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I'm Embarassed...whats an OHM?

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Old Jun 8, 2004 | 04:47 AM
  #1  
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I'm Embarassed...whats an OHM?

Ok I officially don't know ****. What does it mean when an amp can run 1.5-4 ohms per channel. Is it the lower the "ohm" the harder the bass hits? I hate asking these questions but how else can I learn.

I'm sorry...

Ryan
Old Jun 8, 2004 | 07:28 AM
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OHM measures resistence. The lower the ohm, the more power you are going to get from your amp. It depends on your speakers what ohms the amp will see.
Old Jun 8, 2004 | 07:51 AM
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Originally Posted by OpTi
OHM measures resistence. The lower the ohm, the more power you are going to get from your amp. It depends on your speakers what ohms the amp will see.
To add.. The lower the ohm load the less resistance, the less resistance on your amp the more wattage it can push out.. Just think of it like carrying someone and trying the run, the lighter the person the faster you can run (lower the ohm load or resistance the more wattage the amp can push out).. But remember the heavier the person you are carrying the hotter you will get and the harder you will have to work (same with amps, they have to be able to run at that ohm load or else it will overheat and cause damage).. Hope this makes a little sense and helps a little bit..
Old Jun 8, 2004 | 04:55 PM
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Originally Posted by Chitwista
But remember the heavier the person you are carrying the hotter you will get and the harder you will have to work
Also, you can run too fast if the person is too light. You'll get hotter that way too
Old Jun 8, 2004 | 06:12 PM
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Damn... all these analogies are

hehe

Great stuff!!!
Old Jun 8, 2004 | 10:29 PM
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Ill simplify it then

Ohm load is the resistance, if the resistance is lowered (lower ohm load) more power will flow, but it will also run hotter.
Old Jun 9, 2004 | 05:56 AM
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Is there any way to offer the person caryying the other person a cold drink?

Or give them a breather?
Old Jun 9, 2004 | 07:01 AM
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Originally Posted by SleepyMaxxx68
Is there any way to offer the person caryying the other person a cold drink?






Or give them a breather?
Cooling Fans!
Old Jun 9, 2004 | 07:17 AM
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Sleepy wants to give his amp some ice water!!!! There are ways to cheat the system, but not enough to make more power. Just lets you run at the threshold longer. Putting fans across the amp to dissapate heat will help keep it from getting too hot, but not to make more power. If you need more power, buy a bigger amp. An OHM is the measure of resistance, or load on the system. In your house, the light bulb would represent the speaker, as the load on the outlet.
Old Jun 9, 2004 | 02:27 PM
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The equations you guys need to look at to illustrate the power to resistance relationship are the following P=I^2*R and P=V^2/R. If you can maintain the voltage and halve the resistance, you will double the power. Now if you double the power and your resistance is constant your current will need to increase. This is why amps designed to run at really low impedence are often called "high current" amps.

An Ohm is a unit of electrical resistance that has both a passive and reactive component. A voice coil is a complex reactive load that will vary with frequency as well as with the enclosure design.
Old Jun 10, 2004 | 04:14 AM
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I thought the analogies were hard to understand...Dan "The Science Man"
Old Jun 10, 2004 | 06:19 AM
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Amps in their design phase are made to only handle certain loads (Power correction factor). Lower the load resistance does necessarily mean you're gona get more power out of it (there's only so much juice the cute little mosfets and transistors can pump). All that's really gona happen is at the same volume level, the speaker with the lower resistance is going to be pulling more current and playing louder.

If the amp says 1.5~4 OHMS load then it could be one of two things. If its a high end amp, then it might test the load and adjust its output accordingly. If low end amp, don't do anything different and play music and you worry about what happens. In case two it could be 1.5~1000 OHM load....lol
Old Jun 10, 2004 | 07:24 AM
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It's the JL 250/1 1.5-4 OHMs So should I run it at 1.5 or just round it up to 2 ohms?
Old Jun 10, 2004 | 07:50 AM
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Originally Posted by SleepyMaxxx68
It's the JL 250/1 1.5-4 OHMs So should I run it at 1.5 or just round it up to 2 ohms?
You can't just turn a **** or something and tell it to run ar 2 Ohm. What will determine the load are the speakers. That amp is one of the slash series that uses the RIPS system that claims to do the following:

The "Intelligent" portion of the R.I.P.S. System is a circuit that actually monitors output current to optimize the amplifier's output power over a wide range of load impedances (1.5 ohm-4 ohm per channel). Conventional amplifiers are designed to produce optimum power at a particular impedance (2 ohm, for example). When asked to run above that impedance (say, 4 ohm), these amplifiers lose power (half their power from 2 ohm to 4 ohm). This will not happen with a JL Audio “Slash” series amplifier because the R.I.P.S. System detects the actual impedance being driven and adjusts output rail voltages to deliver optimum output. The entire process is seamless, automatic, and results in incredible dynamics for satellite channels and consistent power output for a wide range of subwoofer configurations. It also takes into account the real impedance of the system, rather than relying on often inaccurate assumptions based on a speaker's "nominal impedance".

Took that from JL's site. So the bottom line is that you will have 250W as long as the amp sees anywhere from 1.5-4 Ohm. The next question is what speakers are you going to run?
Old Jun 10, 2004 | 08:54 AM
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I'll be running one 12" W3v2 sub off of the 250/1


&


My interior components VR65CSi 's will be running off of the 300/4 (just a side note)
Old Jun 10, 2004 | 11:03 AM
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MaximaDan..u said "An Ohm is a unit of electrical resistance that has both a passive and reactive component"

i think what you meant to say was this: resistance has a real and an imaginary compnent.... a+b*i ohms...the negative or positive sign of the b*i imaginary part tells us whether the resistance is due to an inductive reactance or a capacitive reactance. even resistors follow this..the b*i part is simply equal to zero though.
Old Jun 10, 2004 | 04:01 PM
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Originally Posted by friendhasmax
MaximaDan..u said "An Ohm is a unit of electrical resistance that has both a passive and reactive component"

i think what you meant to say was this: resistance has a real and an imaginary compnent.... a+b*i ohms...the negative or positive sign of the b*i imaginary part tells us whether the resistance is due to an inductive reactance or a capacitive reactance. even resistors follow this..the b*i part is simply equal to zero though.
OK the use of the word "passive" was wrong. I meant to say that it is a purely resistive element. However, reactance is the proper term to describe either a capacitive or inductive element. An inductor's reactance is described by: X=Omega*L, and a Capacitor's is X=1/Omega*C where Omega is the angular velocity expressed in radians per second.

The basic problem when dealing with Either inductors or capacitors is that the voltage across and the current through are out of phase. For a Cap the Voltage is lagging Current by 90 degrees and for an inductor the Voltage is leading the current by 90.

You end up dealing with polar equations when calculating the total impedence of a system containing multiple RLC elements, but not usually imaginary numbers.
Old Jun 10, 2004 | 05:06 PM
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Man, I think this post is getting a bit too technical for the average .orger to understand I Think that's why I switched majors from EE to Pre-Pharmacy :
Old Jun 10, 2004 | 05:46 PM
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Originally Posted by Maxima Dan
OK the use of the word "passive" was wrong. I meant to say that it is a purely resistive element. However, reactance is the proper term to describe either a capacitive or inductive element. An inductor's reactance is described by: X=Omega*L, and a Capacitor's is X=1/Omega*C where Omega is the angular velocity expressed in radians per second.

The basic problem when dealing with Either inductors or capacitors is that the voltage across and the current through are out of phase. For a Cap the Voltage is lagging Current by 90 degrees and for an inductor the Voltage is leading the current by 90.

You end up dealing with polar equations when calculating the total impedence of a system containing multiple RLC elements, but not usually imaginary numbers.
you're preaching to the wrong person...i'm an EE Major...not usually imaginary numbers? how so? why do you think polar coordinates are used? because this is a complex (imaginary) calculation. you can easily convert from rectangular to polar and vice versa.... if you remember your complex analysis, you must use both rectangular and polar forms of imaginary numbers in order to do calculations...you cannot simply add up two polar numbers, u must convert them to rectangular a+b*i form first, and then convert them back to polar
Old Jun 11, 2004 | 01:36 AM
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Originally Posted by friendhasmax
you're preaching to the wrong person...i'm an EE Major...not usually imaginary numbers? how so? why do you think polar coordinates are used? because this is a complex (imaginary) calculation. you can easily convert from rectangular to polar and vice versa.... if you remember your complex analysis, you must use both rectangular and polar forms of imaginary numbers in order to do calculations...you cannot simply add up two polar numbers, u must convert them to rectangular a+b*i form first, and then convert them back to polar
Buy a better calculator. I am a EE grad and haven't used this stuff for 8+ years. Why? because computer simulations are so good. My HP calc will add up multiple polar values with ease and it is 10 years old. If you are trying to do these calcualtions in rectangualr form you are making your life way too difficult.

The reactance axis is the imaginary axis. We are arguing the same point here.
Old Jun 11, 2004 | 03:57 AM
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Guys um I'm an English Major who wants his sbwoofer to bump....I'm not to worried about loads, equations and what not...

I'll be running a 12W3v2 off of the JL250/1 amp will that bump?
Old Jun 11, 2004 | 04:01 AM
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well dan u said that there are no imaginary numbers involved...thats infact what its all about though, imaginary numbers...we use polar coordinates to represent imaginary numbers...and my calculator is just fine..Ti-89
Old Jun 11, 2004 | 04:10 AM
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you engineers are sick people!
Old Jun 11, 2004 | 06:27 AM
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Originally Posted by SleepyMaxxx68
Guys um I'm an English Major who wants his sbwoofer to bump....I'm not to worried about loads, equations and what not...

I'll be running a 12W3v2 off of the JL250/1 amp will that bump?
Dont expect anything crazy, but yes it should sound good.

Yes they can be very sick people
Old Jun 11, 2004 | 02:20 PM
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Originally Posted by friendhasmax
well dan u said that there are no imaginary numbers involved...thats infact what its all about though, imaginary numbers...we use polar coordinates to represent imaginary numbers...and my calculator is just fine..Ti-89
Ok, you're right about the imaginary numbers. But, I do stand by the terms resistance (real) and reactance (imaginary)...can't believe you made me dig out my old textbooks.
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