Would you upgrade or stop?

[PENG]

The tightly regulated SMPS is not going to have the headroom off a non-regulated linear power supply.
The specs state 29A peak current so that may give a better indication of dynamic power. What does that indicate for watts, say into 4 Ohms?

Typical manufacturer's marketing hypes! It is not possible to comment because one would have to make all sorts of assumptions, such as what I am doing below:

Assumptions:
- the 29A peak current is the peak value of the sine wave, that is, not peak as for maximum rms current.
- and it is not peak to peak, just peak.

In that case, 29A peak = 29/square root 2 = 20.5A
To deliver 20.5A into 4 ohms, the voltage will be 20.5X4 = 82 V rms

I don't believe the rail voltage is as high as 82 Vrms.

So it is not possible to project in terms of "watts". We would have to know the maximum voltage the amp can output and at what load impedance that 29A peak is based on, as well as the phase angle because:

Watts = Voltage X Current X Cosine (phase angle), or Current X Current X the resistance of the impedance based on the following power formula:

Power = VI Cos(ø), ø is the phase angle, or
Power = I^2 R, R is the resistance part of the impedance.

I think it is more important to know the voltage and current. Too bad everyone talks about "watts", when it really isn't as useful for loudspeaker loads.

BTW, there were some posts Audiosciencereview that Revel Performa 228be, due to impedance and phase angle provide a load equivalent to 2.8 Ohms.

Very interesting point, but It depends on what the poster knew at the time. If it was Amir, okay he's an EE and is knowledgeable in audio electronics so I would take his words for it. Otherwise it could be just hearsay, or stated by someone who may or may not fully understand the effects of phase angles. One thing I do know is that phase angle is part of the characteristics of the impedance, but if the absolute value |Z| is 8 ohms at a certain frequency, then it is 8 ohms regardless of the phase angle at that same frequency point.

Note: impedance has two components mathematically speaking, and can be represented as a complex number, with a real term and the imaginary term:

Z = R + jX, where R is the resistance, and X is the reactance (inductive, capacitive, or both).

for details, check out hyperphysics website:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html

The issue concerning phase angle is heat dissipation in the amp, the higher the phase angle typically means the higher the heat the amplifier output stage has to dissipate. How to figure out the equivalency even in that narrow sense is complicated. I don't know how to do it, without spend hours on text books and audio amplifier handbooks. Even then, it still depends on the particular design of the amplifier, some may be less prone to the phase angle effects than others.

That looks to be not a suitable load for an AVR. Even though, the AVR and benchmark both are spec'ed with similar power into 8 (and possibly) 4 Ohms.

I would agree in general sense, but if the AHB2 can do it, then a strong AVR such as the RX-A3080, AVR-X8500H, SR8012, and some of the older Onkyo/Integra/NAD flagship or near flagship AVR should be able to do it, in term of output power. Example: the Onkyo TX-SR805 measured (by S&V) 327.6 W @0.1% THD+N, that's 1 kHz, but the AHB2 measured under the same condition will not come close at all. My Denon AVR did 298 W, not far behind the Onkyo, and @0.0021% THD+N at 167 W (knee point).

Don't get me wrong, I would of course take the AHB2 over any AVR, just that I feel AVR, some anyway, deserves a touch more respect.......

 

[RichB]

Typical manufacturer's marketing hypes! It is not possible to comment because one would have to make all sorts of assumptions, such as what I am doing below:

Assumptions:
- the 29A peak current is the peak value of the sine wave, that is, not peak as for maximum rms current.
- and it is not peak to peak, just peak.

In that case, 29A peak = 29/square root 2 = 20.5A
To deliver 20.5A into 4 ohms, the voltage will be 20.5X4 = 82 V rms

I don't believe the rail voltage is as high as 82 Vrms.

So it is not possible to project in terms of "watts". We would have to know the maximum voltage the amp can output and at what load impedance that 29A peak is based on, as well as the phase angle because:

Watts = Voltage X Current X Cosine (phase angle), or Current X Current X the resistance of the impedance based on the following power formula:

Power = VI Cos(ø), ø is the phase angle, or
Power = I^2 R, R is the resistance part of the impedance.

I think it is more important to know the voltage and current. Too bad everyone talks about "watts", when it really isn't as useful for loudspeaker loads.



Very interesting point, but It depends on what the poster knew at the time. If it was Amir, okay he's an EE and is knowledgeable in audio electronics so I would take his words for it. Otherwise it could be just hearsay, or stated by someone who may or may not fully understand the effects of phase angles. One thing I do know is that phase angle is part of the characteristics of the impedance, but if the absolute value |Z| is 8 ohms at a certain frequency, then it is 8 ohms regardless of the phase angle at that same frequency point.

Note: impedance has two components mathematically speaking, and can be represented as a complex number, with a real term and the imaginary term:

Z = R + jX, where R is the resistance, and X is the reactance (inductive, capacitive, or both).

for details, check out hyperphysics website:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html

The issue concerning phase angle is heat dissipation in the amp, the higher the phase angle typically means the higher the heat the amplifier output stage has to dissipate. How to figure out the equivalency even in that narrow sense is complicated. I don't know how to do it, without spend hours on text books and audio amplifier handbooks. Even then, it still depends on the particular design of the amplifier, some may be less prone to the phase angle effects than others.



I would agree in general sense, but if the AHB2 can do it, then a strong AVR such as the RX-A3080, AVR-X8500H, SR8012, and some of the older Onkyo/Integra/NAD flagship or near flagship AVR should be able to do it, in term of output power. Example: the Onkyo TX-SR805 measured (by S&V) 327.6 W @0.1% THD+N, that's 1 kHz, but the AHB2 measured under the same condition will not come close at all. My Denon AVR did 298 W, not far behind the Onkyo, and @0.0021% THD+N at 167 W (knee point).

Don't get me wrong, I would of course take the AHB2 over any AVR, just that I feel AVR, some anyway, deserves a touch more respect.......

Here is a post by John Siau concerning the AHB2 protection architecture:

https://www.audiosciencereview.com/forum/index.php?threads/review-and-measurements-of-benchmark-ahb2-amp.7628/page-18#post-180791''

I have driven the AHB2 into some intermittent clipping (Clip lights but no Temp lights) that indicated voltage and not current clipping given the above.

There seems to be a trend to using FPGA and microprocessors from Benchmark, Anthem, and even Schitt, to replace analog protection circuits.

- Rich

 

[PENG]

Here is a post by John Siau concerning the AHB2 protection architecture:

https://www.audiosciencereview.com/forum/index.php?threads/review-and-measurements-of-benchmark-ahb2-amp.7628/page-18#post-180791''

I have driven the AHB2 into some intermittent clipping (Clip lights but no Temp lights) that indicated voltage and not current clipping given the above.

There seems to be a trend to using FPGA and microprocessors from Benchmark, Anthem, and even Schitt, to replace analog protection circuits.

- Rich

That's great, so he confirmed my first assumption that the 29 A is the peak value of the sine wave. He also provided the missing link, that is, the minimum load impedance is 1.4 ohm.
V = IZ = 20.5 X 1.4 = 28.7 V
P = VI*Cos(0) = 28.7 X 20.5 = 588.35 W assuming 0 deg phase angle.

So 588.35 W is your answer, at the tripping point..and assuming the 1.4 ohm is resistive/0 degree phase angle. That seems very high for that amp but then in practical term there aren't going to be many loudspeaker that dip down to 1.4 ohm at 0 degree phase angle.

As far as the impedance in ohms equivalency at the phase angles he mentioned, he seems to have done what I thought he (not knowing its John Siau) might have done, that is, simply basing it on the power formula:

Power = VI cos( ) that I mentioned before to calculate the equivalency in terms of watts, and that is not meaningful.

That is a wrong approach imo. Again, not that as long as the impedance is 4 ohms, the amp will still deliver the same current equal to the output voltage V/4, regardless of the phase angle. So what is his impedance equivalency claim based on? If it is based on heat dissipation, or amp stability (that I believe would be a good way), then he should have said so and explained why then he could simply use the power formula.

 

[RichB]

That's great, so he confirmed my first assumption that the 29 A is the peak value of the sine wave. He also provided the missing link, that is, the minimum load impedance is 1.4 ohm.
V = IZ = 20.5 X 1.4 = 28.7 V
P = VI*Cos(0) = 28.7 X 20.5 = 588.35 W assuming 0 deg phase angle.

So 588.35 W is your answer, at the tripping point..and assuming the 1.4 ohm is resistive/0 degree phase angle. That seems very high for that amp but then in practical term there aren't going to be many loudspeaker that dip down to 1.4 ohm at 0 degree phase angle.

As far as the impedance in ohms equivalency at the phase angles he mentioned, he seems to have done what I thought he (not knowing its John Siau) might have done, that is, simply basing it on the power formula:

Power = VI cos( ) that I mentioned before to calculate the equivalency in terms of watts, and that is not meaningful.

That is a wrong approach imo. Again, not that as long as the impedance is 4 ohms, the amp will still deliver the same current equal to the output voltage V/4, regardless of the phase angle. So what is his impedance equivalency claim based on? If it is based on heat dissipation, or amp stability (that I believe would be a good way), then he should have said so and explained why then he could simply use the power formula.

4 ohms at -45 deg = 2.83 ohms - j2.83 ohms -- real and imaginary parts of the impedance, yes (j = sqrt(-1)).
A angB = A cosB + j*A*sinB = x + jy
A = sqrt(x^2 + y^2), B = arctan(y/x)

https://www.audiosciencereview.com/forum/index.php?threads/review-and-measurements-of-benchmark-ahb2-amp.7628/page-23#post-182902

I don't understand these equations but they might explain the thinking behind treating phase as the load equivalence of lower impedance into resistive loads.

- Rich

 

[PENG]

https://www.audiosciencereview.com/forum/index.php?threads/review-and-measurements-of-benchmark-ahb2-amp.7628/page-23#post-182902

I don't understand these equations but they might explain the thinking behind treating phase as the load equivalence of lower impedance into resistive loads.

- Rich

Agreed, unfortunately you don't/won't find much accurate analysis given in detail on this topic. Try googling something like:

Phase angle effects+audio amplifiers+dissipation or any variations you can think of and you won't get too far. I think the reason is that the academic world are not very interested in studies on technical topics only audiophiles/electronic hobbyists may be interested in. Even if there are many research scientists, research students and professors are interested in such topics they would have hard time getting sponsors.

Equivalency between the two effects (impedance vs phase angle) is very complicated to derive and even if someone managed to derive a bunch of formula to calculate the equivalency, a generalize one is likely not very meaningful because of many reasons, and I can immediately think of a couple, such as:

- Both typically varies with frequency, but look very differently, or similar in some cases in terms of the shapes.
- Impedance variations invariably result in load current variations according the Ohms law (we all know what Current = Voltage/Impedance, i.e. I = V/Z, I = V/R right?)
- Phase angle variations alone, does not result in load current variations, but it affects the power dissipation, that in turn results in heat dissipation in the amplifier's output stage.
- Phase angle effects on the output stage heat dissipation would understandably be different among different design of amp, for an extreme example, class D vs class AB..
- Music contents have major effects too, and the effects will be quite different to phase angle vs impedance simply by looking at the typical Phase angle vs freq and Impedance vs freq curves.

Case in point, why don't they (Soundstage networks, Stereophile, Audioholics.com etc.) plot the impedance corrected for phase angle vs frequency graphs for speakers? My guess is, it is not that simple, or even practical. For example: If the combination effects of 4 ohm at a certain phase angle to that amp is 2.8 ohm, that's fine for the AHB2 to do that all day long, but what if it is not the AHB2, but an amp that cannot handle the resulting 40% increase in current. Also, how about providing extra cooling to an amp driving a speaker load that has higher "nominal" phase angles, that should help to an extent I would think, while the effects of high current due to low impedance include not only increase heat, but obviously other issues.

Way back I was probably among the first on this forum who mentioned multiple times the need to consider phase angle, during the days when the only formula you see on this forum on the question of power are P=VI, V=IR, without referencing to the phase angle/power factor i.e. Cos ( ) part of the equation that is applicable to reactive loads such as loudspeakers. @Steve81 was the only person that I know who responded to my point (may have nothing to do with my posts on the topic, too long ago..)seriously, and eventually posted a nice article touching on this topic.

We found a very useful Elliot Sound article on this topic:
http://sound.whsites.net/patd.htm

I also found one that is written by an EE:

Scroll to Sect 14 if you only want to save time
http://www.woodjohn.uk/Introduction to current-drive audio amplifiers Rev. 1.pdf

You may or may not understand all the formula either, but it should give you an idea of the complexity involved in trying to come up with equivalency.

Regardless, it is nice that Mr. Siau's examples gave the readers some ideas, in terms of at least highlighted an effect that often get ignored. I just couldn't help being too curious, and overly critical..

 

[JDM3030]

Just a quick update...

I decided to get another Crown amp. I found a used XLS 1502 in great condition for a really good price. I have the XLS 1002 for my mains now. Moving forward, I will use one of the two channels on that amp to power my center channel. I will then use the new XLS 1502 to power my two front mains. I plan to sell the Dayton APA150 which I previously used to power the center channel in bridged mode. Down the line I may upgrade the subs but that will be a much bigger financial investment so I won't be doing that for quite some time.

Thanks again for all the feedback and help!