Running 4 ohm studio monitors off vintage amp / receiver rated at 8 ohms?

[dualflip]

It means that it will output 160W if using a 8ohm load, it also means that if you use a 4ohm speaker, following ohms law, it will be outputing 320W (although the amp will melt before this happens), this could in turn overstress the amp since there will be twice the power at its output. However the 160W is the maximum power output of the amp when connected to an 8ohm load, so if you connect a 4ohm speaker and dont completely blast the speakers or overload the amp, you should be fine.

Keep it at a moderate level, think that in theory, at full volume you are getting twice the power that you would have if you used 8 ohm speakers...

I must say, long time ago, back before i knew anything, I paralleled 3 4ohm speakers and connected them to 750W amp at full volume so I could have "more blast" , that effectively lowers the total impedance to 1.3 ohms! thats 31AMPs!! at maximum output. After some minutes at full volume the amp melted... good times!

[666666]

Originally Posted by dualflip

...it also means that if you use a 4ohm speaker, following ohms law, it will be outputing 320W

Thanks for the input!

I just found the SX-1250 service manual online and it states the following for this amp:

160W @ 8 ohm
200W @ 4 ohm

This info was not in the owner's manual, only in the service manual. Their 4 ohm rating here does not seem to follow the typical ohm calculation, I'd image there's a good reason. This amp / receiver does have three sets of speaker outs (switchable) and they talk about ways to wire things up so you can run four speakers at once and so on... likely this has something to do with the ratings.

So they ARE giving us a 4 ohm rating here. Not sure if this implies that it's ok to use 4 ohm speakers. There is still no specific mention about whether or not they recommend the use of 4 ohm speakers.

It's a really nice amp / receiver, at least for casual listening purposes. Has nice, useful tone controls too. In a perfect world I'd find a nice pair of vintage period home stereo speakers to go with it (which I'd imagine would likely be 8 ohm typically), but until I do, for now, I may as well make use of some of the good stereo monitors I have laying around (most of them being rated at 4 ohm).

Thanks!

[Richard Crowley]

Originally Posted by 666666

I just found the SX-1250 service manual online and it states the following for this amp:

160W @ 8 ohm
200W @ 4 ohm

That means, at least, that the amp is somewhat safe with a 4 ohm load. Otherwise, I would have recommended against even trying this.

Their 4 ohm rating here does not seem to follow the typical ohm calculation, I'd image there's a good reason.

dualflip gave us the reason. You are correct that simple Ohm's law would suggest 320W @ 4 ohms. But the amplifier (either the output transistors AND/OR the power supply) aren't up to producing 320W

This amp / receiver does have three sets of speaker outs (switchable) and they talk about ways to wire things up so you can run four speakers at once and so on... likely this has something to do with the ratings.

They typically use a complex arrangement of series/parallel (and sometimes build-out resistors) to achieve switching all those combinations while not putting too much load on the amplifier.

So they ARE giving us a 4 ohm rating here. Not sure if this implies that it's ok to use 4 ohm speakers. There is still no specific mention about whether or not they recommend the use of 4 ohm speakers.

The fact that they even state a 4 ohm rating means (at least IMHO) that they imply that it is safe to use with a 4 ohm load (although clearly at reduced power capability).

Remember also that amplifier power ratings, and published speaker impedances are only approximate. Most speakers are only "8 ohms" or "4 ohms" in a relatively narrow frequency band, and a higher impedance elsewhere.

[vince @ speck]

Results depend on the amp. Most manufactuers don't show a Z below published specifications but many amps can handle 4 ohm load just fine.

I have a friend that uses a 1 ohm speaker on a 8 ohm amp...works fine... it get hots, but it works.

[dualflip]

Originally Posted by rcrowley

dualflip gave us the reason. You are correct that simple Ohm's law would suggest 320W @ 4 ohms. But the amplifier (either the output transistors AND/OR the power supply) aren't up to producing 320W

Exactly, the 320W comes from a simple ohms law calculation, in this case, ohms law considers a source capable of providing an infinite amount current (or voltage for that matter) so you will get a linear ohms VS power ratio, but reality is a whole different thing, like rcrowley said, amps are limited either by its components or the power supply, if not it would be so simple as to get the same power from 1 amp rather than using 2.

Just dont do what I did long ago with the three parallel speakers e

[johnnyc]

Originally Posted by dualflip

Exactly, the 320W comes from a simple ohms law calculation, in this case, ohms law considers a source capable of providing an infinite amount current (or voltage for that matter) so you will get a linear ohms VS power ratio, but reality is a whole different thing, like rcrowley said, amps are limited either by its components or the power supply, if not it would be so simple as to get the same power from 1 amp rather than using 2.

Just dont do what I did long ago with the three parallel speakers e

The reason for not having a linear power vs speaker ohms relationship has to do with the output impedance of the amplifier. If you view the amp as a voltage source then ideally its output impedance would be zero, but in reality it's non-zero.

Alternatively you could view the amp as a current source with a non-ideal non-infinite output impedance.

[JohnRoberts]

Originally Posted by johnnyc

The reason for not having a linear power vs speaker ohms relationship has to do with the output impedance of the amplifier. If you view the amp as a voltage source then ideally its output impedance would be zero, but in reality it's non-zero.

Alternatively you could view the amp as a current source with a non-ideal non-infinite output impedance.

Audio amplifiers are basically voltage sources but they have a finite source impedance, and transformer windings have resistance and will cause PS voltage sag with increased current draw, so they will rarely double when dropping load impedance in half.

The probable explanation for the only modest increase at 4 ohms for that amp is likely current limiting to protect the output stage devices.

JR

[johnnyc]

Originally Posted by JohnRoberts

Audio amplifiers are basically voltage sources

I'm not necessarily saying they aren't. But from a black box model perspective you can view it as either, they are equivalent, and one might be more convenient and intuitive for a given analysis. Given that the purpose of a power amp stage is often to increase current drive capability, choosing the current source model approach is not unreasonable.

[JohnRoberts]

Originally Posted by johnnyc

I'm not necessarily saying they aren't. But from a black box model perspective you can view it as either, they are equivalent, and one might be more convenient and intuitive for a given analysis. Given that the purpose of a power amp stage is often to increase current drive capability, choosing the current source model approach is not unreasonable.

Not really... The black box model would be a voltage source, with a very small R in series.

JR

[johnnyc]

Originally Posted by JohnRoberts

Not really... The black box model would be a voltage source, with a very small R in series.

JR

You can transform a Thevenin equivalent into a Norton and vice versa.

[Bill Way]

Chances are the Pioneer will do just fine - I remember that, when it was new, we all wanted one.

But... what speakers? If their nominal impedance is 4-ohms, great. But impedance varies with frequency, so it's a good idea to know what their impedance curve looks like, and specifically where the low points are and how low they are.

And... the SX1250 has a pretty good protection circuit for the outputs. If the load impedance drops too low you should hear a couple relays click inside the chassis and silence through the speakers. Suggest you get the service manual from hifiengine.com. (You'll need it anyway to re-cap that thing.)

WW

[JohnRoberts]

Originally Posted by johnnyc

You can transform a Thevenin equivalent into a Norton and vice versa.

You could, but that doesn't illuminate how an actual audio power amp behaves wrt loads.

If it was predominantly a current source, the power in the load would drop in half when the load impedance was halved. An ideal voltage source, doubles power in the load when the impedance drops in half.

A real world power amp deviates from an ideal voltage source as noted, but that is the best characterization for it's behavior.

JR

[johnnyc]

Originally Posted by JohnRoberts

If it was predominantly a current source, the power in the load would drop in half when the load impedance was halved. An ideal voltage source, doubles power in the load when the impedance drops in half.

True but only if you neglect the internal impedance. Let's go back to the example, by my calculations you get a 2.9 ohm internal impedance. Subsequently you can model the amp as a 48.8V voltage source with 2.9 ohms in series, OR a 16.8A current source with 2.9 ohms in parallel. Both will give you 160W for 8 ohm load and 200W for 4 ohm load.

Granted the above is simplified and there are other factors at work as you mentioned earlier. But given finite impedances, both voltage and current source models remain valid.

[JohnRoberts]

Originally Posted by johnnyc

True but only if you neglect the internal impedance. Let's go back to the example, by my calculations you get a 2.9 ohm internal impedance. Subsequently you can model the amp as a 48.8V voltage source with 2.9 ohms in series, OR a 16.8A current source with 2.9 ohms in parallel. Both will give you 160W for 8 ohm load and 200W for 4 ohm load.

Granted the above is simplified and there are other factors at work as you mentioned earlier. But given finite impedances, both voltage and current source models remain valid.

That is an useful circuit analysis tool for hypothetical sources, but again it does not represent how real power amps actually behave.

Actual power amplifiers use negative feedback to deliver source impedances on the order of tens of milli-Ohms, which neither simplistic model correctly reflects.

The internal resistance model is somewhat useful to model the unregulated PS for the amp, but not output power. Dissipation in amplifier output devices are not linear, so if the the 200W power output is due to current limiting and not voltage limiting, that model is not even accurate for the PS.

JR

[dualflip]

Originally Posted by johnnyc

True but only if you neglect the internal impedance. Let's go back to the example, by my calculations you get a 2.9 ohm internal impedance. Subsequently you can model the amp as a 48.8V voltage source with 2.9 ohms in series, OR a 16.8A current source with 2.9 ohms in parallel. Both will give you 160W for 8 ohm load and 200W for 4 ohm load.

Granted the above is simplified and there are other factors at work as you mentioned earlier. But given finite impedances, both voltage and current source models remain valid.

Im guessing that you are using Norton and Thevenin to get the equivalent Voltage or current with a single impedance, May I ask how did you get to those numbers?

[johnnyc]

Originally Posted by dualflip

Im guessing that you are using Norton and Thevenin to get the equivalent Voltage or current with a single impedance, May I ask how did you get to those numbers?

Yes, (admitedly simplified) Thevenin and Norton equivalents were used

Power=V^2/R

You end up with 2 equations and 2 unknowns and can then solve for the variables.

160W=[(V*(8/(8+Rint)))^2]/8
200W=[(V*(4/(4+Rint)))^2]/4

You can do a similar thing for I or just convert the Thevenin to Norton.

[JohnRoberts]

Now calculate damping factor, using either of your models.

JR

PS: Perhaps useful for modeling batteries and unregulated power supplies. The source impedance of the power amp is independent of the power it can deliver.

[666666]

OP here. Thanks for all the info, guys!

I have one more, related question about power amps, load, etc.

I also have a dedicated amp (Hafler P4000) and I wish to follow it with a speaker switcher so I can have the ability to switch between two sets of passive speakers.

In another thread recently (in a different section), it was suggested to me that this is a bad idea because while switching between the speakers, there will be a short time that the amp will see no load as one set gets turned off and then the other set gets turned on.

Someone else then suggested that as long as program isn't being fed into the amp and the amp is at idle, it wouldn't really matter. But not sure if this is true. If this IS true, then all one would need to do is simply mute the program being fed to the amp before doing the switching.

Another thought.... if it's indeed bad to actually turn off one pair of speakers then turn on another pair, at idle or otherwise (cutting all load from the amp momentarily), I know I've seen speaker switchers that will allow BOTH connected pairs of speakers to run at the same time. So then, while one pair is playing, you could switch on the second pair (thus two sets would be sounding at once, at least momentarily), then just switch off the first pair. In this way the amp would never be in a "no load" situation, it would just see a change in impedance for a short time, and let's say you keep the amp at idle during this switch, perhaps this is ok?

Just trying to figure out a safe way to switch different sets of passive speakers on my dedicated amp... if possible. I wish to be as gentle to my amp as possible so it gives me a lot of years of reliable use.

Thanks!

[Richard Crowley]

The recommendation to avoid switching during operation is a good idea. That is the path I would take.

The devices that allow switching more than one set of speakers often use complex arrangements of series and parallel and even extra resistors to avoid presenting too low a load impedance to the amplifier.

[JohnRoberts]

Running an amp with no load was mainly and issue with tube amp, not solid state which will remain stable and be OK.

For relays it is harder to switch open while passing high current, and contacts can degrade, so generally not good practice to switch while playing at full tilt.

When switching between only two loads it is pretty simple with double throw relay contacts. More complex speaker switchers need to be careful about overlap in multiple relay timing.

JR

[dualflip]

Originally Posted by johnnyc

Yes, (admitedly simplified) Thevenin and Norton equivalents were used

Power=V^2/R

You end up with 2 equations and 2 unknowns and can then solve for the variables.

160W=[(V*(8/(8+Rint)))^2]/8
200W=[(V*(4/(4+Rint)))^2]/4

You can do a similar thing for I or just convert the Thevenin to Norton.

Ohh great I can see the logic, however I cant see from where did you get the 8/8+Rint part of the formula substituded inside V. Somehow if I were to do it using 2 equations, I would have thought of something like this:

160W=(V^2)/[(Rint*8)/(Rint+8)]
200W=(V^2)/[(Rint*4)/(Rint+4)]

[johnnyc]

Originally Posted by JohnRoberts

Now calculate damping factor, using either of your models.

I never claimed it was the best model. Then again some amps can have damping factors in the single digits and sound good doing it, so you never know

[johnnyc]

Originally Posted by dualflip

Ohh great I can see the logic, however I cant see from where did you get the 8/8+Rint part of the formula substituded inside V. Somehow if I were to do it using 2 equations, I would have thought of something like this, although it seems more complex to solve:

160W=(V^2)/[(Rint*8)/(Rint+8)]
200W=(V^2)/[(Rint*4)/(Rint+4)]

The 160W and 200W refer to the power delivered to the load so you need to calculate the actual voltage across the load, hence the voltage divider times V. Also for Thevenin the load would be in series not parallel. It would be in parallel for Norton.

JR will argue that this model is not correct for this amplifier, and there is a good chance he is right, there are probably other factors at play. Even so this still provides a good illustration of how the power delivered to a load can be affected by the internal source impedance of the amp driving it. In the real world some amps don't use negative feedback and don't have milli-ohm source impedances, even if they are in the minority.

[JohnRoberts]

Originally Posted by johnnyc

I never claimed it was the best model. Then again some amps can have damping factors in the single digits and sound good doing it, so you never know

I apologize for being a pest but I spent over a decade at Peavey trying to explain some of the same principles to dealers and music store salesmen in factory seminars. So I worry about a model that gives a user the wrong impression about how his gear works.

====
Yes old tube amps had generally much lower DF than modern solid state amps due to higher natural source impedance of tubes before NF, and output transformers.

This higher source impedance can interact with non-flat cabinet impedance for better or worse. This is often used in guitar amp design for the "better", and I have seen solid state guitar amps, that use feedback to artificially raise output impedance for some "better'. But this can also be worse and that is why back in the old tube days we also saw 16 ohm loudspeakers to reduce this interaction.

JR

[dualflip]

Originally Posted by johnnyc

The 160W and 200W refer to the power delivered to the load so you need to calculate the actual voltage across the load, hence the voltage divider times V. Also for Thevenin the load would be in series not parallel. It would be in parallel for Norton.

JR will argue that this model is not correct for this amplifier, and there is a good chance he is right, there are probably other factors at play. Even so this still provides a good illustration of how the power delivered to a load can be affected by the internal source impedance of the amp driving it. In the real world some amps don't use negative feedback and don't have milli-ohm source impedances, even if they are in the minority.

I understand that you are using the voltage divider formula for an unknown current to substitute for V, and that is great!. I just like math and im obviously deviating from topic, I just want to give this more thought.

You are right, shouldnt have used paralel, dont know why when I pictured the load in my mind, it looked like a paralel resistor.. ohh well

[johnnyc]

Originally Posted by JohnRoberts

I apologize for being a pest but I spent over a decade at Peavey trying to explain some of the same principles to dealers and music store salesmen in factory seminars. So I worry about a model that gives a user the wrong impression about how his gear works.

No apology needed, if you have a differing view by all means point it out. I spent many years doing RF design so it's entirely possible that has caused me to over-emphasize the source load perspective which may not be as useful for audio applications.

Here is another example though with a QSC RMX850 amp, it is speced at:
200W@8ohms
300W@4ohms
430W@2ohms (higher distortion so a bit inflated)

Given these numbers shouldn't we expect more power at 4 ohms given a milli-ohm output impedance? Or is there really that much sag in the power supply?

[johnnyc]

Originally Posted by dualflip

I understand that you are using the voltage divider formula for an unknown current to substitute for V, and that is great!.

I was looking at it more as an unknown voltage. If power=(V^2)/R then we need to find how much voltage is across Rload (the speaker). This voltage would be equal to the source voltage times the voltage divider equation.

[dualflip]

Originally Posted by johnnyc

I was looking at it more as an unknown voltage. If power=(V^2)/R then we need to find how much voltage is across Rload (the speaker). This voltage would be equal to the source voltage times the voltage divider equation.

Yes thats what I actually meant, I meant that the voltage divider formula you used, is great for when you dont know the current, thus by multiplicating it by the total voltage, you get the voltage at the speaker, thats what I meant for unknown current.

On a side note, and following the main (of the now hijacked thread) argument. It is obvious that when doubling the load the expected or theoretical power doesnt match the actual rated power. Following your model, it basically states that a huge amount of power is being wasted by the internal impedance of the amp. In the case of your QSC amp, the expected power at 4 ohms should be 400W yet, its delivering 300W, that means (following your model) that 100W are being dissipated by the internal impedance of the amp, which seems a bit extreme.

Wouldnt it sound more convincing to say that the lower power value is caused by limitations or non linearities of the internal components of the amplifier when connected to different loads, or could it be the PSU mentioned earlier? or a combination of both, and of course some will also be dissipated by the internal impedance.

[JohnRoberts]

The internal impedance model is useful for looking at say voltage sag in a battery with current draw, or transformer windings with current draw, but internal losses in audio power amps are complex and nonlinear.

While this is an over-simplification, the power wasted in an output section is not a fraction of the power delivered to the load, but instead the product of the current delivered, times the difference between output voltage and full power supply voltage. So heat sink (waste) dissipation is actually more than output power at 1/3 power output, and drops proportionately as output power increases. For example lets say a simple class A/B amplifier with +/- 48V amp power supply is putting 16 volts peak into a 4 ohm speaker, At the peak of that waveform, the loudspeaker is getting 4A for 64W instantaneous power. At that same instant the output devices are dissipating 4 amps times 48v-16v for 128W. Of course sine wave power is not the same as that momentary peak power, but you see the general relationship.

This is more complex for power amps using variant class power supplies. Class G/H wastes far less in heat-sink power than class A/B, and newer class D even less waste heat, with class D approaching linear dissipation.

Lets, just say it's complicated.

If you want to characterize your car battery voltage when the starter motor is pulling 100a the internal resistance model is OK. Audio amps, not so much.

JR