Differences between high current amps and normal ones

[gus6464]

This subject has caught my interest as I have been told by some dealers that with certain speakers it's not the amount of wattage you give them that's important, but the amount of current.

My best experience has been with the B&W CM1. I heard them with both a Yamaha receiver and a Rotel RB-1070 high current amp. While at high volumes I would have a hard time telling them apart, it's when I listened to them at low to moderate levels that they really differentiated themselves. They were just 2 completely different speakers, which is what got me thinking that there actually might be something there in regards to high current amps. What do you guys think about this? Is there really something there between high current amps and normal (class D?) ones?

 

[no. 5]

This subject has caught my interest as I have been told by some dealers that with certain speakers it's not the amount of wattage you give them that's important, but the amount of current.

I think it's just marketing.

I'm sure some of the knowledgeable will chime in here, but I have to say, every time I hear that statement it never sounds right.

 

[deedubb]

Part of depends on the speakers you are running. If you have 4 ohm speakers that might drop down below that at times, you'll need something with a fair bit of current to run them decently.

 

[PENG]

The fact is, Power (Watts)=Voltage (Volts)XCurrent (Amps)XPower Factor (the Cosine of the phase angle between the Voltage phasor and the Current phasor). Alternatively stated, Power (Watts)=Current (Amps) X Current (Amps) X Resistance (ohms, the resistive component of the impedance). And remember, Current=Voltage/Impedance.

So watts, volts, amps are all inter-related and are inseparable. Not all sales people in an audio store have a good understanding of electrical theory. They could, sometimes unintensionally mislead their customers, even audiophiles.

From the formula you can easily see why high current capability is important for low impedance speakers. That being said, it is virtually not a factor at low listening volume because even a "low current" receiver can suppply the current needed by a 4 ohm or even 2 ohm speaker at low volume. If you experienced the difference you did, I can only suggest you do it again in DBT.

Let's see an example, to produce 85 dB at 4 meters in a 15X20 room (not open field so the inverse square rule does not apply fully), you should not need any more than 10watts. So if the music produces an average of 85 dB and peaks to 97 dB occasionally, it only require 160 watts/channel at the very most. Any Yamaha RX-V2500 or above should be able to produce well in excess of 200 watts into 4 ohms in 1 or 2 channel mode, and much higher if it is for a short duration and/or 1 channel only.

Again, whether it is a Rotel amp or a Yamaha receiver, if it delivers the same watts into a 4 ohm load, the currents will be exactly the same.

 

[PENG]

I think it's just marketing.

I'm sure some of the knowledgeable will chime in here, but I have to say, every time I hear that statement it never sounds right.

You are absolutely right. It would make sense if it was the other way round. That is, at high volume, amps with high current capability would win in general but not at low volume. Who would otherwise buy a $4,000 25W Lexicon integrated amp?

 

[B3Nut]

A high-current amplifier is simply one in which the power supply and output stage can pass enough current to drive low-impedance loads. The "high-current" label is pretty much a marketing label, it gets tacked onto even entry-level receivers with slightly beefier power supplies than one usually sees at that level, but not much else (there is still only one pair of output transistors per channel, driving low impedance loads with a single pair can push the transistors out of their safe operating area - this is why many receivers have a "4-ohm" mode, it limits the supply rails and engages current limiting in the output stage to keep from blowing outputs.) When an amplifier attempts to swing a given voltage into the load, the load will attempt to draw a certain amount of current, and if the power supply cannot provide it or the output transistors cannot safely pass that current, output either sags (and the amplifier clips, and you can toast your tweeters if the distortion harmonics are strong enough) or you toast outputs (and if the protection circuit doesn't trip, possibly your woofers as well since up to 30-50VDC could pass through whichever transistor shorted to the speakers.)

To deliver large amounts of current, you need a beefy power supply, multiple output device pairs, and GOOD heat sinking. The amplifier can be class A, AB, B, D, whatever...the physics still apply. Also, you really can't divorce power in watts from current in amperes, since P = IE (P is power in watts, I is current in amperes, E is potential in volts.) If an amplifier cannot deliver the current a speaker asks for, the power in watts dissipated into the load will not be there either.

 

[gus6464]

Ahh ok I see. So what is the difference between Class A, B, D, AB, T and whatnot? Also when it comes to headroom in driving speakers, is there an audible difference?

 

[furrycute]

Wow, a lot of people here really seem to know their physics. I am impressed.

Speaking from subjective personal experience only (I am kind of rusty on that physics stuff). I have owned a pair of low impedance low sensitivity floorstanders, a pair of Infinity Kappa 600. I drove them with both a Yamaha integrated and a Yamaha pre/amp combo (CX-1/MX-1). And there was an audible difference between the two power sources.

When the speakers were driven by the integrated, the sound just seemed thin, like the bass wasn't there. When the speakers were driven by the pre/amp combo, the music just seemed fuller, more full bodied. And this difference was present in almost all listening volumes. And oh, that Yamaha MX-1 amp weighs over 50lbs.

Anyways, that's just my experience.


And as to different classes of amplification. Only the really high end amplifiers will work in class A mode. Class A mode gets really hot. Most external amplifiers will work in class AB mode. Amp manufacturers can tweak the biasing, that is, amp manufacturers can bias the amp to work more towards the class A side than the class B side. Class H is how digital amps work. Some sort of switching mode, which I am not too familiar with.

People say class A sound the best. Class AB with high class A bias sound the second best (biased more towards the class A end). Digital amps, well, you need to find the good ones, the early ones didn't sound too good, hence they are mostly delegated to driving subs and not your mains.

Yamaha makes a digital amp. People who have heard it say it's quite good. It's about the only external amp Yamaha makes nowadays. It's quite a shame. Because the MX-1 is a fantastic amp.

 

[PENG]

Wow, a lot of people here really seem to know their physics. I am impressed.

Speaking from subjective personal experience only (I am kind of rusty on that physics stuff).

I have had similar subjective experience a few weeks ago when I had a RX-V659 for about a week. It first sounded a little bright and thin until I went back and forth between it and the Bryston a few times using the same CD and speakers. I still thought the Bryston sounded fuller, it has to be, the RX-V weighs only 27.1 lbs. The fact is, I would have hard time telling the difference in a blind test. The physics/engineering principles presented so far to show the inseparable relationship between watts, voltage, current and impedance are pretty basic, proven, and can be found in any college/university text books or wikipedia.
http://en.wikipedia.org/wiki/AC_power#Basic_calculations_using_real_numbers)
As such, it is more reliable than subjective experience such as the one I had recently.

 

[furrycute]

The power supply in an external amp has to be bigger than that power supply in a mid range receiver. I guess that's where the difference lies.

 

[PENG]

The power supply in an external amp has to be bigger than that power supply in a mid range receiver. I guess that's where the difference lies.

Agree, that's why I thought at higher volume the difference can be explained but not at the low volume Gus was talking about. Whether the power supply is rated for 100W or 500W, if the demand is only 10 watts, even the 100W unit would have enough headroom 99% of the time.

Of course the perceived difference could be real but it would have to be for other reasons, not due only to power/current capability, assuming Gus was talking about mid range receivers such as a RX-V659 or better. By the way, I have seen several popular speaker's impedance vs frequency curve and noticed that the low impedance dips typically lie outside the 30 to 70 hz range.

 

[furrycute]

Another thing to think about are the bass peaks.

Have you ever seen those old styled power amps with analog meters? I had one, and it was quite educational to observe how much power was being used during music playback.

During most playback, the meter would just hover around 1 or 2 watts. But when there are bass heavy passages, the meter would swing all the way to the other end. I guess the larger power supply in an external amp deals better with these bass peaks than the smaller power supply in a mid range receiver.

 

[dilznoofus]

As a long-time Magneplanar owner, I can assure you that for some of us it is ALL about the current.

Not all speakers have the same requirements, but some speakers crave current and will not "light up" until they get it.

 

[mtrycrafts]

... Who would otherwise buy a $4,000 25W Lexicon integrated amp?

Is this correctly printed? 25 watts and $4k?

 

[mtrycrafts]

People say class A sound the best. .

Yep, people say a lot of things and most are incorrect audio is no exception when people speak

 

[mtrycrafts]

Also when it comes to headroom in driving speakers, is there an audible difference?

We did the amp classes just the other day and a wiki was posted that explained it well, how they operated.

Headroom is nice at the peaks in a dynamic music at the limits. Otherwise, headroom doesn't come into play until you demand greater power that the designed continuous amount

 

[mtrycrafts]

Part of depends on the speakers you are running. If you have 4 ohm speakers that might drop down below that at times, you'll need something with a fair bit of current to run them decently.

That is kind of relative to your volume demand, listening distances and speaker sensitivity.

 

[PENG]

Is this correctly printed? 25 watts and $4k?

I was indeed my mistake. This is at least the second time tonight I have to thank people to point out my mistake. Thanks!

It isn't a Lexicon, it is a Luxman L550A MKII Class A integrated amp rated for 20 WPC into 8 ohms (40 WPC into 4 ohms), not 25. List price $4,500, not $4,000.

http://www.avguide.com/news/2007/09/20/luxman-returns-to-us-market-after-25-years/

http://www.onahighernote.com/luxman/?c=8&id=31

I read the review recently in Chapter/Indigo but it is not available on line. The reviewer was impressed and did think more power was needed unless the room was bigger (something like that).

Anyway, I was just trying to make the point that a 20 WPC X2 Class A solid state amp that weighs 48 lbs for sure has high current capability but it is not going to be higher than that of any mid range 120WX7 (e.g. RX-V2600, Denon AVR3805) receiver that have been tested to do well over 200 W into 4 ohms with 2 channels driven simultaneously. The high current thing is real, but it is all relative..

 

[TLS Guy]

We have had this before. The salesman is wrong. The current the amp provides is determined by the load resistance. As the load resistance drops, the source (amp) has to provide more current

I have responded to this before. Here is the complete math, from a previous posting of mine to a similar question.

Let us consider the possibilities for an amp that can provide 200 watts into an 8 ohm load.

Now the power delivered is the square of the current X the resistance.

So if we have the 8 ohm load the square of the current is 200 watts/8 ohms which is 25. The square root of 25 is 5 amps. So the amp has to deliver 5 amps to produce 200 watts into 8 ohms. Now the voltage from ohms law is the current X the resistance which is 5 amps X 8 ohms which is 40 volts

Now lets take the four ohm load. The square of the current is 200 watts/4 ohms. Which is 50, the square root gives a current of 7.071067 amps.
Now the voltage from ohms law is 7.071067 amps X 4 ohms which is 28.284 volts.

Now if the amp did not drop its voltage when driving the 4 ohm load then the power delivered is the square of the voltage divided by the resistance. So we would have 40 volts squared/4 ohms this is 1600/4 which is 400 watts. Now you might say what is the issue. Well if the amp drops voltage supplying the four ohm load it is current limited. The transistors will be stressed and the amp will likely hit a hard clip and in the worst case scenario will clip at five amps, if that is its maximum current output. If you do the above calculations that would be 100 watts into 4 ohms.

The point is that amplifiers that have to deliver more current require larger power supplies and bigger output devices. It is the current output of the amp that has a bearing on cost. The maximum current output determines how much power an amp can deliver to a given load. The voltage an amp can deliver is determined by the operating voltage of the power transistors. Obviously the output voltage can not exceed that.

Now to class.

Class A. The bias is set so that the output devices operate only on the linear part of the curve. The quiescent current is high and the efficiency relatively low. The output devices are subjected to more heat and therefore shorter life is a possibility. Distortion products are even harmonic which are considered less objectionable. Large amps with big power supplies and multiple output devices are required for high output. They are very heavy and expensive unless power output is low.

Class B. Low quiescent current, with high efficiency and large amounts of crossover distortion. Large quantities of odd harmonic distortion. Not suitable for critical listening.

Class A/B. At low power bias is class A. As power increases bias moves to class B. The efficiency is typically 60% or a bit better. Problems of odd harmonic distortion and crossover distortion reduced to very low levels on the best designs. This is by far the commonest arrangement for high fidelity amps.

Class C. Suitable as RF amps. Only one successful, high fidelity design by the late Peter Walker OBE of Quad. This is a class C feed forward design. The class C output dumper transistors are rescued by a very good class A amp as a feed forward error correction. Extremely good amps and my preferred topology.

Class D. These are switching amps using N channel switched MOSFETS in the output. Until recently confined to sub woofer use, because of severe problems with matching load impedance. Recent developments, especially PWM of the input have resulted in improvement. Cheap to build. This class will likely take over the receiver market. Expect more problems than we have now with amp speaker interface. Powered speakers with digital crossovers will likely be the best approach. I expect these amps to take over the budget, if not the entire receiver market, almost immediately, to maintain price points.

Class G. These are class A/B amps. The rail voltage has switch points depending on the input signal to increase efficiency. These are the so called "smart Power supply" amps.

Class H. Extension of G, but the H has an infinitely variable rail supply voltage as it is modulated by the input.

Those are the classes in use in currently available hi fi amps.

 

[PENG]

We have had this before. The salesman is wrong. The current the amp provides is determined by the load resistance. As the load resistance drops, the source (amp) has to provide more current

I have responded to this before. Here is the complete math, from a previous posting of mine to a similar question.

Let us consider the possibilities for an amp that can provide 200 watts into an 8 ohm load.

Now the power delivered is the square of the current X the resistance.

So if we have the 8 ohm load the square of the current is 200 watts/8 ohms which is 25. The square root of 25 is 5 amps. So the amp has to deliver 5 amps to produce 200 watts into 8 ohms. Now the voltage from ohms law is the current X the resistance which is 5 amps X 8 ohms which is 40 volts

Now lets take the four ohm load. The square of the current is 200 watts/4 ohms. Which is 50, the square root gives a current of 7.071067 amps.
Now the voltage from ohms law is 7.071067 amps X 4 ohms which is 28.284 volts.

Now if the amp did not drop its voltage when driving the 4 ohm load then the power delivered is the square of the voltage divided by the resistance. So we would have 40 volts squared/4 ohms this is 1600/4 which is 400 watts. Now you might say what is the issue. Well if the amp drops voltage supplying the four ohm load it is current limited. The transistors will be stressed and the amp will likely hit a hard clip and in the worst case scenario will clip at five amps, if that is its maximum current output. If you do the above calculations that would be 100 watts into 4 ohms.

The point is that amplifiers that have to deliver more current require larger power supplies and bigger output devices. It is the current output of the amp that has a bearing on cost. The maximum current output determines how much power an amp can deliver to a given load. The voltage an amp can deliver is determined by the operating voltage of the power transistors. Obviously the output voltage can not exceed that.

Coming from you (an obvious experts in this field) I am a little disappointed with the math presented. It is not that the math itself is wrong, but your example implies the loads (8, 4 ohms) are purely resistive. The fact is, an 8 ohm reactive load (say that happens at 1 kHz) will demand more current than an 8 ohm purely resistive load for the amp to deliver the same wattage. The VA will be the same but the Watts will be different, higher in the case of a reactive load. This is why I prefer the more complete formula that include the power factor.

I know I am splitting hair, your math is correct.