[Stecchino]

These questions may sound like they're coming from a ignorant newcomer...(ahem), well I guess they are

[size=xx-small](This is kind of a general question but I'm asking it in the context of headphones so I posted in this forum as opposed to elsewhere.)[/size]

Questions
When a headphone has a "low" impedance rating:

1. Does that mean it has a low amount of ohms?
2. If the answer to number 1 is yes, does that then mean it has a low resistance to electric currents (or sound or whatever) and therefore will usually require some kind of amplification because it allows the currents to flow into and out of the headphone?

 

[lindrone]

Yes, low impedance => lower Ohms => lower resistance...

What this means, is that it'll require less voltage to drive the headphone to a louder degree. It also means that it draws more current, which would imply a shorter battery life from portable devices. Although the realistic difference between a low and high impedance headphone drawing battery power really varies on the headphone, battery, source... and other complicated issues that I don't even understand

On the other hand, high impedance headphones require less current, but takes more voltage to drive. Which means that your battery life may be lengthened, but your source might not be able to crank up the headphone loud enough to the point where it's usable.

Of course, there's other factors involved that determines how easily a headphone is driven as well other than just impedance. High sensitivity, for example, will make a relatively high impedance headphone still very easy to drive (case in point, Shure E5c is such a design).

 

[Stecchino]

Quote:

Originally Posted by lindrone Yes, low impedance => lower Ohms => lower resistance... Of course, there's other factors involved that determines how easily a headphone is driven as well other than just impedance. High sensitivity, for example, will make a relatively high impedance headphone still very easy to drive (case in point, Shure E5c is such a design).

Is there are statistical measure for sensitivity or is that a subjective thing or simply not measured in units?

 

[breez]

Sensitivity is quite often expressed as dB per mW (at 1KHz).

 

[pedxing]

Quote:

Originally Posted by breez Sensitivity is quite often expressed as dB per mW (at 1KHz).

Just in case you didn't know:
db - decibels - a way of measuring loudness with sound. I think it actually measures the relative power of the sound.

mW- milliwatts - amount of power.

In other words, the higher the db and lower the mW, the more sensitive the headphone is. However, it does not represent anything about quality or much or anything else.

 

[breez]

Quote:

Originally Posted by pedxing Just in case you didn't know: db - decibels - a way of measuring loudness with sound. I think it actually measures the relative power of the sound. mW- milliwatts - amount of power. In other words, the higher the db and lower the mW, the more sensitive the headphone is.

Yep and the relation is to the threshold of hearing (0dB) in this case.

 

[lindrone]

Either that, or you can get really mad at your headphone and yell at it, and see at which point the headphone starts crying and gets its feelings hurt.

 

[mkmelt]

When a headphone has a "low" impedance rating,

1. Does that mean it has a low amount of ohms?

An ohm is a measure of resistance that a conductor (wire or coil) presents to the flow of an electric current. 1 ohm is defined as the amount of resistance required so that 1 ampere (amp) of current flows when 1 volt is applied.

Low, medium, and high impedance are relative terms. For speakers, low might be considered below 4 ohms, medium could be considered 8 ohms, and high anything over 15 ohms.

For headphones, low impedance generally means up to 32 ohms, medium impedance would be between 60 and 100 ohms, and high would be anything greater than 100 ohms.

2. If the answer to number 1 is yes, does that then mean it has a low resistance to electric currents (or sound or whatever) and therefore will usually require some kind of amplification because it allows the currents to flow into and out of the headphone?

For a given amount of power, the impedance of the headphone or speaker will determine the voltage that must be applied, and the amount of current that must flow. To power the headphone or speaker to acceptable output levels, an amplifier must be able to provide both the needed voltage and current.

Example # 1: Two speakers with different impedances.
The following equation is derrived from Ohm's law (see below).

For a Speaker A rated at 8 ohms, to deliver 1 watt of power the following is true: 1 watt = (V)^2/ 8 ohms, that is the voltage (V) squared, divided by the impedance 8 ohms. In this case, the voltage needed is equal to the square root of 8 or approximately 2.83 volts.

Suppose Speaker B was rated at 4 ohms. To deliver 1 watt of power the following is true: 1 watt = (V)^2/ 4 ohms, that is the voltage (V) squared, divided by the impedance 4 ohms. In this case, the voltage needed is equal to the square root of 4 or exactly 2 volts.

The power is the same (1 watt), but the required voltage is higher for speaker A (the 8 ohm rated speaker).

To determine the current (in amperes), we can use Ohm's law,
where V = I * R.

For the 8 ohm speaker: V(2.83 volts) = I (current) * R (8 ohms). Dividing 2.83 volts by 8 ohms, yields a value for I of 0.35375 amperes (approx. 1/3 of 1 ampere)

For the 4 ohm speaker: V(2 volts) = I (current) * R (4 ohms). Dividing 2 volts by 4 ohms, yields a value for I of 0.5 amperes (exactly. 1/2 of 1 ampere)

The power (1 watt) is the same, but the lower impedance speaker requires more current( 0.5 vs 0.35 ampere) from an amplifier.

Example 2: Two headphones, Headphone A is low impedance (32 ohms) and Headphone B is high impedance (320 ohms).

To determine the voltage needed to provide headphone A with 50 milliwatts of power (a reasonable power level that might be expected from a headphone amplifier), we would compute 0.050 watts = V^2/32 ohms,
and V = 1.26volts.

To determine the voltage needed to provide headphone B with the same 50 milliwatts of power, we would compute 0.050 watts = V^2/320 ohms,
and V = 4 volts.

It is clear from this example that a portable player running on 1.5 or 3 volts will have no difficulty providing 50mw of power for the low impedance headphones, but will be unable to provide the 4 volts needed for the high impedance headphones.

The current required for 50mw is computed using Ohm's law.

For the low impedance phones:
1.26 volts = I * 32 ohms, and I = 0.039375 ampere (approx. 40 milliamps.)

For the high impedance phones:
4 volts = I * 320 ohms, and I = 0.0125 ampere (approx. 13 milliamps.)

The high impedance phones require approximately 1/3 the current of the low impedance phones for the same 50 milliwatts of power. Assuming these headphones are to be powered by a portable device, depending on the type of battery powering the amplifier, the higher current flow required by the low impedance phones would drain the battery 3x times faster than would the high impedance phones.

In practical terms, many real world low impedance headphones and earbuds don't require 50mw of power, so the battery drain will be slower. Also, some well known high impedance phones such as Sennheiser HD-580 and HD-600 (both rated at 300 ohms) actually need more than 50mw of power for peak levels, typically 100~200mw so the current required would be a bit more, and the required voltage (5.47~7.75 volts) would be higher than the 4 volts in the previous example. A portable player or amplifier that uses less than a 6~9 volt battery or an external power supply will not be able to drive these high impedance phones to full output.

 

[Stecchino]

Thanks a million lindrone, breez, mkmelt, and pedxing. Between all of your responses I've got a much clearer conception of how the electricity behind my music works! Granted, I'd still have a hard time explaining it to someone else, but I've got it in my head. I could always just point them to this thread.

To personalize all of this...(please correct me if I'm wrong)

My Etmotic ER-6 phones have a sensitivity rating of 108 dB SPL for a 0.4 volt input and an impedance of 48 ohms. So, there will be less voltage (power) necessary to drive the ER-6 loud enough, however, because it's on the lower end of impedance, it will theoretically draw more current which reduces battery life quicker (on battery-dependent portables/amps). Furthermore, the ER-6 has a relatively high sensitivity, which means that in theory it will be easy to drive to decent listening volumes (also true in REALITY).

Now, the ER-4S which has an impedance of 100 Ohms and a sensitivity of 108 dB SPL for 1 volt input is 4 times LESS sensitive than the ER-6. In other words the ER-4S requires 4 times the amount of power (.25 volts vs. 1 volt) to get 108 dB SPL. The ER-4S has a higher impedance than the ER-6 and will require more power (voltage) to drive it to good listening levels but will draw LESS current (theoretically lengthening battery life if applicable).

The Sennheiser HD-497 which I have on order has a low-impedance of 32 ohms and a sensitivity of 112dB/mW, which seems to me that they are moderately highly sensitive phones.

One thing I get tripped up on...Etymotic states their tech specs for sensitivity in terms of volts and most other companies express it in terms of mW. So, which headphone is technically more sensitive, the ER-6 (108 dB for .25 volts input) or the HD-497 (112 dB/mW)?

 

[pedxing]

Correct me if I am wrong because I only know ohm's law when it comes to electronics, but i think you confused something:

V= I*R so,
V/R = I

at 108 db with er-6:
0.40 volt input/ 48 ohms = 0.008333.... amps

at 108 db with ER-4S
1 volt input/ 100 ohms = 0.010 amps

I think the ER-4S is drawing more current, maybe reducing battery life in comparison to er-6. But I don't know anything about batteries in general, so I can be completely wrong about the correlation between current draw and battery life.

Also, sensitivity measurements can vary by the way the manufacturers derive the numbers. I don't know if Ety and Senn do their measurements and calculations the same way, otherwise it would be like comparing apples to oranges. The fact that senn 112 db and ety uses 108 db makes the comparison already tricky because I don't know if the difference in db can be used to linearly interpolate (or extrapolate, not sure which term) comparable values.

 

[mkmelt]

[My Etmotic ER-6 phones have a sensitivity rating of 108 dB SPL for a 0.4 volt input and an impedance of 48 ohms.]

Don't confuse Power (measured in Watts) with Electromotive Force (measured in Volts).

Power = Voltage^2 / Resistance

With 0.4 volts and an impedance of 48 ohms, the power level required for 108db output from your ER-6 phones is 0.0033 watts of power (3.3mw).

[So, there will be less voltage necessary to drive the ER-6 loud enough, however, because it's on the lower end of impedance, it will theoretically draw more current which reduces battery life quicker (on battery-dependent portables/amps).] - True

[Furthermore, the ER-6 has a relatively high sensitivity, which means that in theory it will be easy to drive to decent listening volumes (also true in REALITY).] - True

[Now, the ER-4S which has an impedance of 100 Ohms and a sensitivity of 108 dB SPL for 1 volt input is 4 times LESS sensitive than the ER-6. In other words the ER-4S requires 4 times the amount of power (.25 volts vs. 1 volt) to get 108 dB SPL.] - Almost correct, see below.

With 1.0 volt and an impedance of 100 ohms, the power level required for 108db output from your ER-4S phones is 0.01 watts of power (10mw). The ER-4S requires approx. 3x the power (10mw vs. 3.3mw) of the ER-6. In terms of sensitivity, for the same 3.3mw of power, the ER-4S will produce approx. 103db. This is approx. 5db less output than the rated 108db output for the ER-6 for the same power level. So it would be correct to state that the ER-4S is 5db less sensitive than the ER-6.

Keep in mind that for every 3db increase in acoustic output level, the power must be doubled. So to raise the output of the ER-4S from 103db back up to 108db, the power must be doubled from 3.3mw to 6.6mw, and then raised again by 50% to reach 10mw for the same 108db output level as the ER-6.

[The ER-4S has a higher impedance than the ER-6 and will require a bit more voltage to drive it to good listening levels, but will draw LESS current (theoretically lengthening battery life if applicable).] - True

[The Sennheiser HD-497 which I have on order has a low-impedance of 32 ohms and a sensitivity of 112dB/mW, which seems to me that they are moderately highly sensitive phones.] - True

[One thing I get tripped up on...Etymotic states their tech specs for sensitivity in terms of volts and most other companies express it in terms of mW. So, which headphone is technically more sensitive, the ER-6 (108 dB for .25 volts input) or the HD-497 (112 dB/mW)?]

For 10mw of power the ER-4S will produce an output level of 108db (for sensitivity measurements, acoustic output is usually measured at a standard frequency of 1Khz). For the ER-6, 10mw of power will produce an acoustic output level of approximately 113db. For the same 10mw, the output level of the HD-497 would increase by approximately 10db over the rated 112db/1mw to reach 122db. Based on those specifications, the HD-497 is 14db more sensitive than the ER-4s, and approx. 9db more sensitive than the ER-6.

Based on the rated sensitivity of 112db/mw, just 1 or 2 mw is all the power needed to drive the HD-497 to full output, while a bit more than 10x this amount of power would be required to drive the ER-4S to the same output levels. As the power levels required by the headphones in these examples are all very low (10mw or less), any portable player should be capable of driving these phones to full output. Battery life would depend more on the type of device (CD player, hard drive MP3 player, or flash player) than on the power required to drive any of these headphones.

 

[Coalescent]

Thanks guys! I love learning new technical information about my new hobby, and this thread is very enlightening. It looks like some of you spent a lot of time writing these posts, and I just wanted to say that one relative noob here says, "Thanks!"

-Chad

 

[Zuerst]

Then one can utilize capacitor for low-pass or high-pass filter to separate signals to lower and higher to high end and low end driver in speakers or maybe in the case of Shure E5 and other dual driver phones???

 

[hedrick]

I admit that I'm a bit surprised by discussions of high vs low impedance headphones. There seems to be a general impression that high-Z phones are harder to drive than low-Z phones. With speakers, low impedance is considered harder to drive. Low-Z speakers tend to require lower voltage and higher current to reach a given volume. Inexpensive amplifiers often have problems with that, running out of current from their power supplies.

Now obviously headphones are a different situation, because their impedance is much higher than speakers, and the amount of power required is much lower. I guess you're less likely to run into current limits based on the power supply. But still, you would think that some of the same effects would occur. Based purely on ohm's law, you would expect that a high impedance headphone

* would tend to require higher voltage to get to a given output (obviously this depends upon sensitivity as well as impedance). You are more likely to run into the upper voltage limit of your device, which would mean either that you couldn't get to as high a volume as you want, or if you try to push it, you might end up clipping

* but given that your headphone output can produce the necessary voltage to get the volume you want and you don't push it into clipping, a high-impedance phone should be easier to drive, i.e. it should require less current and power. While I doubt that you're going to run into power supply current limitations as when driving speakers, I do wonder about the ability of op amps and other ICs to deliver significant current.