[TStewart422]

So, using the spreadsheet and spec sheet of an amp; how would you determine if the amp is up to properly driving a particular headphone?

 

[sanderx]

Quote:

Originally Posted by TStewart422 /img/forum/go_quote.gif So, using the spreadsheet and spec sheet of an amp; how would you determine if the amp is up to properly driving a particular headphone?

If it lists the output power for the given headphone impedance in the spec sheet, you just look in the spreadsheet and see if where it lies wrt the headphone's requirement. If it lists power output at a number of impedances, look at the closest one and then apply I=sqrt(P/R) && E=I*R and look in the correct columns of the spreadsheet (higher is better).

If it only lists an ambiguous max output power rating you are stuffed

You will also need to know the amplifier gain and the line out voltage level of your source, when multiplied, these should give a value somewhat higher than the Vpp for the headphone at highest volume (110db is a good highest volume here).

 

[plonter]

HI guys,and thank you for the helpful comments, but i think that i didn't quite get it yet. sorry..but it's a little hard for me to understand this...
let's put the sensetivity issue asside for a moment, let's say that all headphones have the same sensetivity...than why a higher impedance phones should be EASIER to drive?? it just doesn't make sence to me.

I always thought that: the higher the impedance, the more power you need to drive it. and I know it's true in some cases (many ask about certain amps ability to drive the senn hd650/600 or 800), but in this thread i hear differently. please explain to me like i am a 5 years old..

 

[TStewart422]

Quote:

Originally Posted by sanderx /img/forum/go_quote.gif If it lists the output power for the given headphone impedance in the spec sheet, you just look in the spreadsheet and see if where it lies wrt the headphone's requirement. If it lists power output at a number of impedances, look at the closest one and then apply I=sqrt(P/R) && E=I*R and look in the correct columns of the spreadsheet (higher is better). If it only lists an ambiguous max output power rating you are stuffed You will also need to know the amplifier gain and the line out voltage level of your source, when multiplied, these should give a value somewhat higher than the Vpp for the headphone at highest volume (110db is a good highest volume here).

According to these calculations, the Compass can completely drive the HD650s, k701s, the SR225is, and the DT880s with no issues. I may have to make that space on the desk after all...

 

[Steve Eddy]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif HI guys,and thank you for the helpful comments, but i think that i didn't quite get it yet. sorry..but it's a little hard for me to understand this... let's put the sensetivity issue asside for a moment, let's say that all headphones have the same sensetivity...than why a higher impedance phones should be EASIER to drive?? it just doesn't make sence to me. I always thought that: the higher the impedance, the more power you need to drive it. and I know it's true in some cases (many ask about certain amps ability to drive the senn hd650/600 or 800), but in this thread i hear differently. please explain to me like i am a 5 years old..

Let me take a stab at it.

A higher impedance is easier to drive because it demands less power from the amplifier than a lower impedance.

An amplifier is designed to output a given voltage for a given voltage input to it, based on how much gain the amplifier has.

For example, if an amplifier has a gain of 10, if you put 1 volt into it, it will output 10 volts.

How much current it is asked to deliver depends on the impedance of the load that it's driving.

Ohm's Law for current is I = E/Z where I is current, E is voltage, and Z is the impedance.

So let's take our amplifier that's outputting 10 volts as an example.

If we connect an 8 ohm load to its output, then the amount of current that load will want to draw is 10/8, or 1.25 amps.

If we halve the load impedance and instead connect a 4 ohm load to its output, that load will want to draw 10/4, or 2.5 amps. Twice as much current as the 8 ohm load.

Ohm's Law for power is P = I x E where P is power, I is current and E is voltage.

So for the case of the 8 ohm load where it's wanting to draw 1.25 amps from the amplifier, the power that the load is demanding from the amplifier is 10 x 1.25, or 12.5 watts.

For the 4 ohm load, it is 10 x 2.5 or 25 watts. Twice the power.

In other words, as the impedance decreases, it demands more power from the amplifier, not less, and this is what makes lower impedance loads "harder" to drive and higher impedance loads "easier" to drive.

Rather like lifting a 100 pound weight is harder and requires more "power" than lifting a 50 pound weight. And this is why higher impedance loads are sometimes referred to as "light loads" and lower impedance loads as "heavy" loads.

Does this help?

k

 

[plonter]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif Let me take a stab at it. A higher impedance is easier to drive because it demands less power from the amplifier than a lower impedance. An amplifier is designed to output a given voltage for a given voltage input to it, based on how much gain the amplifier has. For example, if an amplifier has a gain of 10, if you put 1 volt into it, it will output 10 volts. How much current it is asked to deliver depends on the impedance of the load that it's driving. Ohm's Law for current is I = E/Z where I is current, E is voltage, and Z is the impedance. So let's take our amplifier that's outputting 10 volts as an example. If we connect an 8 ohm load to its output, then the amount of current that load will want to draw is 10/8, or 1.25 amps. If we halve the load impedance and instead connect a 4 ohm load to its output, that load will want to draw 10/4, or 2.5 amps. Twice as much current as the 8 ohm load. Ohm's Law for power is P = I x E where P is power, I is current and E is voltage. So for the case of the 8 ohm load where it's wanting to draw 1.25 amps from the amplifier, the power that the load is demanding from the amplifier is 10 x 1.25, or 12.5 watts. For the 4 ohm load, it is 10 x 2.5 or 25 watts. Twice the power. In other words, as the impedance decreases, it demands more power from the amplifier, not less, and this is what makes lower impedance loads "harder" to drive and higher impedance loads "easier" to drive. Rather like lifting a 100 pound weight is harder and requires more "power" than lifting a 50 pound weight. And this is why higher impedance loads are sometimes referred to as "light loads" and lower impedance loads as "heavy" loads. Does this help? k

thanks a lot for the detailed explenation mate! i can't argument with the formulas (I = E/Z and P = I x E) ...and i understand what they are saying, but it doesn't make sence to me logically. i mean...how can it be that a high impedance headphone is driven more easely than a low impedance one? but i guess the formulas don't lie!

I my language (hebrew) impedance means "resistance". that means that the higher the impedance the higher is the headphones/speaker's resistance.
and by what these formulas are saying, this term is just the other way around...
what's that confuses me the most is that sometimes people taliking on high impedance cans as hard to drive (most of the times actually) and sometimes as easy to drive...why is that?
if LOW impedance cans are the harder to drive...than it should always be like that, isn't it? than my RS1 is hard to drive even though i can drive them with my ipod?
am i missing something here?

 

[sanderx]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif I always thought that: the higher the impedance, the more power you need to drive it. and I know it's true in some cases (many ask about certain amps ability to drive the senn hd650/600 or 800), but in this thread i hear differently. please explain to me like i am a 5 years old..

Well, you can input 300Ohm, 97db into the spreadsheet for HD600 and 300Ohm, 103db into the spreadsheet for HD650. That will tell you the voltage and current requirements.

It won't tell you how much the amplifier must cost for some head-fi'ers to accept it as sufficent for driving the phones or if you should select tubes over solid state.

 

[sanderx]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif I my language (hebrew) impedance means "resistance". that means that the higher the impedance the higher is the headphones/speaker's resistance. and by what these formulas are saying, this term is just the other way around... what's that confuses me the most is that sometimes people taliking on high impedance cans as hard to drive (most of the times actually) and sometimes as easy to drive...why is that? if LOW impedance cans are the harder to drive...than it should always be like that, isn't it? than my RS1 is hard to drive even though i can drive them with my ipod? am i missing something here?

You really can't talk about the drive requirements based on impedance only. You must also consider the efficency, see above in the thread for an example with K701.

 

[Steve Eddy]

Quote:

Originally Posted by sanderx /img/forum/go_quote.gif You really can't talk about the drive requirements based on impedance only. You must also consider the efficency, see above in the thread for an example with K701.

I think he's only wanting to set efficiency aside in order to try and keep things simple and less confusing in order to get a better understanding of something more fundamental.

Certainly efficiency plays a critical role when you're talking about specific headphones. But what he's trying to reconcile is something more basic, i.e. how the notion of higher resistance translates into something that's easier to drive.

I think the sticking point for him is that he's looking at "resistance" as something which needs to be "overcome," which is understandable given how "resistance" is more generally thought of outside of electronics.

So the way he's looking at it, something which presents a greater resistance demands a greater effort to overcome it, whereas less resistance requires less effort to overcome it. For example a nut which has been cinched down very tightly presents a greater "resistance" to being unscrewed and requires more effort to unscrew it than a nut that wasn't cinched down so tight.

That's the fundamental hurdle he's needing to overcome and a headphone's efficiency hasn't anything to do with that. That's a finer point that can be discussed only once he's made it over this more fundamental hurdle.

k

 

[Steve Eddy]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif thanks a lot for the detailed explenation mate! i can't argument with the formulas (I = E/Z and P = I x E) ...and i understand what they are saying, but it doesn't make sence to me logically. i mean...how can it be that a high impedance headphone is driven more easely than a low impedance one? but i guess the formulas don't lie! I my language (hebrew) impedance means "resistance". that means that the higher the impedance the higher is the headphones/speaker's resistance. and by what these formulas are saying, this term is just the other way around...

Fundamentally "impedance" and "resistance" mean the same thing. In electronics, "impedance" and "resistance" are used to distinguish between AC conditions and DC conditions respectively, or where a circuit includes inductances and capacitances instead of just resistances.

But for our discussion, you can think of impedance and resistance as synonymous.

And as I explained to sanderx, it seems that the reason you're having troubles with this is that you're looking at "resistance" as something which must be overcome, and that if something presents a higher resistance, it requires more effort to overcome it and as a consequence you're looking at high impedance headphones as being harder to drive rather than easier to drive.

But the impedance or resistance is a resistance to current flow and when combined with voltage, dictates how much power is demanded by the load. So all else being equal, less current means less power demanded and an easier load to drive, whereas more current means more power and a harder load to drive.

Quote:

what's that confuses me the most is that sometimes people taliking on high impedance cans as hard to drive (most of the times actually) and sometimes as easy to drive...why is that?

That's where a headphone's efficiency comes in.

The headphone's impedance isn't inherently a gauge of how efficient it is. Two headphones may have the same impedance, but one may be more efficient than the other.

What this means is that the more efficient headphone requires less power, and hence easier to drive, than a less efficient headphone that requires more power and hence harder to drive.

It all depends on the specific headphone, and not necessarily its impedance.

Quote:

if LOW impedance cans are the harder to drive...than it should always be like that, isn't it?

Not when you consider efficiency.

If a low impedance headphone is efficient enough, you can get the same level of output from it with less power, than you can from a less efficient headphone with a higher impedance. When the amplifier has to deliver less power, it's easier for it to drive the load.

Quote:

than my RS1 is hard to drive even though i can drive them with my ipod? am i missing something here?

Yes. You're missing the efficiency half of the equation.

k

 

[sanderx]

Maybe an analogy to understand the "hard to drive" and "easy to drive" vs high/low impedance thing. Consider a heater and the cord connecting it to the wall socket. The heater has high impedance and is very easy to drive, it gives off heat very easily. The cord that connects it to the wall socket is very low impedance and is ridiculously hard to drive - you need enormous amounts of power before it starts to give off even a fraction of the heat of the heater.

The same applies to headphones, except that the difference in the impedance of the heater and the cord is "slightly" larger than the impedance difference between say 16 ohm headphones and 600ohm headphones, but the principle is the same.

 

[plonter]

thanks you koyaan for the very well worded explanation, i think i got it.
it's always bothered me for not getting to the buttom of this, and now i understand it more.
headphones sensetivity is not something new to me, but i always thought that a headphone with a high impedance will be hard to drive no matter of how much is the sensetivity factor of it. I guess the sensitivity has more influence than i thought.


just one more question than...what headphone will be harder to drive?:
A. 300ohm and 117db
B. 60 ohm and 106 db

what have more influence here...the impedance or the sensetivity?

 

[Steve Eddy]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif thanks you koyaan for the very well worded explanation, i think i got it. it's always bothered me for not getting to the buttom of this, and now i understand it more.

You're welcome. Glad it helped.

Quote:

headphones sensetivity is not something new to me, but i always thought that a headphone with a high impedance will be hard to drive no matter of how much is the sensetivity factor of it. I guess the sensitivity has more influence than i thought.

Well, it certainly CAN have enough influence to turn the tables, which rules out "no matter how much the sensitivity factor" is.

Quote:

just one more question than...what headphone will be harder to drive?: A. 300ohm and 117db B. 60 ohm and 106 db what have more influence here...the impedance or the sensetivity?

Well, if the sensitivity figures are given as dB per a particular power level instead of a voltage level, then the one with the higher sensitivity rating will be the easier to drive as a milliwatt is a milliwatt.

That's why loudspeaker sensitivity ratings can be confusing.

A loudspeaker manufacturer might rate the sensitivity of their speaker referenced to 2.83 volts, which corresponds to 1 watt into 8 ohms, which is fine if the speaker's an 8 ohm speaker. But if the speaker is a 4 ohm speaker, then it would take 2 watts to achieve the same SPL output.

It's best to give sensitivity ratings in decibel milliwatts (dBm) for headphones and decibel watts (dBW) for loudspeakers. That way you can make direct comparisons between models and see how far a milliwatt or a watt will get you.

k

 

[plonter]

thanks. and if you mentioned speakers, i was always confused when i saw the low impedance rating of speakers... (8 or 4 ohm) it was like: how could it be that a speaker has so little impedance and need much more powerfull amp than headphones do?...now i kind of got it.

thanks again.

 

[Steve Eddy]

Quote:

Originally Posted by plonter /img/forum/go_quote.gif thanks. and if you mentioned speakers, i was always confused when i saw the low impedance rating of speakers... (8 or 4 ohm) it was like: how could it be that a speaker has so little impedance and need much more powerfull amp than headphones do?...now i kind of got it.

Quote:

thanks again.

You're welcome.

k