headfones and amp requirements ?

[quadklipsh]

guys ,

im using a fostex t20 rp headfones, they are awesome .

very nice sound.

they do require an amp. as i feel some shyness of loudness and punch when playing it with a laptop or the sound card audigy2 .

though 96 db isnt bad but i now feel they should be used with an amp.

so i used an old marantz 2230 .it has headfones socket on the front .and a sweet sound .

i wanna know if the wattage of the amp for loudspeakers is usually known (30 watts in this amps case).

if i used a 100 watts amp or a 300 watts amp, for headfones , would there be any difference ???

do i need a bigger amp to go louder with these headfones.?????

i do know that the headfone socket is barely giving only a feww hundred milliwatts /milli volts ,to go to ear blowing levels through a 100 db headfones , generally used by youngsters and gamers .

i use it for music purely.

i remember the guy who was selling it to me had me auditioned it on the yamaha 130 watts amp. which he said was notorious for its sonic characteristics and harshness. and cranked up the volume to pretty loud level. and these fostese handled that pretty well, without discomfort to my ears .

[quadklipsh]

looks like no ones in the headfone usa here . i must have posted in the wrong place

[quadklipsh]

correction:

looks like no ones in the headfone usage here . i must have posted in the wrong place

[moray james]

try here

http://community.klipsch.com/forums/81.aspx

[WMcD]

I have not tried such "can" type earphones with a computer. The exception is my noise cancelling Sennheiser which have a built in amp.

I see from the Internet that your units are rated with a 50 ohm input impedance. This should not be too much of a burden for computers (8 ohms would be). OTOH, I doubt that all computer output cards have much available power. Maybe enough for ear buds, but not big drivers like you have.

Therefore it makes sense that you're finding better performance with a traditional amp. Given that these are pro units, they may be expecting an amp with low output impedance, as ordinary hi-fi amps have. Computer cards don't IIRC The bottom line is that there are technical reasons why performance with the computer is okay, but somewhat mediocre.

I would not suggest buying a very powerful amp just for the headphones. 30 watts is more than enough as long as it doesn't have any noise or hum. You're looking for a couple of clean watts, and not more.

Wm McD

[quadklipsh]

thanks william .

i read a couple of topics related to the ohm thing. i dont get what they say .

i mean

heck !

1) which is a difficult load . 500 ohm headfone or a 50 ohm headfone .

what i understand from the loudspeaker world is that the 8 ohm rated speakers are far easier loads than 3.2 ohm speakers .

is it the same for headfones too?

2) ..if the ohm ratings are the same for two HPs , will there be difference of sensitivity in both ?

thanks again if you can explain in english plz . im a doctor of medicine not electronics . lolz .

and whats the role of sensitivity rating of a given headfone?

[Guest " "]

You have to factor the circut design used in the headphone jacks. Some are fairly simple with just a dropping resistor. Others us a transformer. Others use a seperate buffer circut. So basically, the power out put of an amp has very little to do with the power output of a headphone jack. A 100watt amp could have a 600ohm resistor at the headphone jack, a 10 watt amp could have a 60ohm resistor at the jack. Thats why folks buy headphone amps, they want control and predictability. A headphone converter box that connects to yor amp speaker connections would give you control, but based on the output of different amps, the results will vary. Get one that has an autoformer in it instead of dropping resistors.

[WMcD]

Regarding the load thing.

For simplicity, we will have to speak about the load in ohms and assume it is resistive. Going into the closely related concept of impedance is going to make things more complicated than needed here.

Resistance is a quality of a load. Not quality as in high quality or low quality. Just something that descibes somethings. Note the plural.

Resistance is always a ratio. It is the ratio of Voltage applied to the resistance divided by the current through the resistance. Resistive loads of given ratio R always have the same ratio of Voltage over (divided by) Current. We use the units of electomotive force or potential in Volts (named after the Italian scientist Volta). We use the the units of current intensity (I) named after the French scientist Ampere. And the ratio is named after Mr. Ohm, who was German. Often we see resistance in Ohms indicated by an omega.

So what does this mean? Suppose we have any device characterized as a resistance of 10 ohms. It means that if 10 volts is applied across the terminals the current through the device will be 1 ampere. Always, always, always. R = V/I. If 100 volts is applied to the unit, 10 amps will flow through.

Basically, if you have low R, the device is drawing more current. relative to applied voltage. If we had a 1 ohm resistor, the current flow is the same numerical value as the applied voltage. If we had a 100 ohm resistance the current will be 1/100 ampere for every volt..

As you can see, the lower the resistance, the more current flow. Very generally speaking, amplifiers can deliver only so many volts and so much current. So we can generally observe that lower resistances draw more current and are a tougher load.

We see math expressions for this ratio. But it applies in real time too.

Suppose we capture a graph of a music signal over time as applied to an R of 10 ohms. The voltage graph will look varied over time. Suppose it looks like a graph of mountains and valleys in Wyoming. The current will look the same, except 1/10th as tall. this is the ratio again. At high voltage peaks, there are high current peaks.

So what is the problem?

The problem is that amps and other devices supplying power to a load are generally limited by the current they can provide. Otherwise we could use a load of 0.00000001 ohm. Since power = V x I. we could draw enough power to light up all of Pittsburgh. Of course that can't happen.

In the case of computer cards, there is a limit to the power (or any amp). This is in terms of volts it can apply to a load, and current it can supply. When amps run out of current, the voltage output sags too.

A high power amp is like a fully charged automobile battery. You can turn on the dome light before starting the car. These have a high resistance and don't draw much current. Then you lower the resistance overall applied by cranking the starter (switching in a low resistance. The lights may dim slightly, but the high current starter works.

A low power amp is like a weak automobile battery. The dome lights work until you try to crank the engine. Then there is not enough current for the starter and the dome light dims because voltage output sags.

In both these cases, Ohms law for the Rs (here the starter motor and the dome light) still applies. But we see that the weak battery (small amp) can not supply supply current or voltage to the loads when there is a high current demand.

To belabor the point. The low resistance starter motor is like a 10 ohm headphone or speaker. The dome light is like a 100 ohm headphone. If the amp (battery) is weak, it can supply the dome light to full brightness, but not the starter, too.

There is more complicated explanation of how to model a small amp and weak battery. It is that they have a high internal resistance in the circuit and even that fails to account for other issues.

You ask about sensitivity. From memory headphone acoustic output is measured with a dummy head with a microphone where the the human ear is located. Essentially a given amount of electical power is applied to the headphones and the sound pressure is measured at the ear microphone. This also creates another issue of how electrical power is measured. But, here we assume a given amount of electrical power (maybe 1 milliwatt) is delivered to the headphone. Then the acoustic pressure is measured.

These measurements give some impressive numbers compared to speakers. A speaker being driven with 1 elecrical watt will give, say, 80 dB of sound pressure level to a listener (or microphone) at 1 meter distance. But 1 milliwatt of electrical power (1/1000 watt) to a headphone can give the same 80 dB to the ear canal. Look closely at any headphone spec and loudspeaker spec.

I don't know if you are puzzled by the sound pressure level. With sound we are dealing with sources, like speaker diaphragms, which move back and forth. The resulting air pressure is going plus and minus. The ear drum is moving back and forth too in response to air pressure. The long term average is zero. But over small time periods,there is real work and therefore energy transfer.

Wm McD

[quadklipsh]

[]wm mc dermott ,

thanks aplenty for such an excellent account .

sorry i cudnt attend to this post earlier perhaps i forgot.

dont mind my asking,how did you learn all this .

[garyrc]

To William F. Gil McDermott
:

I love your posts!

Here is something I've always wondered about, and you seem like the one to ask: some years ago, some loudspeakers were advertised as having very little variation in impedance across the frequency range. This was touted as being a very good thing. Other speakers are all over the place impedance wise (including the Klipschorn). At least one McIntosh dealer at the time said that a big rise in impedance within certain frequency bands could cause power to poop out at those frequencies, and, in addition, a related issue was that many speakers that had been rated at 16 Ohm nominal impedance were re-rated at 8 Ohms when solid state came in, without change in the speaker design, since they now needed to be connected to the only output most solid state amps had -- 8 Ohms. The McIntosh amps somehow avoided that problem. Just down the street, a Klipsch dealer said that the speakers were at their most efficient at the points of their impedance peaks, so there was no problem.

What do you make of this?

[WMcD]

Thank you for the kind words.

It is difficult to address the comments of others -- we can't ask them why or what their statements are based upon. We also can not gauge the level of their technical knowledge.

One starting point is that speakers are designed and tested using amps which have a constant voltage output across the frequency spectrum -- and this constant voltage at all loads which the speaker presents at those freqs. If you have an amp and speaker hooked up to it, it is fairly easy to measure this with the proper equipment.

It is true that amps without feedback tend to be sensitive to the load and have an output which sags when impedance is low. These are the single ended triodes, which some people say have better sound.

In this case, a speaker with constant impedance versus freq may work better. Actually, I suspect people are just looking for a flat speaker impedance because flat is better, in their mind.

I believe that this flat impedance is achieved by the use of component in the crossover which presents a load in addition to the drivers. But this is a complicated situation. It is similar to when "Zobel" networks are used so that a given driver presents a constant impedance to a crossover output. The crossover output is sensitive to the load.

One thing to discuss is that the acoustic performance of given driver actually becomes more flat because its impedance is uneven. The prime example is a woofer in a sealed box.

At the mechanical resonance frequency the diaphragm is free to move and undergoes greater excursion. This means it has more acoustic output. because it moves farther. However, its electrical impedance is high at that freq. We see that in the curves of electrical impedance. The hump identifies the resonance point. OTOH, the higher electrical impedance means that the input terminals to the woofer driver is absorbing less power from the amp than at other freqs above or below.

As you can imagine, this is a type of equalizing the power absorbed by the woofer driver at resonance. Actually, this mechanical and acoustic resonance hump (somewhat broad) of displacement is used to equalize the bass response.

Sorry, that is going far afield. Suffice to say, variations in electrical impedance and power absorption are "good" in the proper situations.

- - -

I can't say what was going on, back in the day, with nominal speaker impedance and transistor amps, or any amps. I can only think that people were use to having taps on output transformers on tube amps. With the exception of the Mac autotransformer transistor amps, these all went away with transistor amps. People were use to having numbers to match. The popular technical press, and advertizing departments, did not having the engineering smarts to realize what was going on in the first place.

The comment by the Mac dealer was IMHO just more of the lack of knowledge. It is true that as impedance rises (generally) the amp is not delivering as much current, and therefore power, to the speaker. But the speaker designer and tester knew that.

What is more true is the following: If speaker impedance goes lower and lower, the amp might run out of capability to deliver current and applied voltage sags.

Look again about what I wrote above about the weak car battery. .So to that extent the amp "poops out" and delivers less than required voltage and therefore power.

But, when speaker impedance goes up, the load is asking for less current. Now the voltage output of the amp IS what it should be and does not sag as much, if at all. The amp is not "pooping out." It is happy. It is true that power output is less but it is because the speaker is not demanding as much current.

Let's tie that into PWK's autotransformer use. In his designs, because of the autotransformer, the input impedance at the speaker terminals in the range where the midrange is working is high, up to 70 ohms for so. Therefore there is less current being drawn from the amp. The power delivered is lowered. But that is a designed-in feature because the midrange horn is more efficient and it's electrical consumption of power must be reduced. This may lead to lower distortion from the amp.

WMcD

[garyrc]

Thanks, WMcD! I will try to absorb that.