2. amps-preamps
  3. 192831-171531

Question about high current amps versus "not high current amps"

Recently I read a reply to a post about a certain speaker, and the person who replied typed that (and I am going to paraphrase somewhat) the speaker required a high current amp to perform well and it wasn’t the WPC that was important.

Sorry as I am afraid that these are probably going to be  "audio electrical questions for dummies," but here goes:

I vaguely remember being taught the PIE formula, so I looked it up online for a quick review and if I am understanding it correctly,

P (power/watts) = I (current/amps) x E (electromotive force/voltage) .

My first question would be: if I am understanding that correctly, how can wpc NOT matter since watts are the sum of current x voltage? I mean if you have so many WPC, don’t you then HAVE to have so much current?

My next question would be, if I am understanding PIE correctly, is E/voltage going to be a fixed 110 vac out of the wall, or is that number (E) determined by the transformer (so it would vary by manufacturer) and it is that (different transformers that are used in different amps) going to be the difference between a high and a lower current amp?

Or am I completely off base thinking that P is wpc and P is actually the spec in my owners manual that lists "power consumption as 420 watts operate 10 watts stand by"?

And lastly, what would be an example of a high current amp and what would be an example of a low current amp?

Thanks.

 

immatthewj

Showing 1 response by richardbrand

richardbrand's avatar
richardbrand
394 posts
 

Ohm’s law is one of the simplest laws of physics, yet seems to be so hard to apply properly.

What matters is the voltage drop across the load, Multiply this voltage drop by the current flowing, and you have the power (or Wattage) produced at the load.

The current that flows is just given by the relationship voltage drop = current times resistance.

Amplifiers might be considered to be ’straight wires with gain’ to quote Peter Walker of Quad. They try to produce an output voltage which is a multiplier (the gain) of the input signal voltage. So the voltage drop is given by the input signal and the gain, and the resistance is also fixed (at least nominally).

If the rated maximum power into 8-Ohms is say 100-Watts, that must be the voltage drop times the current. The current is the voltage drop divided by the resistance. So 100 = V * V / 8 or V = sqrt (800) or about 28-Volts.  Note that this is derived just from the power and the load resistance.

This is true for Direct Current. When we measure alternating currents, it is conventional to think of them as equivalent to the direct current that would produce the same average power. Turns out this is the square root of the average (mean) of the instantaneous value squared. Abbreviated to RMS (root mean squared).

I feel better now ...

More to discover