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Bi-Wiring, yes, again.

Discussion in 'Audio Hardware' started by Katz, Aug 29, 2019.

  1. Bolster

    Bolster If it ain't broke try harder..

    Location:
    UK
    I think my Wharfedale Diamonds are optimised for bi wiring.
     
  2. Bananajack

    Bananajack Forum Resident

    Location:
    Singapore
    I tried a lot in the past ... maximum I heard a small effect from biwiring. Did I? It wasn’t an A/B comparison.

    The best and very audible thing I heard from biwiring posts was (and is) when I had AudioNote jumpers
    made from a better silver cable (not their prefabricated ones) and silver plated bananas. THAT really
    sounds a lot better. Cutthroat expensive though ... but worth it.

    If you are doing biwiring, at least grab the chance and use a 2nd cable run that improves the tweeter or
    woofer (purists may feel suicidal now lol).
     
    Bolster likes this.
  3. Ingenieur

    Ingenieur Forum Resident

    Here's my explanation.
    2 way speaker
    There are 2 filter sections in parallel:
    A hi-pass (attenuates lo freq)
    A lo-pass (attenuates hi freq)
    These are bridged or paralleled

    When you bi-wire (or bi-amp) you remove this bridge or jumper connecting the hi & lo filters.

    hi-filter Z ~ 1/freq
    As freq goes higher Z decreases
    As freq goes lower Z increases
    It presents a low Z to highs and a high Z to lows

    Low-filter Z ~ freq
    As freq goes higher Z increases
    As freq goes lower Z decreases
    It presents a low Z to lows and a high Z to highs

    So when you bi-wire the high filter cable blocks or rolls off lows, ie, low freq power is attenuated.
    The wire carries the high signal only (mostly)

    And the low filter cable blocks or rolls off highs, high freq power is attenuated.
    The wire carries the low signal only (mostly)

    This should provide lower IM distortion, lower Z since you have 2 cables in parallel.
     
    Big Blue and Ripblade like this.
  4. Ingenieur

    Ingenieur Forum Resident

    You can think of it in terms of a Thevenin equivalent circuit.

    You have 'slid' the jumper connecting the hi and lo filter sections down to the amp terminals while extending the wires from the jumper to each filter section.
     
    F1nut and Ripblade like this.
  5. Bananajack

    Bananajack Forum Resident

    Location:
    Singapore
    I would never dare to quarrel with an Ingenieur, but ...
    Both speaker cables carry the same until they reach the crossover, it’s only filtered out afterwards

    Otherwise tell me how that should be possible that a simple crossover influences backwards (by resistance)
     
  6. Ingenieur

    Ingenieur Forum Resident

    how can I clarify....hmmmm
    There is a Law KCL, Kirchhoff's Current Law
    It states the current entering a node must equal the current leaving the node, expressed mathematically the sum of currents at a node must equal 0.

    consider the filter input the node.
    We can't have hi AND lo currents on one side of the node and only one (lo OR hi) on the other.
    The attenuation is at the amp terminals and the respective current does not travel down the wire.
     
  7. Bananajack

    Bananajack Forum Resident

    Location:
    Singapore
    I happen to know Kirchhoffs law ;)
    But doesn’t stay the current the same everywhere, whilst a high pass e.g. basically works with variable
    resistance (impedance change due to capacitor behavior), so the current simply becomes heat?

    Low pass is more or less the same ...
     
    timind likes this.
  8. Ingenieur

    Ingenieur Forum Resident

    a C or L generates no heat or power, only reactive power which does no work
    The real power drives the cone like a motor. Electrical power is converted to mechanical work.

    Z = R + j(Xl -Xc)
    Xl = j 2 Pi f L (lo pass primary Z component)
    Xc = 1/(j 2 Pi f C) (hi pass)
    j = sqrt(-1)

    So Z does vary with freq.
    It blocks (attenuates) the hi or lo

    visualize the speaker wire getting bigger towards infinity and shorter towards 0
    Z goes to 0
    Now the amp is connected directly to the filters and the hi lo splits there.
    As it does when you have 1 cable with the jumper installed.
    Assume the wire is connected to the low filter which is jumped to the hi, only hi will travel thru the jumper.
     
    Ripblade likes this.
  9. Ingenieur

    Ingenieur Forum Resident

    Power perspective
    P = V I
    We know V is the same at the hi and lo filters since they are in parallel. Assume 10 V

    We set the amp to a 10 W average level
    Let's say hi and low sections (tweeter and woofer) each get 5 W.
    If 1 cable it carries all 10 W
    So I= P/V = 10/10 = 1 A

    now bi wire
    Each section still gets 5 W and 10 V
    So each filter sections respective cable must carry:
    I = 5/10 = 0.5 A
    But the total is still 1 A and 10 W
    The difference in current is because each cable only carries the hi or lo, not both
     
    Ripblade likes this.
  10. Bob_in_OKC

    Bob_in_OKC Forum Resident

    Location:
    Dallas
    With biwiring: At the point where the wires leave the amplifier they are joined and have not passed through a crossover. So just to be clear - Those with expertise in the matter say one cable carries the high and one cable carries the low? Is this a consensus?
     
    Ingenieur likes this.
  11. Davey

    Davey like smoke from a lithium dream

    Location:
    SF Bay Area, USA
    No, not a consensus, depends on the crossover design.
     
  12. Ingenieur

    Ingenieur Forum Resident

    For most 2 way system using a first or second order filter, yes.
    The 'impedance' actually reactance of a C or L was given above, but simplified here:

    Xc ~ 1/f the C reactance is INVERSELY proportional to frequency.
    If freq = 100 Hz Xc ~ 0.01
    If 10,000 Hz, 0.00001
    It blocks lo and allows hi thru

    Xl ~ f the L reactance is proportional to frequency.
    If freq = 100 Hz Xc ~ 100
    If 10,000 Hz, 10,000
    It blocks hi and allows lo thru

    the hi pass C blocks the lows (hiZ for low f) and the L (hiZ for hi f) forces them thru the speaker
    A freq dependent current divider

    the low pass is the inverse
    the lo pass L blocks the highs (hiZ for hi f) and the C (hiZ for lo f) forces them thru the speaker

    Very simplified but the basic concept is there
    A 2 way will have one of each in parallel.
    The jumper removed separates them.
    And you run wire to each filter and it only carries the hi or lo. Actually some of each gets thru but is greatly attenuated. The slope of the filter determines how fast the lows roll off in a hi pass and vice versa for the lo pass.

    if you have a high pass with a 6 dB slope and 400 Hz cutoff.
    At 200 Hz -6 dB
    100 -12 dB
    And so on
    -6 dB = 10 log (power ratio)
    So 100 Hz will have 1/16 the relative power of 400, ~6%

    [​IMG]
     
    Last edited: Nov 21, 2020
  13. Ingenieur

    Ingenieur Forum Resident

    Sorry if this has already been posted.
    In the link the writer did actual testing.
    He sent a signal with 5 low freq pulses and 5 high. He measured the current for 3 scenarios
    1 cable
    Bi-wired
    Low freq cable
    High freq cable

    One can see in the low pass cable the high pulses were attenuated and in the high pass the lows were attenuated.

    excerpt from his conclusions:
    We theorised and proved by measurement, the counter-intuitive notion that the high and low frequencies travel only in their designated cables. We have also shown by measurement that the theory that this would reduce intermodulation distortion...

    Bi-wiring Speakers: An exploration of the benefits -
     
    timind and Big Blue like this.
  14. tubesandvinyl

    tubesandvinyl Forum Resident

    Speaker jumpers definitely make a big difference. It's a cheap experiment.

    Upgrading the jumpers from the stock Tannoy jumpers made a very nice improvement!!!
     
  15. Ripblade

    Ripblade Forum Resident

    Location:
    The Six
    Current through the cables varies according to the impedances in their respective filters. It's Ohm's Law. No consensus required; anyone who disagrees doesn't understand the physics.
     
    Ingenieur likes this.
  16. Bob_in_OKC

    Bob_in_OKC Forum Resident

    Location:
    Dallas
    I’m not asking about a difference in current. That part is simple. It’s the question of only high frequencies going to one cable and only low frequencies to the other that doesn’t seem straightforward.
     
  17. Ripblade

    Ripblade Forum Resident

    Location:
    The Six
    Current is inversely proportional to impedance, yah? So how does bass current flow through the cable connected to the tweeter? And how does treble current flow through the woofer cable?
     
  18. Bob_in_OKC

    Bob_in_OKC Forum Resident

    Location:
    Dallas
    Note that the author of the article linked above called this idea counter-intuitive and seems to be making a case for it to be so. That means to me it is being presented as not necessarily understood to be true at the outset. That’s what I’m trying to explore.

    Next, I’ll say this is the first time I’ve ever heard the term bass current. This is news to me.
     
  19. Ripblade

    Ripblade Forum Resident

    Location:
    The Six
    Current at bass frequencies. Is that better? I'm trying to keep the language simple because most people don't understand the equations @Ingenieur so aptly posted.

    It's not counterintuitive to me, but I have training in electronics, and a college certificate commemorating it. I'm no genius by any stretch, but this is the easy stuff.
     
    Ingenieur likes this.
  20. Ingenieur

    Ingenieur Forum Resident

    The difference in current is DUE TO frequencies being attenuated.

    a musical signal can be thought of a series of sine waves of different frequencies. Almost any signal can be constructed out of sine waves, even a square wave. See Fourier
    If a transform is done from the time/temporal domain to the frequency domain you'll have a bunch of spikes/power at varying frequencies.

    This is not consensus nor debatable, it is mathematically proven ,fact.
    So is the attenuation of freqs or signal in each cable due to the filters.

    below is an Fast Fourier Transform FFT VS time plot
    FFT shows all power at 1 frequency

    [​IMG]
     
    Last edited: Nov 21, 2020
    Ripblade likes this.
  21. Ingenieur

    Ingenieur Forum Resident

    Counterintuitive is not the word I would use, not sure what fits.
    But you are correct, without having studied the subject you would naturally assume ALL frequencies flow together and how in the heck would you separate or filter them when all mixed up. A instant of music will have dozens of frequencies combined.

    You can think of a frequency having power, so therefor also V and I (P= V I)
    Engineers talk in terms of these variables, not the bass notes, but the low freq signal which is described by its power, V and I.
     
    Last edited: Nov 21, 2020
  22. Ingenieur

    Ingenieur Forum Resident

    Is it worth it? imho yes, I've done it.
    And this is why I think it helps.
    Every cable has 3 basic properties:
    R the predominant factor
    C (small in our case)
    L (very small in our case)

    assume one cable carries hi and the other lo
    The negative effect of C and L are phase shift or time distortion in opposite fashion.

    The negative effect of C is 1/freq
    Meaning the higher f the less negative effect or distortion, (or the lower f the more effect)

    The negative effect of L is proportional to freq
    Meaning the higher f the more negative effect or distortion, (or the lower the less the effect)

    so by separating the freqs you have minimized or negated the C distortion in the high freq cable and the L distortion in the low freq one.
    In the high you reduced C
    In the lo you reduce L
    It's like having better cables with lower C and L, also R since you have cables in parallel.
     
    Ripblade likes this.
  23. Ripblade

    Ripblade Forum Resident

    Location:
    The Six
    I liken it to the cheapest upgrade one can make, assuming the cables are inexpensive. Those who disagree are usually using expensive cables and don't want the added expense, or are ppl who doubt that cables make any difference to begin with. Others agree but simply don't like it.

    To each their own, but that doesn't negate the science.
     
  24. Bolster

    Bolster If it ain't broke try harder..

    Location:
    UK
    Wow, at lot of these recent discussions went over my head at great speed.

    However can I ask what you guys made of the Q Acoustics bi wiring analysis? I thought it was quite accessible for the layman.

    Also is there some consensus from the group that bi wiring speakers that have sympathetic crossovers improves the sound or at the very least sounds different?

    On the subject of cables I'm using budget friendly Fisual S-Flex cable Fisual S-Flex Studio Grade White Bi-Wire Speaker Cable 4 x 2.5mm - Price Per Metre - Speaker Cables - AV Online - UK Home Cinema and Hifi Specialists
     
  25. Ingenieur

    Ingenieur Forum Resident

    The Q Acoustics analysis is excellent with actual testing.

    I can't hear a difference although I haven't done a blind test. It will be subtle. imho the crossover has very little to do with the difference when bi-wiring, it is the cables.
    The closer the cable parameters get to the jumper parameters the less difference. I could see long, small cables having a negative effect and possibly having a drop or 'suck out' at the xover freq.
    For example some jumpers, the bar type, have essentially no L or C and very little R.
    My KEF R3 have an internal switch to separate filters.
     
    Bolster likes this.

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