# ohms

Discussion in 'Mobile Phones / Entertainment' started by Mark*L, Sep 23, 2003.

1. ### Mark*L Auditioning

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I have a ? I would like to purchase an amp but dont quite understand the whole ohm thing. You supposedly get more power at different ohms...Could someone please explain this? I have heard that it has to do with the resistance...

2. ### Frank Grimes Second Unit

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What exactly are you needing to power? A driver with a lower resistance (say a 2 ohm driver) can draw more power than one with a higher resistance (say 4 ohm), but it really has to do with the amplifier being used. A driver with a lower resistance can damage an amplifier if not properly paired. Explain your situation more specifically and what products you are looking at and we will help you out.

Also, give a gander to http://www.eatel.net/~amptech/elecdisc/caraudio.htm
for more information. JLaudio.com also has a fairly simple tutorial on wiring/mathing resistance of speakers to amps properly (with many pictures! YIPPEEE HOOORRAAYYY!)

3. ### Mark*L Auditioning

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I bought two mtx 4000's from a friend for \$50 with inclosures and i would just like to hook then up to my stereo.

4. ### Tom Jr Extra

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Ohms Explained:

Let's start in the DC (direct current) domain. It's easier, and it will make the AC (alternating current) stuff easier, too.

First, we'll define some variables:
V = voltage (volts)
I = current (amps)
R = resistance (ohms)
P = power (watts)
Don't ask me why we use 'I' for current.

With these, the following relationship holds:
V = IR
I = V/R
R = V/I
They're all the same, re-arranged using basic algebra.

Now, let's say you have a 9 volt battery and a 100 ohm resistor. If you connect the resistor to the battery, you get a current of (9 volts)/(100 ohms) = 0.09 amps.

So what is power? By definition, instantaneous power is voltage multiplied by current:
P = VI (watts)

From this, you can derive other useful equations:
P = (V^2)/R
P = (I^2)R

In our 9V battery example, power comes out to (9 volts)*(0.09 amps) = 0.81 watts.

Now, onto the AC realm. Technically, we now have to deal with impedance, which is a voltage/current ratio that varies as frequency varies. Fortunately, we can simplify this for the audio realm and just use resistance. In this respect, impedance = resistance. Voltage and current are also fairly easy to deal with. The equations above are still valid, if you are talking about RMS voltage and current. RMS stands for 'root mean squared' and is a measure of the average voltage or current in an oscillating (alternating) signal. Forget peak. RMS is the measure of constant, sustained voltage, current, and power.

So, let's say that you have an amp rated at 200 watts into 2 ohms. What is the RMS voltage necessary to achieve this? We can re-arrange P=(V^2)/R to get:

V = sqrt(P*R)

So, V = sqrt(100*2) = 20 volts. This will be the RMS voltage your amp is sourcing to the speaker at full capacity. This voltage is directly related to the signal voltage coming from your head unit, typically about 2V rms. So the signal has been amplified by 20 (2V -> 20V rms).

Also note that the current is:

I = V/R = 20/2 = 10 amps rms

Now, what happens if you hook up a 4 ohm speaker to the same amp with the same head unit? You get:

P = (V^2)/R

P = (20^2)/4 = 100 watts

and

I = V/R = 20/4 = 5 amps rms

As you can see, you get less power using a 4 ohm speaker than you do when you use a 2 ohm speaker. So, we should all use the lowest speaker impedance we can, right?

Not always.

Your amp will be rated for a certain load. Load = impedance = resistance (in the audio case). There should be a specification for this with the amp. If it says it's rated for a 4 ohm load, and you hook up a 2 ohm load, you'll be pulling twice as much current from your amp than it's designed for. This tends to fry electronic components.

I bet that got your attention.

With that, the recommendation is to use the lowest impedance speaker you can, while still within the load limits of the amp.

There is one more issue, since you have multiple speakers. Since these are subwoofers, you'll likely be buying a mono amp. You don't need stereo at low frequencies since bass is non-directional. So, how do you hook up 2 speakers to one output. There are two possibilities:

- Serial
- Parallel

If you hook up your speakers in serial, you add their impedances.

example: 2 + 2 = 4 ohms
example: 4 + 4 = 8 ohms

If you hook up your speakers in parallel, you add their inverses.

example: 1/(1/2 + 1/2) = 1 ohm
example: 1/(1/4 + 1/4) = 2 ohms

Since you have two identical speakers (and not more), this can be simplified as follows:

Serial => multiply impedance by 2
Parallel => divide impedance by 2

How to hook them up:
Serial:
Amp + to Speaker A +
Speaker A - to speaker B +
Speaker B - to Amp -

Parallel:
Amp + to Speaker A + and Speaker B +
Amp - to Speaker A - and Speaker B -

So, sorry for the book, but armed with a calculator, you should be able to figure out speaker impedance possibilities, as well and powe needs. Then you can buy an appropriately specified amplifier.

Hope this helps.

5. ### Mark*L Auditioning

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Thank you very much!

6. ### scott>sau Stunt Coordinator

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Mark, Tom did a A+ technical explanation of "Ohms Law" (a guy with the last name Ohm, a German developed Ohms Law), anyway it is impedance. It is what causes a speaker (load) to create resistance to an amplifier (voltage generating device). If you have two 12" subwoofers that are 8 Ohm each and bridge them, they will become a 4 Ohm load. The amp will see half of that (2 Ohms). Some amps can remain stable into 2 Ohms, some heat up, shutdown, or blow outputs.