Impedance and current. Technical question.
Mar 22, 2006 at 1:33 AM Thread Starter Post #1 of 8

slinger1182

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I read this this thread following the link Jahn linked in this thread. Nomad in the first thread said that Earmax original is good for high impedance headphones but not low impedance headphones.

Well from my high school Electrical Engg course I remember that:

i = V/R
So it seems to me that to get enough current to drive a headphone, you need a high voltage or low impedance (or both). So this stumps me that if Earmax original can swing enough volts to drive high impedance headphones, why should it not preform as well for low impedance headphones?

Confused. Can someone clear it?
 
Mar 22, 2006 at 2:08 AM Post #2 of 8

Blitzzz

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Quote:

Originally Posted by slinger1182
I read this this thread following the link Jahn linked in this thread. Nomad in the first thread said that Earmax original is good for high impedance headphones but not low impedance headphones.

Well from my high school Electrical Engg course I remember that:

i = V/R
So it seems to me that to get enough current to drive a headphone, you need a high voltage or low impedance (or both). So this stumps me that if Earmax original can swing enough volts to drive high impedance headphones, why should it not preform as well for low impedance headphones?

Confused. Can someone clear it?



You have to remember that V and R are given and fixed. For example, if your amp runs off 10V wallwart then it will supply 10V, and if your headphones are 100ohm then R = 100 and you can't change either of these. Therefore it all depends on if the amp is able to supply the amount of current needed.

Using the above example, you would need i=10/100=0.1A to drive the headphones. As R decreases (lower impedance), i will increase (they are inversely proportional) to a point where the amp is no longer able to support that much current.
 
Mar 22, 2006 at 3:50 AM Post #3 of 8

slinger1182

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Quote:

Originally Posted by Blitzzz
Using the above example, you would need i=10/100=0.1A to drive the headphones. As R decreases (lower impedance), i will increase (they are inversely proportional) to a point where the amp is no longer able to support that much current.


Aah, ok, but the doubt still remains.

Suppose the amp has a current limit of X amperes. Now if this X amperes still drives a headphone of higher impedance effectively, it should still drive a headphone of lower impedance as effectively, shouldn't it?

Or am I missing some other factors here? Sorry, but I am a noob afterall.
 
Mar 22, 2006 at 4:06 AM Post #4 of 8
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Well let's take a look at this issue. I will compare the HD650 and Grado RS1 (though specs are the same for all Grados), because though both have the same efficiency at 98dB/mW, the former is high impedance (300), the latter low (32).

To play at a sound level of 98dB, both headphones need 1mW of power. However, due to the differing impedances, the require different voltage and amperage:

HD650: 0.56V, 1.87mA
RS1: 0.18V, 5.59mA

Now let's say, purely for speculation's sake, that you have an amp that can output 1V and 4mA. In this situation, both headphones would be able to have enough voltage to reach 98dB (and beyond), however, due to the amperage limit, the Grado would not be getting enough current, however the Senn would. I believe the result of this would be clipping (someone please correct if I'm wrong) on the Grado.

Reverse the situation, say your amp can output 0.4V and 10mA. Now the Grado has plenty of voltage and current to reach 98dB, but the Sennheiser doesn't have enough voltage, so it is limited to the volume produced at 0.4V. As such, there is no clipping or distortion, the headphone just won't play as loud as you want.

The first situation is closer to a tube amp, the latter closer to a solid-state amp. Of course ANY decent amp should be able to drive these headphones well beyond 98dB, I just used that figure for simplicity's sake.
 
Mar 22, 2006 at 4:08 AM Post #5 of 8

kin0kin

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I suggest you bring the question somewhere else, specifically the DIY community/headwize.

Given that a 32ohm impedance headphone (Grado), in order to determine the current that the phone will draw, you will need to determine the voltage of your listening level. Say you listen at 2V, and given that R = 32 the amp will need to be able to deliver:

V = IR
2 = I X 32
I = 0.625A (62.5mA)

however, say your amp is a single stage cmoy, with a chip that can only deliver up to +/-40mA, then the chip does not have the capcbility to drive the phone properly.

High impedance phone is less current dependent, to illustrate this, assuming that you have the senn 650, @ 300ohm. In order to get the samue volume as the low imp phone, you may require to pump the volume to say 5V, so given V= 5, R = 300 you can get the current that the phone draws:

V = IR
5 = I X 300
I = 0.167A (16.7mA)

this is why high impedance phone doesnt require current as much as low impedance phone, and virtually can be driven by just a cmoy with high operating voltage.

please note that current is dependent on V and R, not the other way round. Current do now flow if there is no load or when the impedance is very very very high.
 
Mar 22, 2006 at 4:25 AM Post #6 of 8

slinger1182

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Got it! Thanks.
 
Mar 22, 2006 at 7:10 AM Post #7 of 8

Ferbose

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Many of the above discussions assume output impedance of the amp is zero ohms, which is not true for most amps, especially tube ones.
Looking at Earmax specs, it can deliver 100 mW per channel, which is certianly enough for any Grado headphone.
http://www.audioadvancements.com/headphones/earmax.html

I suspect Earmax has very high output impedance, due to its tube design.
If the headphone's impedance curve is not flat and varies a lot across different frequencies, then amps with high output impedances will create serious frequency response uneveness. Frequencies with impedance peaks will sound too soft if the amp's output impedance is high.

For speakers, this kind of problem is often described as having a low damping factor. For speakers, low damping factor generally creates weak bass because most spekaers have impedance peaks in the bass region. Google dampling factor and you will find very good explanations of this problem.

Amp output impedance is a serious issue in headphone amps because headphones have wildly different impedances (16-600 ohms). Most speakers today have 4 or 8 ohms (nominal) impedance.
 
Mar 23, 2006 at 1:26 AM Post #8 of 8

nikongod

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the ideas of voltage swing to drive the headphones only aply if the "amp" can suply the necessary current.

a tube can easuly swing near the rails in the audioband, but thats with miliamps of current. there is a VERY rigid limit on the current suplied. the "way around this" is tu use a transformer, but that is not the case with the amp mentioned above.

similarly (well the opposite) a ss amp can usually provide substantial current, but the voltage swing is often very limited.

in the case of a capacitor coupled (or direct coupled, if the case is such) tube amp, with no output buffer you are limited to the current that the tube can "suply." trying to get 5ma out of most tubes is like trying to get a product worthy of their generally perceived reputation out of the boze factory. it happens every now and then, but its the exception to the rule.
 

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