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

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 ...

A lot of good information here.  The bottom line for me  is to make sure the amp you choose can drive the speakers you choose.  A lot of the manufacturers information is misleading because it’s marketing instead of science.  Note some of the online reviews that show a speaker’s impedance.  The manufacturer will state it’s 8 ohms and when it’s tested It may dip as low as 2 ohms.  That means you would want to select an amp that shows it has been rated and has power at 2 ohms.  Some amps will only show specs at 8 ohms.  If the amp can handle 2 ohm loads the manufacturer will usually publish it because it’s a selling point.  That’s one of the reasons I usually only buy equipment I can buy and try in my system and return for no or little cost if it doesn’t sound right to me in my listening room.  

@atmasphere 

Thank you for the lesson, very useful. I was a Ground Radio Repairman in the USAF. My impedance training was focused on matching antennae to transmitters (along with other factors). A visual "test" of radios, like a big SSB, was the size and number of capacitors. We paid attention to impedance matching. Along with the characteristics of the antenna termination... open, capacitance, resistance, inductive , etc. I've forgotten 90% of what I learned (it was so very long ago). But for my humble needs, if the manufacturer lists power output to the speakers as doubling, or nearly so, from 8 ohm to four and again to two ohms, it is a high current amplifier. If you open the hood, large transformers, and large capacitors are a visual clue. Do not get across the terminals of a large cap and ground without discharging the cap. The potential (voltage) can kill, which explains the "No User Serviceable Parts"  stickers.

Not just any speaker will perform its best with just any amp.

I’ve always thought the higher the capacitance reserve an amplifier has, the more amps it can deliver when needed. This isn’t correct?

 

@thecarpathian - kind of sort of. Think of the amplifier as a weight lifter, and the power supply as the stage the athlete stands on. The better the power supply, the stiffer the stand and the better the weight lifter is able to dead lift up to the same point.

The capacitance reduces power supply ripple and attempts to feed the amplifier constant DC  voltages. That is, how much the DC rails vary based on incoming AC and amplifier demand.

Many things go into what the ideal capacitance levels are. Amplifier feedback, speaker load, etc. Generally speaking though, if you want to design a high current amplifier you will need to add more capacitance to ensure it can perform to spec even during high demand.

Should also point out that most amplifiers use linear, unregulated supplies, meaning they are subject to the long term voltage stability coming out of your wall.  Seasons, your neighbors, your HVAC and refrigerator can cause instability even with dedicated lines. 

@erik_squires ,

To your weight lifter analogy;

I would liken the weight lifter to the amplifier, the weight he's lifting to the speakers, and the power supply to how much muscle (power reserves) he has.

Theoretically, the more muscle he has (capacitance), the more weight (low impedance speaker) he can lift with minimal strain.

Is this reasonable? I hope so, because now I need an aspirin...