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50-AMP remote meter needed

Discussion in 'Electronic Basics' started by ltj, Nov 14, 2003.

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  1. ltj

    ltj Guest

    We have a large industrial motor that draws up to 50 AC amps that we need to
    monitor in the control room 100 feet away. Looks like a current
    trransormer takes our 50A and drops it to a 5A (full scale) level, which we
    would run to a panel meter. I'm not sure this is a good way to go. Perhaps a
    "permanent" clamp-on amp meter that outputs something besides mA-A signals
    would be better? Looking for suggestions...thx
     
  2. Tom Woodrow

    Tom Woodrow Guest

    you could use a meter with both a clamo-on amp meter and an RS232
    interface and set it up to poll the meter up on a regular basis.

    Tom Woodrow
     
  3. There are some pretty inexpensive Hall-effect-based current to voltage
    converters available. They tend however to be a little inexact. You did not
    tell what you need the data for. If it is just to see if some known and
    constant level(s) has (/have) been exceeded, the easiest solution would
    probably be such a sensor connected to a (/multiple) comparator(s) with the
    inverse input on the needed fixed level(s). When the limit is reached, the
    comparator will output series of 50Hz pulses that can easily be used to
    trigger some sort of monitoring circuitry. The duty cycle of these pulses
    will also provide a rough overview of how far the limit has been exceeded.
    More exact current measurements are also possible, but you will need
    converters that do not 'drift' too much. A current transformer will also
    work, but with considerably higher losses due to heating.

    Dimitrij
     
  4. Steve

    Steve Guest

    If all you need it a method to read the current on an anolog meter, there
    are pick up loops that you just put on the motor lead and the loop windings
    measure the field going through that lead.. Very easy to hook up to a remote
    meter..

    Just check with a industrial electrical supply house.

    Steve
     
  5. Luhan Monat

    Luhan Monat Guest

    Running large amounts of current over that distance is not usually a
    good idea. Put a 'current to voltage' transformer (little xformer with
    large hole in the middle) over one of the AC wires and connect a
    voltmeter remotely. Now you have only low voltage, low current running
    back to the control room. Use a voltmeter and the specified load for
    the transformer to read the current.
     
  6. The low tec way to do this is just as you have described, using a pair
    of 14 ga wires. If you want to go higher tech, get an AC current to
    4-20 ma transmitter and read that signal with a scalable numeric
    display. Yokagowa makes a display that includes the 24 volt supply to
    power the transmitter over the signal wires (called a two wire
    transmitter) with an AC power connection at the meter. But lots of
    companies make the transmitters and displays. If you suspect that you
    might ever want to connect this signal to a DCs or other industrial
    control system, its input can be driven in series with the panel
    display.
     
  7. John Fields

    John Fields Guest

    ---
    There is no such thing as a "current to voltage converter" transformer.

    What you're describing is using what's called a "current transformer"
    and a "burden resistor" to generate an AC voltage which is proportional
    to the current flowing in the primary, which is the single wire running
    through or looped through the center of the toroidal core.

    Using it as you've described is _not_ a good idea in that if the
    secondary circuit is broken anywhere along the 200 feet of cable high
    voltage will appear at the break.

    Nor would it be a good idea to place the burden resistor directly across
    the secondary unless the resistance of the volmeter and the run of wire
    is taken into consideration.

    More appropriate would be the use of a 4-20mA transmitter and receiver
    as John Popelish has suggested.
     
  8. There is no need to run currents over that distance. If the shunt is
    located at the motor, then 49.999 amps go thru the shunt, and only 1
    mA (assuming a 1 mA meter movement) has to travel remotely to the
    meter. If the meter is insulated and the wiring to the meter is
    capable of handling the voltage, then there is no problem because the
    I squared R voltage drop is negligible at that current. Just make
    sure that the circuit is fused with low current fuses at the shunt.

    The transformer solution does have the advantage of isolation, so it's
    a more attractive, even tho much more expensive, solution.


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  9. Fred Abse

    Fred Abse Guest

    Sorry, John, beg to differ here. Any transformer having current in the
    primary will induce a voltage in the secondary. There is no need for a
    burden resistor. OK, I know there's no such thing as an infinite impedance
    measuring device, but a 200 meg input DVM is reasonable approximation, and
    I've used one of those with Rogowski coils before now .......

    The modern type of feedback-linearized Hall effect current converter
    *does* need a burden resistor, since you're actually measuring the
    feedback current.

    ISTM that's as near current to voltage as dammit.
     
  10. John Fields

    John Fields Guest

    ---
    If a shunt is used at the motor, then the meter and the wire going to it
    from the shunt will appear as three series resistors in parallel with
    the shunt and the current going through the meter will be I = E/R, where
    E is the voltage dropped across the shunt and R is the resistance of the
    meter and the wiring connecting it to the shunt. Since the shunt will
    look like a very low impedance voltage source it will only be capable of
    supplying current into the meter depending on the voltage which it drops
    with the motor current flowing through it.

    For example, if we have a 50A 100mV shunt it will supply 1mA into
    50mV/1mA = 50 ohms.

    I believe most 1mA AC meters will have an internal impedance of around
    1000 ohms, so that makes it a non-solution.
     
  11. John Fields

    John Fields Guest

    ---
    I suspect that if you were to use a current transformer with an unloaded
    secondary in this application you'd be in for a nasty surprise!

    Let's take, for example, a current transformer with a 1000 turn
    secondary and a single turn primary (an unlooped wire going through the
    center of the toroid) with 50A going through it. Since the turns ratio
    of the transformer is 1000, 50A flowing through the primary will try to
    induce 50/1000 = 50mA in the secondary. So, since E = IR and the
    unloaded secondary looks like an open circuit, the voltage induced in
    the secondary will be E = IR = 0.05A*infinite ohms = infinite Volts!

    Looking at it a little more realistically, with your 200 megohm load on
    the secondary the voltage would only rise to 5.0E-2A * 2.0E11 ohms =
    1.0E11V = 100,000,000,000V, but it might still be a good idea to load
    the secondary down a little more heavily to get the voltage down to
    where your meter can measure it on a more conveniently available range.
    ;)
     
  12. Luhan Monat

    Luhan Monat Guest

    Putting the resistor directly accross the secondary would work fine. The
    wire resistance is too low to effect the voltmeter reading, and the
    voltmeter's effective resistance is way to high to effect the
    transformer's load.
     
  13. John Fields

    John Fields Guest

    ---
    Sloppy work.

    You should cite the wire size and the internal impedance of the
    voltmeter before running off half-cocked. And, it's 'affect', not
    'effect'.
     
  14. Luhan Monat

    Luhan Monat Guest

    You must be using 'voltmeter' in a diffent sense than originally
    proposed. I'm talking about a voltmeter as you would use as a test
    instrument, not just some 'meter' that reads out in 'volts'.

    Most commercial voltmeters present at least 1 megohm to the circuit they
    are measuring. Hence, wire size and loading are not significant.
     
  15. Luhan Monat

    Luhan Monat Guest

    Also, to be fair, given that you were using the other meaning of
    'voltmeter', your original comment is entirely logical.
     
  16. John Fields

    John Fields Guest

    ---
    I think it's much more likely that a passive meter would be used as a
    full-time monitor, reducing that 1 megohm considerably.
    ---
    ---
    Depends totally on the voltmeter, the wire size, and the current
    transformer.

    Take a look at a 0-5VAC iron vane panel meter (Simpson 2153) wired to a
    current transformer with 24AWG wire.
     
  17. John Fields

    John Fields Guest

     
  18. Don Kelly

    Don Kelly Guest

    ----------
    Sorry, I don't follow that reasoning.

    A 50A, 100mV shunt will produce 100mV at 50A and have a resistance of 0.002
    ohms unless it is tailor made for a specific meter as some older Weston
    shunts were. As long as the meter and connecting wires have an resistance
    which is orders of magnitude higher than that of the shunt, there is no
    problem.
    It is not designed to drive a specific current in the meter and, in fact,
    the lower the meter current, the better. Use a millivoltmeter and label the
    scale so 100mV =50A.
    The only problem may be that a shunt is not normally used for AC but as this
    appears to be for indication only, that should not be a problemWhether AC or
    DC, in an industrial environment it might be a good idea to use shielded
    and/or twisted leads.
     
  19. John Fields

    John Fields Guest

    ---
    Actually, there is.

    A 1mA meter with an internal resistance of 1000 ohms will require 1V to
    drive it to full scale, which a 100mV shunt will be unable to produce.

    Watson's suggestion to use a 1mA movement, therefore, won't work unless
    the meter has an internal resistance of R = E/I = 100mV/1mA = 100R.

    Looking around for solutions, though, a Simpson 05400 iron vane 50mA
    AC milliammeter

    http://www.simpsonelectric.com/pdf/webpdfe/Round AAC.pdf

    with an internal impedance of 80 ohms might work with a 20 ohm series
    resistance, which could be made up by the wire run. 200' of 30AWG comes
    out to about 20 ohms, but it's kind of flimsy so a better solution would
    be, say, 200' of #20 (or #18, or #16...) and a 50 ohm rheostat.
    ---
    ---
    The problem with going that way seems to be that there seem to be no
    passive AC panel meters rated for 100mVAC full scale. At least I've had
    no luck at all in finding one. I suppose putting a step-up transformer
    primary across the shunt and measuring the voltage on the secondary
    would work.
     
  20. ltj

    ltj Guest

    Great feedback! ... I'm learning a lot
    thx
     
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