Thread of Basic Questions

[tomb]

I make that assumption because the maximum output voltage is almost always mentioned in the spec sheet of headphone amps. I have no doubt that the amps could reach the claimed voltage. Something that's not always mentioned is the impedance of the load (and the actual power drawn) when they measured the max output voltage. An amp might be able to output 10V into 1kOhms but not into 10Ohms because it lacks the power to do so.

The way I ballpark what amplifier a headphone needs is by looking at the sensitivity of the headphone which is typically in dB/mW (at 1kHz) and it's impedance (which is just an average value). I pick the desired SPL let's say 100dB SPL average is pretty loud already. Since most music is dynamic reaching an average of 100dB SPL requires to reach even higher values (let's just say 20dB headroom) depending on how dynamic the music really is. This would allow for a loud listening experience even with fairly dynamic music.

From the sensitivity, impedance and desired SPL I calculate the voltage the headphone needs to reach the required max SPL level. If the amp has a higher output voltage than I calculated there is already a chance the pairing will work (it will be loud enough for sure). So in this case one of the questions is can the amp actually output their claimed voltage into the headphone I want to use it with? The higher the impedance, the more likely it actually can. The question could be rephrased as well: I know the amp I'm looking at will have enough voltage to drive my headphones, the question is, will it have enough power to do so?

I understand that there are other requirements for the amps like low noise, low distortion but these usually aren't described in the spec sheets in a way it could be useful to me so after checking if the output voltage is enough I just bite the bullet, order the amp and hope I will like the sound.

Interesting. I've never really heard of anyone selecting headphone amps in this manner. Amplifiers are usually selected on the ability to deliver power at a specified impedance, with distortion and noise capabilities that are commendable.

Voltage is a very limiting factor, I would think in selecting an amplifier, because - that voltage will be different for loads at different impedances. If you state that the impedance of the load is very seldom cited for that specified maximum voltage, what have you achieved?

I'll try to use V instead of U. Hmm, didn't you miss that equation as well? If I wanted to visualize P = V x I (power equals voltage times current) in a simple V-I graph I would think that the power is the rectangle that can be drawn after plotting the (V;I) coordinate. I think the product of two simple numbers, V and I, can be most easily visualized by the area of a rectangle (so the area corresponds to power which equals to V x I) where the sides' length are V and I because the area is calculated by multplying the sides' length anyways. If the area (power) is the given number, there are of course an infinite pair of V-I values that would draw a rectangle with the correct area just like how a certain load can correspond to an infinite pair of V-I values depending on the load's resistance.

So I think we are on the same page just looking at it differently. However there's one thing I still don't understand.

JMO, but a rectangle is a poor way to represent a vector. Vectors have a magnitude and direction. The magnitude is power, the direction is the slope of the line on an X-Y graph with voltage and current as axes. If you want to draw location lines to represent the end of the vector where it crosses each axis (as opposed to the origin), then I guess you could say you have a rectangle, but two of your sides are superfluous to the physical representation.

I'm fairly sure headphones are meant to be driven with the same voltage gain across the frequencies, and definitely not the same power gain across the frequencies. Is this where we disagree?

Here again, is a limitation in understanding/agreeing on the definition of words. There is NO power gain. Gain is the measure of the amplification factor. If your amplifier is amplifying voltage, then it operates with voltage gain.

Power is the energy (over time) required for the load to respond to the amplified output. Given a voltage gain, the amplifier must also supply an amount of current to equal the power that the load requires to respond to that voltage gain. That ratio of voltage to current is dictated by the load's (headphone) impedance. The ability of the amplifier to supply that current, with the voltage gain that headphone needs, is based on the output impedance of the amplifier. If the impedance matching between the amplifier and the headphone do not result in an advantageous combination, the amplifier will fail to deliver the necessary current (assuming voltage gain).

If we don't disagree on that, then the idea of a headphone amp not being able to swing enough voltage at the frequencies where the impedance is high is still looking nonsense to me.

Sorry, but we disagree again on how you presented the scenario.

Why would it be harder for the amp to swing the same output voltage into a higher impedance compared to a lower one? Since it's actually easier to have a good output voltage into a higher impedance because that requires less power.

Here we go again ... "easier to have a good output voltage into a higher impedance because that requires less power."
NO, NO, NO. A higher impedance only represents a different combination of voltage and current is required to deliver the power the headphone requires.

If you have a 300 ohm Sennheiser that is rated at 98dB of 1mW input, it is exactly the same amount of power as a 32 ohm Grado rated at 98dB of 1mW input, PERIOD. Both examples are still at 1mW. The difference is the 300 ohm Sennheiser requires 0.55V and 1.8mA for that 1mW. The 32 ohm Grado, however, requires 0.17V and 5.5mA for 1mW.

Please don't go running off with those numbers and think a Sennheiser only needs an amp with a voltage swing of a half-volt. I did not look up the specifics on a Senn or Grado. Also, those voltages are RMS values and not peak-to-peak, but peak-to-peak are actually more important within the amplifier (MHO). You also burn up voltage in all the parts on the way from the power supply to the output. The same can be said of the current. dBs are also not consistent, because transient peaks may require many, many more dBs.

The point being, an amp does not swing the same output voltage into different impedances, PERIOD. Some amps are better at swinging voltage (tubes), some amps are better at pulsing current (solid-state). Since power at different impedances requires different combinations of voltage vs. current, an obvious corollary is that some amps are better at some impedances than others. Those are very general statements that can easily be obscured by altering details of the circuit, using transformers (hint: "TRANSFORM" is a magic word), etc., and we haven't even broached the subject of efficiency (sensitivity).

Try your own numbers on this little calculator here:
http://www.sengpielaudio.com/calculator-ohm.htm
(Don't get too confused about the diatribe that follows after that little calculator at top - or something called, "Power Factor." In actuality, that is what I've been trying to explain to you. You'll see they refer to something called the Power Factor Angle. Another way of stating it is a vector with magnitude and direction (angle). He also chooses to start a semantic argument on "RMS." Try not to be swayed on that one.)

I'm sorry to say that I have to do other things, now.

 

[sander99]

I'm fairly sure headphones are meant to be driven with the same voltage gain across the frequencies, and definitely not the same power gain across the frequencies.

Yes.

Since it's actually easier to have a good output voltage into a higher impedance because that requires less power.

Yes.

From the sensitivity, impedance and desired SPL I calculate the voltage the headphone needs to reach the required max SPL level. If the amp has a higher output voltage than I calculated there is already a chance the pairing will work (it will be loud enough for sure). So in this case one of the questions is can the amp actually output their claimed voltage into the headphone I want to use it with? The higher the impedance, the more likely it actually can.

Yes.

 

[tomb]

Yes.

Yes.

Yes.

Gah! More heresy!

How does a higher impedance require less power? If the impedance is high enough, does that mean it requires no power at all?

 

[sander99]

How does a higher impedance require less power? If the impedance is high enough, does that mean it requires no power at all?

It requires less current (hence less power) to keep the voltage (or voltage swing) up at a specific value over a higher impedance.
And indeed, if the impedance is infinite (the conduction is zero) then 0 ampere and 0 watts are needed to keep the voltage up. (But that has no practical meaning of course, you won't be able to create any sound with that.)

P = I x R, period! You cannot create or remove the amount of energy required for the headphone.

Moreover, the Senns have a pretty big impedance bump around 100Hz. On the HD800, the impedance goes from about 350 ohms at 10Hz, all the way to 650 ohms at 100Hz! It only averages 300 ohms from 20-20kHz. What does that mean? It means that all around the areas of mid-bass, where "warmth" generally comes from, power disappears with many solid-state amps, because they simply can't swing the voltage needed at that high impedance. A typical solid-state amp could be clipping at this point, but it depends on the gain. It might just sound bass-light or tinny.

If you have a 300 ohm Sennheiser that is rated at 98dB of 1mW input, it is exactly the same amount of power as a 32 ohm Grado rated at 98dB of 1mW input, PERIOD. Both examples are still at 1mW. The difference is the 300 ohm Sennheiser requires 0.55V and 1.8mA for that 1mW. The 32 ohm Grado, however, requires 0.17V and 5.5mA for 1mW.

The Sennheisers sensitivity of 98 dB of 1 mW input holds for one specific frequency (1kHz?).
If at another frequency (you mentioned 100Hz) the impedance is 650 ohm, that doesn't mean that we need 1 mW there to reach the same 98dB, or that we need 1 mW at all there.
At that frequency the same 0.55V will result in less current and less power compared to the situation at 1 kHz.
(And what SPL we reach depends on the frequency response. If for example the response is equal at both frequencies, then apparently the headphone's efficiency is higher at the frequency where the impedance is 650 ohm so with the same voltage and less current and less power we reach the same SPL.)
(If the frequency response has a dip at 100 Hz and we want to EQ it up, or in general if we want to use EQ, then we may need more voltage.)

 

[castleofargh]

Maybe let's look at it in a simpler form. We ask one thing from an amp, it's to be able to output a signal up to a given amplitude.
Obviously any amp will have limits in how much power it can output without turning into a distortion machine. Some hard limits come from the power supply, input signal, amp design(let's take a basic op amp, it's good up to a given voltage and then, enjoy clipping). After that, the load itself may decide whichever gives first between voltage and current, but each amp might have different limits.

At the same time, if we didn't purchase some headphone with ultra extreme specs, and if we did bother to read half a spec on the amp's datasheet, it's likely that all is well and that most listeners will achieve pretty loud volume level without putting the amp on its knees. Not all headphones need 15V, not all listener would use such a headphone at those levels anyway. Some do exist and some listeners do play awfully loud music. If those guys rely on purely subjective experiences, I wouldn't be surprised if they tend to look for a lot of power and avoid SS amps. Because of you do reach some limit for the amp, an SS amp will tell it to you in a very clear and audible way^_^.

How does a higher impedance require less power? If the impedance is high enough, does that mean it requires no power at all?

This as you said before, comes with assumptions. IF the voltage remains the same then P=U²/R and P goes down as R goes up. The reasoning is clear enough and they both insisted on same voltage at one point. But would the voltage remain the same when we change the load presented at the amp? If I plug different resistors into my amp, I get different voltage readings. In the headphone it's going to be the same thing, the impedance ratio between amp output and headphone at a given frequency will give us a voltage, when at another frequency the headphone has a different impedance, meaning a different impedance ratio that will affect the voltage, and as a result the FR.

Probably an issue somewhere when trying to juggle between the point of view of the load and that of the amp(or am I the one doing that?). or maybe just the fact that like me they tend to think voltage=loudness(because twice the voltage for a headphone should give a solid +6dB), but then forget that the sensi gives a relation between power and dB SPL, where the voltage=loudness stands only so long as the impedance is fixed. from the same P=U²/R, if P=1mW and R goes up, why would V stay the same?
QED.

 

[VNandor]

The reasoning is clear enough and they both insisted on same voltage at one point. But would the voltage remain the same when we change the load presented at the amp? If I plug different resistors into my amp, I get different voltage readings.

This is the point that should be adressed properly.
Before I go into ramble mode again are the resistors you tested have a "high enough" resistance? The point I was trying to make is that my line of reasoning only works if the resistance of the load is high enough. By high enough I mean something like 8 times more than the tested amp's output impedance. If so, I wouldn't expect the measured voltage change more than 10% but if it does, then that's where I am wrong.

If I simulate an amplifier circuit (with a low output impedance) the output voltage stays largely the same (for a given input and gain) for any resistor until their resistance becomes too low.

 

[castleofargh]

This is the point that should be adressed properly.
Before I go into ramble mode again are the resistors you tested have a "high enough" resistance? By high enough I mean something like 8 times more than the tested amp's output impedance. If so, I wouldn't expect the measured voltage change more than 10% but if it does, then that's where I am wrong.

If I simulate an amplifier circuit (with a low output impedance) the output voltage stays largely the same (for a given input and gain) for any resistor until their resistance becomes too low.

Oh, now I get it.
Sure, the variation in voltage should remain small, so long as we do follow an idea of impedance bridging.
But then there are some multidriver IEMs with stupidly low to high impedance. And also tube amps with high impedance, the type @tomb mentioned a few times to distinguish them from SS amps.
If the core disagreement is the amount of voltage change that can be disregarded, I think we'll find an understanding ^_^.

 

[VNandor]

I'll try to recap where I disagree with him but it looks like he is fed up with me anyways.

Moreover, the Senns have a pretty big impedance bump around 100Hz. On the HD800, the impedance goes from about 350 ohms at 10Hz, all the way to 650 ohms at 100Hz! It only averages 300 ohms from 20-20kHz. What does that mean? It means that all around the areas of mid-bass, where "warmth" generally comes from, power disappears with many solid-state amps, because they simply can't swing the voltage needed at that high impedance. A typical solid-state amp could be clipping at this point, but it depends on the gain. It might just sound bass-light or tinny.

He insists the reason the power "disappears" where the impedance is higher, is some shortcoming of the amplifier, specifically that it "can't swing the voltage needed at high impedance".

So I try to go back in his line of reasoning to the point we are still agreeing and then go from there because he is clearly wrong.
So far I think we agree on Ohm's law and that power equals current multiplied by voltage so that's good at least. I think we are disagreeing on what the job of an ideal voltage amplifier is, and that a headphone amp should work like an ideal voltage amplifier.

I will get the boring math stuff out of the way first.

JMO, but a rectangle is a poor way to represent a vector. Vectors have a magnitude and direction. The magnitude is power, the direction is the slope of the line on an X-Y graph with voltage and current as axes. If you want to draw location lines to represent the end of the vector where it crosses each axis (as opposed to the origin), then I guess you could say you have a rectangle, but two of your sides are superfluous to the physical representation.

A rectangle is indeed a poor way to represent a vector. I think a vector can be best represented with a line (or "arrow") that points from point A to point B. However I do think that a vector is a poor way of representing power in this context and an area of a rectangle is both more hands on and simpler. Calculating an area of a rectangle is elementary school level. Vectors are a more abstract concept and I believe its being tought at secondary school.


You keep saying that the magnitude of the vector would represent the power. Imagine a vector that represents 1V and 1A. 1Volt times 1 Ampere obviously equals to 1 Watt. What would be the magnitude of this 1V and 1A vector though? If you use Pythagoras you can find that the magnitude is square root of 2. If you do the calculations while keeping the units, at least the unit would still be correct at the end but the actual number is incorrect as you can see. The magnitude of this vector would never describe the power correctly (well, unless the power is 0).

Here we go again ... "easier to have a good output voltage into a higher impedance because that requires less power."
NO, NO, NO. A higher impedance only represents a different combination of voltage and current is required to deliver the power the headphone requires.

A higher impedance at the output of the amplifier represents a different load (different power drawn) for a voltage amplifier.

Consider the following and assume an ideal voltage amplifier:
First case:

• The input signal of the amp is a 1Vrms sine wave at 1kHz. The gain is 10. The impedance that is put to the amplifier's output varies with frequency. It happens to be 300ohms at 1kHz. What should be the output voltage and how much power this impedance draws?
• The input is 1Vrms, the gain is 10 the and since this is a voltage amplifier, this means the output must be 10Vrms. So what's the actual load (power drawn) in this case? It is 10V*10V/300ohm which equals to 333mW.

Second case:

• Use the same configuration as before but change the input frequency to 100Hz. The input signal of the amp is a 1Vrms sine wave but at 100Hz. The gain is still 10 (there are no changes to the volume knob). The impedance that is put to the amplifier's output varies with frequency. It happens to be 650ohms at this particular frequency. What should be the output boltage and how much power this impedance draws in this second case?
• The input is 1Vrms, the gain is still 10 (didn't change to volume knob) and since this is a voltage amplifier, the output must be 10Vrms. What's the power drawn in this case? It is 10V*10V/650ohm which equals to 154mW.

So you aren't correct to assume that different impedances present the same load (to voltage amplifiers). You can see that a higher impedance always draw less power from a voltage amplifier if you don't change the gain and the input signal for the different impedances.

Again, my point is that if the input amplitude stays the same and the voltage amplifier has to output X amount of volts at 1kHz into 300ohms, it also has to output the same X amount of volts at 100Hz where the impedance is 600ohms. Since the output voltage did not change while the impedance went up that means the power the amplifier outputs has to decrease to make sure the gain stays the same at both frequencies. If a voltage amplifier would output the same amount of power at 300ohm impedance and 650 ohm impedance (at the different frequencies) the voltage gain would not be the same accross the frequencies which would mean it's not functioning correctly as a voltage amplifier. I think we can agree on that?

 

[VNandor]

If you have a 300 ohm Sennheiser that is rated at 98dB of 1mW input, it is exactly the same amount of power as a 32 ohm Grado rated at 98dB of 1mW input, PERIOD. Both examples are still at 1mW. The difference is the 300 ohm Sennheiser requires 0.55V and 1.8mA for that 1mW. The 32 ohm Grado, however, requires 0.17V and 5.5mA for 1mW.

Let's go through this one as well.

In this imaginary scenario, connect the DAC to the input of the amp and the Sennheiser to the output of the amp. Let's use a 1kHz sine wave test signal with a 0.1Vrms. Set the gain to 10. That way, if the amp works as a proper voltage amplifier,it will swing 1V into the Sennheiser headphones. The 300ohm Sennheiser would draw 3.3mW in this case and the headphones would be around 103dB loud.

If you switched the Sennheiser with the Grado, and didn't change anything else, the amplifier would swing the same 1V into the Grado headphones if the amplifier worked correctly. The 32 ohm Grado would draw 31mW and it would be around 113dB loud.

Notice that the voltage swing did not change. The drawn power however did change, and because of that, their loudness aren't the same despite having the same sensitivity. If you wanted the headphones to have the same loudness you would have to change either the input signal's amplitude or the gain of the amplifier.

 

[tomb]

Maybe let's look at it in a simpler form. We ask one thing from an amp, it's to be able to output a signal up to a given amplitude.
Obviously any amp will have limits in how much power it can output without turning into a distortion machine. Some hard limits come from the power supply, input signal, amp design(let's take a basic op amp, it's good up to a given voltage and then, enjoy clipping). After that, the load itself may decide whichever gives first between voltage and current, but each amp might have different limits.

At the same time, if we didn't purchase some headphone with ultra extreme specs, and if we did bother to read half a spec on the amp's datasheet, it's likely that all is well and that most listeners will achieve pretty loud volume level without putting the amp on its knees. Not all headphones need 15V, not all listener would use such a headphone at those levels anyway. Some do exist and some listeners do play awfully loud music. If those guys rely on purely subjective experiences, I wouldn't be surprised if they tend to look for a lot of power and avoid SS amps. Because of you do reach some limit for the amp, an SS amp will tell it to you in a very clear and audible way^_^.


This as you said before, comes with assumptions. IF the voltage remains the same then P=U²/R and P goes down as R goes up. The reasoning is clear enough and they both insisted on same voltage at one point. But would the voltage remain the same when we change the load presented at the amp? If I plug different resistors into my amp, I get different voltage readings. In the headphone it's going to be the same thing, the impedance ratio between amp output and headphone at a given frequency will give us a voltage, when at another frequency the headphone has a different impedance, meaning a different impedance ratio that will affect the voltage, and as a result the FR.

Probably an issue somewhere when trying to juggle between the point of view of the load and that of the amp(or am I the one doing that?). or maybe just the fact that like me they tend to think voltage=loudness(because twice the voltage for a headphone should give a solid +6dB), but then forget that the sensi gives a relation between power and dB SPL, where the voltage=loudness stands only so long as the impedance is fixed. from the same P=U²/R, if P=1mW and R goes up, why would V stay the same?
QED.

I cry foul. Nowhere was that stipulated that I remember. If it was, then the entire conversation is stupid.

It's stupid, because the most simplest P = I x V equation will tell you that if one variable is constant and the other variable changes, then the Power will vary. Why the hell would it NOT?

The idea that I meant anything about keeping Voltage constant is absolutely ridiculous.

This is a silly game and is not accomplishing anything if fundamentals like that can be misconstrued. It's the same as saying I said 2 + 2 = 5.

 

[tomb]

I'll try to recap where I disagree with him but it looks like he is fed up with me anyways.


He insists the reason the power "disappears" where the impedance is higher, is some shortcoming of the amplifier, specifically that it "can't swing the voltage needed at high impedance".

So I try to go back in his line of reasoning to the point we are still agreeing and then go from there because he is clearly wrong.
So far I think we agree on Ohm's law and that power equals current multiplied by voltage so that's good at least. I think we are disagreeing on what the job of an ideal voltage amplifier is, and that a headphone amp should work like an ideal voltage amplifier.

I will get the boring math stuff out of the way first.



A rectangle is indeed a poor way to represent a vector. I think a vector can be best represented with a line (or "arrow") that points from point A to point B. However I do think that a vector is a poor way of representing power in this context and an area of a rectangle is both more hands on and simpler. Calculating an area of a rectangle is elementary school level. Vectors are a more abstract concept and I believe its being tought at secondary school.


You keep saying that the magnitude of the vector would represent the power. Imagine a vector that represents 1V and 1A. 1Volt times 1 Ampere obviously equals to 1 Watt. What would be the magnitude of this 1V and 1A vector though? If you use Pythagoras you can find that the magnitude is square root of 2. If you do the calculations while keeping the units, at least the unit would still be correct at the end but the actual number is incorrect as you can see. The magnitude of this vector would never describe the power correctly (well, unless the power is 0).



A higher impedance at the output of the amplifier represents a different load (different power drawn) for a voltage amplifier.

Consider the following and assume an ideal voltage amplifier:
First case:

• The input signal of the amp is a 1Vrms sine wave at 1kHz. The gain is 10. The impedance that is put to the amplifier's output varies with frequency. It happens to be 300ohms at 1kHz. What should be the output voltage and how much power this impedance draws?
• The input is 1Vrms, the gain is 10 the and since this is a voltage amplifier, this means the output must be 10Vrms. So what's the actual load (power drawn) in this case? It is 10V*10V/300ohm which equals to 333mW.

Second case:

• Use the same configuration as before but change the input frequency to 100Hz. The input signal of the amp is a 1Vrms sine wave but at 100Hz. The gain is still 10 (there are no changes to the volume knob). The impedance that is put to the amplifier's output varies with frequency. It happens to be 650ohms at this particular frequency. What should be the output boltage and how much power this impedance draws in this second case?
• The input is 1Vrms, the gain is still 10 (didn't change to volume knob) and since this is a voltage amplifier, the output must be 10Vrms. What's the power drawn in this case? It is 10V*10V/650ohm which equals to 154mW.

So you aren't correct to assume that different impedances present the same load (to voltage amplifiers). You can see that a higher impedance always draw less power from a voltage amplifier if you don't change the gain and the input signal for the different impedances.

Again, my point is that if the input amplitude stays the same and the voltage amplifier has to output X amount of volts at 1kHz into 300ohms, it also has to output the same X amount of volts at 100Hz where the impedance is 600ohms. Since the output voltage did not change while the impedance went up that means the power the amplifier outputs has to decrease to make sure the gain stays the same at both frequencies. If a voltage amplifier would output the same amount of power at 300ohm impedance and 650 ohm impedance (at the different frequencies) the voltage gain would not be the same accross the frequencies which would mean it's not functioning correctly as a voltage amplifier. I think we can agree on that?

Pure semantics. You are trying to analyze gain from the amp's perspective. Draw your control volume (as in thermodynamics, not sound pressure) somewhere around a realistic world scenario.

The point is, you are trying to achieve a volume to the end user and the headphone is the one drawing the power. That power is the result of a desired volume - the amount of gain you are citing as an example is superfluous, because it can't be held constant, depending on the source signal and the type of headphone. The end result is volume to the ear and a particular headphone will draw the same amount of power to produce that volume, period.

Whether one amp can supply that same power, depends on the impedance of the headphone, thereby commanding the combination of voltage and current that must be applied. That impedance also dictates whether one amp or another has a better chance of supplying it, depending on the amp's topology.

 

[bigshot]

I'm always impressed at how seemingly basic things can be transformed into being incredibly convoluted and complicated so quickly.

 

[tomb]

I'm always impressed at how seemingly basic things can be transformed into being incredibly convoluted and complicated so quickly.

Amen to that.

 

[megabigeye]

I'm glad that I asked this question and that there's a discussion going on in this thread, but it seems like blood pressures are rising.

If you're getting hot under the collar, please just back off rather than feeding the fire. It's really not that important.

 

[bigshot]

I'm not upset. I just would like to see someone give some general practical advice that people can actually use. Pure theory and small exceptions are great, but they don't always help people assemble a good sound system.