need help sensing a disconnected ground

Discussion in 'The Projects Forum' started by jcbeck84, Aug 29, 2009.

  1. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    I am going to start by saying I'm a mechanical person and not and electrical, but I am really confused on this one. I have a valve control that uses a simple 5 kOhm pot to control the desired speed, however if the ground from this pot to the controller gets disconnected anywhere between the signal the controller sees from the pot is a full tilt give it all you got. seeing as how this valve controls hydraulics to some very heavy and expensive machinery this could potentially be a very bad situation. Basically I need to know if there's a way I can sense if that ground wire is still connected to the pot. The supply wire is 5V 10 mA max, the sweeper is 0-5V around .18 mA and the ground in question has a constant 1mA at a potential so low I can't even use it to signal a transistor. Is there any way I can electronically tell if this ground is connected without corrupting my sweeper signal? Please give me any ideas you've got.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Monitor the current in the wire on the ground side of the potentiometer.

    Since your supply to the high side of the pot is 5v, you should always have 1mA current flow through the ground side of the pot. If the ground side current falls much below 1mA, you know that there is an open circuit.
     
  3. eblc1388

    Senior Member

    Nov 28, 2008
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    You don't.

    Use redundancy, e.g. multiple cables, different cable routes, screwed cable connection panel etc... to prevent the ground wire from ever being disconnected while the plant is in service.
     
  4. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    The hydraulic powerpack is a mobile sort of system and the potentiometer is in a pendant connected to the main box via a cable. So there is only one route from the controller to the potentiometer. How do you go about monitoring the current in the ground without applying any resistance in the line and choking off the current? I do have that 1 mA but the potential on it is extremely low.
     
  5. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    If you can stand losing a bit of the maximum setting, you could add a PNP transistor in line with the +5V feed to the top of the pot.

    Connect the emitter to the +5V supply and the pot circuit to the base, with a resistor across base to emitter. If you used 1k5 that would take roughly 0.5mA before the transistor started to turn on, so about 0.5mA base current with the 5K pot as a load.

    You could then have the LED of an opto isolator (eg 4N32) with a 470 Ohm resistor in series from the transistor collector to 0V/Ground.

    That would effectively give you a volt-free contact that could feed PLC input be or used to drive a small relay with the 'OK' signal.

    The whole 5V circuit should be within your 10mA current allowance.

    Generally with any variable control like that, the 'maximum' will have some sort of calibration and you may be able to reset it so the 'new' maximum setting of the pot gives the same result in the equipment - or possibly the abolute maximum is never used and the slight change won't matter.
     
  6. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    The controller does have a calibration procedure so I could recalibrate it to get my top end back. I will give this a shot and let you know how it goes. Thanks.
     
  7. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    I gave the circuit a test run on the breadboard and it ran just as expected. I'm using this optoisolator: http://www.toshiba.com/taec/components/Datasheet/4N29DS.pdf
    and wondering if my understanding of it's limitations are right. Basically I need someone to explain the difference between the Collector-Emitter voltage and Emitter-Collector voltage? Maybe this is one of those obvious things... At any rate, if I am limited to 5 volts through the detector I will need to boost that to operate a relay and take action on a disconnected ground. This could be done with an NPN transistor using the positive signal from the optoisolator at the base, relay ground to the collector, and actual ground to the emitter correct??? If no then maybe someone could point me in the right direction with these transistors.
     
  8. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    Hi,
    the 30V rating is the relevent one, the 5V rating is the reverse polarity breakdown voltage.
     
  9. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    You may want to consider all disruptions for a practical approach. Say you just loose common, or maybe just power, or both, or your power and common are shorted together.

    If your concerned about top end runaway, then maybe consider increasing the control voltage to the pot, add a series resistor to bring the control back to working range, and add a higher than working voltage trip. This way, an open return and a shorted power to control is covered.

    Last but not least, is don't forget your E stop
     
    Last edited: Sep 3, 2009
  10. CDRIVE

    Senior Member

    Jul 1, 2008
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    Have you considered something like this. RS sells some very low voltage/current Piezos. Did you check to see if there are any unused conductors in the pendant cable? The Piezo can also be physically located at the controller.
     
  11. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    I apologize for not getting back to the previous posts. Apparently I don't get sent any more update emails after so long or something.... At any rate, the disconnected ground is the fault that results in something very bad happening as the controller see's full speed and goes for it. Everything else results in zero speed or shut down. The only fault that isn't covered is short from +5V to sweeper which is unlikely enough that I'm going to not worry about it. The buzzer is a good idea except it wouldn't do anything to stop the system from ramping up and out of control.

    I would like to know rjenkins how this little circuit works exactly, which we know it does. My understand of transistors is that when you apply a lower voltage to the base current flows from emitter to collector on a pnp transitor. But according to this circuit only a small amount is flowing that way while the rest is going to the base???
     
  12. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    Hi,
    transistors are effectively current operated, the base-emitter junction appears as a diode so the voltage drop does not change much once it starts conducting.

    You are correct that normally the base current would be some fraction of the collector current, but in this application the current through the pot circuit is quite low and well within the capability of the transistor.

    It's probably got high enough base drive to switch a relatively heavy collector load, but having a smaller load does not affect whether the transistor is on or off, it's just a minimal-component way of doing the job.
     
  13. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    I think I got it. The minimal difference in voltage between the emitter and base allows only a small amount of current to flow through the transistor preserving as best as possible the current and voltage for the pot??? And the resistor after the transistor is to lower the voltage so the isolator doesn't get ruined???
     
    Last edited: Oct 8, 2009
  14. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    Was I right?
     
  15. CDRIVE

    Senior Member

    Jul 1, 2008
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    I believe that this is close to what Robert was describing to you. The switch is not required and was used only to simulate an open ground return.
     
    Last edited: Oct 9, 2009
  16. CDRIVE

    Senior Member

    Jul 1, 2008
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    I just realized that the circuit that I posted could have supplied you a 5V control voltage directly off the Collector of Q1 connected to a 10K to ground. Here's a variant circuit using a relay and dry contacts. I modified that schematic at the time and date of this edit.
     
    Last edited: Oct 9, 2009
  17. CDRIVE

    Senior Member

    Jul 1, 2008
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    As per jcbeck84's request via PM this is the theory of operation for the CurrentSensor schematic.

    Condition: Ground Return Is OK.

    In order for current to flow through the Pot it must pass through the Emitter-Base Diode junction of Q1. You will loose ~ .6V across this junction, leaving ~4.4V across the Pot and ~ 880uA flowing through it from ground. Bipolar Transistors are current sensitive devices so this very small current will turn the Emitter-Collector junction on, similar to a switch. This, in turn, allows current to flow through the Opto LED which then lights it. The light from the LED strikes the photo sensitive Base junction of the Opto's NPN Transistor which turns the Emitter-Collector junction of the Opto Transistor on. Because this transistor is on current flows through the coil of Relay K1, which closes it's contacts.

    Condition: Ground Return Is Open:

    No current is flowing in the Emitter-Base junction of Q1 causing Q1's Emitter-Collector junction to also switch off. This causes the LED to go out, causing the Opto Transistor to shut off and thus depriving the K1 of coil current, which opens the contacts.
     
  18. jcbeck84

    Thread Starter Active Member

    Dec 20, 2008
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    Regarding Q1 as this is where my confusion is: Starting with 0V everywhere in the circuit, you turn it on and 5V comes in the supply line. The supply line via the 1.5K resistor applies a small voltage and current to the base of the transistor. Once this voltage (or current?) is applied current is allowed to flow from the emitter to the base across the diode junction. This is where my understanding drops off. This current the flows OUT(?) of the base and through the pot at 4.4V and 880uA allowing current to also flow through the base collector connection. Why is this current/voltage going from emitter to base and through the pot as opposed to emitter, base, and collector??? Also, what is the function of R7?
     
  19. CDRIVE

    Senior Member

    Jul 1, 2008
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    The 1.5K (R5) does not aid in forward biasing the Base-Emitter junction. It does the opposite in fact. Base to Emitter resistors are used to give BJT's stability over temperature, which is the same reason I put one on the Base to Emitter junction of the Opto Transistor. They are not absolutely necessary in switching operations (I use them as habit) but they are a must when a transistor is biased as an analog amplifier.

    In the electronics community we subscribe to the theory that electron current flows from negative to positive. Some industries, like electricians, subscribe to the positive to negative theory. I matters little which you follow, as long as you don't mix and match them. ;)

    There would be little point in me posting graph plots which depict Bipolar Transistor action, as we have sections for that on the forum. We have information showing transistors as switches (non linear) and as linear amplifiers. Can anyone please point JC to some links?
     
  20. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    It may be simpler to think of it in a more conventional context.
    Turn my original circuit upside down so the transistor emitter is connected to the 0V side and imagine it's an NPN so all polarities are reversed, the supply to the pot and collector load is +5V.

    You now have a simple transistor switch, controlled by the base current and with a load in the collector side.

    The base circuit (current through the pot) turns it on & breaking that circuit turns it off.

    The base-emitter resistor ensures it does not turn on until a small but significant current flows, so leakage such as moisture in a damaged cable cannot keep the system on.

    The actual operation is exactly the same, but using a PNP transistor so all polarities can be reversed to suit the application (sensing loss of ground rather than loss of power).
     
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