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Beyerdynamic Custom one pro - impedence clarification needed!

post #1 of 2
Thread Starter 

Hey guys,

Audiophile noob here. I am on the verge of buying Beyerdynamic's COP given the it almost have everything im looking for. Just a couple of questions that I just can't find a definitive answer to be completely sure about my purchase.

Basically, Impedence.

Since im new to the Audiophile world I have no accesories. (dac, amps, etc) I will be using it:
-90% desktop PC

-10% on the go with my iphone.

Basically the general consensus is that since these cans have a 16ohm impedence, amps are not requiered/needed to have the best possible experience. However the only thing I want from these cans is to be able to have virtualized surround sound 5.1/7.1. And so for this I need a sound card(internal or external). 

HOWEVER,

I read a thread I can't recall explaining that most internal soundcard have a minimum 10ohm output impedance that will potentially degrade it sound especially by making it bass muddier. Is this true?

If true, is there a solution to achieve a virtualized sound for gaming ONLY? Perhaps an external DSP like the Mixamp have no issue with the impedance? 

Thanks in advance!

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post #2 of 2
Quote:
Originally Posted by happyhour13 View Post
 

Hey guys,
...

Basically the general consensus is that since these cans have a 16ohm impedence, amps are not requiered/needed to have the best possible experience. However the only thing I want from these cans is to be able to have virtualized surround sound 5.1/7.1. And so for this I need a sound card(internal or external). 

HOWEVER,

I read a thread I can't recall explaining that most internal soundcard have a minimum 10ohm output impedance that will potentially degrade it sound especially by making it bass muddier. Is this true?

If true, is there a solution to achieve a virtualized sound for gaming ONLY? Perhaps an external DSP like the Mixamp have no issue with the impedance? 

Thanks in advance!

 

Well...16 Ohms is a comparatively low impedance. Keep in mind that speakers can and do have impedance values around a few Ohms (though to be fair, headphones, just like speakers, do not present a uniform impedance to the source driving it - it varies as a function of frequency).

 

Anyway...

 

The thing to remember is this, and it applies universally:

 

1) A low-impedance source can adequately drive a high impedance load.

 

2) A high impedance source cannot adequately drive a low impedance load.

 

The thing to remember here is that 'high' and 'low', as used here, are relative terms.

 

When you break it down to a simplified model, the output amplifier (sound card, whatever) can be modeled as a perfect source (zero output impedance, purely real, and capable of any sane voltage swing, as well as sourcing the required current). After that is the internal resistance of the device (which is the output Z of the source). This is what the source and its resistance (more properly, impedance) 'look like' mathematically.

 

In other words, if you draw a (ideal) source with its internal Z as a series impedance, and then draw the headphone load going from that to ground, you have the basis of a voltage divider. That is, the ratio of Zi (internal device impedance ) and Zl (load resistance) determine how much VOLTAGE will be developed across the load (the headphones). Note this does not speak to the current available to the load (and consequently, not the power sourced to the load), just the voltage.

 

Example: Let's suppose that the device Zi is 16 Ohms, and let's suppose that the load (headphones) is also 16 Ohms. To simplify the example, let's assume the load is purely real (100% resistive)). This isn't reality, but it makes the example easier.

 

The voltage develop at the headphones will be given by VLoad = V out * (ZL / (Zi + Zl)).

 

Thus, VLoad = V out (16 / (16 + 16))

 

VLoad = V out (1/2)

 

So in our example, one half of the voltage developed by the output will be developed across the headphones, but also, 1/2 will be developed internally (a voltage drop internal to the device). That is, you're not going to get even close to what the 'ideal' source is capable of sourcing, because its own internal resistance does not allow for that.

 

Now, let's re-work the equation and assume that the output Z is 0.1 Ohm, and the load is still 16 Ohms.

 

In this case we get VLoad = V out * (ZL / (Zi + Zl)) just as before, but now we get (substituting values):

 

VLoad = V out * (16 / (16.0 + 0.1))

 

VLoad = V out (16/16.1)

 

So you can see with with a driving impedance that is 'much lower' than the load impedance, nearly all of the voltage is developed into the load (the headphones) whereas when the device's output Z was 16 Ohms, you'd be losing 1/2 the available voltage. Mind you, you can look at this example in the complementary fashion; in the case where a headphone was 0.1 Ohm (which would never happen, but bear with me) and the output Z was 0.1 Ohm, then the same thing that happened in the 16 & 16 Ohm case would happen again (voltage dropped across the load would be 1/2 the total available voltage).

 

Mind you, there are many other factors at play here, not the least of which is the amount of current that the driving amplifier can source. That is, it's possible that a device can swing a hefty voltage open-circuit or with a very high Z load, but not when a more typical load is seen by the device. This is why headphone amplifiers will usually state the amount of power (usually stated in mW) that can be developed into the load (the headphones). Also, I'm not taking into account the reactive portion of the impedance, but these are good general concepts (the voltage divider in particular) to keep in mind when wondering whether a certain load is 'friendly' to a certain device.

 

Again, there are many factors to consider, but usually output Z is specified, but if it is not, given the stated power available and  stated load impedance, you can figure out just how much current can be sourced.

 

Hope this helps

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