Before I go on I thought I’d point out a couple of other things involving power and power supplies. I stated that going from 8-ohm to 4-ohm speakers can allow an amplifier channel to deliver more power within the constraints of the designed current limits. I also said that the power supply could only reliably deliver power up to a given limit and when that limit is reached, the power well runs dry. If, as is the case of multi-channel surround sound (5.1, 7.1, etc.), 4-ohm speakers are used for the fronts and 8-ohm speakers are used elsewhere and the fronts end up demanding more power at similar volumes then that dips deeper into the power well leaving less for the remaining speakers. This is easy to do and is in some cases hidden due to the way specifications for speakers are written to tell us how loud a speaker will play. Speaker specifications will usually state this as either Efficiency or Sensitivity. It is measured in db and in this case db is the measure of the Sound Pressure Level or SPL. It’s the same SPL the safety guy at work would measure to determine if hearing protection is required in a given work environment. The sticky point comes with the conditions under which the specification is measured. It’s usually stated by the following:
1 kHz @ 1 Watt @ 1 meter
..or..
1 kHz @ 2.83 Volts @ 1 meter
Meaning they run a 1kHz tone at a power of 1 watt (2.83 volts) to the speaker and measure the SPL from 1 meter away. I prefer the former and here’s why. Assume the conditions above are applied to an 8-ohm speaker. By the equation wheel we can see that for 8-ohm speakers they say the same thing.
Let’s run 1 watt to that speaker and see what voltage develops.
V = √(P*R)
V = √(1*8)
V = √8
V = 2.828427125 ~ 2.83
Now do the same for the 4-ohm speaker.
V = √(P*R)
V = √(1*4)
V = √4
V = 2.00 volts … a 41% difference that gives me an idea of how much headroom I’ll have – db for db.
Now let’s go the other way – let’s apply those 2.83 volts to the 4-ohm speaker and calculate the power being delivered.
P = V²/R
P = 2.83²/4
P = 8/4 = 2 watts … a 2-for-1 difference that betrays the otherwise hidden fact that, db for db, going by this spec I’ll be dipping twice as deep into my power supply well.
To me the first definition is more – apples to apples – than the second. It relates the Efficiency/Sensitivity more directly to the power supply spec of watts which is more useful in my opinion.
Another thing that may be of interest is using the specs to do a little detective work to ESTIMATE the limits of the power supply. I’ll use the specs for my AVR since I have them handy. By my signature you can see that’s a Pioneer Elite SC-71. You can go by the manufacturer’s spec but I prefer using independent lab tests if I can find them as I tend to believe they have less bias. My favorite is Sound and Vision. They did a full review of the SC-71 with a nice lab workup you can find here:
http://www.soundandvision.com/conten...ver-test-bench
First let’s figure out how deep my power supply well is. We see that the power limit into 8-ohms for all seven channels driven (Hey, chinzo, LOOK we’re finally getting to your question – sorta) @1.0% THD (I’ll be getting to that) is 106.8 watts.
106.8 * 7 = 747.6 watts (in the manual the spec is 560 watts – under spec’ing or conservative spec’ing is a good thing to find in a manufacturer in my opinion)
It’s deep enough for me (for now anyway). That’s the power supply total output but what about the voltage limits. When will this thing begin to clip? For that I need to push a single channel (or just two) up to its limit without draining the power supply well. When I find the limits in those conditions I’ll know a little more about upper voltage. Higher speaker impedance yields higher voltages so a limit pushing spec into an 8-ohm speaker is what I’m looking for. And I find it in the first paragraph under the Harmonic Distortion graph. That paragraph reads as follows:
Quote:
This graph shows that the SC-71’s left channel, from CD input to speaker output with two channels driving 8-ohm loads, reaches 0.1% distortion at 120.6 watts and 1% distortion at 146.0 watts. Into 4 ohms, the amplifier reaches 0.1% distortion at 204.3 watts and 1% distortion at 247.3 watts.
The spec is most likely to be pushing the envelope at the 1.0% point and beyond so I’ll use that. For 8-ohms that’s 146.0 watts. Again from the equations:
V = √(P*R)
V = √(146.0 *8)
V = √1168
V = 34.18 volts
But that’s RMS and I want the value for peak (I’ll discuss the difference later but peak is where clipping occurs) which is RMS*1.414 which comes to 48.33. So it’s safe to assume the power supply voltage limits are at least +/- 48.33 volts. An estimate of power supply current can now be made by using the max power we found before and this voltage value in our calculations.
{I have to do an edit here. The power supply peak voltage calculation is mixing RMS and peak values which is a NoNo. } {Corrected using a value of 1057.1 watts peak vs. the 747.6 watts RMS , sorry for the mix up.}
I = P/V
I = 1057.1/48.33
I = 21.87 amps peak or 15.47 amps RMS
Just for grins let’s calculate the voltage using the specs for a 4-ohm speaker. At 1.0% distortion the channel was pumping 247.3 watts.
V = √(P*R)
V = √(247.3 *4)
V = √989.2
V = 31.45 volts RMS or 44.47 volts peak
That’s less than what was calculated for the voltage limit for the power supply and we haven’t reached the power supply max power limit so the best safe assumption is this limit is driven by the current limitations of the amplifier stage itself.