Minimum frequency on Cat.5

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.

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:

'Most Category cables require a minimum bend radius of "four times'

'Bend Radius: 25.4mm'

'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.

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?

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

Correct - although the practical effects are much more prevelent at high frequency - crosstalk failures when testing to CAT6 and sometimes 5e

--
Hal

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.

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

Cheers! Rich

kHz.

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.

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.

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

w"

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'.