So I'm looking at this ($149.99 at Sears)
Manual says 8-16 ohm impedance, and I'm wondering how bad will it be if I connect my 4-ohm subwoofers to it?
Are we talking Blue Smoke here?
Michael
No blue smoke, but you wont be sending the maximum power to the speakers (and thus hurting yourself in the long run).
Check out "maximum power transfer", it basically says that impedance in must equal impedance out for power to be transferred effectively.
The old "impedance matching" thing almost certainly doesn't apply here because the amp specs are not based on output impedance. (Unless this amp has an output transformer... which pretty much went out with vacuum tubes.) Modern solid state amps have output impedances that are near zero at low frequencies.
The "8-16 ohms" rating is more likely based on power handling capability of the output stage. If you use a lower impedance load, it takes more current at lower voltage to get the same power, which means the output stage has to dissipate more power. Like a pair of voltage regulators with changing setpoints, the positive and negative output devices have to pass the output current while dropping the voltage difference between the supply voltage and the output voltage.
Let's take a simple example: Say the supply is 40 VDC and at some instant the positive output device of the amp has to deliver 50W into its load. (No RMS here, just instantaneous power.) So with an 8 ohm load the output is 20V to give 20^2/8 = 50W, and the device itself has
40-20 = 20V across it, while the current through it is the same as the current through the load = 20/8 = 2.5A. So the device has to dissipate 2.5 * 20 = 50W.Now consider a 4 ohm load. First of all, you need to turn the volume control down to get the same 50W into the load, since now the output voltage must be 14.14 since 14.14^2/4 = 50W. So the device now has
40-14.14 = 25.86V across it. But its current is 14.14/4 = 3.535 and the power it dissipates is 25.86*3.535 = 91.4W.So if the power rating of the amp is based on the power that the output devices can handle, you'd need to derate it significantly to use a 4 ohm load. Some amps list separate power ratings for 4 ohms (and sometimes 2 ohms). If the 4 ohms rating is less than the 8 ohm, the output devices are the limiting factor. If the amp is more sturdily built (a pro amp, or better grade home amp) the 4 ohm rating will be higher, indicating that the limit is more likely just the supply voltage. (+/-40V gives 28VRMS, which into 8 ohms gives 100W, or into 4 ohms gives 200W.)
But this amp wasn't rated for 4 ohms, so you may have to derate it. And you'd be taking a chance that someday somebody didn't crank up the volume too high.
Best regards,
Bob Masta DAQARTA v3.50 Data AcQuisition And Real-Time Analysis
Ok, thanks for the detailed explanation.
If I place my two 4-ohm subs in series, that would give an 8-ohm impedance, right?
Would sound quality suffer if I put two speakers in series?
Michael
Get some big 4ohm resistors, and put them in series. That will protect the output drivers from excess dissipation.
The resistors need to be rated at 100W, of course...
kind of pricy...
so use 2 50W 2 ohm resistors for each instead:
Regards, Bob Monsen
Wrong! Modern amplifiers just like your wall plugs are constant voltage. Impedances are NEVER matched. The output impedance of a modern amplifier like your wall plugs is near zero ohms, a few hundredths of an ohm.
Matched impedances do transfer maximum power but they also dissipate as much power in the source as is delivered to the load. That condition is very wasteful and unnecessary. As an example consider operating a 100 Watt light bulb on a matched impedance basis. The impedance would be: Z = V/(sqrtP) =
120/10 = 12 ohms and the wall plug would also have to be 12 ohms if it was matched. A 100 Watts would go to the light and another 100 Watts would be wasted in the 12 ohms at the generator. That's absurd! The same thing is true of amplifiers. You are trying to deliver power to the speakers not dissipate just as much in the amplifier.Matched impedances are ONLY used to avoid reflections of energy in RF work and to properly load vacuum tube plates. Delivering power is best done constant voltage.
Yes very likely.
And please don't post in HTML.
Graham
Whjy do you think not ?
Graham
Yes. Reduction of damping factor.
Graham
That's plain silly.
The right answer is to buy an amp suitable for 4 ohm loads.
Graham
I didn't. I pasted a link.
What did it look like on your end?
M
Or just buy new speakers...
M
There was an HTML tag which screws up the editing for the reply. That'll often happen when posting links. With Netscape it warns me about that and I can turn off the HTML (which I do when posting links).
Graham
That too. I'd buy a more comfortably rated amp myself though.
Graham
If you can buy new speakers, that is the way to go. If you can't buy new speakers, or just love those old 6 foot tall woofers, you can use the resistors to protect the amp.
BTW, isn't Eeyore a stuffed donkey?
Regards, Bob Monsen
Yes and no.
Technically, the damping factor is the rated load impedance divided by the actual output impedance. It's just a way of specifying the output impedance of the amp, which (as has been pointed out already) for modern amps is very low. So if an amp has a damping factor of (say)
100 with an 8 ohm load, then the output impedance is 0.08 ohm. (Note that damping factor almost always falls with frequency, but the OP is talking about subwoofers so there is no need to worry about that here.)Putting 2 drivers in series doesn't affect the amp's output impedance. However, each speaker is now being driven through an additioanl 4 ohms (nominally), so from the speaker's point of view the effective damping factor has dropped substantially. Nevertheless, this may or may not be a problem, depending on the particular speakers. If the OP already has the amp and speakers, he should try it. Otherwise, why buy this amp? The stated price doesn't look all that attractive.
Best regards,
Bob Masta DAQARTA v3.50 Data AcQuisition And Real-Time Analysis
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