Curing Audiophilia Nervosa

Discussion in 'Sound Science' started by watchnerd, Jan 18, 2016.
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  1. castleofargh Contributor
    then you only need acoustic, psychoacoustic, marketing wouldn't hurt. the rest is but a walk in the park ^_^.
    the other option is to embrace ignorance so long as we can find gears to test ourselves and pick by taste. sadly this requires to live in special places or to have plenty of money to waste on trial&error.
     
    Killcomic likes this.
  2. bigshot
    All you need is one cheap amp that plays well with your cans. Then plug whatever you want into it using line out. Or find cans that work with anything. That's what I did.
     
    Last edited: Nov 25, 2017
  3. Killcomic
    I thought that the lower impedance the headphone/IEM, the more likely it will play nice with everything.
    Am I wrong in that assumption?
     
  4. castleofargh Contributor
    you're wrong, definitely. but take comfort in the fact that way too many audiophiles think like you. it mostly comes from assuming sensitivity and impedance are always linked(they aren't), and just false assumptions all around from trying to get electrical behaviors without looking at electricity.
    without going back to electrical stuff in detail when we have plenty of topics that would be a better fit, there is one idea simple enough to make sense: if you get headphones/IEMs with extreme specs(sensitivity and/or impedance), it's less likely they'll work best with the majority of sources. if only because the majority of sources are designed to be plugged into headphones/IEMs with average specs. or it doesn't matter which wheel diameter is the best for traction, or power, or consumption, if it doesn't fit on standard cars, and standard cars are what you're going to use.
    historically designers were trying to make do for load within the 16 to 300ohm range. didn't mean a 16ohm IEM would bring out the best performances from the amp, but it meant that using IEMs reaching as low at 5ohm somewhere, could be a Star Trek adventure.
     
  5. Strangelove424
    To put it simply... that is a wrong assumption. I think you might be thinking of amp output impedance, where lower resistance ensures better matching with more variety of headphones.
     
    ev13wt likes this.
  6. Killcomic
    Indeed! I really need to educate myself on output impedance, sensitivity and such.
     
  7. bigshot
    You really only need to do it once. Find an amp and cans that work together and you're set.
     
  8. Redcarmoose
    image.jpeg
    I used to have a Thorens TD 160.
    http://www.theanalogdept.com/thorens_td_160_dept_.htm


    I read about them and finally picked a used one off EBay around 2006. They had a great sound, but I was surprised how much color the TD160 added. Just a fullness and body, maybe due to the spring suspension? It was like an acoustic instrument in comparison to tables I had before or after. But it did have it's purpose adding body to thin digital fast sounding new wave records. That's the thing, at times the romance is not from accuracy but well matched distortions.

    It even made 1200s sound less musical? Though that would be a great shoot out, to get an old 1200 and a old TD160 and try to hot rod them up as much as possible, then find out which one the group thought was better. They are coming from two opposing directions. Though the TD160 was always like a midrange audiophile legend, and the 1200s were DJ workhorses which could allow DJs to match Beats. The 1200 were and are very strong and resilient against any form of abuse, where the TD160s had a funny rickety way about them. If you ever take the bottom panel off one you will see what I mean. Almost looks like they are a combination of cheap wires, glue and rubber bands!

    Just add years of dust and this is what they look like inside!
     
    Last edited: Nov 25, 2017
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  9. 71 dB
    Sensitivity/watt tells how much sound you get per watt. For example 100 dB/mW (mW = 0.001 W).
    Sensitivity/voltage tells how much sound you get per volt. For example 110 dB/V.
    Power [W] = voltage [V] * current [A]
    Impedance/resistance [Ω] tells the ratio of voltage and current: If power is 1 W, and voltage is 2 volts, then current must be 0.5 amperes because 2 V * 0.5 A = 1 W. Resistance here is voltage/current = 2/0.5 Ω = 4 Ω. So, if you know power and impedance, you can calculate voltage and current.
    Amps have limited power, voltage and current. Bad load impedance means either voltage or current limit is reached long before power limit. Let's say our amp can give 0.3 volts and 0.02 amperes max so that the max power is 0.3 V * 0.02 A = 0.006 W = 6 mW. In what impedance can it give this power? Well, 0.3 volts / 0.02 amperes = 15 Ω How much can the same amp give to 100 Ω ? Well, the voltage limit is reached first and at max voltage the current is 0.3 volts / 100 Ω = 0.003 A so that the power is only 0.0009 W = 0.9 mW. How much can we get to 10 Ω ? Well, now the current limit is reached first: At 0.02 A the voltage = 0.02 amperes * 10 Ω = 0.2 volts. The power is 4 mW.

    This is a very simplified presentation which excludes the effect of output impedance and the complexity of the load impedance, but if you understand it completely, you understand the issue quite well.

    Output impedance is a complex issue, the main thing to know is that the voltage the amp produces is divided over it an the load. So, if the amp produces 1 volts, the output impedance is 1 Ω and the load is 9 Ω, the voltage over load is 0.9 volts and the rest 0.1 volts is over output impedance. So, the bigger the load impedance is compared to the output impedance is, the more you get the amp voltage over your load. If your load above was only 1 Ω, the 1 volt would be divided in half between the "equal" impedances. Because of this, if your load isn't constant in frequency, the voltage is divided differently in different frequencies and you get frequency responce errors. The larger the load impedance is compared to output impedance, the smaller these errors are. Large output impedance means also less control of the load for the amp, because the amp just puts out some voltage and output impedance and load uses it up the way they "want." Less control means more distorted and less accurate sound, but also more "relaxed" which people may like, but if your load is purely resistive and constant in frequency, output impedance only attenuates the sound and the control of amp isn't compromized at all.
     
    Killcomic and Alcophone like this.
  10. Whazzzup
    There is no cure, just give me more cowbell.
     
    Deftone, Killcomic and sonitus mirus like this.
  11. bigshot
    It was probably because it had a different cartridge than you were used to.
     
  12. Redcarmoose
    I did use the same cart on the table before the TD160 but a different one after. Good question as the constantly of the cart was not the same.
     
  13. bigshot
    Whenever you hear a difference the acoustic part is always the one to suspect. Cartridges have a response curve. A turntable can only add noise.
     
    ev13wt likes this.
  14. Killcomic
    Thank you for the explanation but it really went over my head. I can build you a computer blindfolded and troubleshoot it while passed out drunk, but this stuff is a bit beyond me.
    I'm saving it for later study, because if there's one thing that bugs me, it's not understanding something.
     
  15. 71 dB
    Everything is over our head before me understand it. Understanding takes time and effort. If understanding was easy, everyone would understand everything. I spend years in university to understand this and much more complex stuff, but this isn't impossible for anyone. All math you need here is multiplication and division. Primary school math.

    Start with Ohm's law: current = voltage / resistance => I [A] = U [V] / R [Ω].

    I = current and it's unit is A (ampere)
    U = voltage and it's unit is V (volt)
    R = resistance and it's unit is Ω (ohm)

    Now what Ohm's law tells us is that voltage is the potential for current. The more voltage, the more electric current you have. Sounds intuitive, right? However, the current is partly "blocked" by resistance. Resistance makes it harder for electrons to move. You can't swim as fast as you run, because water has greater resistance than air on your body. That is why resistance is denominator in the equation. Increasing it decreases current (while voltage stays the same). If you feed 0.1 volts to 25 Ω headphones, you get 0.1 V / 25 Ω = 0.004 A of current, but if you change to 100 Ω headphones, the current drops to 0.001 A.

    Now simply doing arithmetic you have 3 equations, all describing Ohm's law if you remember just one:

    1. current = voltage / resistance
    2. voltage = current * resistance
    3. resistance = voltage / current

    So, you have quite a lot if you remember and understand just current = voltage / resistance. Absorb Ohm's law and when you feel you get it we go forward and study power.

     
    Killcomic likes this.
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