Ohm Level on Speakers

[daveo]

How does this Ohm rating work. I know it's something to do with resistance but what should I be looking for? Is there a big difference between 8 Ohms and 6 Ohms?

I'm guessing that the lower the Ohms the easier it would be to blow the speaker up, am I right?

[bobwindy]

The ohm rating is the resistance of the speaker coil. The higher the resistance the less current is allowed to flow, hence less strain on the speaker.

[homerjhandley]

You can have an equivelent Ohm rating or higher speakers on the amp, but the higher the Ohm rating the less noise it will have (ie the amp will have to work harder to produce the noise).

If you put a lower rated speaker on an amp then you run the risk of blowing up the speakers or amp or both.

From what I have seen most HC speakers are rated at 8Ohm (my kenwood amp supports 6-8)

Car speakers are rated @ 4 so i could damage a component if i use them.

speakers used for most home stereo units are at 6Ohm.

I think this is correct.

Also dont forget, it also depends how you wire them up.


if you use 2 8 Ohm speakers and attach them both to one connection (ie both - to the - out and both + to the +) it halfs the Ohm so this will be 4Ohm. (series)

If you wire them parallel (ie one speaker - to the output -, the + to the 2nd speaker - and the 2nd speaker + to the output +) this will be 16Ohm, as it doubles.

Also @ 0 Ohm this is no resistance so the amp would probably not like it very much

;)

do a search for this subject on the net, the place is full of electrical things like this.

[lynx]

Have a look here (http://www.the12volt.com/ohm/ohmslaw.asp) . This is a very informative site.

[MikeK]

Actually, if you wire in parallel, you reduce the total resistance.
In series, you increase it.

For series circuits

Rtotal = R1 + R2


For parallel

Rtotal = (R1 x R2) / (R1 + R2)


Don't forget though that the rating of a speaker is the nominal impedance, not the resistance.
The coil resistance of the driver is only part of that, and may actually be misleading, as the impedance of the pair (woofer and tweeter) plus the crossover, can, at certain freqs, actually be less than this DC resistance, which is really little more than the coil resistance of the woofer. That impedance will also vary with frequency.

[lynx]

Originally posted by MikeK
I'll only bother posting that if anyone's interested.
Please do.

[MikeK]

First of all, I've got mixed up posting on two similar threads
(there is another asking about 16ohm speakers).

Assume for this example, that a loudspeaker is a pure resistance, as it gets much more complex if you treat it as a reactive impedance (which is what it really is). The theory is basically similar (but the calculations are much more complex, can only be done if you know all the component values and speaker parameters, and in any case, the results vary with each frequency you calculate for).

Assume an amplifier has a maximum output voltage of 32V, and a maximum output current of 2A.
Roughly speaking, this means a max output power of 32V x 2A = 64VA, or 64W.
To ensure that the amplifier does not get overloaded, the minimum load resistance should be Vmax/Imax or 32V/2A, which is 16ohms.


Now if you put an 8ohm load onto this amp, it will reach it's max output current at only 16V, as 16V/8ohms = 2A (the max current).
This isn't a problem, as long as you make sure the output voltage never goes over 16V, but 16V at 2A is only 32W, or half the rated output power of the amp.
If you drive the voltage over 16V, the amp will either cut-out (due to it's internal protection) or start to "clip" the current. Basically a clipped AC waveform has a large DC component, and speaker coils do not like this at all - they burn out.
Also, the amp's output transistor may not appreciate being driven at max current, at DC, for very long either. (although the max current is just as likely to be dictated by the amps power supply - if it can only supply 2A max, then that's all the final output transistor can deliver - likewise with voltage, if the power supply is only 32V, then the output voltage can't exceed that (with common designs that is - as is often the case, there is a way around most things))
Clipping is easily audible, as it sounds like horrendous breakup distortion. This is basically why low impedance (nominal 4ohms or less) speakers are often described as difficult loads.

Now to bring the total load up to 16ohms, you put an 8ohm resistor in series with it, so the total load is now 8ohms + 8ohms = 16ohms.
So 32V (max voltage) across a 16ohm load will draw 2A (max current), and 32V x 2A = 64W, which from the amps point of view is fine.

But - the load is now two 8ohm loads in series.
So of the 32V, half will be across the resistor, and half across the speaker, ie 16V.
Now 16V across an 8ohm speaker means 2A current (which we already knew) - and 16V x 2A is 32W.
So, again you have only half the power output being dissipated by the speaker.

In either case, all you can deliver to the speaker itself is 32W (and with or without the resistor, you are running the amp at max current)

If 32W is too much, then fine, turn the volume down.
But once you do that, there's really little need for the resistor at all, as output voltage is well within limits (less than 16V, when the max is 32V), and output current is less than the 2A max limit.


Of course, you could compromise and use a 4ohm resistor, for a 12ohm total load.


That's the basics - just to get the basic point across.
It gets a lot more complex in real life, as a speaker isn't a pure resistance. It's a reactive load, and reactance varies with frequency (the reactance is due to the inductance of the driver coils and crossovers, and the capacitance in the crossover).
It causes AC voltage and current in a circuit to go out of phase (and the amount of this phase shift also varies with frequency, as it's a product of the reactance). Also, a music waveform contains multiple frequency components.
The impedance quoted on a speaker is also only a nominal value, say 8ohms. This could vary between say 4ohms and 16ohms at various frequencies (if you don't take into account any impedance spike caused by the driver approaching resonance), but usually a speaker with a quoted nominal imp. of 8ohms, will usually have a minimum impedance somewhere around 5ohms.

Another difficulty is the way that manufacturers quote the amps performance, in watts/channel. They don't always measure it the same way, and some are more optimistic in any case than others.
It would be a lot more useful if they also quoted the max constant output voltage, the max constant output current as well as max power delivery and quoted them into a purely resistive load as well as into a real reactive load.
(I say constant, as music contains transients which can cause instant peaks way in excess of these figures).


For instance.

Take a typical AV amp, quoted at 100W/channel into 8ohms. (does this mean that all 5 channels can be driven at 100W simultaneously?)

Using ohms law, we can derive some ball park figures for Vmax and Imax.

P=I2R (ie I squared).

100= "I2" x 8, therefore I2 = 12.5, so I= approx 3.5A.

P= VI

100= V x 3.5, therefore V= approx 30V

So, we can now work out a safe power level for driving a 4ohm speaker. The actual level could be far higher than this, but we don't know, for an 8ohm load, if it's the voltage or the current which is the limiting factor. So we have to assume that current is the limiter (as with a reduced load resistance, it's that limit we'll hit first).

so V=IR

V=3.5 x 4 so V is about 14V

P=VI

P=14 x 3.5 which is approx 50W.


So, if an amp can deliver 100W into an 8ohm load, then it should be OK for it to drive a 4ohm load at 50Wmax. (and a 2ohm load at 25W using a similar calculation)

Likewise, for a 16ohm speaker.

V=IR

30 = I x 16 , so I is approx 1.8A

P=VI

P= 30 x 1.8 so max power into a 16ohm load is again 50W.

This time, we are assuming that the voltage cannot exceed 30V
(which it may be able to without problem, but we don't know)
As you increase load resistance (which is the same thing as reducing the load), the voltage increases and the current decreases - and vice versa when you lower the load resistance (increase the load)

Remember though, that these figures are what we know is safe, based on the figures derived from the quoted specs for power delivery into an 8ohm load.
To be safe, I'd probably assume that the 100W quoted is a bit optimistic, and assume 70% of the quoted figure, or 70W/channel.
It may also be the case that the limiting factor is neither the voltage nor the current - it could be the amount of heat which the output stage can dissipate - heatsinks can only be made a certain size if they are to fit inside an amp. In other words, given a large enough heatsink, the power delivery capacity may be 200W, but is limited in the design by the cut size heatsink.

[lynx]

Both informative and well conveyed,well done. :)

[Roy]

Nice touch Mike, very well presented.

It gave me horrible flashbacks of my Electronic Engineering HND though... ;)

All of which I can't remember in anything like the detail you do, do you use that knowledge daily in your job?

[daveo]

Thanks for taking the time to respond as you have done :). I was really after a laymans guide but I think I get at what you are saying!