Need help understanding a basic relay-oscillator circuit

Discussion in 'General Electronics Chat' started by stormBytes, Sep 1, 2010.

  1. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
    41
    1
    I'm following an intro to electronics book and working with the attached diagram. I'm trying to understand how this circuit works. For simplicity, I'll refer to the current as flowing from positive-to-negative poles. I'll start by describing what I observe when breadboarding the circuit.

    If I press the button momentarily and release it:

    - the green LED lights up for a (barely perceivable) instant

    - I hear the relay engage (click) as the yellow LED lights for an instant

    - The yellow LED goes out, the Green LED lights up and fades out, and I hear the relay disengage

    I'm trying to understand, step-by-step (on a 4-year-old's level) what is happening in my circuit. Here's what I can gather so far:

    - When the button is NOT depressed, there is zero current flowing through the circuit, and so, nothing happens.

    - As soon as I depress the button, current flows from 12V source, through the button, through the relay's pole, to terminal-1 (t1), through path (b) - energizing the coil, and simultaneously through path (c), lighting the green LED for an instant

    - Once the coil is energized, the relay engages severing the connection between pole-and-(t1) which causes the green LED to go out, and engaging (t2) which lights the yellow LED.


    What I don't understand is what's happening hereafter and more importantly, why. Any insight would be greatly appreciated, keeping in mind my knowledge/skill levels.

    [​IMG]
     
  2. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    It's as you've stated, but add;

    When you push the momentary, it begins to charge the capacitor. Once the cap is charged to a sufficient voltage the relay switches. The cap then begins a discharge through the green led and it's current limiting resistor.
     
  3. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    The Cap is also discharging through the Relay coil always including in parallel withe the green LED and resistor.

    You should also understand relays have a switching current and aholding current that is about half the switching current.

    That allows a delay before switching with the capacitor charging to a higher voltage to allow the switching current.

    Then there is another delay while the capacitor discharges until the relay current falls below the holding current.

    This hysterisis is important in lowering frequency. Otherwise the relay would burn out quickly switching so fast that it would look like both LEDs were always lit.
     
  4. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
    41
    1
    Sorry I'm not the sharpest pencil in the box -

    Why does charging of the capacitor affect the relay? The cap wired parallel to the relay and should click-off (engage) as soon as power is applied, regardless of what other components (caps, led's, etc) are also being fed by the voltage source - as I understand.

    The way the cap discharges I don't get at all -

    Here's a step-by-step of what I understand to happen here, forgive the redundancy:

    1. I press the momentary switch for an instant, bridging the positive voltage source and the pole of the relay.

    2. Current flows through the pole of the relay through path (b) to the cap & relay coil, and through (c) to the green LED

    3. The relay-coil is energized and the pole is disengaged from terminal (t1) and engages terminal (t2)

    4. As the voltage source has been disconnected (positive) the circuit is now powered by the capacitor.

    As I understand it -

    (i) The relay should REMAIN engaged to (t2) for as long as power remains in the capacitor

    (ii) Once the cap voltage drops below the minimum required voltage to sustain the relay coil's energized state, the relay should revert to its open state
     
  5. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    If your supply to the coil / cap was infinite and instantaneous, then yes, but that isn't real world. You will always have some series resistance, which results in an RC time constant. The coil, as mentioned, requires a certain torque to pull in. This torque is current related, voltage induced. The RC time constant dictates when that voltage is reached.

    It was well noted by Potato Pudding in regards to hysterisis on the discharge side.
     
    Last edited by a moderator: Sep 1, 2010
  6. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
    41
    1
    I was hoping someone would actually provide a step-by-step explanation of whats happening in the circuit.
     
    Last edited: Sep 1, 2010
  7. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
    41
    1
    Well anyhow, I gave up on understanding this circuit for the time being, I guess I'll get back to it when I have a better understanding of what's going on.
     
  8. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    The diagram is drawn incorrectly. The top set of contacts (t1) are normally closed, and and should be on top. They also power the relay coil.

    [​IMG]

    It is not complex. When the relay coil is energized the common is pulled down, which breaks the contact and deenergizes the relay coil. The common contact bounces up, and powers the relay coil again. Cycle repeats as long as there are power.

    The capacitor slows everything down.
     
  9. stormBytes

    Thread Starter Active Member

    Jan 26, 2010
    41
    1
    Hey, Thanks for that -

    The diagram is taken right out of the book, so... I guess maybe the author decided to draw it that way!? who knows..

    Point is -

    I'm sure there's a perfectly sensible reason for how things work here. I'm trying to tie my observations with the theory, and so far that hasn't happend. It requires looking at each electrical interaction step-by-step and making sense of it. I can't really dwell on this one circuit forever so I decided to shelf it for now and hopefully will get back to it when my brain can make some sense out of what I'm seeing, or when someone's nice enough to take the time and explain it methodically.
     
  10. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    Reading is your friend, but you have to go through the process. If you can't understand this why would you understand the next step. Knowledge builds on previous knowledge, so you need to have the patience and perseverance to work through problems.

    I was building this circuit when I was 13 from my 300 in 1 electronics kit. You don't need the LEDs or the switch, just the relay for a buzzer.

    BTW, it makes a dandy tinglier across the coil.

    Try redrawing the schematic the way I suggested, it will be much more obvious. Only two contacts (the normally closed set, which is to say they only make contact when the relay is not energized) and the coil is needed for the circuit.

    Since the capacitor takes time to charge and discharge it slows things down, but it isn't really needed either.

    Simpify, then understand the simplification.
     
    Last edited: Sep 4, 2010
  11. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Bill, hate to add this but if the cap wasn't there the contacts might not stay in the NO position for long enough that the corresponding LED would ever light up long enough to notice. Wouldn't hurt to have a diode across the coil either to help protect the cap.

    Regardless this is obviously just meant to be a learning circuit, not anything you'd actually use for anything I can think of.
     
  12. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    Actually it has a lot of uses. It was how they created high voltages for the old tube radios, since it was a cheap source of AC. I remember the old truck my Dad and I worked on that had a working tube AM radio. You turned it on and could hear the buzz from the mechanical setup before the tubes warmed up enough to work.

    It is also a small scale shocker. The coil on the relay acts as a small flyback transformer. Similar arrangements were used a lot for practical jokes such as shocking gum packs.

    Then there is the obvious use as a noisemaker. It is the basis for most older buzzers, and still works well for the application.

    The original comment about the schematic stands, who ever drew that circuit isn't very professional. How you draw a schematic goes a long way towards understanding it. There is often a reason why schematics are drawn in a particular manner. The layout needs to correspond to the physical layout to some extent.
     
    Last edited: Sep 4, 2010
  13. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    The capacitor also snubs the CEMF spikes. A Diode across the coil is more common and effective for snubbing.
     
  14. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
    92
    Bill makes a good point to mention the use for power supplies.

    I think that buzzing relay oscillator power supplies are the reason that certain types of switching power supplies are called vibrators even though they have used transistors for more than 50 years, and tubes for many years before transistors became common.
     
  15. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    I've worked on many of those old radios, most that I encountered were made by Motorola.

    Even after the early transistor units came out with the doorknob PNP Germanium output the older tractors were still 6V and some farmers had a tendency not to want to change from something that's served them well for years.

    I remember one model they must not have made many of. Motorola used to be very good at keeping parts around for ages after a model became obsolete but not in this case. First IF transformer I ever had to rewind by hand because physical limitations didn't leave enough room to sub anything we could find in its place.
     
Loading...