What does Current mean in a power amp???

Ray Dall, Electronics Theory.com site here is good for basics on ohms law, circuits about as briefly as is possible. About 50 pages (big print.) An overview of tubes begins on page 37.

I remain,

Look for an amp that has a high power rating into 4 ohms. An idea high current amp will double it's rated power into 4 ohms. Most amps can't do this but my Bryston goes from 240 wpc to 400 wpc from 8 to 4 ohms. Look for my posting talking about the difference between "Sony watts" and "good" watts.

If your speakers impedance is rated (nominally) 4 ohms or less, or if certain frequencies dip much below 4 ohms, you will need a power amp that is rated into 4 ohm loads. The lower the impedance, the more current being drawn from the amplifier. Amps put out volts, the load draws current. I'm being very general here for the sake of brevity because there are many factors to consider, such as room size, how loud you listen, etc. There are other things to know about your speakers, such as minimum impedance dip, phase angle,
electrical sensitivity (sometimes referred to as efficiency) and such. Ask the speaker manufaturer.

Rwd,
What kind of speakers are you intending to drive?

"High Current" is a relative term. For a load with a given, fixed impedance, a higher powered amp will always deliver higher current INTO THAT LOAD. In other words, for a fixed 8 ohm load, a 200W amp will supply more current than a 100W amp. Therefor, based solely on current requirements, it could appear that using an amp with a higher rated power output would be the answer.

However, the rated power of an amp into an 8 ohm load is not really a good indication of the MAXIMUM current delivery capability of that amp. Most speakers do not have a constant impedance over the entire audio frequency range, and many have less than 8 ohms nominal impedance. As described in the post above, using ohm's law, into an 8 ohm load, a 100 watt amp would deliver 3.54 amps of current (assuming a 0 degree phase angle between the voltage and current waveforms, for you hardcore engineers :). If instead of an 8 ohm speaker, we were driving a 4 ohm speaker, then to deliver the same 100 watts, the amplifier would have to output 5.00 amps of current. If the impedance dropped to one ohm, to provide 100 watts of power to the load, the current delivered would have to be 10 amps, almost 3 times the current at the 8 ohm rated power output. The ability of an amp to deliver higher current into low impedance loads is influenced primarily by the design of the power supply, and the type and number of output devices. The ability to dissipate larger amounts of heat also becomes an issue as current delivery increases.

If you have difficult to drive speakers, such as Martin Logans, where the impedance drops as low as one ohm at 20kHz, it is important to have an amplifier that is designed to drive low impedance loads.

Therefor, it is probably more helpful for you to consider an amplifiers performance into low impedance loads, rather than zero in on the term "high current". As described above, it is possible to have an amplifier with a large power rating into 8 ohms that will have "high current" at 8 ohms, but may not be suitable for driving low impedance speakers. It is also possible to have an amp with a modest power rating into 8 ohms, and thus a modest current delivery at 8 ohms, that is capable of delivering much higher current, or even much higher power, into lower impedance loads.

In general terms, if you have speakers with an unusual characteristic impedance plot, you probably want to look for an amp with the ability to drive a 2 ohm load continuously. If a manufacturer won't put into writing that their design is capable of driving a 2 ohm load, then it probably isn't.

That said, if your speakers sound good to you with the amp that you are using today, and the amp does not appear to be running abnormally hot, don't mess with a good thing.

A minor clarification to my earlier response:

Ohm's law does not provide for a definition of power in an electrical circuit, rather it defines a relationship between the DC voltage, DC current, and DC resistance. When I stated "using ohm's law", I omitted that I was applying ohm's law, along with the basic power equation, to obtain the relationship of P=I^2*R, where P=power, I=current, and R=resistance.

Sorry for any confusion that may have caused,
Mike