[nnck]

As the title says - I'm looking for info about how much current both of these amplifiers put out. Cant seem to find the info online. Does anyone have the inside scoop? Dan, can you answer this one for us?

[nnck]

Not sure if this question has been seen or not. But still interested in an answer if possible.

Or is this information just not known? Thanks

[Æ]

Not sure if this question has been seen or not. But still interested in an answer if possible.

Or is this information just not known? Thanks

You should be able to locate the power output of the amps in watts. Then it is quite easy to use Ohms law to determine the current and/or voltage. The amount of current will depend on the load. Typically a solid state amp will deliver twice as much current into a 4 ohm load as an 8 ohm load. If you don't know Ohms law, I can do the math for you or provide you with a link.

[nnck]

You should be able to locate the power output of the amps in watts. Then it is quite easy to use Ohms law to determine the current and/or voltage. The amount of current will depend on the load.

How about an example? Just using a nominal 8ohm load. The KWA100SE is rated at 100 watts and the KWA150SE is rated at 150 watts, correct?  So what would be the current produced?

Current = Square Root of (Watts/Impedance)

So the KWA100SE current = 3.5 amps.
KWA150SE current = 4.3 amps.

But, still a little confused by what you say - the wattage ratings of 100 and 150 for these 2 amps are generally not what they are outputing at any given time. It's typically a lot lower than this I think. Depends on how loudly it's being played. Are the 100 and 150 watt values the maximum outputs? So usually the current is much less?

[dminches]

I would suggest you contact Dan directly.

[nnck]

Also, what you are saying doesnt fit in general with other information I am getting elsewhere, so there must still be some additional confusion. Let me explain - I am looking at speakers which the manufacturer suggests work best with amps which can deliver high current (i.e., a 20 watt amp with high current will outperform a 200 watt amp with low current.) If what you are saying about the method of calculating current is true (using the fixed load and wattage of 2 amps), there is no way that a lower wattage amp could produce more current than a higher wattage amp.

What am i missing? What does "a 20 watt amp with high current will outperform a 200 watt amp with low current" mean?

[dminches]

What amps do the speaker manufacturer recommend?  I would start there and also ask Dan if he thinks his amps are a good match for those speakers.

[nnck]

What amps do the speaker manufacturer recommend?  I would start there and also ask Dan if he thinks his amps are a good match for those speakers.

dminches-
I already know that the speakers I have in mind should be compatible with the Modwright amps. They are often paired as such at audio shows, etc. I'm just trying to gather some info independent of the amp and speaker manufacturer recommendations. And trying to learn something in the process. So it's mostly out of curiosity that I ask.

I realize that current is not a spec that is often given for amplifiers. But I'm more curious than ever about what a statement like "a 20 watt amp with high current will outperform a 200 watt amp with low current" means? According to AE and Ohms law, there is no way that a lower wattage amp could produce more current than a higher wattage amp. There is a good chance I am missing something. If someone could help me understand this I'd appreciate it.

[Æ]

dminches-
I already know that the speakers I have in mind should be compatible with the Modwright amps. They are often paired as such at audio shows, etc. I'm just trying to gather some info independent of the amp and speaker manufacturer recommendations. And trying to learn something in the process. So it's mostly out of curiosity that I ask.

I realize that current is not a spec that is often given for amplifiers. But I'm more curious than ever about what a statement like "a 20 watt amp with high current will outperform a 200 watt amp with low current" means? According to AE and Ohms law, there is no way that a lower wattage amp could produce more current than a higher wattage amp. There is a good chance I am missing something. If someone could help me understand this I'd appreciate it.

Current only "flows" when there is a voltage to push it. And then the amount of current flowing is dependent on the impedance of the loudspeaker. Typically a 4 ohm loudspeaker will draw twice as much current as an 8 ohm loudspeaker for the same given amplifier voltage.
Some amps are capable of even more current into lower impedances. Whereas some equipment such as receivers cannot give up more current into 4 ohms and are only safely rated for 8 ohms.

What you are probably thinking of, is "dynamic headroom." The ability to give a brief blast of extra power when called for. An amplifier with a very heavy duty power supply. A power supply having power (current) in reserve, the ability to exceed the specs. A large transformer and plenty of filter capacitors.

[kc8apf]

Current only "flows" when there is a voltage to push it.

Actually, it's the other way around.  Voltage exists when a current flows through a resistance.  Many power supplies and amplifiers are voltage regulators which adjust the output current such that the voltage between the output terminals equals a set voltage. 

And then the amount of current flowing is dependent on the impedance of the loudspeaker. Typically a 4 ohm loudspeaker will draw twice as much current as an 8 ohm loudspeaker for the same given amplifier voltage.
Some amps are capable of even more current into lower impedances. Whereas some equipment such as receivers cannot give up more current into 4 ohms and are only safely rated for 8 ohms.

Yes, the impedance of the loudspeaker comes into play in terms of average current, but where high current matters is when the desired output voltage is changed.  In a square wave, the edge transition is extremely sharp and is basically an infinite slope.  If you fed that to an amplifier, the amplifier will take some time to completely make the change and thus the output will be a square wave with a rounded leading edge.  How fast the transition occurs is called the slew rate of the amplifier which is expressed as a rate of voltage over time. 

Slew rate is governed by many factors, one of which is how much current the amp can provide to the loudspeaker.  Remember that a voltage regulated amp works by controlling the output current such that the voltage across the loudspeaker meets a set point.  When the set point is changed, the amplifier will quickly open up the output current to try to match the new set point.  The drivers and crossover components within the loudspeaker provide inductance and capacitance that will need to charge sufficiently before the output voltage will change.  If the amplifier merely adjusted the output current such that it would hit the voltage set point assuming an optimal loudspeaker impedance, this charging would cause a lag in the output and the voltage rise or slew rate would be fairly slow.  Instead, it opens up the current well beyond what is necessary to maintain the set point which allows the driver and crossover components to charge very quickly.  Once the set point is reached (or preferable before to avoid overshoot), the current is backed off such that the voltage across the loudspeaker once again matches the set point.  An amplifier that can source a lot of current during that surge means that it can maintain a very fast slew rate even with a large amount of inductance and capacitance within the loudspeaker.  This mainly shows up in highly dynamic sources where the input signal has large, sudden swings.  A high-current amp will be able to change the output quickly enough to accurately reproduce the large swings in the input while low current amps will potentially cut off some of the sharpness of the signal.

What you are probably thinking of, is "dynamic headroom." The ability to give a brief blast of extra power when called for. An amplifier with a very heavy duty power supply. A power supply having power (current) in reserve, the ability to exceed the specs. A large transformer and plenty of filter capacitors.

If you have an amplifier that can put out 150 watts into 8 ohms, you know that it can deliver 4.3 amps of current.  If you moved up to a 250 watt amplifier, it can do 5.6 amps.  You'll likely won't need 250 watts of constant power at normal listening volumes, but when a large, sudden bass note happens, the ability to deliver extra current will make the leading edge of that note much more sharp and accurate.

[Æ]

Actually, it's the other way around.  Voltage exists when a current flows through a resistance.  Many power supplies and amplifiers are voltage regulators which adjust the output current such that the voltage between the output terminals equals a set voltage.

So, you never have to switch your amplifier on? It just starts pushing current magically all by itself? Are we talking electron flow or conventional current flow?

Not the way I learned it. When I insert my meter test leads into a live wall outlet, I can measure voltage with no problems. Now on the other hand if I were to switch the meter to current, it would fry immediately, at a minimum it would certainly blow the fuse.

Regardless of which one comes first, you cannot have current without voltage. Resistance cannot be negative only positive. But you can have voltage without current, batteries and capacitors are two examples that I am familiar with.
I can turn on my amplifier and measure voltage at the output with no load connected. No load no current. OK maybe some leakage current. . . . And I hope we aren't including super conductors into the equation, I like my environment to be at a nice comfortable temperature.

Anyway if you have some links to updated information, please do.

[Phil A]

I believe there is a bit of info here: http://www.audiocircle.com/index.php?topic=64421.90;imode

[Phil A]

I can tell you that the KWA150SE driving my Thiel 3.7s runs tons cooler than my old 14BSST (which was at least 900W into 4 ohms) and drives the load much better (the 3.7s although rated at 4 ohms run between 2-3 ohms on a large part of the audio band).  The 14BSST could get so hot to the touch on one side at times it could burn your hand.  The KWA150SE (which I have since mid-June) has never gotten more than moderately warm to the touch.

[nnck]

Before this thread gets too beyond my capabilities, I'd like to take the technical level down a notch.

It seems we've already established with ohms law that a 100 watt amplifier into an 8ohm load delivers 3.5 amps of current (4.3 amps for a 150 watt amp). And it sounds like what you folks are saying is that good amplifiers are also probably capable of storing excess power (in capacitors?) and delivering extra current when needed.

So does anyone know how much current amplifiers like the KWA100SE and KWA150SE are capable of putting out when needed? Even a rough idea? Can they put out 25 amps of current? 30 amps? more? If you dont know about these amplifiers specifically, how about for amplifiers of a similar design?

[kc8apf]

So, you never have to switch your amplifier on? It just starts pushing current magically all by itself? Are we talking electron flow or conventional current flow?

Of course you have to turn your amplifier on.  But all the power switch does is complete the circuit on the primary windings of the transformer.  You get zero current flow when there isn't a complete circuit.

Not the way I learned it. When I insert my meter test leads into a live wall outlet, I can measure voltage with no problems. Now on the other hand if I were to switch the meter to current, it would fry immediately, at a minimum it would certainly blow the fuse.

Voltmeters work by connecting a meter and a high-valued resistor (generally 1Megaohm or greater) in series with the test leads.  That completes a circuit such that current flows through the meter's coil and across the resistor.  The resistor's value is known so the amount of current flowing through the meter's coil is proportional to the voltage being maintained by the power supply.  For a wall outlet, the power company acts as a giant power supply that adjusts the current flow from their generators such that the voltage across the grid is ~120VAC.

Ammeters, on the other hand, complete the circuit by putting just the meter's coil in series with the test leads.  The current flowing through the load also flows through the meter's coil and causes it to behave as a small motor.  A spring forces the needle against the motion of the motor such that the needle to move proportionally to the current.   If you connect an ammeter across a live wall outlet, you are effectively putting a small piece of wire in the socket and causing a short circuit.

Regardless of which one comes first, you cannot have current without voltage.

True.  Even a trace on a PCB has an intrinsic resistance that causes a voltage to be created from
the current.

Resistance cannot be negative only positive.

Technically there are cases where resistance becomes negative but it's more of an artifact of the math than anything else.

But you can have voltage without current, batteries and capacitors are two examples that I am familiar with.

For both batteries and capacitors, with nothing attached to the leads, you have a lot of charge but no voltage or current.  As I explained above, attaching a voltmeter completes a circuit and causes a small current that is used to determine the voltage.  The battery's chemistry acts as a voltage regulator and adjusts the current so that the voltage generated across the load matches a specific value.

I can turn on my amplifier and measure voltage at the output with no load connected. No load no current. OK maybe some leakage current. . . . And I hope we aren't including super conductors into the equation, I like my environment to be at a nice comfortable temperature.

Except for that pesky 1MOhm resistor inside the voltmeter that creates a load.

Anyway if you have some links to updated information, please do.

http://en.wikipedia.org/wiki/Voltmeter
http://en.wikipedia.org/wiki/Ammeter

[kc8apf]

For clarity, electrostatic potential (confusingly also called voltage) is actually a measure of how much work per unit charge _would be required_ to move a charge from infinity to the charge at a given point.  This is really just an alternative way to express the strength of the electric field.

Voltage w/o current only exists in static electric fields which generally have very little actual charge in it.  I've ignored this in the discussion above because static electric fields are rarely used to power devices.

Nearly all power generation is done instead by generating a current by moving wire through a magnetic field.  This means that all voltage in such a system originates from a current.

[Æ]

For clarity, electrostatic potential (confusingly also called voltage) is actually a measure of how much work per unit charge _would be required_ to move a charge from infinity to the charge at a given point.  This is really just an alternative way to express the strength of the electric field.

Voltage w/o current only exists in static electric fields which generally have very little actual charge in it.  I've ignored this in the discussion above because static electric fields are rarely used to power devices.

Nearly all power generation is done instead by generating a current by moving wire through a magnetic field.  This means that all voltage in such a system originates from a current.

Yes, that was all very helpful to the original poster. You thoroughly answered his question.

Now if you'll excuse me, I need to go and flick the needle on my Galvanometer.

[kc8apf]

Yes, that was all very helpful to the original poster. You thoroughly answered his question.

He asked why high current matters.  That deserves a reasonable explanation.  As for his other question about how much current these particular amps can deliver, only the engineer who designed them can say.

Now if you'll excuse me, I need to go and flick the needle on my Galvanometer.

*sigh*

[dminches]

The answer to the OP's question should be directed to the person who designed and built the amp, Dan.  Frankly, I wouldn't rely on an answer from anyone other than him (not that people aren't trying to be helpful).