I have an Ipod Shuffle. It's supposed to last 10 hours between charges. The stock headphones drew about 16 ohms of power. I bought some Sony headphones (that make me look like Princess Leia, circa 1977) but they have good bass. But they draw 24 ohms. Can I expect the Ipod to have less battery life?
I'm not an electrical engineering geek, but I figured some of you just might be.
Headphones' Power Draw Question
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Sales Dude McBoob
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My guess is that the more power consumptive drivers in your larger headphones will draw more juice from the iPod battery.
If you think about it, it makes sense.
Also, the volume that you listen to stuff at will affect the battery life. If you crank it, you also squeeze more juice out of the battery.
And the material that you listen to could also eat more battery. If you played a glitched out breakbeat remix of Flight Of The Bumblebee on one iPod, and played a simple 1000k tone on another, my guess is that the Bumblebee iPod would die first.
In short, if you want your pod to last, only listen to sine waves.
If you think about it, it makes sense.
Also, the volume that you listen to stuff at will affect the battery life. If you crank it, you also squeeze more juice out of the battery.
And the material that you listen to could also eat more battery. If you played a glitched out breakbeat remix of Flight Of The Bumblebee on one iPod, and played a simple 1000k tone on another, my guess is that the Bumblebee iPod would die first.
In short, if you want your pod to last, only listen to sine waves.
Well, here's a definitive answer:
If you want to buy a smaller, portable set of headphones, you should pay attention to the headphones' impedance. Impedance indicates how efficiently headphones use electrical power and is measured in ohms. The lower the impedance, the more efficiently headphones use power. Headphones typically draw power from their audio source, such as a portable CD player or a home stereo amplifier. If you intend to use your PC headphones for portable audio devices, you want a pair of headphones with low impedance, such as 40 ohms or less. Most portable headphones hit this mark, but larger headphones intended for home stereo use may have higher (in some cases much higher) impedance.
Source: http://www.smartcomputing.com/editorial ... 5390&guid=
If you want to buy a smaller, portable set of headphones, you should pay attention to the headphones' impedance. Impedance indicates how efficiently headphones use electrical power and is measured in ohms. The lower the impedance, the more efficiently headphones use power. Headphones typically draw power from their audio source, such as a portable CD player or a home stereo amplifier. If you intend to use your PC headphones for portable audio devices, you want a pair of headphones with low impedance, such as 40 ohms or less. Most portable headphones hit this mark, but larger headphones intended for home stereo use may have higher (in some cases much higher) impedance.
Source: http://www.smartcomputing.com/editorial ... 5390&guid=
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Sales Dude McBoob
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The real drag about using high impedance headphones with a consumer device is that there is usually a low ceiling to how loud you can turn up the volume. I used to own a pair AKG 240M headphones, and those things needed a lot of juice. On a portable CD player you couldn't crank it up enough to efficently rock out. I wish I had bought the 240S because they're 55 ohms. Actually, I wish I had bought neither because the rings pop off of both of them.nebulae wrote:Well, here's a definitive answer:
If you want to buy a smaller, portable set of headphones, you should pay attention to the headphones' impedance. Impedance indicates how efficiently headphones use electrical power and is measured in ohms. The lower the impedance, the more efficiently headphones use power. Headphones typically draw power from their audio source, such as a portable CD player or a home stereo amplifier. If you intend to use your PC headphones for portable audio devices, you want a pair of headphones with low impedance, such as 40 ohms or less. Most portable headphones hit this mark, but larger headphones intended for home stereo use may have higher (in some cases much higher) impedance.
Source: http://www.smartcomputing.com/editorial ... 5390&guid=
Kinda crazy. I just looked up the 240M and they've been discontinued. I bet a lot of producers are pissed
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Michael-SW
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Michael-SW
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The 240M is a high impedance headphone (600 ohms), which mostly isimportant if you are going to connect a lot of phones to the same output.Sales Dude McBoob wrote: Kinda crazy. I just looked up the 240M and they've been discontinued. I bet a lot of producers are pissed
The low impedance version K240 is still being made.
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Sales Dude McBoob
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The topic of impedance is a super-sized can of worms.nebulae wrote:I thought the higher the impedance the more the resistance, and therefore the more power required. Please un-confuse me
I wish I had the know-how to lay it all down on the table, but I don't.
I believe the impedance is, in essence, the resistance in the cable transfering the audio. The impedance rating on the specific headphone itself is only a rating on how that headphone's drivers take in the signal flow. The actual resistance in the cable is dependant on what the headphones are plugged into, and the make-up of the cable itself. The resistance in the cable will vary from device to device, as will the preformance of the headphones.
Am I on point?
I am one of the geeks you're looking for... 
Basically, as the resistance goes up, the power consumption goes up.
McSalesBoob - yes, you're on point and have a grip on this stuff.
Power is measured in watts
P = I * V power equals current times voltage
then there's Ohm's Law
V = I * R
so... substituting to combine those equations you can look at their relationships differently:
P = V^2/R
or
P = I^2 * R
The way headphone output drivers work (devices call operational amplifiers), they're always going to put the same voltage waveform for a given song, regardless of the load (headphones), it's the current that will change, so
P = I^2 * R
applies. As the resistance goes up, the power consumption goes up.
What's impedence...
A resistor is just a thing that turns voltage into current, nothing tricky happens, the math is really simple. It gets weird when you consider the capacitance and inductance of a device.
Capacitors are just two plates separated by a small space, then they're wrapped up really tightly into a little cylindrical package. Electrons go in, go onto a plate and can't get to the other plate, there's a space there. BUT when they're active, when they move around, when it's an AC current or a current made by an ipod they can coax electrons on the other plate to move (this is due to electromagnetic fields). A side effect is that in a capacitor the current of a signal will LEAD the voltage, a phase difference is produced, which messes with ohm's law and the power equation above. Capacitors are also found measured between the shield of a cable and the conductor. 75 Ohm coax cable, for example, or your speaker wire.
Inductors... Think of an electromagnet, a piece of iron with a wire wrapped around it, pass a current through it and it becomes a magnet. That's all an inductor is, electrons go in, they go around and around the iron core, create a magnetic field, which store energy. A speaker driver is a big ol' inductor. A side effect is that in an inductor the voltage of a signal will LEAD the current (opposite from a cap), a phase difference is produced, which messes with ohm's law and the power equation above.
So that's all weird and fucked up, so they call it an impedence for short. Impedences also change with frequency.
You want to match impedences to get the best power transfer between devices. If the impedences don't match you get reflections in the line, exactly like waves in a bathtub, this leads to distortion and loss of line level, basically it's inefficient.
Think of a series of rapids in a river, the water is the electron, the water's potential energy (how high it is from the GROUND) is its voltage, how fast the water's moving (kinetic energy) is current. The rocks in the water are like resistors.
Just buy an extra charger, it's much simpler.
hth, geek out.
Basically, as the resistance goes up, the power consumption goes up.McSalesBoob - yes, you're on point and have a grip on this stuff.
Power is measured in watts
P = I * V power equals current times voltage
then there's Ohm's Law
V = I * R
so... substituting to combine those equations you can look at their relationships differently:
P = V^2/R
or
P = I^2 * R
The way headphone output drivers work (devices call operational amplifiers), they're always going to put the same voltage waveform for a given song, regardless of the load (headphones), it's the current that will change, so
P = I^2 * R
applies. As the resistance goes up, the power consumption goes up.
What's impedence...
A resistor is just a thing that turns voltage into current, nothing tricky happens, the math is really simple. It gets weird when you consider the capacitance and inductance of a device.
Capacitors are just two plates separated by a small space, then they're wrapped up really tightly into a little cylindrical package. Electrons go in, go onto a plate and can't get to the other plate, there's a space there. BUT when they're active, when they move around, when it's an AC current or a current made by an ipod they can coax electrons on the other plate to move (this is due to electromagnetic fields). A side effect is that in a capacitor the current of a signal will LEAD the voltage, a phase difference is produced, which messes with ohm's law and the power equation above. Capacitors are also found measured between the shield of a cable and the conductor. 75 Ohm coax cable, for example, or your speaker wire.
Inductors... Think of an electromagnet, a piece of iron with a wire wrapped around it, pass a current through it and it becomes a magnet. That's all an inductor is, electrons go in, they go around and around the iron core, create a magnetic field, which store energy. A speaker driver is a big ol' inductor. A side effect is that in an inductor the voltage of a signal will LEAD the current (opposite from a cap), a phase difference is produced, which messes with ohm's law and the power equation above.
So that's all weird and fucked up, so they call it an impedence for short. Impedences also change with frequency.
You want to match impedences to get the best power transfer between devices. If the impedences don't match you get reflections in the line, exactly like waves in a bathtub, this leads to distortion and loss of line level, basically it's inefficient.
Think of a series of rapids in a river, the water is the electron, the water's potential energy (how high it is from the GROUND) is its voltage, how fast the water's moving (kinetic energy) is current. The rocks in the water are like resistors.
Just buy an extra charger, it's much simpler.
hth, geek out.
In my life
Why do I smile
At people who I'd much rather kick in the eye?
-Moz
Why do I smile
At people who I'd much rather kick in the eye?
-Moz
if you need to boost the power you could try getting a cmoy headphone amp...
google or ebay for 'cmoy' and around $30 - you can get a very portable entry level heaphone amp , built in to small containers ( like sweet tins) - i use one with my AKG k701's which benefit from the extra upmhhh...or you can build them yourself
http://search.ebay.com/search/search.dl ... title=cmoy
j
google or ebay for 'cmoy' and around $30 - you can get a very portable entry level heaphone amp , built in to small containers ( like sweet tins) - i use one with my AKG k701's which benefit from the extra upmhhh...or you can build them yourself
http://search.ebay.com/search/search.dl ... title=cmoy
j
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robbmasters
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As power consumption goes up, so does the power output (i.e. the perceived volume level). So for the same perceived volume level, headphones of different resistances will consume roughly the same amount of power (you'll just have the volume knob on different settings). Therefore, the impedance of your headphones has no effect on battery life. The efficiency of the headphones will make a difference, but this is not the same as the overall impedance.
However (since P=V^2/R) the larger the resistance of your headphones, the less power output (perceived volume level) you'll get for a specific voltage - for example, the maximum voltage across your headphone outputs. Therefore, higher impedance headphones won't go up so loud - so you might want lower impedance ones anyway.
However (since P=V^2/R) the larger the resistance of your headphones, the less power output (perceived volume level) you'll get for a specific voltage - for example, the maximum voltage across your headphone outputs. Therefore, higher impedance headphones won't go up so loud - so you might want lower impedance ones anyway.
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