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Monitoring a Live, 80W, 12VDC Load

  1. Dec 21, 2009 #1
    Greetings, folks!

    I need easy, visual confirmation of the continuity of a resistance wire-based heating element circuit (~80W, 12VDC): it is simply a loop of Nichrome resistance wire (with a fuse!) connected to a 12VDC battery bank.

    If the circuit breaks somewhere, I need to know ASAP--that heat is essential! As I am completely off-grid on a solar-panel system, I also need to minimize the additional loading of the circuit. Incorporating a single LED would be ideal: cheap, tiny (size and draw), reliable, and visible in all lighting conditions. (The LEDs I currently use are each 12VDC, in series with a 750-Ohm 1/4W resistor.) I don’t need any other information; I only need to know if current is flowing through the circuit!

    My challenge is:

    - the LED must be on one side or the other (high or low) of the load. If it were to span the load then it would simply be in parallel with the load and would be illuminated even if the resistance wire of the heating element breaks, and

    - I can’t put the LED directly in series with the load! ;-)

    A cheap DC multimeter would tell me the current going through it, but it doesn’t light up and I only need to know if any current is going through it! Thus I assume some form of shunt is necessary. I am considering a 100-Watt, 750-Ohm resistor in parallel with the LED (and its resistor) on the low side of the load as the only practical solution.

    My assumptions regarding the resistor:
    The 100 watts is to allow for the full current passing through the heating element.
    The 750 ohms is to force enough current to pass through the LED (and its 750 ohm resistor) so that it illuminates.

    Is this a functional solution?

    Is it the most practical and reliable solution?

    Is there a cheaper solution?

    Thoughts?

    Please forgive my ignorance and thank you for being such a valuable resource of information!

    Thanks!!
     
  2. jcsd
  3. Dec 21, 2009 #2
    Wind about 10 turns of wire (in series with the heater) around a small magnetic compass, and point the coil East-West. If the compass points North-South, the current is off.
    Bob S
    [added] A 10-turn, 6-amp, 4-cm diameter coil will produce about 6 Gauss (6 x Earth's field).
     
    Last edited: Dec 21, 2009
  4. Dec 21, 2009 #3

    vk6kro

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    :) Hi Bob.

    LotusDome:

    I wonder if you could afford to lose 0.6 of a Volt?

    If so, something like this may be OK:

    LED indicator.PNG

    There are other ways (opamps and microprocessors) that would allow you to have less of a voltage drop, but this one is pretty simple.
    The transistor could be a BC548 or similar low power, high gain device.
     
  5. Dec 22, 2009 #4
    LotusDome,
    I think that the other contributors have given some promising and relatively simple :smile: suggestions for an indicator. Perhaps it would also be helpful for you to think a bit more about your original solution. Possibly I have misunderstood your description. but it does not appear to be correct.

    In particular, think about why you thought that a 100W 750 ohm resistor would be required.
    How much power can be developed in a 750ohm resistor with only 12V supply? How much current could pass through it? Remember I=V/R, P=V2/R

    A shunt for monitoring current usually has a low value to mimimise power loss. The voltage across it should be as small as possible. In your case, developing the full LED voltage across a shunt would be very wasteful, so it would be better to use an amplifying switch such as a transistor, which needs less input voltage. Alternatively, an electromagnetic indicator like the compass in a coil would drop very little.
    In days gone by, one might have used a commercial analogue ammeter!!!
     
    Last edited: Dec 22, 2009
  6. Dec 23, 2009 #5
    All of my old cars prior to about 1960 had analog ammeters with a low voltage drop. You could possibly find one in an old junkyard.
    Bob S
     
  7. Dec 23, 2009 #6
  8. Dec 23, 2009 #7
    Thanks, folks, I really appreciate your input!

    vk6kro’s schematic looks promising and I expect(ed) that Adjuster’s concerns about my (limited) understanding of resistors are correct. Because of the storm that has rolled through, I cannot yet add any more info. as I must get offline, but I wanted to express my appreciation ASAP.

    When I can get time online again, I will investigate and let you know how it goes!

    My deep appreciation and Happy Holidays to you all!

    Blessings.
     
  9. Dec 23, 2009 #8

    sophiecentaur

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    How about using a reed relay with a very low resistance coil? There is plenty enough current to operate a reed switch via a hand wound field coil with a few tens of turns. The switch contacts could operate a signal light and give you an audible buzzer alarm too. Suck it and see! Totally fail safe.
     
  10. Dec 25, 2009 #9
    I've been watching this one for awhile, and I'm surprised that no one suggested a comparator chip, say an LM139, with a low value resistor (i.e. a 6" piece of wire) in series with the load. That would keep the drop minimal, and require little power.

    - Mike
     
  11. Dec 25, 2009 #10
    Yes, it's obviously a good and cheap way to do it, for someone familiar with op-amps etc. However, the content of the initial post suggests that the writer may not have this background, and so might need quite a bit of help with it. (LotusDome, please comment if you think I have got this wrong!)

    For instance, if you take advantage of the comparator's high sensitivity and use a very small sense resistance, the wiring layout needs to be considered carefully. Voltages dropped in unfortunately arranged wiring can easily upset this sort of thing. The arrangement of the sensing and reference connections to the comparator would have to be carefully described.
     
  12. Dec 26, 2009 #11

    vk6kro

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    Thanks adjuster. Well said.
    Using a comparator with inputs near its supply rails would be full of traps for a beginner.

    Another way which I wouldn't suggest would be this one:
    Hall probe in toroid.PNG
    A small gap is cut in a powdered iron or ferrite toroid and a Hall probe chip is inserted in the gap. A few turns of wire are wound on the the toroid. The magnetic field is concentrated on the chip which can then be used to switch a LED on or off.

    I exaggerated the gap size. It should be a tight fit for the Hall chip.
    And the toroid would normally be circular.
     
  13. Dec 30, 2009 #12
    Greetings, folks, and thank you all for you kind assistance!!

    I have much technical training and experience, but no training in electronics, thus my vast ignorance in this regard. Also, due to an extreme accident (I am told I fell about 200 feet, head-first into an ice-wall), I currently have very limited resources. In a week I will have the only chance to visit a Radio Shack for the next few months; thus I most-humbly request some further information regarding resistors.

    Specifically:

    When a resistor is rated at, say, 5W, does it not mean that the resistor can only allow 0.42A of current (at 12VDC) to pass before it risks failure (as I perhaps incorrectly assumed)?

    In the circuit that vk6kro was VERY kind to provide (thank you!), there are two resistors used to divert a controlled amount of electricity to the transistor, used as a switch, that would then allow the flow of electricity through the LED--if there is any current going through the main load. (I do need some form of illumination as I must be able to check the continuity easily at night and it must be small.) Without the specified 0.1-ohm, 5W resistor, there wouldn’t be any resistance to force some electricity to the transistor; it would simply follow the “route of least resistance.” Perhaps I am incorrect here as well!

    If so, then a 0.1-ohm, 5W resistor in a 12.5-15.5V circuit (depending on the battery bank being charged by the solar array or not) with an ~80W load should be a small fraction of the necessary capacity (again, as I perhaps incorrectly assumed!).

    What I do not understand then, is how the choice of resistors is determined; thus please allow me to ask:

    What are the calculations behind the choice of the 1K ohm resistor and the .1 ohm, 5W resistor, given that the circuit will be drawing 80W (or more, given the voltage at the time--thus my belief that I needed a 100W resistor to be safe from overheating)?

    In vk6kro’s schematic, I assume that the “1K” represents a 1K-ohm resistor; I am using 750 ohm, .25 watt resistors for single red LEDs. Would .25W then be a proper value for this resistor as well?)

    Blessings to you all for your patience and efforts!!
     
  14. Dec 30, 2009 #13

    vk6kro

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    If you look at the current flowing, 80 watts at 12 volts means that 6.66 amps is flowing.

    (12 volts * 6.66 amps = 80 watts)

    So, 6.66 amps through 0.1 ohms is 0.666 volts.
    (V = I * R...... so .....V = 6.66 * 0.1 = 0.666 V).

    It takes 0.6 volts to turn on a silicon transistor so there is 0.066 volts across the 1 K resistor.
    So, there will be a base current of about 66 uA.

    With a high gain transistor with a gain of about 200 this will give a collector current of about 13 mA.
    This will be enough to light the LED, and you can change the 1 K for something smaller if you want it brighter. (Yes, 1 K = 1000 ohms). Maybe 470 ohms would be a better choice.

    Having a LED on when it is working is also a check that the power supply has not failed. If the LED only came on when the circuit failed, you wouldn't know if the power had gone off.

    EDIT: The power dissipated in the 0.1 ohm resistor is as follows:
    Power = Voltage * Current = 0.6666 volts * 6.666 amps = 4.44 watts so a 5 watt resistor would be OK, but it will get hot. 10 watts would be OK too, if that was available.

    The 1 K can be anything but 0.25 watts would be fine. It is dissipating 44 uWatts.

    .
     
    Last edited: Dec 30, 2009
  15. Dec 30, 2009 #14

    sophiecentaur

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    Guys. We have Eight Amps to dtetct here. Why are people insisting on solutions that will detect milliAmps?
    A simple 40 ampere turn reed switch will operate reliably with, say, 20 turns of wire, wound into a solenoid, with a ball point pen body as a former and held together with Araldite. The switch can be put in parallel with an LED and the two can be put in series with a 1kohm resistor. As long as there is current, the LED will be off. When continuity fails, the switch will open and the LED will light. That solution is almost FREE!
    It's at least worth a try- old technology is good for some things and easy to understand.
     
  16. Dec 30, 2009 #15
    Good ..... except when the 12 V battery is discharged, the LED will also be off. The LED has to be ON when the current is flowing. Then the LED is OFF whenever EITHER the load is disconnected OR the battery is dead.

    I just placed a single wire carrying 3 amps on top of a compass needle oriented N-S, and the compass needle rotated 45 degrees. So a couple of turns around the compass will work adequately, unless a remote readout is needed.
    Bob S
     
  17. Dec 30, 2009 #16

    sophiecentaur

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    Ok. Have the switch in series with the LED,then. No current, no light.
     
  18. Dec 30, 2009 #17
    Lol.
    Good thing there are no psychologists replying - you'd hear about how to regress your amperage through hypnosis or something.

    Why not...
    wrap one of the power leads around a toilet paper derder?
    Then, turn on the power, set a cake pan under the derder and dump a handfull of paperclips into your electromagnetic coil?
    If power goes off, the clips will clatter like madness into the cake pan below.
    If you want a visual alert, put a paperclip on the bottom of a red straw. The straw will fall out of the coil if power cuts off.

    MacGyver's kid sister was consulted for these suggestions.
     
  19. Dec 30, 2009 #18
    For the transistor based alarm, would it be worth adding a very weak pull-up resistor from the transistor base to +12V, (say about 47kohms if the suggested 1k base feed is used). The idea is to increase the base voltage slightly, although not enough to turn the transistor on with the element failed.

    I'm concerned that the voltage might not be enough to guarantee the transistor on if it is exposed to cold weather, or perhaps if the battery voltage is low. Alternatively, raise the 0.1 ohms a bit, but that would waste power.

    Finally a suggestion for LotusDome, if mobility problems or any other reason mean you can't easily get to a dealer, can't you get components by mail/internet order? Actually, I've just had a thought: ice-wall accident...storm rolling in - are you at a very remote location?
     
    Last edited: Dec 30, 2009
  20. Dec 30, 2009 #19
    Thanks for the response, BenchTop! You are the only other person I know, besides the person who told me, what a derder is! ;-)

    Thank you, Adjuster, for your intuitiveness. I am a monk, living alone in the vast, UT wilderness after the accident; my production company is “dormant.” (The details of the interim R & D work I am doing, and thus why I need these indicator lights, is described on the Status Updates page of www.LotusDome.com ) Temperature here is a big concern, thus this thread (see below).

    The pull-up resistor sounds like a wise choice, given the conditions, Adjuster. Where would it go in the schematic? FYI, Digi-Key may be necessary as Radio Shack doesn’t seem to have any resistor in the 0.1-ohm, 10W range; I just wanted to spare the shipping and retrieval costs as I am a long way from a dirt road.

    Given vk6kro’s kind (and patient!) advice:

    > If you look at the current flowing, 80 watts at 12 volts means that 6.66 amps is flowing.

    I learned it long ago as the West Virginia law, W = V*A. Although I haven’t yet mentioned it, as it is running off a battery bank fed by solar panels, the voltage can get up to 15.5V when charging. This is why I asked for the equations, to not impose any more than I already am.

    > So, 6.66 amps through 0.1 ohms is 0.666 volts.

    This is what has me so baffled. Why would there suddenly only be 0.666 volts in a 12V circuit? Did it get “absorbed” by the ~80W load and thus there is no “pressure” on the low side?

    In closing as the temperature plummets, a brief elucidation of the conditions here:

    It does get “a bit chilly” here so I appreciate your concern regarding exposure, Adjuster; it has rarely gone above freezing during the day this past month and gets down to well below zero degreesF at night. One of the indicator lights--the one being discussed--will be to tell me if the composting toilet is freezing up solid . . . again. (I have just built and installed two aluminum, resistance wire heating plates inside the drum to compensate, thus this thread) Another indicator light is to tell me if the cat’s 48W 12V electric blanket is functional--they get the electric blanket not me; it’s a compromise they both grudgingly accept. ;-) FYI, it can get over 100 degreesF during the summer.

    Blessings to you all for your kindness and patience (and God bless satellite modems!). I am deeply appreciative, and humbled, by your help.

    Richard Fairbanks
     
  21. Dec 30, 2009 #20

    vk6kro

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    I am a monk, living alone in the vast, UT wilderness after the accident;

    WHAT ACCIDENT?

    > So, 6.66 amps through 0.1 ohms is 0.666 volts.

    This is what has me so baffled. Why would there suddenly only be 0.666 volts in a 12V circuit? Did it get “absorbed” by the ~80W load and thus there is no “pressure” on the low side?

    If you like, you can imagine your 80 watt 12 volt load as a 1.8 ohm resistor made up of 18 resistors all in series with 0.1 ohms marked on each one.

    Each one has 6.66 amps flowing in it and 0.666 volts across it.

    Can you see that all these 0.1 ohm resistors would equally share the supply of 12 volts and all their 0.6666 volts would add up to 12 volts?

    Adding another 0.1 ohm resistor reduces the overall current slightly because you now have 19 0.1 ohm resistors in series and 6.3 amps flowing, so each resistor in the series string would have 6.3 amps flowing in it and 0.63 volts across it, including the recently added external 0.1 ohm resistor.

    All these precise calculations change when the battery voltage rises to 15 volts, though.


    However, I do like the reed switch idea. I'm not comfortable with using 5 watts just to turn a LED on.
    Since it was Sophiecentaur's idea, maybe he could draw up a detailed picture of how to do it.
    It is an easy and efficient way of doing it.
     
    Last edited: Dec 31, 2009
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