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Ammeter internal current draw

Discussion in 'Electronic Basics' started by Ken C, Mar 2, 2006.

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  1. Ken C

    Ken C Guest

  2. John Fields

    John Fields Guest

    ---
    This is really a digital millivoltmeter reading the drop across the
    shunt and, typically, the input impedance to these voltmeters is
    about 10 megohms, so the current in the leads (and, thus, the drop
    across them) will be negligible compared to what's going through the
    shunt.
     
  3. Ken C

    Ken C Guest

    Thank you. I still don't think I have enough information to avoid
    trial & error.

    The voltage drop along the meter leads will indeed be very small in
    absolute amount, but not as a fraction of the voltage drop across the
    shunt -- which will also be very small.

    I think I need to know either the typical current through the meter
    leads or the typical voltage drop across this kind of shunt.

    I don't have instruments accurate enough to measure such small
    variables. My only recourse is to try different lead lengths and to
    see what happens to the readings.

    Ken C
     
  4. John Fields

    John Fields Guest

    ---
    Not true. Let's say the shunt drops 300 millivolts with 30 amps
    across it.

    With 30 amps going through the shunt, your circuit will look like
    this:


    30A--->
    +V---------+
    |
    [RL] 300mV
    | /
    +--------+
    | |
    [SHUNT] [METER]
    | |
    GND--------+--------+

    and if your meter impedance is 10 megohms, the circuit will look
    like this:

    30A--->
    +V---------+
    |
    [RL] 300mV
    | /
    +--------+
    | |
    [SHUNT] [10MR]
    | |
    GND--------+--------+

    Now, 300 millivolts across 10 megohms will cause:


    E 3.0E-1V
    I = --- = --------- = 3.0E-8A = 300nA
    R 1.0E7R

    to flow through the wires connecting the meter to the shunt.

    Now, just for grins, lets say that you've got the meter located 1000
    feet away from the shunt and you've got them connected up with #24
    AWG copper wire. That's a total length of wire of 2000 feet, and a
    total resistance of 51.4 ohms, so the voltage that'll be dropped
    across the wire will be:

    E = IR = 3.0E-8A * 51.4R ~ 1.54E-6V = 1.54µV,


    which is:


    1.54E-6V
    ---------- * 100 ~ 0.00051% of full scale.
    3.0e-1V

    the meter will have, best case, a resolution of 1 part in 2999,
    which is:


    100
    ------ = 0.0333%
    2999

    so you can see that the effect the length of the wire is going to
    have on the system will be miniscule.
     
  5. John Fields

    John Fields Guest

     
  6. Ken C

    Ken C Guest

    Thank you very much. You are a gentleman and a scholar (no kidding!).

    Ken C
     
  7. Ken C

    Ken C Guest

    Will I minimize the antenna effect if I use coax for the signal leads,
    like RG-174? There will be RF radiation (1.8-432 mHz) in the
    immediate vicinity. RF currents in conductors is a mystery to me and
    your experience would be invaluable.

    Ken C
     
  8. Chris

    Chris Guest

    Hi, Ken. The input impedance of your ICL7107-based digital panel meter
    is almost certainly 10 megohms. Mr. Fields is correct -- a longer line
    might act like an antenna, picking up electrical noise. It sounds like
    he's been there before. I have, too.

    The ICL7107 is very good at cancelling out 60Hz noise, but less good at
    other frequencies. If you've got a longer wiring length from the shunt
    resistor to the DPM (say, over 6 feet or so), you've got a close source
    of RF energy, or if your wiring is going through wiring duct with high
    current lines (could cause inductive pickup), you'll want to provide
    yourself some protection for your DPM input.

    Note here that you will need a separate, floating power supply for the
    DPM -- they work that way. Unless special circuitry is included on
    board, the 7107 can't measure its own power supply.

    Omitting the power supply, here's what your basic circuit should look
    like (view in fixed font or M$ Notepad):

    |
    | ___
    | .---o-o---|___|---o-o----.
    | | |1 milliohm | |
    | | | | |
    | | | _ | |
    | +| | + / \ | .-.
    | --- '---(DPM)---' | |
    | - \_/ | |1 ohm
    | 12VDC| '-'
    | | |
    | '------------------------'
    |
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    Your "30 amp" shunt is almost certainly a 1 milliohm resistor, set so
    the DPM will read 29.9 for 29.9 amps. One of the easiest ways to
    reduce the effect of electrical noise is to provide a load resistor
    across the DPM to directly provide a lower input impedance:

    |
    | ___
    | .---o-o---|___|---o-o----.
    | | |1 milliohm | |
    | | | | |
    | | | | |
    | | | | |
    | | | ___ | |
    | | o---|___|---o |
    | | | 220 ohm | |
    | | | _ | |
    | +| | + / \ | .-.
    | --- '---(DPM)---' | |
    | - \_/ | |1 ohm
    | 12VDC| '-'
    | | |
    | '------------------------'
    |
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)


    This simple step helps most problems. But for higher energy
    interference, you can either use protection diodes or a load capacitor
    here.

    | ___
    | .------o-o---|___|---o-o----------.
    | | |1 milliohm | |
    | | | | |
    | | 100 ohm| | |
    | | .-. | |
    | | | | | |
    | | | | | |
    | | '-' | |
    | | | 2 X | |
    | | | 1N4001 | |
    | | | | |
    | | o-----|<----o |
    | | | | |
    | | o----->|----o |
    | | | _ | |
    | +| | / \ | .-.
    | --- '---(m V)---' | |
    | - \_/ | |1 ohm
    | 12VDC| '-'
    | | |
    | | |
    | '---------------------------------'
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    The diodes here are valuable, because they limit any voltage excursion
    to 600mV or so, which is overrange for the DPM, but won't cause damage.
    This setup is good for inductively coupled noise, which could be
    higher energy.

    Here's another setup, which is probably better for "antenna-type"
    noise:

    | ___
    | .------o-o---|___|---o-o----------.
    | | |1 milliohm | |
    | | .-. | |
    | |220 ohm| | | |
    | | | | | |
    | | '-' 0.1uF | |
    | | | || | |
    | | o----||-----o |
    | | | || | |
    | | | _ | |
    | +| | / \ | .-.
    | --- '---(m V)---' | |
    | - \_/ | |1 ohm
    | 12VDC| '-'
    | | |
    | '---------------------------------'
    |
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    Make sure you use a good ceramic cap which is made to be low impedance
    at your frequencies of interest. And attach the resistor right at the
    DPM.

    Or use a combination of these. Component selection kind of depends on
    additional information, like the length of your wire run, other
    potential sources of noise, &c.

    By the way, coax won't hurt, but you should be able to use plain
    twisted pair for just about any application.
    don't have the equipment to track down problems here. It might be best
    just to build in protection, test it well, and then test it better. If
    it doesn't cause problems, it's probably good.

    Many times, people forget that there's another solution to this type of
    problem. If you have problems with other sensitive electronics in this
    installation, it might be better to go with a time-tested solution
    that's impervious to RF energy and doesn't require an external power
    supply -- namely, an analog meter with external shunt. Your total
    installed cost will be competitive, even if you didn't have any
    problems with your DPM. If you have to spend hours of fiddle-factor
    time on your DPM, the analog meter comes out way ahead. Assuming
    everyone who's using the meter can actually read analog meters, you can
    just install it, enjoy a quick win, and move on to the next problem.

    Since your DPM is on the low end of its range (you're only going to be
    using 299 counts of the 2000 count full scale range), a good analog
    meter would probably be just about as accurate as your DPM, too.

    Good luck
    Chris
     
  9. Rich Grise

    Rich Grise Guest

    I wouldn't use coax cable for something like this - especially since
    the meter is isolated (i.e., floating) I'd personally be more comfortable
    with twisted pair. But, if you have a pile of RG-174 just lying around,
    then by all means, give it a try!

    Good Luck!
    Rich
     
  10. Ralph Mowery

    Ralph Mowery Guest

    Unless you can ground one side of the coax, it may be worse than other
    methods. It will unbalance the wires as far as RF is concerned. In that
    case the twisted pair would be beter if needed.
     
  11. John Fields

    John Fields Guest

    ---
    If your circuit looks like this:

    Vin>-------+
    |
    [RL]
    |
    +--------+
    | |+
    [SHUNT] [METER]
    | |
    GND>-------+--------+

    which it should,

    Then what I would do would be to decouple the meter from the coax
    line like this:

    Vin>-------+
    |
    [RL]
    | R1
    +----------------+--[1kR]--+---------+
    | | | |+
    [SHUNT] [0.1µF] [0.1µF] [METER]
    | |C1 |C2 |
    GND>-------+----------------+---------+---------+

    Where the 1000 ohm resistor and the 0.1µF caps are mounted at the
    meter end of the cable, as close to the meter as possible.

    At 1.8MHz the reactance of the caps will look like:

    1 1
    Xc = --------- = -------------------------- ~ 1 ohm
    2pi f C 6.28 * 1.8E6Hz * 1.0E-7F

    So even if you wind up with something horrible like, say, 100
    millivolts of RF across C1, your circuit will look like this:


    0.1V
    |
    +------+--E1
    | |
    | [1000R]
    | |R1
    [1R] +-----+--E2
    |X1 | |
    | [1R] [10M]
    | |X2 |R2
    +------+-----+


    Where X1 and X2 are the reactances of C1 and C2, and R2 is the
    resistance of your meter.

    The [AC] voltage at E2 can be determined from:

    E1 X2
    E2 = ---------------
    R1 + (X2||R2)

    Since 1 megohm in parallel with 1 ohm is essentially 1 ohm, we have:

    0.1V * 1R
    E2 = ------------ = 100µV
    1000R + 1R

    Now, assuming that the sensitivity of your meter is 200mV full scale
    and is a 2000 count meter means that its resolution will be:


    200mV
    ------------- = 100µV
    2000 counts

    Voila! we have reached the point where 100mV of RF at 1.8MHz on the
    center conductor of the cable could cause a reading error of 1 LSD
    on the meter.

    As frequency increases, the reactance of the caps will decrease,
    decreasing the amount of RF presemt at the meter, further decreasing
    the reading error.

    One thing: Be sure that the caps you get have a self-resonant
    frequency (SRF) higher than 432MHz, asince if its lower than that
    they'll start looking inductive and could cause trouble.

    Finally, what about that 1000 ohm resistor and the DC drop across
    it? OK, Here's that circuit:


    Vin E1
    |
    [1000R]
    |
    +-----E2
    |
    [10MR]
    |
    GND

    Since:

    E1 R2
    E2 = ---------,
    R1 + R2

    if we set E1 to 1V we'll have:


    1V * 1E6R
    E2 = ------------- = 0.999V
    1E3R + 1E6R


    which is an error of:


    1.0V - 0.999V
    --------------- * 100 = 0.1%
    1.0V

    of the reading, which isn't bad and, since the meter accuracy is +/-
    0.5% should be fine.
     
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