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 . 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. -- http://www.piclist.com hint: To leave the PICList mailto:piclist-unsubscribe-request@mitvma.mit.edu