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Calculating power output for impedance, need help

post #1 of 11
Thread Starter 

Heya,

 

So I'm pretty retarded here as I can't seem to figure this out, so I'm needing some help.

 

I'd like to see simply how one can calculate the 6 watt rated power output into 32ohms that the Schiit Lyr advertises based on whatever specs they have listed on their website. My understanding is that the output impedance of the Lyr is less than 1ohm (0.2 if I remember). If someone could put it into "I can count to potato" form for me to understand, I'd greatly appreciate it.

 

Very best,

post #2 of 11

The specs say: "Maximum Output: 40V P-P into 32 ohms typ". This means that:

 

Peak voltage = peak to peak voltage / 2 = 20 V

RMS voltage (for sine wave) = peak voltage / sqrt(2) = 14.142 V

Power = voltage ^ 2 / resistance = 14.142 * 14.142 / 32 = 6.25 W

 

post #3 of 11
Thread Starter 

Heya,

 

Perfect thanks. I was messing up on dividing the peak voltage by 2 initially.

 

I can now count to potato.

 

Very best,

post #4 of 11

One should also note that there is absolutely no correlation between output impedance and output power. 

 

You can have a very powerful amplifier with a very high output impedance, or a very low-powered amp with a freakishly low output impedance. 

 

Nay-sayers can state it like ohms law or the power equation. 

post #5 of 11
Thread Starter 

Heya,

 

Hrm.

 

So let's see if I'm counting to potato correctly. The Schiit Asugard is 20V P-P.

 

So peak is 20v / 2 = 10v

RMS voltage is 10v / (square root of 2) = 7.071v

And power = 7.071v ^2 / 38ohm (assumed) = 1.36 watts

 

Sound about right?

 

Very best,

post #6 of 11

Quote:
Originally Posted by MalVeauX View Post

So let's see if I'm counting to potato correctly. The Schiit Asugard is 20V P-P.

 

So peak is 20v / 2 = 10v

RMS voltage is 10v / (square root of 2) = 7.071v

And power = 7.071v ^2 / 38ohm (assumed) = 1.36 watts


Maybe.  They're not very specific about it.  Many amplifiers are current limited into lower impedances, so just because they can do 20V P-P into 600 ohms or 80 ohms does not mean they can do the same 20V P-P into 38 ohms or so (they might start clipping before reaching that high).  However, it's a discrete FET-based class A design, so it can probably handle a lot of current and actually do 20V P-P into 38 ohms, not like any headphones other than a few planar magnetics are really going to be operated with that kind of power output.  I would sure hope it can do so, at the listed power consumption.

post #7 of 11
Quote:
Originally Posted by nikongod View Post

One should also note that there is absolutely no correlation between output impedance and output power. 

You can have a very powerful amplifier with a very high output impedance, or a very low-powered amp with a freakishly low output impedance. 

Nay-sayers can state it like ohms law or the power equation. 

That's very wrong because there's even a theorem that says that the "maximum power transfer occurs when the load impedance is equal to the complex conjugate of the source impedance".
Output impedance has a very big effect on the output power depending on the load.

For example, an amp driving a 600 ohm load could produce 78 mW, but with a 16 ohm load the same amp only produces 75 mW due to a 100 ohm output impedance. The same example with 0 output impedance: 106 mW into 600 ohms and 4000 mW into 16 ohms (theoretically).
Edited by xnor - 2/15/12 at 2:41am
post #8 of 11

 

Quote:
Originally Posted by xnor View Post

That's very wrong because there's even a theorem that says that the "maximum power transfer occurs when the load impedance is equal to the complex conjugate of the source impedance".
Output impedance has a very big effect on the output power depending on the load.
 

 

There should be an equation to explain this similar to:

V=I*R or P=V*R

There is no law relating output power an output impedance. 

While you could write an equation to estimate the maximum output power of a given amplifier into a given load if you define unloaded voltage, max output ucrrent, load impedance, and output impedance there is no reliable way to estimate the maximum output power of an amplifier by looking at just its output impedance and the load impedance. 

 

I do agree that at some point the realities of the situation do create limits, but you can not in any way say that since an amplifier has a 1ohm output impedance it can put out a bunch of power - which is the correlation that the OP looks like it was leaning towards. The info for output voltage or current are not given - only output impedance and load. 

 

Maximum power transfer does not apply to almost any audio application. 

http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

It requires a damping factor of 1.

In the case of a speaker & an amplifier - Measurements guys freak out when you suggest a damping factor of 1 between the power amp and the speaker. Id certainly try it with headphones or single driver speakers as I find I often like the changes this makes to the sound but thats kind of a personal thing.

post #9 of 11
Quote:
Originally Posted by nikongod View Post

 

 

There should be an equation to explain this similar to:

V=I*R or P=V*R

There is no law relating output power an output impedance. 

While you could write an equation to estimate the maximum output power of a given amplifier into a given load if you define unloaded voltage, max output ucrrent, load impedance, and output impedance there is no reliable way to estimate the maximum output power of an amplifier by looking at just its output impedance and the load impedance. 

 

I do agree that at some point the realities of the situation do create limits, but you can not in any way say that since an amplifier has a 1ohm output impedance it can put out a bunch of power - which is the correlation that the OP looks like it was leaning towards. The info for output voltage or current are not given - only output impedance and load. 

 

Maximum power transfer does not apply to almost any audio application. 

http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

It requires a damping factor of 1.

In the case of a speaker & an amplifier - Measurements guys freak out when you suggest a damping factor of 1 between the power amp and the speaker. Id certainly try it with headphones or single driver speakers as I find I often like the changes this makes to the sound but thats kind of a personal thing.



 



My I totally understand why they would do so. It is defined as distortion after all, not like everyone hates distortion. Audio signals are considered AC after all and for good regulation, low impedance/high damping factor are (mostly)desirable in amps. As for changes, after that, it is as you said, up to the user :)

 

As for no laws, are you sure? Impedance's relationship with power dives into complex numbers and with all the mathematics and phasor/vector diagrams that go into it, I won't(or most for the matter) will be able to explain it in detail with formulas at all. It still starts with I = V/R though the R is subbed with Z and Z can be further expressed in reactive and non reactive loading laws/math that I will not pretend to understand. In short, its REALLY complicated.

 

Also to quote a wiki article on Impedance Matching(its kind of relevant),

 

"Impedance matching is not always necessary. For example, if a source with a low impedance is connected to a load with a high impedance the power that can pass through the connection is limited by the higher impedance. This maximum-voltage connection is a common configuration called impedance bridging or voltage bridging, and is widely used in signal processing. In such applications, delivering a high voltage (to minimize signal degradation during transmission or to consume less power by reducing currents) is often more important than maximum power transfer."

post #10 of 11

 

Quote:
Originally Posted by firev1 View Post

As for no laws, are you sure? Impedance's relationship with power dives into complex numbers and with all the mathematics and phasor/vector diagrams that go into it, I won't(or most for the matter) will be able to explain it in detail with formulas at all. It still starts with I = V/R though the R is subbed with Z and Z can be further expressed in reactive and non reactive loading laws/math that I will not pretend to understand. In short, its REALLY complicated.

 


Quite sure. 

There is no way to reliably say that a given amplifier can output a given amount of power given just output impedance and load impedance. 

 

Given output impedance, load impedance, and unloaded output voltage or current you can work it out. BUT that process starts with using output impedance and load impedance to figure out what the LOADED output voltage will be after that its the basic power equation. 

 

I have a power amp with a 2ohm output impedance, how much power can it put into an 8ohm speaker? This question is unanswerable. 

post #11 of 11
Quote:
Originally Posted by nikongod View Post

There should be an equation to explain this similar to:
V=I*R or P=V*R
(P=V*R ? It's either P=V*I or using ohm's law: I^2*R or V^2/R)

There is, since it works like a simple voltage divider: Vout = Vin * Zload / (Zload + Zout) so 8 Vrms with 100 ohms output impedance and 600 ohms load (see my previous example) results in 6.857 Vrms --> 78 mW into the headphones.

(Btw, if we take a look at the impedance curve of the headphone we can calculate the influence of the output impedance with the same formula. Let's say the headphone has 700 ohms at 100 Hz and 600 ohms at 1 kHz. --> 20*log10(7 / 6.857) = 0.18 dB bass "boost")
Quote:
There is no law relating output power an output impedance.
Actually if there was no relation you could not calculate the output impedance using math like this:
Zout = (Zload * (Vnoload - Vload)) / Vload

But I guess what you're trying to say is that one cannot reason output power from output impedance or vice versa, which of course is right. Though it's still less likely that a headphone amp with high output impedance will deliver lots of power into low impedance loads. wink.gif
Quote:
While you could write an equation to estimate the maximum output power of a given amplifier into a given load if you define unloaded voltage, max output ucrrent, load impedance, and output impedance there is no reliable way to estimate the maximum output power of an amplifier by looking at just its output impedance and the load impedance.
No but some manufacturers provide specs with a high and a low impedance load. You can use both values to interpolate and even estimate the power outside that range. Extrapolating output power into higher impedance loads should be fairly accurate.
Quote:
I do agree that at some point the realities of the situation do create limits, but you can not in any way say that since an amplifier has a 1ohm output impedance it can put out a bunch of power - which is the correlation that the OP looks like it was leaning towards. The info for output voltage or current are not given - only output impedance and load.
The op has specified the output power into a 32 ohms load and close to zero output impedance. That's enough information to calculate e.g. the min. power the amp can output into loads ranging from 32 up to thousands of ohms.
Quote:
Maximum power transfer does not apply to almost any audio application.
I know, I was just saying that there's a well defined relationship between output power and output impedance. As firev1 wrote, impedance/voltage bridging is the preferred configuration for headphones but it's also a matter of choice as you wrote. I'm sure some headphones sound better with higher output impedance.
Edited by xnor - 2/15/12 at 1:47pm
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