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Using 250ohm headphones with amp and computer
starbux58
post Sep 5 2012, 11:58
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I have found a reseller for Beyerdynamic DT990 headphones that is selling the 250ohm model for half that of the other models 32ohm and 600ohm. So the 250ohm model is within my financial reach as its $200 US lower than the others. I am also buying an integrated stereo amplifier from NAD electronics in Canada to use the Beyers with primarily. Does anyone know if the standard headphone jacks on amplifiers and receivers will adequately drive or power these? I can buy some AKG's that I know will work but have wanted to try these Beyers for awhile now. I know too that companies like FiiO make affordable (for my budget)desktop amps too that could work with the Beyers.
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pdq
post Sep 5 2012, 14:29
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Some amplifiers might have problems driving 32 ohms, but I doubt there are any that have difficulty driving 250 ohms.

Are you sure this model is similar to the others other than impedance? The low price sounds suspicious.
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dhromed
post Sep 5 2012, 15:25
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QUOTE (pdq @ Sep 5 2012, 15:29) *
Some amplifiers might have problems driving 32 ohms, but I doubt there are any that have difficulty driving 250 ohms.


Shouldn't that be the other way around? 250 is a higher resistance than 32, requiring more power from the amplifier. Maybe I'm mixing up the sequence of events here.
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saratoga
post Sep 5 2012, 16:12
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QUOTE (dhromed @ Sep 5 2012, 10:25) *
QUOTE (pdq @ Sep 5 2012, 15:29) *
Some amplifiers might have problems driving 32 ohms, but I doubt there are any that have difficulty driving 250 ohms.


Shouldn't that be the other way around?


No, its correct.

QUOTE (dhromed @ Sep 5 2012, 10:25) *
250 is a higher resistance than 32, requiring more power from the amplifier. Maybe I'm mixing up the sequence of events here.


Higher resistance doesn't require more power. P=I^2*R. So if you increase the resistance, less current is required from the amp at a given power, and hence the amplifier likely performs better.
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godrick
post Sep 5 2012, 16:40
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QUOTE
Higher resistance doesn't require more power. P=I^2*R. So if you increase the resistance, less current is required from the amp at a given power, and hence the amplifier likely performs better.


For my clarity, in the higher resistance situation, the assumption that the amp may perform better would be because one assumes that it is easier for the amp to produce the higher voltage required to produce a sound of a certain loudness than it is for the amp to produce the higher current required in the lower resistance situation - correct? And across both cases, we'd need to assume that we are talking about speakers of the same efficiency so that the equal power assumption makes sense, correct?

In other words, for speakers of the same efficiency and the same target power output, a higher resistance speaker requires relatively higher max voltage and lower max current capability from the amp, and a lower resistance speaker requires relatively lower max voltage and higher max current capability from the amp. If I'm missing something or stated something inaccurately, please correct me.

I think it used to be that bipolar junction transistors (BJTs) used for amps were better at producing high current compared to metal oxide semiconductor field effect transistors (MOSFETs), and MOSFETs were better at higher voltage operation than BJTs, but that situation may very well have changed in the last decade with integrated circuit advances, and the actual circuit design significantly affects voltage and current limits. I used to use graphs of safe operating areas to determine the combined effects of all of that and see if my amp would work well with my load, so I wonder if such a graph is available for the amps in question.

This post has been edited by godrick: Sep 5 2012, 16:51
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xnor
post Sep 5 2012, 17:03
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Especially with receiver headphone jacks you're better off with a higher impedance headphone. The reason for this is that those jacks usually have a high output impedance which will act as a voltage divider. A dynamic headphone driver typically has a higher impedance at the resonant frequency and a lower impedance in the midrange to treble. Therefore, the voltage will be divided unequally (=> bass boost, "loose" bass). This effect becomes smaller if the headphones' impedance is high compared to the output impedance.
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pdq
post Sep 5 2012, 17:15
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There is another factor to consider. Very often the headphone amplifier uses a single-ended power supply, and because of this it needs a blocking capacitor in the output to remove the DC component and pass just the AC.

A capacitor of sufficient size to pass very low frequencies can be large/bulky/expensive, so often the manufacturer will cut corners and use a capacitor that is large enough for high-impedance headphones, but has low-frequency roll off for very low impedance headphones, such as 32 ohms.
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Porcus
post Sep 6 2012, 09:01
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QUOTE (saratoga @ Sep 5 2012, 17:12) *
QUOTE (dhromed @ Sep 5 2012, 10:25) *
250 is a higher resistance than 32, requiring more power from the amplifier. Maybe I'm mixing up the sequence of events here.


Higher resistance doesn't require more power. P=I^2*R. So if you increase the resistance, less current is required from the amp at a given power, and hence the amplifier likely performs better.


To supplement:

P = U^2/R as well, so it requires more voltage. In principle, the issue would be whether the current or the voltage is the limiting factor. In practice, there are other issues; the headphone outlet of an amp runs the speaker outlet through a resistor. Now impedance isn't constant in frequency. And then there's pdq's cap argument.

In practice the max voltage isn't an issue except some portable devices, which cap the output voltage in order to save the listeners' ears (and themselves from lawsuits ... and I can only assume that some brands are happy to use any excuse to cause trouble for users who want to connect 3rd party earphones).


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Arnold B. Kruege...
post Sep 6 2012, 16:17
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QUOTE (saratoga @ Sep 5 2012, 11:12) *
Higher resistance doesn't require more power. P=I^2*R. So if you increase the resistance, less current is required from the amp at a given power, and hence the amplifier likely performs better.


The exposure that you pick up is that high impedance phones are more likely to require more voltage for a given level of loudness.

250 ohm phones may require up to as bit more 3 times more voltage than 25 ohm phones for a given SPL at the ear.

Many portable music players are voltage-limited, so this can be a real issue.
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pdq
post Sep 6 2012, 16:38
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Lighter loading means the output can be driven a little closer to the supply voltages, so this partially compensates for the higher voltage needed for equal loudness with higher headphone impedance, but this effect is probably small.
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Arnold B. Kruege...
post Sep 9 2012, 04:13
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QUOTE (pdq @ Sep 6 2012, 11:38) *
Lighter loading means the output can be driven a little closer to the supply voltages, so this partially compensates for the higher voltage needed for equal loudness with higher headphone impedance, but this effect is probably small.


I would expect a small and ineffective benefit from a well-designed active-driven headphone amp. Typically their source impedances are on the same order or even much less than low impedance headphones.

OTOH, the passive (series resistor) headphone jacks on some receivers and integrated amps might come close to providing enough extra voltage due to reduced loading to do some good. Their impdeance is typcially much, much more than low impedance phones and may even come close to matching high impedance phones.
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