proper setup), but I've been warned too many times to even attempt it.

Once you get your alt you can run .5 no problem on your electric ... That 2k won't be that bad .. I've ran some big amps at .5 for the electric I had and was just fine .. People forget about rise it seems

 

[Chriszle 10+ year member]

Man you guys should really do a little research before informing someone.... .5Ω/2Ω, will draw the same amount of amperage if the power is the same. Look up Ohm's law and read about it. Apply it to all questions about electrical/ ohm load.

Thank you. I think I just became a little more stupid by taking the time to read this post. Maybe you should take your own advice and look up ohm's law.

 

[basswiigee]

Once you get your alt you can run .5 no problem on your electric ... That 2k won't be that bad .. I've ran some big amps at .5 for the electric I had and was just fine .. People forget about rise it seems

was thinking that, average on clamp was .5Ω 2200 -2500w rise to 1.5- 1.9

 

[TaylorFade 10+ year member]

I usually observe in the neighborhood of 85% or better on "good" amps at 4ohms or so. ~ 80% at 2 ohms and it usually gets to 70% or so by the time I get to 1 ohm. Below that... it gets retarded.

Consider a 2,000w output from a 14v source. A 100% efficient amplifier (not possible) would draw ~ 143A. Assume a real world rated 2,000w@1ohm amp with 70% efficiency (which is common). That 2,000w now comes at a cost of 204A.

Now consider a 5k amp that is capable of 2,000w @ 2 ohms with 80% efficiency. That's 179A. Same 2,000w, but less draw. But at the expense of a relatively gigantic amp.

But don't get it twisted. I'm talking about 1 ohm amps here. Amps with lowest rated impedance of 2 ohms aren't inherently more efficient at 2 ohms than a 1 ohm amp at 1 ohm.

I used this to my advantage at finals actually. I only "needed" about 8k or so. But I was severely battery restricted. So I ran about 15kw worth of amps at a higher load to facilitate that power with less draw. I could have gotten the same power from a single amp at a lower load if I had the reserve, but I was limited.

 

[argenis69 10+ year member]

Thanks Guys, looks like you were right basswiigee . //content.invisioncic.com/y282845/emoticons/cool.gif.3bcaf8f141236c00f8044d07150e34f7.gif

 

[Jepalan]

Time for some theory. First let's look at what the output of the amp is doing when driving the speaker load at 2400 watts RMS.

For a 2 ohm load the RMS voltage at the speaker is V = sqrt(2400 x 2) = 69.2 V RMS

For a 0.5 ohm load the voltage at the speaker is V = sqrt(2400 x 0.5) = 34.6 V RMS

For the 2 ohm load the amp has to put I = 69.2 V / 2 ohms = 34.6 Amps RMS into the speaker

For the 0.5 ohm load the amp has to put I = 34.6 V/0.5 ohm = 69.2 Amps into the speaker

So you see for 2400 W into the 0.5 ohm load the amp is putting 69.2 amps RMS into the speaker at 34.6 V RMS

for 2400 W into the 2 ohm load the amp is putting 34.6 amps RMS into the speaker at 69.2 V RMS

If the amp was 100% efficient (could convert 2400 W at input to 2400 W at output in any V/I combination) then it would pull I = 2400W/13V = 184.6 Amps in *both* cases.

But the amp is not perfectly efficient and in general because it has to output more current to run 0.5 ohms at 2400 W it is less efficient at converting input power into output power - if for no other reason than ohmic losses in the internal power devices that have to supply the higher current.

If the amp is 90% efficient then it will pull 186.4 x 1.1 = 205.4 Amps from the alt/batt system to supply 2400 W to output

If the amp is 80% efficient then it will pull 186.4 x 1.2 = 223.7 Amps from the system to supply 2400 W to output

 

[keep_hope_alive Premium Member]

the simple answer is lower impedance loads result in more current in the output stages which increase heat which decreases efficiency. the decrease in efficiency is what you are concerned about.

 

[tommydh 10+ year member]

Time for some theory. First let's look at what the output of the amp is doing when driving the speaker load at 2400 watts RMS.
For a 2 ohm load the RMS voltage at the speaker is V = sqrt(2400 x 2) = 69.2 V RMS

For a 0.5 ohm load the voltage at the speaker is V = sqrt(2400 x 0.5) = 34.6 V RMS

For the 2 ohm load the amp has to put I = 69.2 V / 2 ohms = 34.6 Amps RMS into the speaker

For the 0.5 ohm load the amp has to put I = 34.6 V/0.5 ohm = 69.2 Amps into the speaker

So you see for 2400 W into the 0.5 ohm load the amp is putting 69.2 amps RMS into the speaker at 34.6 V RMS

for 2400 W into the 2 ohm load the amp is putting 34.6 amps RMS into the speaker at 69.2 V RMS

If the amp was 100% efficient (could convert 2400 W at input to 2400 W at output in any V/I combination) then it would pull I = 2400W/13V = 184.6 Amps in *both* cases.

But the amp is not perfectly efficient and in general because it has to output more current to run 0.5 ohms at 2400 W it is less efficient at converting input power into output power - if for no other reason than ohmic losses in the internal power devices that have to supply the higher current.

If the amp is 90% efficient then it will pull 186.4 x 1.1 = 205.4 Amps from the alt/batt system to supply 2400 W to output

If the amp is 80% efficient then it will pull 186.4 x 1.2 = 223.7 Amps from the system to supply 2400 W to output

Finally someone actually showed the actual calculations and not just droppping ohms law. Not to add more crap on the topic anyone mention heat created from inefficiency at the lower load which = distortion of the wave which is only visible by scope but distortion is never the best thing