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Project Help!!

Discussion in 'General Electronics Discussion' started by Marsell4k, Jun 6, 2012.

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  1. Marsell4k

    Marsell4k

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    Jun 6, 2012
    Hi,

    I am trying to make something that I think will be basic for most electricians but not for me and need help/advice (in layman's terms please).

    The project involves a low powered circuit with a sensor/switch that when pressed will power a solenoid, unlocking something. When the sensor/switch is not touched the solenoid will lose power and revert back to its default "locked" state.

    I have the switch/sensor and I'm sure I can find the correct type of miniature solenoid, but what else would I need to use exactly and what would connect to where?

    Sorry for what is probably a completely basic or stupid question, but I'm likely to bodge this up with getting advice first!

    Cheers.
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    It depends on the characteristics of your "sensor/switch" and the solenoid.
    If you already have the sensor/switch, can you describe it, or give a part number, or ideally a link to a data sheet?
    Can you explain what you want to do, in more detail? So far, it sounds to me like you want to remotely unlock something, using a switch. When you turn the switch ON, the solenoid will activate and unlock the thing. When you turn the switch OFF, the solenoid will deactivate and return to its rest position, which will lock the thing.
    If it's as simple as that, you just connect the switch and solenoid in series across a supply voltage that's suitable for the solenoid. When the switch is ON, it completes the circuit and the solenoid is energised from the power supply.
    The power supply must be able to provide the voltage and current needed by the solenoid. The switch must be rated to switch at least that much voltage, and to switch and carry at least that much current. This shouldn't be an issue except for tiny switches.
    If this isn't a complete description of what you want, please give much more detail. This will save a lot of time and effort. Especially you need to describe how the solenoid will lock and unlock your "thing'.
     
  3. Marsell4k

    Marsell4k

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    0
    Jun 6, 2012
    Thanks for getting back to me so quickly, unfortunately I haven't had time to respond until now.

    I know that it is difficult to provide help when I haven't disclosed the purpose for my project, but unfortunately that's very difficult for me to do. I can however provide the link to the sensors data sheet here . I do not yet have the solenoid but it will be around 2-3cm in size and what it locks/unlocks will be fairly small.

    The principle is as simple as it sounds, with the sensor being pressed to power up the solenoid and release the lock and vice versa. Nothing else is connected or running off this.

    It's only a small circuit but knowing virtually nothing about electronics I need help with which parts to source, what to consider or know and what to connect to where. Once I know the right parts to buy and connect I can do the rest!
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    OK, the sensor you're using is a "force sensing resistor". It behaves like a resistor whose value varies depending on the pressure applied.
    It can't drive a solenoid directly. Even with 10 kg pressure applied, its resistance is about 250 ohms, which does not allow enough current to operate a low-voltage solenoid; in any case, the sensor is not designed to have high currents passing through it, or high voltages across it - it would probably overheat and be damaged. So a simple circuit is needed to buffer the sensor and drive the solenoid.
    I've attached a simple circuit that should do what you want. I've assumed the solenoid is operated by 12V DC and that it draws no more than a few hundred milliamps. The MPSA14 transistor is rated for 500 mA maximum collector current.
    The circuit is a simple comparator. As you put pressure on the sensor pad, its resistance will decrease, raising the base voltage because of current flow through the sensor and R1. The sensor and R1 together, and VR1, form a voltage divider. When the voltage at Q1 base reaches about 1.4V positive (relative to the emitter, which is connected to the common rail), Q1 conducts and passes current from collector to emitter, activating the solenoid. D1 is needed to prevent damage to Q1 when Q1 turns off and the solenoid generates a "back EMF" voltage pulse.
    VR1 is a "trimpot" or "preset potentiometer". Turning it clockwise increases the pressure threshold of the circuit.
    This circuit is very simple and is missing a characteristic called hysteresis. This means that if you hold the sensor around the threshold, the solenoid will activate and deactivate rapidly and randomly, according to small variations in pressure on the sensor that cause the circuit to quickly alternate between its two states (below threshold, solenoid off, and above threshold, solenoid on). If you press and release the sensor firmly and cleanly, this may not be a problem; if it is, you'll need a slightly more complicated circuit with hysteresis.
    Anything more, just ask.
     

    Attached Files:

  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Here's a better design with hysteresis, as I explained in the previous post.
    It uses three transistors but it will have a cleaner behaviour.
    I'm not sure whether you'll want to tackle this circuit, being a newbie to electonics, but you seem pretty smart and I'm sure you can do it if you give it a go.
    You can build it on stripboard (aka "veroboard" in the UK). The components should be available from your local electronics shop.
    Let me know if I can help.
     

    Attached Files:

  6. Marsell4k

    Marsell4k

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    Jun 6, 2012
    Wow! Thats a great response, very fast and sounds like exactly what I need. Thank you so much. I am away from tomorrow so I will unlikely be back on this forum for a little while, but I will definitely have a couple of things that I'd like to clarify/have explained in my poor laymans terms!

    Just seen you've posted a second response. Yes I will look to get the parts and set this up. I'm really not as good with electronics as you give me credit for, but I like to learn and pick things up pretty quick. I'm hoping to use standard batteries for this I take it that wouldn't be a problem?
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Cool :)
    Battery power will work but both circuits have idle current, which will run the batteries down over a period of time. The first circuit has lower idle current than the second, although the second will only draw about one milliamp. Your choice of battery will also be affected by the amount of current drawn by the solenoid, even though it will only be activated infrequently. You need a battery with enough current capability to supply the solenoid without the terminal voltage dropping too much. I'd suggest eight D-cells or two 6V lantern batteries (not sure if you have them where you are... wherever that is... they're made from four E-cells and have two springs on the top). Mains-derived power would definitely be better.
    You need to choose your solenoid (aim for one with the lowest power requirement that will do the job reliably) and decide how often you want to replace the batteries.
    Here are some links I found for stripboard construction.
    http://en.wikipedia.org/wiki/Stripboard
    http://www.kpsec.freeuk.com/stripbd.htm
    There is even software available to help design stripboard layouts! Google "stripboard layout software".
     
  8. Marsell4k

    Marsell4k

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    Jun 6, 2012
    There's definitely a lot to digest there, but I have already browsed some of that before so that's a minor headstart I suppose. The lower the power consumption and longer the battery life, the better for this, but it's not essential as I'm only trying to build a working prototype. If sucessful then I'm sure I could find ways (with professional help and money) to streamline this and improve efficiency.

    I'm very much looking forward to making a start on this as once I have it made and working (even if crudely for now) then I can proceed to testing which is going to be a massive step for my project.

    Thanks again, I really appreciate your help and when I back online again I'll be in touch.

    Ps. If you can think of any pointers or ideas for a novice then feel free to email them or post on here and I'll pick them up when I can.
     
  9. BobK

    BobK

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    Trying to figure out why you would not want to use a comparator IC.

    Bob
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Simplicity. A circuit with a comparator will need about the same number of resistors, perhaps one or two more, and a buffer transistor, so it's the choice between two transistors or an IC.
    The need for battery power changes things though. I would want less than 100 uA quiescent, which eliminates my transistor Schmitt trigger and standard comparators. I could replace the second transistor with a FET and increase the pullup resistor value to a few hundred kilohms, that would reduce the quiescent current, but a comparator IC might be better overall.
    I found the LTC1440/1/2 which look great but are only rated up to 11V supply voltage.
    Do you have any suggestions for suitable low-current comparators?
     
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    This version has a quiescent current below 0.2 mA.
     

    Attached Files:

  12. BobK

    BobK

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  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    LM393 supply current is a bit high, typically 400 uA, maximum 1 mA or 2.5 mA at 30V supply. It's a lot lower than other standard devices though.
     
  14. BobK

    BobK

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  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    LTC1541/2 looks like a nice IC. Maximum supply voltage 12.6V is a bit close for my liking (absolute max is only 13V). Also not ex stock digikey or element14.
     
  16. Marsell4k

    Marsell4k

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    Jun 6, 2012
    I've just come back from being away and wanted to pick up where I left off in regards to my force resisting sensor and solenoid help post that you were so kindly helping me with.

    Whilst away I was giving this some thought and have decided that I will be better to get some help with making a working prototype from someone in the trade who can put together this circuit for me. I think it will be quicker and easier than trying to tackle this myself. Before I do that though I had a couple of questions about the last circuit you had mentioned on the message board;

    For practical reasons I am now thinking of using a small electromagnetic plate (instead of solenoid) to hold the lock shut and then de-magnetise it when a current is run through it, following the FSR being being activated. Nothing else is different. I've been browsing the web this morning for one, but can you suggest any specifications I would need to bear in mind when selecting one? Obviously low power comsumption is important as I want to be able to use small and/or fairly inexpensive batteries and not need to change them every week. Plus what other considerations do I need to bear in mind with batteries? Would something like this work?

    Oh, also the electromagnet doesn't need to be particularly strong. Can't give a measurement in newtons, but I'd estimate enough to stop around 10-15kg's of force being applied.

    As usual I really appreciate your advice!

    Thanks again.
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    I don't think a normally energised solenoid will be workable with a battery supply. The amount of current needed to hold the plate in the locked position will be prohibitively high. Perhaps a motor and gear system to move a locking bolt? Run the motor one way to lock it, the other way to unlock it, and it remains in its current position when the motor is not driven.
     
  18. Marsell4k

    Marsell4k

    8
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    Jun 6, 2012
    Maybe I'm confused or just haven't explained it properly, but I am thinking of an electromagnet that in its natural state (no current) is magnetised and keeps the door locked. Then when current paseses through it, (from sensor being activated) it is released and unlocks.

    Can i still get by with a low voltage/current etc with this?
     
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I've never heard of an electromagnet that's permanently magnetised and becomes unmagnetised when energised. It would have to have a permanently magnetised core and be designed so that the electromagnet's field "cancels out" the permanent magnetism in the core. I don't know if it's even possible. I suspect not. Try googling it.
     
  20. Marsell4k

    Marsell4k

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    Jun 6, 2012
    Check this link. It shows magnets that 'energise-to-release'.
     
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