4 ohm vs 8 ohm speakers

[3db]

I don't see much point in even trying to calculate it though degree electrical engineers (obviously self taught individuals could do that too if they are good) should know how to do it. If one is to do it properly one would likely have to spend some time reviewing their old text books and/or notes, not just electrical circuit analysis but also advanced math including Laplace transforms as minimum. One would also need the crossover circuit diagram complete with values of every single components used. So except for DIY people such as yourself, it will be much easier to either search for lab measurements (e.g.Stereophile and Soundstage websites) or take their own measurements and plot their own graphs if they have the equipment.

In practice, I would agree for most reasonably well designed speakers, phase angles shouldn't be too bad, while there are certainly no shortages of good sounding speakers that are truly 4 ohm or less nominal. I think 3 dB's point is that it is a factor that should not be ignored, but I guess in doing so his choice of word ("meaningless") appeared a little strong for some.

After all, we were taught at University that impedance needs both a magnitude and a phase angle.

 

[PENG]

After all, we were taught at University that impedance needs both a magnitude and a phase angle.

Its easy to calculate it for one freq. To plot a graph you are going to use a transfer function and Laplace, that's all distance memory for me.

 

[TLS Guy]

Its easy to calculate it for one freq. To plot a graph you are going to use a transfer function and Laplace, that's all distance memory for me.

True, but in practice you can spot a major problem with dead reckoning, and most likely be correct.

If you see a plot of a speaker with an impedance of 3.1 ohms in the major power band and a phase angle around -60, then that is pretty much a 100% certain an incompetent design with a ringing crossover. I don't care how many zillions it cost, how impressive it looks and how much the audiphools fawn over it, it's a dud pure and simple.

 

[PENG]

True, but in practice you can spot a major problem with dead reckoning, and most likely be correct.

If you see a plot of a speaker with an impedance of 3.1 ohms in the major power band and a phase angle around -60, then that is pretty much a 100% certain an incompetent design with a ringing crossover. I don't care how many zillions it cost, how impressive it looks and how much the audiphools fawn over it, it's a dud pure and simple.

I agree with you 100%. You mentioned few members would know how to calculate impedance phase angle, and I simply cited one reason. That reason being, the impedance of a passive crossover network in loudspeakers will typically be complex; and it is not only frequency dependent, but also time dependent. So accurate calculations to cover the frequency spectrum will involve advance math that includes complex numbers, transfer functions and some mathematical transform such as Laplace, among others.

 

[Johnny2Bad]

You mentioned in your post that a 8 ohm load provided a higher damping factor than a 4 ohm load. Phase angle alters what the amplifier sees as impedance and simply stating 8 ohms without qualifying a phase angle is only half the story. If you would have qualifed 8 ohms at a specified phase angle, then I would have not brought up point. I suggest very strongly that you assess the accuracy of your statements and include relevant parameters.

Im very well versed in impedance. One cant have impedance with out a frequency dependent component such as a capacitor or an inductor from which phase is derived. Anything else u care to add about impedance?

I specifically stated (and you quoted the text) that "There are some other differences that come up with different loudspeaker impedance loads. ..."

Maybe (okay, obviously) you don't read "other differences" as referring to what had not been discussed previously, such as Phase Angle. I do, and I am going out on a limb here, but I believe most people who comprehend English do as well.

An 8 ohm load DOES provide a higher damping factor than a 4 ohm load, and does so regardless of the Phase Angle. So, in order to satisfy you, here we go:

An 8 ohm load provides a higher damping factor than a 4 ohm load, regardless of the Phase Angle.

Happy now?

 

[TLS Guy]

I specifically stated (and you quoted the text) that "There are some other differences that come up with different loudspeaker impedance loads. ..."

Maybe (okay, obviously) you don't read "other differences" as referring to what had not been discussed previously, such as Phase Angle. I do, and I am going out on a limb here, but I believe most people who comprehend English do as well.

An 8 ohm load DOES provide a higher damping factor than a 4 ohm load, and does so regardless of the Phase Angle. So, in order to satisfy you, here we go:

An 8 ohm load provides a higher damping factor than a 4 ohm load, regardless of the Phase Angle.

Happy now?

There is no damping factor at all with a speaker having a passive crossover. This is only a consideration for active designs.

 

[highfigh]

An 8 ohm load DOES provide a higher damping factor than a 4 ohm load, and does so regardless of the Phase Angle. So, in order to satisfy you, here we go:

An 8 ohm load provides a higher damping factor than a 4 ohm load, regardless of the Phase Angle.

How does a lower impedance load increase the damping factor? Inquiring minds want to know.

http://www.crownaudio.com/media/pdf/amps/damping_factor.pdf

 

[TLS Guy]

How does a lower impedance load increase the damping factor? Inquiring minds want to know.

http://www.crownaudio.com/media/pdf/amps/damping_factor.pdf

That is all BS from Crown, because the DC resistance of the inductors dwarfs the output impedance of the amp! To make damping a factor you must have an amp in the speaker case directly connected to the speaker with a short stout wire. Unless that is the case then the amp damping the speaker is a complete red herring.

 

[PENG]

How does a lower impedance load increase the damping factor? Inquiring minds want to know.

http://www.crownaudio.com/media/pdf/amps/damping_factor.pdf

If you look at the definition, Johnny2Bad is right mathematically speaking. In real world, SS amps typically have output impedance so low that 4ohm or 8ohm speaker impedance won't make any audible difference. Tube amps could be a different story, depending on the design.

 

[Johnny2Bad]

How does a lower impedance load increase the damping factor? Inquiring minds want to know.

http://www.crownaudio.com/media/pdf/amps/damping_factor.pdf

Damping Factor is the total impedance of the load (loudspeaker, crossover elements, and both legs of the speaker cable, usually) divided by the output impedance of the amplifier (very low, typically, in Solid State amplifiers, and low but much higher than SS with Vacuum State amplifiers).

It is a purely resistive calculation, although it varies with frequency, it is unaffected by Phase Angle.

For example the Audio Research Reference 75 (Stereophile Review; Measured Values):
" ...
The output impedance was relatively low for an amplifier using a single pair of output tubes, at 0.74 ohm at 20Hz and 1kHz, and 0.95 ohm at 20kHz from the 4 ohm transformer tap.
From the 8 ohm tap these impedances were 1.25 and 1.56 ohms, respectively. ..."

From the above you can see that the Damping Factor is also frequency dependent, and as is typical, is lowest at low frequencies (where it's actually needed, according to theory) and highest at high frequencies (where it plays no, or almost no, role). This is true for Solid State amplifiers as well.

A typical output impedance for a Solid State amplifier is perhaps 0.10 ohm or less, often much less.

So, @ 20 Hz the AR R75's Damping Factor with a 4 ohm load connected to the 4 ohm transformer tap is 5.4; a Solid State amplifier (using the typical value I presented above) with the same load connected would have a DF of 40.

The formula to calculate is:

DF = Zload / Zsource

Manufacturers sometimes specify the DF but usually only testing reveals the values at various frequencies; manufacturers typically specify their DF value at 1 Khz (where the DF has no genuine useful function). So you might see DF specified at perhaps 200 or 500, but this is the 1 Khz value, not the value available below 100 Hz, where the benefit is supposed to be evident.

In theory, the Damping Factor helps the amplifier resist Back-Electromotive Force (Back-EMF) which again, in theory, should result in tighter control of the woofer by the amplifier. There are those who advocate high DF's as a benefit (particularly, these days, in Car Audio applications).

Since the SS amplifier has an inherent advantage in the DF specification, during the 1960's, when SS amps were battling for market share versus Vacuum State amplifiers, the SS manufacturers claimed that the DF of their amplifiers were an inherent advantage, resulting in "tighter" bass.

In practice this rarely was evident, and the SS amps of the 1960's are not known for particularly good sonic qualities in the first place ... Solid State didn't really come into it's own until the 1970's. That is why tube amps from the 60's are sought after items on the used market, while you can only sell SS amps of that vintage to the naive buyer of "Vintage" gear.

High Damping factors in Solid State amplifiers can be achieved by the Circuit Designer by increasing the amount of Global Negative Feedback, so in reality you can "Dial In" almost any DF you want in a SS amp (and high GNF improves all distortion specifications as well).

Since high GNF is associated with poor sonic qualities, you can use the DF specification as a kind of warning that the amplifier you are considering may not be the bargain it claims to be, despite the fact that it's specifications "blow the competition out of the water".

 

[PENG]

It is a purely resistive calculation, although it varies with frequency, it is unaffected by Phase Angle.


The formula to calculate is:

DF = Zload / Zsource

You are not correct this time. As shown in the formula you listed, it is not "purely resistive" as such, Z is impedance. Phase angle is automatically in the equation when you are dealing with impedance that is represented by a complex number. A complex number has magnitude and phase that are directly related.

If purely resistance is involved, phase angle will be 0 degree.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html

Here's a link for you, or you can read up on wiki (better IMO in this case) if you prefer.

 

[3db]

I specifically stated (and you quoted the text) that "There are some other differences that come up with different loudspeaker impedance loads. ..."

Maybe (okay, obviously) you don't read "other differences" as referring to what had not been discussed previously, such as Phase Angle. I do, and I am going out on a limb here, but I believe most people who comprehend English do as well.

An 8 ohm load DOES provide a higher damping factor than a 4 ohm load, and does so regardless of the Phase Angle. So, in order to satisfy you, here we go:

An 8 ohm load provides a higher damping factor than a 4 ohm load, regardless of the Phase Angle.

Happy now?

No not really.

 

[Johnny2Bad]

Well. all I can say is I'm not playing in your P*ssing Contests. Have fun.

Thank the Lord that there are people who read these topics and get something out of them.

Otherwise there would be no point in participating whatsoever. And it's very very close to that as it is.

 

[izzzzzz6]

Hi Guys. I ended up here whilst looking for an answer to a question.
I would like to buy a new driver for an active speaker that has obtained some damage to the cone.
The speaker is an active PA speaker, an RCF ART 500A usually it drives a 15" bass and i believe a 1 or 2" compression driver. 400W to the 15" and 100W to the compression driver. I can find a re-cone kit but at the price difference i would prefer to fit RCF drivers that i have found from blue aran.
However the original driver is 4 ohm and any 15" drivers available to the public from RCF or their distributors are all 8 ohms. I have found a driver that appears to use the same chassis or at least very similar to the original. The original 500A driver must probably be available to OEM or dealers only.
So i was looking at 2 options.
1. A RCF L15P540 500W AES 1000W program power.
Or
2. A RCF L15P530 350W AES 700W program.
Since these are 8 ohms rated not 4 ohms as the original was then should i go for the 350W driver as the amplifier probably wont be able to drive the 8 ohm so hard?
Or should i just go for the 500W driver as there is plenty of headroom on the volume control on the back of the amp. Will the amp find it's own comfort zone as such?
Hopefully this scenario will help some people understand better their choice of speaker. because i am a bit stumped right now. I know i can probably pay 2 times as much and just buy the correct driver but in theory i am thinking that one of these 8 ohm drivers is going to be well suited as an equivalent.

 

[TLS Guy]

Hi Guys. I ended up here whilst looking for an answer to a question.
I would like to buy a new driver for an active speaker that has obtained some damage to the cone.
The speaker is an active PA speaker, an RCF ART 500A usually it drives a 15" bass and i believe a 1 or 2" compression driver. 400W to the 15" and 100W to the compression driver. I can find a re-cone kit but at the price difference i would prefer to fit RCF drivers that i have found from blue aran.
However the original driver is 4 ohm and any 15" drivers available to the public from RCF or their distributors are all 8 ohms. I have found a driver that appears to use the same chassis or at least very similar to the original. The original 500A driver must probably be available to OEM or dealers only.
So i was looking at 2 options.
1. A RCF L15P540 500W AES 1000W program power.
Or
2. A RCF L15P530 350W AES 700W program.
Since these are 8 ohms rated not 4 ohms as the original was then should i go for the 350W driver as the amplifier probably wont be able to drive the 8 ohm so hard?
Or should i just go for the 500W driver as there is plenty of headroom on the volume control on the back of the amp. Will the amp find it's own comfort zone as such?
Hopefully this scenario will help some people understand better their choice of speaker. because i am a bit stumped right now. I know i can probably pay 2 times as much and just buy the correct driver but in theory i am thinking that one of these 8 ohm drivers is going to be well suited as an equivalent.

You need to replace it with the original driver. The impedance is the least of your problems. The replacement driver must have identical Thiel/Small parameters or the speaker will not sound the same at all.

 

[billmiller]

Hi,
will you guys please explain to me what the difference is between a 200 watt 4 ohm speaker and a 200 watt 8 ohm speaker and what one needs a more powerfull amplifier than the other?

It all has to do with an amplifiers ability to source current to a load. With all things equal, a 4 ohm speaker requires twice the current draw than a similar 8 ohm version.



An amp rated to deliver 200 watts into a 4 ohm speaker needs no more power than an amp rated to deliver 200 watts into an 8 ohm speaker. The amp power needed in both cases is the stated 200 watts.

When your amp is rated to deliver power into 4 ohms, and you choose to deliver it to 8 ohms, or vice versa,
(rated at 8 and delivering to 4) then you have to understand the relevance of Jacobi's impedance matching theorem.
matching maximizes the transfer of power in a circuit. The best a circuit can do is to deliver 1/2 of its power to its load. That occurs when the internal resistance (IR) of the power source is equal to the (RL) load resistance of the circuit.. If a circuit has an IR of 4 ohm, max power is delivered to a 4 ohm speaker load. (and 1/2 is dissipated by the IR as a loss). If RL (speaker imedance) were to increase, the speaker load will receive a larger share of available system power (as compared to a static IR which now receives a lesser share), and efficiency is improved. However, total circuit current has been reduced with a resulting exponential loss of circuit power. (Watts = I squared R). Alas, the higher ohm speaker is now sharing more of LESS, with the end result being negative yardage. There has been a loss of load power compared to the matched circuit. Using simple series DC circuits, I calculate the power reduction from 4 ohm to 8 ohm speakers to be from 200 to 173 watts. A lot of "specmanship" is used in stereo component systems. If your stereo is a Pyle, then your "200 watts" may actually indicate peak power, and not Root Mean Squared (RMS) or DC equivalent power. Now if you are intending on using the 8 ohm speakers, and your amp is rated for 4 ohm speakers, you must multiply the 173 watts by .707 for a result of 122 real watts of power delivered to your 8 ohm speakers. You get only 61% of what has been advertised. (which is why they do it this way). The addition of Inductance and Capacitance doesn't alter the premise of impedance matching. This stuff has been known since 1870. Give a shout out to Jacobi, who wrote the theorem on impedance matching in 1870.

 

[TLS Guy]

We have been down this road many times before.

Power is voltage X amps. So the 8 ohm drive needs half the current and twice the voltage of the 4 ohm load. So the four ohm load is double the current.

That double current load has to pass though the output devices. These devices all have a resistance and it will be pretty much the same whether 4 or 8 ohm

Now this is were the problem is you have overlooked. The heating effect of that current is the square of the current times the resistance. So heating goes up by the square of the current and has to be dissipated from the output devices. Heat kills.

Just in case you think that upping the impedance would be a good idea, you are wrong again. This means increasing the voltage. High voltage and semi conductors is a bad idea, because those nasty high voltages spark and discharge through the semi conductor material and makes holes in it, destroying the device.

So there you have it.

 

[Irvrobinson]

Well. all I can say is I'm not playing in your P*ssing Contests. Have fun.

Thank the Lord that there are people who read these topics and get something out of them.

Otherwise there would be no point in participating whatsoever. And it's very very close to that as it is.

It would help your credibility a lot if your clueless misinterpretations of engineering and technology didn't so often closely follow the pattern of a Wikipedia page:

Damping_factor

 

[Matthew J Poes]

We have been down this road many times before.

Power is voltage X amps. So the 8 ohm drive needs half the current and twice the voltage of the 4 ohm load. So the four ohm load is double the current.

That double current load has to pass though the output devices. These devices all have a resistance and it will be pretty much the same whether 4 or 8 ohm

Now this is were the problem is you have overlooked. The heating effect of that current is the square of the current times the resistance. So heating goes up by the square of the current and has to be dissipated from the output devices. Heat kills.

Just in case you think that upping the impedance would be a good idea, you are wrong again. This means increasing the voltage. High voltage and semi conductors is a bad idea, because those nasty high voltages spark and discharge through the semi conductor material and makes holes in it, destroying the device.

So there you have it.

Most semiconductors are operated close to their voltage limit anyway. To make a higher impedance and higher voltage amplifier possible (without going BTl type circuits) would far exceed the voltage the devices are designed around. The last couple amplifiers I designed and built ran a bit under +/- 100 volt rails and that was very close to the limit of the output devices. If we suddenly decided we wanted to make 16ohm speakers standard (for whatever reason) and run the same power, you would be looking at 200 volt rails! That’s really hard to design around, including the special power supply needs.

A German pro company known as SPL has designed such an amplifier that operates very high voltage rails. I’m not clear if their amplifier runs the same rails on the current output stage. I hope to review their amplifier in the future simply to answer this curiosity.

I really feel that once we get into these kinds of power needs, designing an amplifier with high current capabilities, keeping rail voltages manageable, and just moving to 240 lines makes more sense. If someone is already at that extreme, what’s a couple dedicated lines (I have 4 in my room).

 

[Verdinut]

Most semiconductors are operated close to their voltage limit anyway. To make a higher impedance and higher voltage amplifier possible (without going BTl type circuits) would far exceed the voltage the devices are designed around. The last couple amplifiers I designed and built ran a bit under +/- 100 volt rails and that was very close to the limit of the output devices. If we suddenly decided we wanted to make 16ohm speakers standard (for whatever reason) and run the same power, you would be looking at 200 volt rails! That’s really hard to design around, including the special power supply needs.

A German pro company known as SPL has designed such an amplifier that operates very high voltage rails. I’m not clear if their amplifier runs the same rails on the current output stage. I hope to review their amplifier in the future simply to answer this curiosity.

I really feel that once we get into these kinds of power needs, designing an amplifier with high current capabilities, keeping rail voltages manageable, and just moving to 240 lines makes more sense. If someone is already at that extreme, what’s a couple dedicated lines (I have 4 in my room).

By the way, the Outlaw 976 review is well presented and comprehensive.

When I have to replace my Marantz SR5010 which I'm using as a pre-pro at present, the then available Outlaw processor shall definitely be on my list for comparison with the available products in the reasonable price range.