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Minimum frequency on Cat.5

Discussion in 'Electronic Design' started by [email protected], Dec 16, 2008.

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  1. Guest

    Will using a single pair in cat.5 cable as an open collector bus for "slow"
    multidrop serial communication work ..?
    I have read that cat.5 cables have nasty complex impedances below 100-200 kHz.

    The voltage level I will use is likely to be 3.3V. Output I/O from an PIC
    or AVR GPIO pin in threestate mode. Active or FPT termination to be used in
    the cable ends. So in essence it should work like a single line scsi bus.

    Also what's the correlation between cable bend radius and maximum frequency
    (or bitrate) used ..?

    The bend radius hint so far is 4 times the cable diameter. But no
    clarification in respect to frequency.

    Any hints on joining three cat.5 cables into an T-junktion..? Such that
    two maincable pairs pass through, and the remaining two goes into a
    patch cable.
     
  2. Guest

    The 1000's of metres of CAT5 I've measured looks perfectly 'normal'
    from DC to way beyond 100MHz. (as any old bits of wire should). At low
    frequencies you only need to bother with the pF's per metre adding up.
    Yes, technically no reason why a multi drop open collector won't work
    but you'll have to be careful about driving the accumulated line
    capacitances
    God only knows where the "bend radius" thing comes from. Electrically
    no concern. Maybe it's just a mechanical suggestion to avoid damage.
    Personally I wouldn't dream of running single line links over any
    distance (eg beyond 10mtr). Just add an inverted version of the signal
    and run 'differential'. 100's of times better performance.
     
  3. Guest

    Don't you get an nasty electrical circuit of RC-RC-RC-RC characteristics at low
    frequencies ..?

    (and mostly CL at high)

    I had a another second thought. And that is that an open collector network
    will need a pull-up (working as a terminator too?) at the ends.
    The bus length will be like ~40 meters (132 ft) long. So when a node release
    the line it will need to wait for the pull-up perhaps 20-40 m away to pull
    the voltage upp again.
    Bend radius sources:
    http://www2.electronicproducts.com/...d-pair_cable-article-belden-nov2005-html.aspx
    'Most Category cables require a minimum bend radius of "four times'

    http://www.bhphotovideo.com/c/product/314294-REG/GGI_UTP5E25_UTP_Cat_5_350MHz.html
    'Bend Radius: 25.4mm'

    http://www.tilcom-bg.com/en/Catalogue/TelecommunicationCables/
    'min bending radius: 4 times cable diameter'

    The theory I read were that the bend has to be shorter than some part of
    the wavelength. Somewhat similar to the characteristics of fiberoptic cables.
    The reason to not go for differential is to keep node complexity and price
    down. If there's a bad problem there's always the option to divide the
    multidrop network into electrical segments with repeaters inbetween.
    The original idea was along the line of a "string of MCUs on wires" with as
    little extra as possible.
     
  4. Jasen Betts

    Jasen Betts Guest

    it needs a pull-up somewhere

    you could use a current-source as the pull-up for faster recovery

    +12
    |
    [3k]
    |
    |/
    VCC+0.6 --|
    |>
    |
    |
    what sort of bit rate are you hoping to get?
     
  5. oopere

    oopere Guest

    If you terminate the cable at the far end your driver will always see
    the characteristic impedance of the cable at any frequencies. At least,
    as a first order approximation. If the "drops" have high enough input
    impedance, they will not change this.

    The driver, however, has to be able to drive a low resistive load. This
    may require some thoughts. Perhaps even paralleling some outputs
    In microstrip lines, bends introduce somo parasitic reactance. In other
    kinds of transmission lines I guess that it is the physical structure of
    the transmission line than changes. For instance, in twisted pairs you
    may end up with nonuniform twists which translate into an impedance jump.
    At hight frequencies or, equivalently, with fast signals, a T-junction
    will introduce reflections. If you are working at low frequencies, a
    short patch line terminated in a high impedance will behave capacitively
    (some hundreds of pF for a couple of meters). These capacitance will be
    effectively loading your main line. This may or may not be important
    depending on the app. From another point of view, an open circuited
    patch line will introduce delayed versions of the signal onto the main
    line. Since the delay is length-dependent, this may or may not be
    important depending on your app.

    Pere
     
  6. Guest

    it needs a pull-up somewhere
    I assume the signal wire goes between the 3k resistor and the NPN-transistor.
    And that Vcc is the minimum voltage level required for an "1" (2.85V for SCSI).

    Any means to place these in the middle of the line without severe mess?
    Assuming a relay needs to be switched faster than a human can notice, say ~5ms
    and that you need to send 4 bytes to accomplish this. And get an response.
    It would mean on the order of 12800 bps. Preferebly slightly faster due
    many devices and maybe longer messages. But in that ballpark.
     
  7. Rich Grise

    Rich Grise Guest

    He's afraid that at higher frequencies, the waves will trip over the
    loop and fall off the cable. ;-)

    Cheers!
    Rich
     
  8. Tim Shoppa

    Tim Shoppa Guest

    I don't think that is true for the cable, but the line + transformer
    at the far end of an Ethernet cable meets that definition.

    Tim.
     
  9. Guest

    I hate interleaving stuff but just have done :)

    No. It just -looks- like something interesting might be happening if
    you put a number of CR segments up on a spice screen. I.e it looks
    like a multipole low pass filter.
    If you plug some real R+C numbers in you'll see this would have a
    break frequency up in the THz area, at this frequency you'd anyways
    have to include the series inductor bits and yet many more LCR
    sections and then you'd notice the whole thing starts to look just
    like a resistance (Ro) with a bit of a time delay. Ro= Squareroot(L/C)
    Low frequencies and wiring looks just like a capacitor (say 100pF per
    mtr). HF it acts like a parallel resistance.
    Yes. The pull ups can be terminators. 100ohms say will drive a fair
    bit of cable.
    Those quoted bending specs' are just mechanical aspects. Don't worry
    about 'em.
    The theory is correct. Electrons don't like being herded around tight
    corners, they start bleeting and running all over the place. Like cars
    on a test circuit they need a banked track to keep cornering forces
    under control and is why PCB traces are better with a curved fillet
    rather than a straight corner. But and a big BUT, this only becomes of
    interest up in the GHz area and does not apply to us mere mortals who
    use normal data rates.
    [Damned internet is full of info, most of it provided by theoreticians
    with no clue as to the context of a real world.God knows how many
    people they've turned away from engineering subjects.]
    Yes. It'll work fine.
    If you ever get a chance in the future, try out the RS485 type chips
    and balanced pair wiring.
     
  10. Jasen Betts

    Jasen Betts Guest

    no, at the bottom of the transistor) also it needs a resistor in
    series with the base (10 should be good)
    do you have any nodes there? a half-powered one at each end is probably superior to
    one in the middle)
    I think you may need to boost the signals a bit (eg: use rs-232
    line drivers) to get that sort of bit-rate over 40m
     
  11. whit3rd

    whit3rd Guest

    Yes, of course, if the baud rate is low enough for the
    length of wire used. SCSI signalling at 1 MHz and up to
    six meters length used similar wiring with OC drivers.

    Cat-5 cable has 110 ohm characteristic impedance, so a
    near-ideal operating condition requires 110 ohm terminators at
    the two cable ends; few gates can drive a 55 ohm load,
    so drivers/receivers are usually employed.
    That's called a 'stub'; it is not recommended. Can you just
    drive a second bus instead?

    To save power, it can work out that instead of true terminators, a
    snubber (110 ohms
    in series with a capacitor) can be placed on the ends (at high
    frequency, this is a 'correct' terminator), with a current source
    pullup
    anywhere in the middle of the bus. As far as a 110 ohm line
    is concerned, a simple 2k ohm pullup resistor is a 'current source'.
     
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