Good idea, Wagner.  Sounds like we're making this WAAAAY too complicated by
going TDR - or even a PIC!!!

Reminds me of my college days in EE class.  We measured lengths of wire runs
by shorting the two wires on the opposite end and using the unknown length
as one leg of a resistance bridge.  Very, very simple math.  I'm sure most
introductory EE textbooks would have the circuit if you really need it.  Or
email me off-line & I'll elaborate (rather than take up everyone's time up
on non-PIC related issues).


----- Original Message -----
From: "Wagner Lipnharski" <wagner@USTR.NET>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Thursday, February 20, 2003 5:42 PM
Subject: Re: [PIC]: How would I build a reflectometer?


> Robert Rolf wrote:
> > If all you want is to measure is cable length, then just measure the
> > cable capacitance (or cable pair C).
> > Knowing your pf/ft it's simple math after you measure C.
> >
> > Many DVM's can measure C directly. Alternatively use the cable C in
> > an RC osc cct and measure frequency or period. The cheap TDR URL
> > posted earlier uses a 74HC14 as the oscillator. Just use the cable
> > for the C and you're done. KISS! Remember to substact out the
> > connecting lead C.
> >
> > Using a 555 and current source for charging C would give you a linear
> > read out that you could display on an analog meter. (One half of 556
> > oscillates using cable C, the other half generates a fixed width
> > pulse which you low pass filter and use to drive meter IOW FtoV). No
> > PIC required <G>.
>
>
> Nothing wrong in measuring capacitance, but most multimeters can't get
> below 100nF.
>
> Other possibility is perhaps build a milliohms meter, simpler to built.
> Most multimeters can give you a resolution of 1mV, so, if you measure the
> resistance of the whole cable spool (500 or 1000 ft) before you use it,
you
> will find out the relative resistance for smaller cut pieces easily.
>
> A simple circuit like that can tell you magic numbers:
>
>
>                        50 Ohms
>              .------.
>    +9VDC-----| 7805 |---R---.
>              '------'       |
>                 |           |
>                 o-----------'
>                 |
>     >-----------o--------> (+)
> Wire                          V meter
>     >-----------o--------> (-)
>                 |
>                _|_
>                GND
>
> The above circuit generates a constant current about 100mA.
>
> An AWG 26 solid gives you 43.6 ohms per 1000 ft.
> Suppose this is your wire.
>
> Connecting both ends of 1 wire of the 1000 ft cable at the [WIRE]
> connectors at the above circuit, the mV meter will indicate 43.6 x 100mA =
> 4.36V
> If this is what you get at the V meter, then the following formula could
be
> used when measuring pieces of this same cable:
>
> Wire Size in ft = mVolts x 1000 / 4360
>
> If you read 0.36V, then it will be
>
>     360 x 1000
>     ----------  =  82.56 ft
>        4360
>
> So, your multimeter has a 1mV resolution, it means you will have a wire
> size resolution of;
>
>     1 x 1000
>    ----------  = 0.22 ft
>       4360
>
> pretty good for a $5 solution, right?
>
> To build your own formula, just measure how many mV you get in the whole
> spool of 1000ft, and replace the 4360 in the above formula by this new mV
> found.
>
> Of course, once the cable is layed, you can't get both sides, so, using a
> simple loop in one pair will solve your problem, but then, replace the 50
> Ohms resistor in the above circuit by 100 ohms to keep the wire V below
5V,
> and remember that you would be measuring the wire distance in double, but
> the formula still telling you the cable length, not wire(s) length,
because
> you are using the 2 wires loop for the spool and for the actual cable
under
> measure.
>
> Suppose using the wire loop, 100 ohms resistor (50mA constant current),
now
> the 1000ft spool (actually 2000ft wire being measured) results in 87.2
ohms
> x 50mA = 4.36 V.
>
> If measuring a smaller cable using the 2 wires connector you find out
0.36V
> (360mV), then
>
>                  360 x 1000
> cable length = ------------- = 82.56 ft.
>                     4360
>
> The tricky is to find a resistor (50, 100 ohms or other value) that gives
> you the biggest number below 5V when measuring the 1000 ft spool.  Of
> course, if your total cable lenght will have more than 1000ft, then select
> the resistor to be able to measure this whole length.  As smaller the
> resistor, better accuracy and resolution at your measurement, but of
course
> will cause overvoltage in longer cables, so, longer cables, higher
> resistors, but don't go below 5 ohms. Always calculate the resistor power
> at least as 35/R in watts.
>
> You would want to take notes of voltage measured at the whole spool using
> diferent resistors, just in case. 20, 50, 100, 150, 200 ohms for example.
> Take notes, then choose a better resistor when measuring the cable and use
> the correspondent values.  Perhaps you could build a rotary switch with
> soldered resistors... multi-range?  a different formula for each position.
>
> Check this link for some idea of Copper AWG vs resistance:
> http://www.thelenchannel.com/1wire.html
>
>
> Wagner.
>
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