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Need help with electronic circuit (Make: Electronics Experiment 8)

Discussion in 'General Electronics Discussion' started by It'snorocketscience, Feb 20, 2015.

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  1. It'snorocketscience

    It'snorocketscience

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    Feb 20, 2015
    Hello there,
    When I put together everything for Experiment 8 in Make Electronics and connected a 12v power supply, the LEDs blinked at the same time when I held down the pushbutton.
    Have a few pictures:

    272795-1.jpg

    My power supply.

    272795-2.jpg

    The relay.

    272795-3.jpg

    The circuitry

    Now, here are the details:
    The relay clicked every time the LEDs turned on.
    Both LED's blink at the same time (one LED should turn on when the other is off).
    The LEDs on the dim out after turning on.
    The relay appears to function like the relay used in the book (p. 59).
    I've checked many times to see if I've wired my circuit correctly, and it seems I have (but just in case, assume I haven't checked in case I've missed something).

    Note: This circuit is for a school presentation project. It is not a quiz or question-related assignment, I've created this circuit to explain how it works to my classmates as part of a free-choice assignment. The assignment is due this Monday.

    (Moderator note: edited to include photos in post and reduce their size from 2.2M each to 50k each -- KrisBlueNZ)
     
    Last edited by a moderator: Feb 20, 2015
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Welcome to electronicspoint.

    Unfortunately neither do I have acess to the book in question, nor is the circuit well recognizable from the photos. Could you attach the schematic diagram from the book for reference?
    I assume that the LEDs are suppose dto go to different contacts of the relay, one normally open, the other normally closed, so you have a changeover contact arrangement (this will ensure only one LED is on at a time). I suspect that your trouble is at this point. Either the relay used is not suitable (tell us which relay you use so we can find a datasheet) or you have made a wiring mistake.
     
  3. KJ6EAD

    KJ6EAD

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    Aug 13, 2011
    Here are the schematics for the basic relay oscillator and the improved version with the capacitor added.

    Since it's an oscillator, when wired correctly the LEDs will alternate at a high enough rate to both appear to be lit simultaneously, but dim. If you intended a slower alternation, a much larger capacitor or a separate timing circuit will be needed.
     

    Attached Files:

    Last edited: Feb 21, 2015
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  4. It'snorocketscience

    It'snorocketscience

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    Feb 20, 2015
    "LEDs will alternate at a high enough rate to both appear to be lit simultaneously, but dim" I used a 1000μ capacitor, and I can see the LEDs dim for half a second before returning to full brightness
    Thanks KJ6EAD for supplying the schematics for me. "mkel_02_066.pdf" is the schematic I've breadboarded.

    "I assume that the LEDs are suppose to* go to different contacts of the relay, one normally open, the other normally closed, so you have a changeover contact arrangement (this will ensure only one LED is on at a time)."
    Both LEDs are connected so that they share the negative power supply from the resistor and one is connected to the normally closed terminal and the other is connected to the normally open terminal. Both terminals lead to the positive end of the power supply.

    It could be the way my cheap multimeter is, but the relay coil had a resistance of around 750Ω where as another latching (latching relays wouldn't work with this circuit, right?) relay had a resistance of around 150Ω.
    Does the resistance of the relay coil make a difference?

    Here is the data sheet for the relay: http://www.datasheetlib.com/datasheet/453231/hfd27-012-h_hongfa.html#datasheet

    [​IMG]
    S1: Relay
    S2: Pushbutton
    D1, D2: LEDs
    R1: Resistor (680Ω).
    C1: Capacitor (1000μf)

    Thanks.
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Your relay has a non-standard pinout, different from the relay used for that breadboard layout drawing. On your relay, the changeover contact (which connects to the pushbutton) is on the third pin down, and the normally closed contact is on the second pin down. So you have to swap the connections to the second and third pins, counting downwards from the top.

    To be exact:
    • The top end of the link wire just below D1 needs to move downwards by two rows, so it connects to the third pin on the relay, not the second;
    • The bottom lead of D1 and the top end of the link just to the right of S2 both need to be moved upwards by two rows, so they connect to the second pin on the relay, not the third.
    There are also problems with the circuit design, though it will work. I admire what Make are trying to do, but they should be more careful.
     
    Last edited: Feb 21, 2015
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  6. KJ6EAD

    KJ6EAD

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    Last edited: Feb 21, 2015
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  7. It'snorocketscience

    It'snorocketscience

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    Feb 20, 2015
    Alright! the circuit works now! Thanks so much, I've spent so many hours working on this, and now it works!

    Just one question, the left hand LED (D1) only stays on for a very brief time. D2 stays on the whole time the button is held down, then when the relay clicks, D1 turns on for the duration of the click of the relay (an extremely brief flash). Is there a way to make D1 stay on longer?
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Great!

    That circuit will never flash the two LEDs evenly, but you can lengthen the duration of the flash on D1. Replace the link that goes from D1 to the relay coil and the positive side of C1 with a resistor of about 100Ω.
     
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  9. It'snorocketscience

    It'snorocketscience

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    Feb 20, 2015
    I tried to do what you suggested, but then the left hand LED stays on (it's not dim, its bright) and the relay makes a buzzing noise (sounds more like steam, and I don't think the relay is switching it's contacts very fast).
    What should I do to fix this?
    Even though the project is due monday, It's not very important to have the LED's have a similar flash duration time, so take your time when you respond--if you want to, of course. ;).
    So I removed the changes you suggest and the circuit is back to normal.
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK, I thought that would work, but apparently it doesn't. I don't have any other suggestions for simple changes that would fix that problem, sorry. Do you have a transistor? For example, a 2N3904 or 2N3906? If so, I can suggest another thing to try.
     
  11. It'snorocketscience

    It'snorocketscience

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    Feb 20, 2015
    Yes, I have 4 2N2222's
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK, here's a circuit that should work. It uses both sets of contacts on the relay - one to make it oscillate, and the other to light the LEDs. It needs an extra resistor, and a transistor.

    272795.001.GIF

    R1 and C1 determine the flash rate. If you use your 1000 µF capacitor for C1, it will flash fairly slowly. Use a lower value to make it flash faster.

    I use the European circuit symbol for capacitors. The hollow plate, at the top in this drawing, is the positive side.

    The duty cycle won't be exactly 50% (i.e. the durations of the flashes on LED1 and LED2 won't be exactly equal) but it should be between 25% and 75%.

    Normally when a transistor drives a relay coil, a diode must be connected across the relay coil, to protect the transistor from being damaged by the "inductive kickback" from the relay. This diode isn't needed here because of how the transistor and relay are connected. I mention this only because you may see other description which say you always have to put a diode across the relay coil.

    Here's a diagram of the relay showing how the circuit symbol translates to the terminal layout. This diagram matches your HongFa HFD27 relay which has a non-standard pinout.

    272795 relay pinout.gif

    The coil is not polarised and can be connected either way round. The two sets of contacts (one set on each side) are also interchangeable.

    Good luck! Feel free to ask any questions here in the thread.
     
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  13. It'snorocketscience

    It'snorocketscience

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    Feb 20, 2015
    OK, thanks for the help, but I must admit I don't know how this electronic circuit works (the original design that worked, not the one just given). I need to write an explanation on how it works (again, its not a quiz or question related assignment), and I don't understand how the circuit works... I'm sorry if it bothers you to explain, but I'm an electronics noob and could use some help (assume that I know what the components are and how they work). :)
    Thanks for your help!
     
  14. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    The original design (with a large capacitor across the relay coil) depends on relay coil hysteresis to oscillate. Hysteresis in a relay coil means that the current (or voltage) required for the coil to actuate (move) the contacts is greater than the current (or voltage) required for the coil to de-actuate (release) the contacts.

    When the pushbutton is pressed, +12 VDC is applied to the relay coil through the pair of normally-closed contacts on the relay, quickly charging the capacitor across the coil sufficiently to allow the coil to actuate its contacts. With the coil actuated, the normally-closed contacts open and +12 VDC power is removed from the coil. However, the charged capacitor now begins to discharge through the coil, so the coil remains actuated until the charge on the capacitor is depleted enough to cause the coil to de-actuate the contacts. With the coil now de-actuated, the normally-closed contacts close again and, if the push button switch is still closed, the cycle repeats.

    The circuit not only depends on coil hysteresis but also on an energy storage component, the capacitor, to provide a delay that determines the length of the cycle. If the capacitor is omitted, the circuit will still oscillate but at a much higher frequency, generally heard as a high-pitched buzz coming from the relay. In this case, the energy storage component is the mechanical inertia of the moving contact along with the spring that restores the contact to the de-actuated position in the relay. The moving mass and the restoring spring constitute a resonant mechanical system whose frequency is approximately the same as the relay buzz frequency.

    Note that the normally-open contact may not close without the energy storing capacitor connected to the coil. In other words, the moving contact may move far enough to open the normally-closed contact, thus de-energizing the coil, but not far enough to close the normally-open contact since the coil is then no longer energized. For this reason (among others) it is preferable to drive relay coils with well-defined signals instead of relying on the relay and its contacts to perform a timing or oscillatory function.

    @KrisBlueNZ circuit uses a transistor and a resistor-capacitor timing circuit to ramp the coil current up and down in a roughly triangular fashion. This would be one of the least complicated ways to create an oscillating relay coil drive, although it still depends on coil hysteresis for the timing. Better would be a 555 timer and use the transistor only as a relay coil driver. You might need a "flyback" diode across the relay coil, depending on how the transistor is connected, but that's a topic for another Learning Electronics experiment.

    Good luck with your classroom presentation. I hope the explanation above helps.
     
    KJ6EAD likes this.
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