Connect with us

Constructing a "Power on time delay relay" with 5 sec. delay. max 6V

Discussion in 'Electronic Basics' started by [email protected], Jul 10, 2005.

Scroll to continue with content
  1. Guest

    I'm trying to build an autonomous "sumo"-robot for competition in
    sumo-robot contests.
    One of the rules state that the robot must be inactive for 5 sec from
    the time it is switched on. Now I know that it probably would be much
    easier to buy a basic-stamp processosor and program it, but my budget
    is quite thin and I think it would be quite fun to construct a robot
    without software.

    I have found a circuit-drawing of a time delay relay, but it demands
    12V and has a delay of roughly 7 sec.

    http://ourworld.compuserve.com/homepages/Bill_Bowden/page2.htm#delay.gif

    Since my motors run on 6V I would like the relay to do that as well.

    QUESTIONS:
    1. How do I calculate the time delay? In the text the author states
    that the delay can be reduced by reducing the R or C values (R=47K,
    C=100uF)but is there a formula that can be used?

    2. Will this solution turn the robot on at "full power" after the delay
    or will it start softly (when I push the button or when the time is
    up?)?

    3. How do I have to change the circuit to reduce the voltage to 6V?

    4. Is there a better alternative that wouldnt cost me more than perhaps
    20$?

    As you can see I have'nt got much experience of electronics, so pelase
    be patient with me :)
     
  2. Tom LeMense

    Tom LeMense Guest

    [...]
    The time delay will depend on the reverse breakdown voltage of the
    base-emitter junction in the '3904 used as a Zener. You can figure that the
    2N3904+2N3053 pair (in a "darlington" configuration) will be fully 'ON' when
    the voltage at the base of the 2N3904 (of the pair) is 1.2V. So, as the
    100uF cap charges through the 47K resistor, its voltage rises. This voltage
    is reduced by (according to the web page's author) 6-9V due to the "Zener"
    and then applied to the base of the "darlington".
    If you use a relay to switch the robot power, it will be ON or OFF. I
    wouldn't use this circuit to control the motor directly, as the motor
    current will "ramp up" somewhat during turn-on. Stick with a relay to
    switch the motor current ON or OFF.
    a) Use a relay with a 6VDC coil, making sure that 'pull-in' votlage is no
    more than 5V.

    b) Replace the "zener" 2N3904 with something that will drop less voltage.
    The 6-9V dropped by the '3904 clearly won't be compatible with a 6V supply
    rail. Some people use an LED which will drop between approx 2V and 4V
    depending on its color, others use several 1N4148 or 1N4001 diodes in series
    to obtain the desired drop. If you were to use 3x 1N4148 diodes (very
    common "switching" diodes) you'll end up with the darlington pair saturated
    (ON) when the cap voltage is approximately 1.2V + 0.6 + 0.6 + 0.6 = 3V.

    c) Adjust the 47K resistor to change the delay time, keep it above 1K ohm
    to keep excessive current from flowing into the base of the darlington.
    This shouldn't cost anywhere even close to $20.

    This circuit that you found will not behave so well over large temperature
    extremes, or if the 6V supply voltage is going to vary quite a bit (the
    threshold of the transistors doesn't track the supply votlage). Nor will it
    be that repeatable from unit-to-unit -- so if you're going to build a whole
    army of sumo robots, they'll all have slightly different turn on delay
    times. For an army of one that only has to work indoors, however, it'll
    probably work just fine.

    I would recommend a means to help discharge the 100uF cap when the power is
    interrupted - the web page author put a 10K from V+ to ground, but the 100uF
    cap will have to discharge through the 47K resistor in SERIES with the 10K,
    that's a pretty long discharge time. What this means is if the power were
    applied briefly, then disconnected, then reapplied briefly, then
    disconnected, &c, the cap will charge up a little bit each time, and
    eventually the relay could turn ON before 5s elapsed when the power was
    applied "for good" If you place a diode across the 47K resistor with the
    anode at the 100uF + terminal, and the cathode at the +6V point, the
    capacitor will "discharge" across the power supply rail when power is
    removed.

    A more precise turn on delay could be generated with the venerable LM555
    timer circuit - a Google search will reveal lots of references to this very
    common (and cheap) IC. While you might not find the circuit you need
    exactly, it ought to give you lots of good ideas.
    Have fun!

    TJL
     
  3. Guest

    Thanks a bunch.
    I can´t say that I grasp it all at first sight,
    but hopefully I´ll figure it out :).

    /David
     
  4. John Fields

    John Fields Guest

    ---
    See below.
    ---
    ---
    Try this: (View in a non-proportional font like Courier.)


    +6V>---> |
    S1 |
    O
    |
    +-----------------------+------+
    | | |
    | | |
    | | |
    +-------+-------+ | |
    | | | +-------+ |
    | +---+---+ [1M] |K | O COM
    | |_ _| | [DIODE] [COIL]- -|
    +--O|R D|O--+ | | O--> |<--O--->6VOUT
    | |_ | | +-------+ NO NC
    +--O|T TH|---+ |
    | | | C
    | OUT|---|--[100R]-----B
    +-------+ |+ E
    7555 [4.7µF] |
    | |
    GND--------------+---------------+
    _____ _______
    When you close S1, the 7555 will come out of RESET before the TRIGGER
    input goes high, initiating an output pulse which will energize the
    relay for about five seconds. This will pull the common terminal away
    from the normally closed contact, disconnecting it from the switched
    6V coming from S1. After about five seconds the 7555 will time out,
    de-energizing the relay, and 6V will appear on the normally closed
    contact. The nice part about this circuit is that the battery only
    has to supply current to the relay for five seconds, but the maybe bad
    part is that S1 has to be able to supply all the current needed by the
    6V load.

    Will that work for you, or would you like something else?
     
  5. Guest

    That sounds great.
    If I have understood it correctly that is ;).
    Correct me if I´m wrong: When power is switched on the the current
    will flow through the 7555 (and therefore through the base of the
    transistor) and the coil. The coil will act as an electromagnet
    "shutting of" the rest of the circuitry by pulling away the common
    terminal (COM?).
    When the 7555 times out power will stop flowing through the coil and
    start flowing through the rest of the circuit!

    Sounds like just the thing ive been looking for.
    A couple of questions arise:
    1. How do i set the 7555 to time out after 5 sec?
    2. The closest thing i found att my components shop is something they
    call "ICM 7555" would that be what your thinking of? (do you know what
    ICM stands for?)
    3. What is the diode doing in the circuit?
    4. Is the "[4.7µF]" a capacitator?
    5. What does NO and NC to the right of the circuit stand for?

    Phew, I think I´m running out of questions for the moment ;).
    Thanks a lot (and again for your patience, if this project works I sure
    will have learned a lot from it :).

    /David
     
  6. Guest

    OK, I found out what NO and NC stands for (Normally Open and Normally
    Closed), but then another question arose... Can a relay be both NO and
    NC?
    Do you buy a specific NO or NC relay or is it just a question of how
    you turn it when hooking it up to the rest of the circuit?

    /David
     
  7. One day got dressed and committed to text
    It's usually the norm to have a relay with 2 contacts and a moving 'arm'
    that will move between the contacts and connect to 1 of the contacts when
    energised and to the other when de-energised. Thus it can be connected as NO
    by using the open contacts when de-energised or NC by using the NC contacts
    when de-energised.

    ----0\ 0-------
    |
    |

    Hope that comes out but no doubt you will get many alternative explanations
    to consider !
     
  8. Randy Day

    Randy Day Guest

    Yes. Look for a DPDT (double pole, double throw) relay.
    That way you'll have two sets of contacts which can be
    used either as NO or NC. Assume the relay below is at
    rest (not powered). Contacts a and c are NC, e and f
    are NO.

    o- b
    a -o--__
    | o- c
    |
    |
    |
    | o- f
    e -o--__
    o- g
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
    You can buy SPDT, DPDT, 4PDT, etc. relays. It's in how
    you hook them up...


    HTH
     
  9. John Fields

    John Fields Guest

    ---
    It's already set. The 1 megohm resistor and the 4.7µF capacitor set
    the pulse time. Get the data sheet at:

    http://pdfserv.maxim-ic.com/en/ds/ICM7555-ICM7556.pdf

    to find out all about it.
    ---
    ---
    Yes, that's the part. "ICM" is MAXIM semi's prefix
    ---
    ---
    It's clamping the high voltage pulse generated by the relay coil when
    it turns off to about a volt over the supply, keeping it from
    destroying the transistor.
    ---
     
  10. ehsjr

    ehsjr Guest

    I'll post a simple circuit - but first, there is something
    *critical* that no one has mentioned. The relay you use must
    be able to handle the current that the sumo draws, and that
    is likely to be fairly large. I have specified a relay that
    should work. Ok, now to the circuit:

    +------+--------+-----+
    | k| | |
    | [Diode] [Coil] o COM
    | | | |
    +6V----+--------+ +--------+ | <--o--->6VOUT
    | | | NO
    | [R3] e|
    [R1] | /
    | +-----[R4]-----b Q1
    / k| \
    P \ ----- c|
    O /<-----r/ \ TL431 |
    T \ --- |
    | a| |
    +---+ | |
    | |+ | |
    [R2] [C1] | |
    | | | |
    Gnd----+---+----+---------------+

    C2 is a 100 uF 16V electrolytic capacitor
    Diode is a 1N4001 diode
    POT is a 20 K linear potentiometer
    Q1 is a 2N5401 transistor
    R1 is a 56K 1/4 watt resistor
    R2 is a 100K 1/4 watt resistor
    R3 is a 330 ohm 1/4 watt resistor
    R4 is a 100 ohm 1/4 watt resistor
    R5 is a 47K 1/4 watt resistor
    TL431 is part number 497-3035-5-ND at Digikey. It is
    a programmable (by resistors R1, R2 and the pot) shunt.

    Relay: Part # 255-1109-ND from Digikey is a 6 volt single pole
    normally open relay with a 68 ohm coil. The contacts can
    switch and carry 16 amps.

    The potentiometer (POT) allows you to adjust the timing
    to exactly 5 seconds.

    k
    Diode: ----|<---- (Banded end at left)

    TL431 2N5401
    ----------- -----------
    | | | |
    | Flat Side | | Flat Side |
    | | | |
    | | | |
    | | | |
    ----------- -----------
    | | | | | |
    r k a e b c

    See the datasheets at:
    http://www.fairchildsemi.com/ds/TL/TL431A.pdf
    http://www.fairchildsemi.com/ds/2N/2N5401.pdf

    It will be great to hear how you make out with
    your project.

    Ed
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-