Sorry for the stupid question:( I know watts need amps. Is that what current is about? If so, how can you tell how many amps an amp is pushing? What defines a high current amp?
I posted here for I am thinking about both SS and tubes.
thanks

The term 'watt' is simply a convention adapted to refer to an amount of work being done. The simplest form is to multiply the current through a load (I) times the voltage across a load (E). This indicates the amount of 'work' being done. Power(work)=Voltage (E) x Current (I).

If you have 10 volts being measured across a load (resistor, speaker, whatever), and you know the resistance (actually it is 'impedence', but more on that later), then you can calculate the work being done, and express that work in Watts. Since Ohm's Law says I=E/R, you can substitute E/R for 'I' in the P=E x I formula, and have P=E^2/R. Simple algebra. You can also sub I x R for 'E', and have P=I^2 x R.

For more basics, take a trip to this guys page: http://www.williamson-labs.com/

With speakers as a load, things are not so simple. Speakers do not present a pure resistive load. In AC elecronics, loads are referred to as an 'impedence', which is the polar sum of its inductive reactance, capacitive reactance, and resistance. I'm not going to go too deep here, but what it means in a nutshell is that a complex load (highly reactive) is much more difficult for an amplifier to drive than a simple resistance. It is possible for a load to be so reactive that it requires an output stage to deliver all the available current the supply can muster, while a considerable voltage is still present across the output stage, since in a complex load current and voltage are no longer in phase. Since P=VxI, then you can see that if an amp is asked to deliver 3 amps to a load while there is still 25 volts across the output stage (not so far-fetched), then 75 watts are being dissapated by that output device..power that is not going to the load, but trying hard to destroy your amp. Takes a damn well built amp to handle some exotic loads.

As for your question about 'high current', I'll dance around this one a bit too.:banana:

An audio amplifier is designed to act as a voltage source. In other words, if it was a perfect amplifier, it would deliver whatever current was necessary into whatever load was connected in order to generate the necessary voltage across that load. Read that again...an amp is a voltage source. It's not supposed to care what load is connected, it's just supposed to give us the voltage we want in order to perform work.

In the real world this isn't going to happen. High currents through the transformer generate large voltage losses, and high ripple currents through the power supply capacitors cause the output voltage to sag, resulting in less power.

Since P=E x I, the more current through the internals of the amp, the more power is lost. Power eaten-up inside an amp is that much more power that is unavailable to the load. While the perfect amp could deliver 100 watts to an 8-ohm load, and 200 watts to a 4-ohm load, and 400 watts to a 2-ohm load, this is never the case. The idea behind a 'high-current' amp is to keep the internal loss of power as small as possible, or have so much power in reserve that internal losses are not important. This is, as you might guess, expensive, since the heart of the ability to deliver large currents originates in the most expensive part of the amp...the power supply. Those big transformers and big filter caps are not cheap.

So, now that I've tossed in more background than anyone really wants, I'll give you a simple answer to the 'high-current' label...a high current amp is supposed to be an amp that is capable of driving complex loads (as we touched on a paragraph or two ago), and/or able to generate a fair amount of the same voltage across a low impedence load that it can across a higher impedence load. It's not exactly like there is some set definition of a high-current amp, it's more of a marketing thing than anything else. If you want to know if an amp is actually 'high-current' read the specs or check out a review.

Sorry you asked?;)

Simple answer wanted?

If RMS power ratings are equal, one amp could be said to have a higher current capacity if it has a higher damping factor specification then the other. This rating is a ratio spec of an amps internal source impedence compared to the nominal load impedence (typically rated at 8 ohms). Example if an amp has a 1/10 amp internal impedence then it would have a damping factor of 80 at 8 ohm load. The things that an amp needs to attain a higher damping figure (lower internal impedence) are the things that Echo talked about, larger power transformer, larger main filter caps, as well as output devices that can operate at higher current drains while staying in their safe operating range.

Too bad many manufactures don't publish this specification. Another somewhat indirect was to infer current would be to publish RMS output wattage at various nominal speaker impedences instead of just 8 ohms. Some do, many don't.

As has been stated before the key is to try and know the true power requirement of your speakers and buy an amp that matches or exceeds that requirement. Buying a 'designer' high current 100 WPC amp for say $1000 wouldn't have anything over a $1000 200 WPC amp with 'normal' current capacity. And both may be overkill if your speakers max out at 50WPC before overheating and distoring.

So beware the marketing tactics made to think you need more then you really do. Such tactics are effective because I see how many 4 wheel drive SUVs are on the road that will never see dirt in their lifetime.....

Lefty

OK, I'll bite.

My SX-1280 claims 185 Watts per channel at 8 ohms. Nowhere do I find any information on other ohm possibilities.

The owners manual states,

NOTE:
If you want to use two sets of speaker systems, make sure that the impedance of each system is 8 ohms or more. If the impedance is less than 8 ohms, the protective circuitry will be actuated when the volume is turned up and you will not be able to enjoy proper stereo performance.

As many of you are aware I did indeed have protective circuitry issues when I hooked up 4 Ohm speakers while I did not experience that problem with 8 ohm speakers. Certainly, that was in large part due a DC offset issue but the fact remains protective circuitry did not engage with an 8 ohm load.

Why would Pioneer employ protection with lower ohm ratings? If I understand correctly, resistance is halved resulting in the potential for (but in the real world never actually) twice the power. Is it opening a wider possibility of clipping? (another term I only partially understand) and if so is that what EW stated earlier, "The idea behind a 'high-current' amp is to keep the internal loss of power as small as possible, or have so much power in reserve that internal losses are not important."
Is it possible that the 1280 does lose signifigant internal power and when pushed no longer has reserve to handle it. If so, where does it go?

I'll go eat more fish now. :D

No, not sorry. I think I partially understand, but answer DB's question first cause it's part of my next one.;)

DB;

Well you ask several questions that all probably relate to the same basic subject. What your Pioneer manual is trying to say is that there is indeed a minimum load (speaker) impedence value that will trigger the protection circuitry if egnored.

Running two sets of 8 ohm speakers in parallel creates a 4 ohm load as seen by the receiver. If you hooked two sets of 4 ohms speakers you would end up with 2 ohms of load impedence. A 2 ohm load at some predetermed output voltage (hence power level) is designed to activate the protection circuit in your Pioneer. If one knew they exact value of the protection trigger circuit it should be possible to show mathamaticly that the circuit is limiting at the same POWER level no matter what the speaker load actually is. Make sense? Not 100% sure as the only protection circuits I have studied closely only dealt with detecting a DC voltage on the output to protect the speakers, not any kind of power limiting function. Maybe EC has more experiance with multimode protection circuits.

My Kenwood KR-9600 receiver (and some others) tries to deal with this case in a fundamentally different method. When you select 2 speaker output mode, it hooks the two speaker sets in series instead of parallel. This means that two sets of 8 ohms speakers will be seen as a 16 ohm load by the receiver. This of course halves the current at any given output voltage level. Just a different simpler method to limit power leaving the amp.

It is all best described with math via ohms law as there is no way to get hung up with just a current or voltage or ohms value alone. It's power limits that we are dealing and concerned with here.

I think it's safe to say and that I'm on firm ground to state that any run of the mill amp could be hooked up to a 1 ohm resistor and as long as you have the volume low enough the amp would would be happy. But as soon as you raised the volume up high enough either a protection circuit, fuse or lacking such, some component would fail due to exceeding it's power dissipation specification.

This help? If not it's EC turn :p:

Lefty

THe speakers I got are rated at 6ohms. I would get better performance out the amp if I connected them to the 8 ohm taps, right?
Rather than try to figure out the current, better just to over power? By the way, the speakers are rated at 1000w, but no way I am I getting into Thor land:cool:

Lefty,

thanks for the reply. I don't believe I'm saying this, but after rereading your post a couple times, I understand it! :banana:

I also started to read the info on the link EW left, that is good stuff. It all brings up more questions but....

ok, MD, you're up! Hope you didn't mind my :butt1:ing in.

Ask away DB!
That's the only way we'll ever learn anything:D
This is just starting to get interesting......

Originally posted by millerdog
THe speakers I got are rated at 6ohms. I would get better performance out the amp if I connected them to the 8 ohm taps, right?
Rather than try to figure out the current, better just to over power? By the way, the speakers are rated at 1000w, but no way I am I getting into Thor land:cool:


Hummm taps hummm, those are the thingees on the back of tube amps right? Tubes are not my thing so others could probably best answer your question. However if your speakers are indeed rated at 1,000 watts (is that possible? real RMS watts?) I think you would want the tap that will supply the most power it can, whatever one that is :p:

Tubes, geeze ;)

Lefty

Originally posted by DingusBoy
Lefty,

thanks for the reply. I don't believe I'm saying this, but after rereading your post a couple times, I understand it! :banana:

I also started to read the info on the link EW left, that is good stuff. It all brings up more questions but....

ok, MD, you're up! Hope you didn't mind my :butt1:ing in.


That's funny, 'cause I reread my post and now I'm confused :p:

Current, voltage, ohms, watts, how come it has to be so complicated? There is just one volume control knob right? How come it just can't be --not enough, just right, too much--? No it has to have math and stuff.

EC writes a book about it and still I don't see why we can't just wire the speakers to the wall power outlet and save a whole lot of decisions and choices and money :confused:

Lefty

I dont think there is any need to worry about connecting a receiver such a Pioneer SX-1280 to 4 ohm speakers, or even a 45W SX-780 for that matter- or just about any receiver that has two pairs of speaker terminals in the back- if the receiver were not capable of handling 4 ohm loads, the manufacturer wouldnt've put the extra set of spkr terminals in the back of the unit to begin with. The instructions either in the manual or printed on the unit itself next to the terminals will usually state the requirement that if both sets of speakers are to be used (A+ B) then both speaker pairs should each be 8 ohms so that the parallel combination will be the equivalent of 4 ohms. (Exceptions being Lefty's Kenwood which utilizes a series configuration).

I also think that even if you did connect two pairs of 4 ohm speakers to the receiver, it would still play ok, though an eqivalent 2 ohm load will certainly draw more current, in which case, when I say it would still be ok, it would be for doing so at low volume levels, not THOR type listening levels! :) And probably an SX-1280 would allow for more leeway when used in this more demanding manner than a smaller SX-780, for example, would.

You all are making the mistake of assuming that your speakers are a resistive load. Re-read the bit about reactive loads in my first post. Speakers are NOT resistors, and their impedence varies according to the frequency of the signal applied to them. A 'nominal' 4-ohm speaker may very well dip down to 2-ohms at some frequencies.

About all of our amps also have a circuit at the output commonly called a 'EI' (voltage-current) limiter. This circuit is designed to reduce the drive current to the output in case of a unacceptable low impedence load, or if the voltage across the load begins to approach the supply rails. The point at which this circuit triggers is totally up the the designer (a lot of amp tweakers disable this circuit...personally I prefer it left alone).

A Pioneer 1280 should have no problem driving any 4-ohm nominal impedence speaker. Two sets of 4-ohm speaks at the same time may be a quite different case, depending on how they are wired when both 'A' and 'B' are selected.

At this point I'm not sure what ya'all are asking...so ask again. Try not to make it more complicated than it really is...the basics are simple.

This is my view of what limits the current in the amp. This is off the top of my head without pulling out my slide rule or is it the slip stick to check the figures. :)

I = E / R. The E or supply voltage is a fixed value in any amp. The amps we run across the E is probably somewhere between 40 to 100 volts. Pick a mid power system with E = 60 volts. A designer has to choose output component to work reliably. Lets say he designs for an R = 5 to 9 ohms speaker system, then

Power = E * E / R.
I = 60/5 = 12 amps.

For SS amplifier, he would pick transistors to handle at least 12 amps in all conditions. A single power transistor can handle about 10 amps max It turns out with power dissipation and current consideration, this turns out to be 2 transistors in parallel. You will find most mid power (80 or so watts) amps has 2 transistors per leg or 4 trs per channel.

Now if I come along and use speaks R = 2 to 5 ohms,
I = 60/2 = 30 amps. This will be ouside the design. Because in the best case, 2 transistors can handle only 20 amps. So to handle lower impedance speaks will need more trans and $$$. Note if R = 0, I = infinity!!! Designer can find no trs to work here even if you have infinite $$$$ to spend. :(

A good designer would protect the amp by putting in the prtection relay circuit to sense when the over current condition is present and click the relay off. In better designs there is a circuit to self limit the max current even if the output is shorted to prtect the amp and speaks.

By the way, EW's explanation was good and since I majored in EE, the stuff about the 'polar sum' isnt difficult to understand, though for others not familiar with the term, 'polar sum' will only draw a blank expression on the readers face.

IMO, Electronic theory would be alot easier to understand if analogies were used more often to explain things. Formulas and fancy terminolgy are easy to forget but understanding the basic concepts will usually always be retained in memory.

Granted, EW doesnt have the time to write a whole simplified essay on the subject (and neither do I- there are other books for that and Online resources) but just to give a short demonstration on using analogies- instead of saying polar sum (which invloves a branch of mathematics using polar and rectangular coordinates and vector quantities), it can be said that with components such as resistors, whether it be in an ac or DC circuit- if you add two of them together that are connected in series, if Resistor 1 (R1) = 5 ohms and resistor 2 (R2) has a value of 3 ohms, then the total resistance is simply R1 + R2 = 8 ohms. HOWEVER, if you now keep R1 but replace R2 with a different component, such as an inductor with a resistance value of 3 ohms, the total resistance (or ac resistance which is called impedance) is NOT simply 5 + 3 = 8 ohms anymore. An inductor behaves differently than a resistor in that an inductor has the property of opposing changes in current and the simplified result of this charteristic is that you CANNOT directly add the 3 ohms of the inductor to the 5 ohms of the resistor. The answer is actually 5 + j3 - which is in the polar (actually rectangular form that EW mentioned).

The basic concept is that you cannot always add two different characteristics as you think that you simply can such as 2 + 2 = 4. The resistor and inductor combine together in a way akin to how two forces might kick a soccer ball in two different directions. 10 + 10 doesnt always = 20 ---> if one person pushes a heavy refrigerator with a force of 10 lbs in one direction (say east) and another person pushes with 10 lbs of force in the exact opposite direction, West, then its obvious that fridge aint going anywhere- and so electrical circuits with various different components also have properties such that you simply cant add two different things together.

I dont know how well I expressed my thoughts into words for the reader (and being its very late now, I cant spend forever trying to come up with the perfect explanation) but I did want to point out my feelings that I think using anaolgies to explain technical jargon makes thinhs alot easier to learn & understand- I wish my EE teachers did more of that instead of just throwing a whole bunch of math equations on the board.

EW- I am aware about speakers not being a purely resistive load. However, unless the speakers happen to be a pair of Infinity Kappa 9's or huge RSIB's for example (which have reputations for being nasty loads), I dont think there's much to worry about in connecting any typical vintage receiver to most 4 ohm speakers. It might be more difficult to judge the true performance now because that vintage receiver isnt a new or young unit anymore- its possible the performance might be somewhat less due to the 25 year old transistors and caps inside which date back to the Carter Administration (Peanuts anyone ? :)- and more stress being put on them in doing a 4 ohm or 2 ohm speaker load test that might make it more sensitive to going into protection mode.

Regards, B/F.

Okay, how's this: My speakers are rated at 500w at 6 ohms, so if I want to get more current to my speakers it's better to hook them up the the eight ohm taps? That way I am under loading the amp right?
I know what EW was saying about speakers. I have read about how some don't measure correct load across the board. My assumption about these speakers is that since the recommendation that I use I high current amp is that they do NOT measure that well across the freqs.
Oh, Lefty, I just found out that there are different variations of my speakers, thus the difference in power.

A good designer would protect the amp by putting in the prtection relay circuit to sense when the over current condition is present and click the relay off. In better designs there is a circuit to self limit the max current even if the output is shorted to prtect the amp and speaks.Most all relay circuits protect from turn-on/off thumps and DC at the output and not much more (Pioneer big-dawgs are a different story, their protection is a little more complicated). The limiter circuitry I spoke of in the post before yours is supposed to protect the output transistors in case of over current or short circuit (pretty simple stuff), but for a dead short I wouldn't trust these circuits as far as I could throw my KR-9600 (i.e., not too damn far).

Your example is valid to a point, but does assume that the supply does not sag under load, and all of the unregulated supplies do. It also assumes that the output is able to swing rail-to-rail, which it can't since voltage is lost across the output devices. (as an aside, the output devices are not usually the limiting factor in a properly designed amp.)

As an example, my little Kenwood KA-5500 has a +/- 45V power supply at idle. I don't recall how much the voltage sags under load, but we can calculate it to a ballpark figure.

If the supply did not sag at all, and under full load there was no voltage lost through the output transistors, then the output could swing the full +/- 45V. A full +/- 45V swing calculates to a RMS voltage of 31.82 V (RMS=PP voltage/(2*sqrt2)). With an 8-ohm load, this would translate to 126 or so watts. But the amp is rated at 55 watts....what gives?

To help solve the mystery, I have measured the actual power output at clipping at 68 watts. So lets work backwards and see what voltage that represents. I had a dummy load on the amp at the time...a pure 8-ohm resistive load. If we know power, and impedence (in this case, a true resistance), we can calculate either voltage or current with ease.

E=sqrt(PxR)
=sqrt(68 x 8)
=23.32VRMS

OK, we know volts RMS. What does it say about our supply voltage?

Volts peak= volts RMS x sqrt(2)
=23.32 x 1.414
=32.98V peak

Now, since I didn't actually measure the voltage at the supply under load, we'll make an assumption of about 2.5 volts lost through each output device (talking solid state here boys). So our +/- 45V supplies at idle dropped to about +/- 35V under a fairly easy 8-ohm purely resistive non-reactive load. We done lost 20V (10 volts from each rail)!!!!! Complicate matters by increasing the load (lowering the impedence) and adding reactive components to it, and you will see that what I describe here is a best-case scenerio. And in actuallity, this amps power supply performance is considered excellent. Gimme a cheap POS amp and I'll give you something to giggle at. :D

If you are interested in why the voltage drops so much (and a 20V loss isn't so bad), we have to delve into transformer regulation and supply filtering..a whole 'nother subject.

Another quick example...my KR-9600 has +/-67V supplies at idle. Yet it is rated at 160 watts at 8-ohms.

V= sqrt(P x R)
=sqrt(160 x 8)
= 35.77VRMS = 50.6V peak

So here with this amp we have lost (theoritically) almost 35 volts from the +/- supplies!! Actually it is not that bad, as I know the 9600 can output a lot closer to 200W..it's simply rated for distortion figures at 160W 8-ohms.

I used to have a little wheel chart for voltage, current, power, and resistance. I could not find it to save my life, so I whipped this up for ya'all. If you need it resized, let me know...it might be too big.

Edit: No argument here Beatlefred. Any of our amps should be able to handle about any 4-ohm speaker on the market. And old transistors don't bother me too much, but ya'all know how I feel about 25-year old electrolytics.

PS. I didn't know you were an EE. Kept it awful quiet...

That is an excelent chart. I shall use it in my physics exam next week.

Good discussion.

But what I think you may really want in this discussion is a graph of your speakers impedance curve. As mentioned, the speakers may be rated at 8 or 4 ohms...but dip into the low 3's...or 2's.


I also don't think a "rating" of (let's say) 500watts is all that important...'least not to me. This same speaker that has a 500 max wattage rating may also have a sensitivity rating in the high 90's or over 100db. I'm not saying ignore it...but it really don't tell you much about how the driver will perform in a speaker system...or how well it will perform with any given wattage amp. What it does tell you is that this thing can take some abuse (;) ). Many pro drivers are rated just like this...with a high sens and high max continuous power rating. For some reason...I just trust the rating on pro drivers more so than I do normal "audiophile" speakers. If I see a pro driver rated at 600 W continuous program power capacity and an audiophile speaker rated at 600W max power (no Homers)...I'm gonna wonder a bit about the audiophile speakers.





Originally posted by millerdog
Sorry for the stupid question:( I know watts need amps. Is that what current is about? If so, how can you tell how many amps an amp is pushing? What defines a high current amp?
I posted here for I am thinking about both SS and tubes.
thanks

But what I still am not sure about...millerdog, what's your concern here? Is there something your not sure about in your system? or is this just a general question?


take care>>>>>

MD- let me give you my erroneous info here on your speaker tap question.

You want to use the four ohm tap for your six ohm speakers.

An output transformer is used to match the high impedance out of the tube to the low impedance presented by your speaker load. The speaker load is reflected back to the tube. The taps on the OPT change the winding ratios, therefore changing the reflected load.

Since a tube is a high volt/low current source, you want the higher load reflected back to the tube output.

Of course, you really want an optimum transfer. Mismatches cause reduced power out and higher distortion than optimum would, but too much current would saturate the transformer leading to worse problems.

If this is wrong, I'm sure one of the tubies will set me straight:)

Oh how I love these discussions .... more please :)

>If you are interested in why the voltage drops so much (and a 20V loss isn't so bad), we have to delve into ......

I would be interesting to measure the supply voltage here. I suspect the droop is not very much... at the most a few volts. Any more the design is pretty crappy.

My thought is a volt or so is lost on the emitter resistor, a couple volt is power supply droop, and the rest ( majority) is set by the driver circuit ( base drive) of the output trans.

>Since a tube is a high volt/low current source, you want the higher load reflected back to the tube output.

Also you want the proper load reflected back to the tube so it will be happy with it. Wrong load ( one that increase plate current) will shorten tube life. However, tube may put out more light in a dark room which is a good thing ??? :D

Geeze we were having such a nice educational discussion and then the tubies dropped in...:p:

I think I'll go research how to make fire with sticks and rocks and things.....

:D

Lefty

C'Mon Lefty, get out of the dark ages!

Everyone knows that tubes produce the best fires, with a great mid-range heat that you may enjoy all night.

Your stick and rocks produce a harsh, uncomfortable fire that leads to a nervous reaction, reducing pleasure and making it unbearable for more than a short time.

Get with the program, move up to the 20th century.

Yeah, yeah, I'm still behind...:p:

Originally posted by SolderIron
I would be interesting to measure the supply voltage here. I suspect the droop is not very much... at the most a few volts. Any more the design is pretty crappy.

My thought is a volt or so is lost on the emitter resistor, a couple volt is power supply droop, and the rest ( majority) is set by the driver circuit ( base drive) of the output trans. The droop is substiantial in the capacitors.

You want an actual measurement? Alrighty then...I opened up old Kenny just for you... ;)

Supply voltage at idle: 44.4V
Power at clipping: 69 watts (23.5VRMS or 33.23V peak)
Supply voltage at clipping: +/- 38.3V

Admittedly not as bad as I thought..a loss of about 6V (12 total). I did not measure CE or emitter resistor voltage (call me lazy). Reason being that..the clipping is obviously due to the EI limiter kicking in, and not a voltage rail limitation.

Still, don't miss the point. Voltages sag. The larger amps sag more than most because of the amount of current being supplied, so they start with a higher voltage to accout for the expected sag.

So let's see if I'm still only two steps behind:

since Vrms = SQRT(p*r) (p = Wrms; r = ohms)

if the RMS is 185: sqrt(185*8) = 38.47 Vrms

And Peak Vrms is:

=Vrms*SQRT(2)
=54.41V peak
=370 Watts Peak (54.41*54.41)/8

So I can now proudly sell my 185 Watt receiver on ebay claiming it is 370 Watts
p and become an instant cheeseball. :D

Now for another ignorant question: How do you determine "Supply voltage at idle?"

>a loss of about 6V

Thanks EchoWars. It is more than I thought. Never measured one before by myself.

6/44 is about 14%, I guess that is in line with 10% resistors and 20% caps.


:)

SolderIron...I hadn't either...odd eh? Just one of those basic things you take for granted. This old Kenwood actually has a bit more filtering than most for it's size...20,000µf for a 55WWPC amp isn't too bad.

Now for another ignorant question: How do you determine "Supply voltage at idle?"Look at the schematic, or open it up and whip out the voltmeter. No way to actually calculate it simply knowing the output power, although you can make a guess.

Now for another ignorant question: How do you determine "Supply voltage at idle?"

Best place I think is right across one of the big main filter caps. For math purposes you only need to deal with one of the power caps, either plus or minus, as the opposite one should be a mirror of the other. Keep in mind it's just not the DC RMS voltage that one might want to monitor but also the AC ripple as it will increace as more energy is removed from the capacitor then can be replenished from the 120HZ pulsating DC rectifier.

A scope is probably the best instrument to observe power supply droop Vs clipping level as the DC rail will start to show a higher AC ripple content as max rated power is approched. The DC rectifiers can only replenish the power to the caps as the rectificer voltage raises higher then the exisiting voltage charge in the cap, and it can only do this at a 120hz peaks. A standard full wave rectifier is a very non-linear device and Ohms law alone does not define well all the power limits in this kind of system...

Bottom line: Nice big heavy transformers, big large value filter capacitors and rugged rectifer diodes ( reletive to the wattage class) are a first priority in a 'current friendly' amp. Other stuff counts too but you can't get more out then you can put in...

Lefty

GF,
When I researched the Polks I came across people mentioning "high current" amps. While I have run across the phrase before, I just never bothered to think about it until now.
Us high efficiency speaker guys don't have to worry about such things;)