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Need help with current sensing circuits

Discussion in 'Electronic Basics' started by Patrick, May 31, 2005.

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  1. Patrick

    Patrick Guest


    I was wondering if anyone has or knows where I can find a simple, cheap
    current sensing circuit that I could use to sense AC currents. I am
    currently working on a power control project that will monitor 120 VAC
    outlets with a total maximum current of 15 amps. We are currently
    working with a current sensor that outputs an AC voltage (actually a
    varying DC voltage since it does not drop below zero) from an AC input
    current from 0-50A. When the input current is zero the output is about
    2.5V. At the full rated input current of 50 amps the sensor output has
    a peak voltage swing of about 2.5V which ranges from zero to 5V at the
    full 50 amps (swings around the 2.5V q-point).

    The current sensor works great, but the problem is that we need to find
    out what the actual current is and this poses somewhat of a challenge
    since the output of the sensor is a varying DC (sinusoidal) voltage.
    We've tried some software techniques to capture the peaks, but it takes
    up too much processor time and is therefore not feasible.

    Anyone have any ideas or circuits I might be able to try? Any help
    will be greatly appreciated.

  2. Patrick,

    What about the electronic skills of your team? A capacitor is enough to
    separate DC from AC although it might be a huge one for 50/60Hz. If that's
    not applicable a simple opamp suffices to subtract the 2.5VDC from the
    signal leaving the AC-part for further processing. Of course you can find
    complete measuring systems on the net. Tektronix sells some great ones but I
    bet they are pretty expensive in relation to your budget. Current sensing
    devices with Hall effect sensors often has a half Vs offset. So if you buy
    one you will have the same problem. AMPLOC makes them for instance. Some of
    them however have integrated electronics to remove the offset.

    petrus bitbyter
  3. neil

    neil Guest

    dragging up my limited analog knowledge ...
    how about demodulating the signal, and smoothing/processing the resultant dc
    level ?
    - use a second ac signal as a reference, using it to control a two-way
    analog mux, using that to route either the true or inverse (via inverting
    opamp) ac signal.
    No doubt there are suitable circuits available on web.
  4. kinyo

    kinyo Guest

    There's an opamp circuit called precision rectifier. Use the rectified
    voltage to charge a capacitor to read a DC voltage instead of AC. You
    will need of course a resistor across the capacitor to discharge it to
    a lower DC voltage when the current being measured drops.
  5. Bob Monsen

    Bob Monsen Guest

    I'm not sure I understand your problem. However, take a look at the
    AD636. It is a device which will give you a DC output equal to the RMS
    value of a waveform. Thus, pass the 2.5V through a capacitor, then run
    it into this thingy. You'll need to calibrate it, but it should give you
    a good view of the current when coupled with your current sensor.
  6. Chris

    Chris Guest

    Hi, Patrick. From your description, I would guess you might have an
    Allegro ACS750SCA-050, which has a zero ohm current shunt and attached
    Hall effect circuit to detect current going through the shunt. When no
    current is going through the shunt, its output is Vcc/2, or 2.5V. When
    current is going through the shunt, a voltage is superimposed on this
    DC level of +/-40 mV per amp up to +/-50 amps (which gives an amplifier
    output of 0.5V to 4.5V). It has a -3dB response at AC current up to
    13KHz, and a combined guaranteed accuracy of +/-2% at room temperature.

    You have been very unclear on exactly what you want. You've said
    you're using some kind of PC, SBC or uC to read the voltage, but the
    repetitive readings are taking up too much processor time. You'd like
    a simple, inexpensive means to do what you're doing now, but you
    haven't been clear on just what you're doing now, how simple, and how
    inexpensive. "Measuring current" can mean just measuring peak current,
    measuring average current or RMS current. You haven't indicated
    exactly what information you need, or in what time frame you need it.
    Do you need information every half cycle, every cycle, a certain number
    of times a second? Oh, and by the way, what kind of accuracy do you
    need? Your error budget is already over 2%. Do you need a combined
    accuracy of 2.1%, 3%, or 12%? The key to a good answer is a well-asked

    Your lowest cost solution for doing just what you're doing now would be
    using a PIC with built-in ADC. You can read the voltage output off the
    ACS750 many times for every AC cycle (generate an interrupt off the
    power supply transformer secondary?), and then either figure out the
    peak current (if that's your pleasure) or add 'em up to integrate total
    current. There are many ways to communicate that information to
    whatever you've got as a processor. And there's no law against having
    two processors in a piece of electronics.

    If you only need information on peak current on only half the cycle,
    you're only taking a few readings a second, and you can live with
    several percent accuracy, the simple and cheap answer is to use a
    schottky diode feeding a cap with a bleeder resistor, like this (view
    in fixed font or M$ Notepad):

    | ACS750 |
    | |
    | |
    | |
    | |
    | | 1N5817 To high Z-in uC ADC
    | |\ |
    | -| >---o----->|-o----o--->
    | |/ | | |
    | | --- .-.
    | | --- | |
    | | 0.1uF | | |1M
    | | | '-'
    '--------------' | |
    | |
    === ===
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    By using software compensation for the diode drop, you can achieve
    accuracy within several percent. The obvious flaw here is that when
    the current decreases, it will take half a second to get an accurate
    reading. If that's OK with you, that's the simplest. You can juggle
    settling time and decay time by changing resistors to get something
    closer to what you want.

    There are a lot of solutions to your problem. The Analog Devices chip
    mentioned in another post being an elegant one chip solution, and while
    not the least expensive, will give you the best, most accurate answer.
    The precision rectifier is a good answer which solves the issue with
    the voltage drop of the diode above. There are many other answers.
    But you could give a little more information if you want to get the
    best answer.

    Good luck
  7. Chris

    Chris Guest

    Small error. Try this:

    ` From ACS750 To ADC
    ` 1N5817
    ` >--->|-o-----o--------->
    ` | |
    ` --- .-. VCC
    ` 0.1uF--- | | +
    ` | | |1M |
    ` | '-' .-.
    ` === | | |1K
    ` GND | | |
    ` | '-'
    ` | |
    ` '------o 2.5V
    ` |
    ` .-.
    ` | |1K
    ` | |
    ` '-'
    ` |
    ` ===
    ` GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    Sorry. Less haste, more speed.

  8. Chris

    Chris Guest

    Sorry-- the first circuit is the one you should use. The multitasking
    threader is defective tonight -- long week.

  9. Will the leakage through the big power Schottky be a problem here? The
    current through the diode is going to be very small, microamps. The
    datasheet for the 1N5817 says that the reverse leakage current is around 100
    uA at 4 V though. My simulation doesn't rectify; the diode drop is
    negligible, in both directions... (but of course it works fine with a 1N4148
    or something, or I guess you could make C bigger and R smaller).

    I think R*C should be bigger anyways, because with a not-leaky diode there's
    still about 0.3 V of ripple on the output for a 1 V amplitude input. I think
    the ripple is roughly given by

    dV/dT*T = (I/C)/f
    = ((Vout/R)/C)/f
    = Vout/(R*C*f)
    = 3/(1meg*0.1u*60) = 0.5 V which is kind of close

    (assuming that the current through R is constant, and assuming that the
    capacitor spends almost all the cycle discharging and hardly any time

  10. mike

    mike Guest

    I know people are tired of hearing me ask, but "What are you trying to
    If you have a purely resistive load and a pure voltage sinewave, you can
    measure about any parameter of the current waveform and know all the rest.

    Problem is that most applications are not even close to that ideal.
    In most cases, you have to actually measure the parameter of interest.
    Some typical parameters might be peak, average, RMS values. Phase
    may or may not be important.

    If it's a current sine wave and you can determine the zero crossing
    with a comparator and you know the frequency, you know exactly were
    the peaks are and you can sample there.

    If it's not a current sine wave, you can peak detect and subtract off
    the 2.5V either with an op-amp or software.

    What do you want to do about noise?
    How fast do you need your loop response?

    All comes back to, "what do you really want to know about this current

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  11. Bob Monsen

    Bob Monsen Guest

    <snip suggestion>

    I agree that a simple scheme may work properly. One thing to factor into
    the discussion is that a microprocessor pin is a good way to zero a peak
    detector. He can use a shottky diode and cap, like you suggest, but skip
    the drain resistor. Before each sample, just use another pin to drain
    the cap. Then, make the pin an input, wait a few cycles, and take the

    Another thing to consider is that some micro ADCs want fairly low input
    impedance; the PIC 12 and 16 series, for example, want 10k.

    Yet another issue is that this won't get him the right value if that
    2.5V center voltage drifts around. He can fix that by using a big cap to
    make the 5VAC signal be centered at zero. Then, his answer will be
    between 0 and 2.5V, minus the drop of the diode.

    If he wants to add a couple of cheap opamps, he can eliminate that diode
    drop as well by building a simple peak detector. However, that will
    involve more external hardware.

    None of this really works, however, if the current isn't a sine wave.
  12. William

    William Guest

    make a drawing at ( 100 dpi will do fine with a scanner)
    ; go to the mailbox the ID is
    'schetsen' and the password is 'diverse'. send the drawing to the same
    mailbox ''
    and I'll see what I can do for you ( for free of course )
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