"Watts" Versus "Current"

Forgive the intrusion, but I've got a pratical question about the use of the two different amps, i.e. low power high current and high power low current.

When driving a speaker at say 10 watts max (including peaks) and your two amps are, say 200wt, one which has 'high current' and doubles down to 2 ohms, and the other which can only produces say a 50% watt increas at 4 ohms, will you hear a difference because of the difference when you play speakers with a mean impedence curve, or does that only happen when you start to approach the power limits available at any particular frequency/impedence demand?

Just curious.

Current is a component of watts (power) as Dan just indicated. Therefore, all watts are not created equal.

1 amp (of current) X 10 volts = 10 watts of power. But 10 amps X 1 volt also = 10 watts. Further, the basic W =A x V really works for direct current (DC) only. The formula is a bit more complicated for AC (as in: music signal). But leaving that aside for now, how the power is created will either be better or less good for certain kinds of speakers.

Why? Because every kind of driver generates a reactive signal BACK TO THE AMP. When the signal from the amp moves a cone back and forth, for example, the action of the cone's voicecoil in the magnetic field actually GENERATES a reverse electric current BACK to the amp! This reverse current adds to the impedance (load) the amp "sees".

What's CRITICAL (in amp selection) IMO, is HOW how that (let's call it 'phantom') impedance is created. If the amp sees a capacitive type of load (an electrostatic speaker) it needs for its available watts to consist of lots of amps (current). If it sees a resistive type load (like cones, domes, ribbons and planar magnetics) it needs for its available watts to consist of lots of volts (to overcome the reverse voltage created by those kinds of drivers.)

I'm going to generalize here, but 1 tube amplifier watt (a typical push-pull circuit with an output transformer, nothing exotic!) will be rich in the current it can make available (for that 1 watt of output.) 1 solid state amplifier watt (a typical push-pull circuit with the output transistors directly coupled to the speaker) will be rich in the voltage it can make available (for that 1 watt of output.)

You can guess where this is headed ;-) Tubes are better than silicon for capacitive loads, and silicon is beter than tubes for resistive loads. All right -- this is a sweeping generalization! HOWEVER, because the impedance of all speakers varies widely depending on frequency (the rated impedance of speakers is only a nominal figure) there could be times when you'd need a HUGE tube amp to drive those dynamic drivers (God forbid acoustic suspension woofers!) properly. Or if you are driving electrostatics (panels only, we're talking about) you might need a HUGE solid state amp to give those stats the current they need when a tube amp of far less output would sound as good (and likely better IMO ;-)

People into fine audio are generally at the 3-digit Stanford-Binet level. But many (probably most) have had little occaision to become familiar with the nature of (yikes!) circuit design. But if you will take a little time (good bathroom reading for instance as it takes several passes to sink in ;-) and go through this excellent and (reasonably) clear, easy to understand online tutorial, you will find yourself more informed than many of the self-appointed/annointed 'know-it-alls'/'experts' that haunt the halls of audio-underworld ;-)
http://www.allaboutcircuits.com/vol_1/index.html
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The confusion with amplifier ratings is that amps do not "produce" watts. An amplifier is a voltage source, just like the 120 volt receptacle outlet on your wall. An easy analogy is using a light bulb. What is the difference between a 100W and a 60W light bulb? And, why is a 100W bulb brighter than a 60W bulb in the same lamp plugged into the same 120 volt outlet?

You cannot say that with a 100W bulb in the lamp that the wall outlet "produces" 100 watts. The outlet supplies a constant 120 volts - that's the key. So, how does one design a light bulb with 100W and one of 60W?

Ohms law: Voltage = Current x Resistance. Watts = Voltage x Current (simplified).

The 60 Watt bulb in a 120 volt outlet can tell us the current. Rearranging the Watts equation gives

Current = Watts/Voltage or 60/120 = 0.5 amps

We need to pass 0.5 amps through the bulb to get 60 watts. We control amperage by resistance. To get that 0.5 amps we now go to the Voltage equation and rearrange to solve for Resistance:

R=V/I or R= 120/0.5 = 240 Ohms.

So, to design a 60W bulb requires an element with a resistance of 240 Ohms. That gets us 60 watts in a 120V outlet.

For a 100W bulb: I = 100/120 = 0.833 Amps, R=120/0.83=144 Ohms.

So, to design a 100W bulb requires an element with a resistance of 144 Ohms. That gets us 100 watts in a 120V outlet.

The difference is the resitances of the tungsten elements WHICH CAUSES A GREATER CURRENT TO FLOW from the receptacle. That extra current is the extra wattage (W=Volt x amps).

Ok, great, you say. What about amplifiers and current? Same thing. The only difference is that speakers, unlike light bulbs, have different resistances at different frequencies. To the amplifier, that presents a varying load. That is like constantly installing different wattage light bulbs in the same socket. That presents a similar varying load to the wall outlet.

But the wall outlet has an infinite store of current, that is, you can lower the resistance and the outlet will supply the greater current to a maximum of 20-amps (the circuit breaker) or 10,000 amps (the maximum short circuit current of the utility transformer).

But your amplifier does not have 10,000 amps of current at its disposal. The amperage that it can supply is limited by its power supply transformer (VERY simplified statement, there).

If you have an amp that is "100 watts", you have a meaningless rating. If you have a 100 watt amp at 8 ohms, now that means everything. To design a 100 watt amp at 8 ohms you will need a 100 watt transformer for the power supply, so we buy a 100 VA transformer. Since the amp is rated for its power at 8 ohms, that means the transformer will have the following properties:

120 volts on the primary, to match the wall voltage. For the secondary voltage: Watts = Voltage(V) x Current(I). We're stuck because we do not know the secondary current. BUt we do know that the load is 8 ohms and Current(I) = V/R. So the power equation becomes:

W= V x I = V x (V/R) = V x V/R

Knowing R=8, the secondary voltage is 100 = VxV/8 = 28 volts. We specify a winding of 28 volts. The current "available" from the transformer is:
Current = Watts/Voltage or I = 100/28 = 3.5 amps

So the 100 watt amplifier is like a wall outlet at 28 volts with a maximum amperage capacity of 3.5 amps (not 15, not 10,000). The speaker is like a light bulb at 8 ohms. Now we have a "100 watt" amp.

When the speaker drops to 4 ohms, the current required is I = V/R or I = 28/4 = 7 amps. That's twice the current. But wait - the transformer only gives us 3.5 amps. That means we cannot maintain 28 volts across a 4 ohm load. We need more current. But since Watts = VxV/R, at 4 ohms that power is 200 watts. Makes sense: twice the current means twice the watts.

What to do? Well, we put in a bigger transformer - a 200 watt transformer at 8 ohms (Or a 200 VA transformer is more accurate). That gives us the doubling of current when the speaker load drops to 4 ohms. But wouldn't that be a "200 watt" amp? No, because at 8 ohms the secondary needs to be at V=40 volts for a 200 watt rating (200=VxV/8). In this case we keep the same 28 volts. That still makes it a 100 watt amp at 8 ohms with a 200 VA (volt-ampere) transformer, meaning you get 3.5 amps at 8 ohms still and 7 amps at 4 ohms. This is key: with a 200 VA transformer you either have a 100 watts at 8 ohms and 200 watts at 4 ohms when wound with a secondary of 28 volts OR an amp with 200 watts only at 8 ohms when wound with a secondary of 40 volts.

Same reasoning if you want the power to double at 2 ohms. Use a 400 VA transformer, secondary at 28 volts, and that gives a 14 amp capacity to take care of the 2 ohm load.

Basically, the power supply determines the current available and the speaker load at a frequency determines the power. A high current amp is an oversized beast that has a bigger transformer wound down to a lower secondary voltage to get the extra current. Kinda long winded, but this is not an easy answer.

You're quite right, nothing really happens until those watts are called for, but when they are, the device appears to the amp as a certain kind of load. And it's the nature of the speaker load (or more accurately, the nature of the speaker's reactance) that will determine mix of amps x volts the amp must supply.

So it's not just that speakers present different impedences at different frequencies, but that those impedances can be resistive or capacitive. Remember that current flows easily through a voice coil (so volts are called for), but almost not at all across the [air-space + step-up-transformer] assembly of an electrostat (so current is called for.)
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Gs5556, great post - I found your bulb analogy very enlightening (pardon the pun). Thanks for sharing your knowledge on this!