Wednesday, February 27, 2013

Make this Simple Delay ON Timer Circuit - Application Note Included

The post explains a simple delay ON circuit which allows the connected load at the output to be switched ON with some predetermined delay after power switch ON.


The explained circuit can be used for all applications which calls for an initial delay ON feature for the connected load after the mains power is switched ON.

The shown diagram is pretty straightforward yet provides the necessary actions very impressively, moreover the delay period is variable making the set up extremely useful for the proposed applications.

The functioning can be understood with the following points:

Assuming the load which requires the delay ON action being connected across the relay contacts, when power is switched ON, the 12V DC passes via R2 but is unable to reach the base of T1 because initially, C2 acts as a short across ground.

The voltage thus passes through R2, gets dropped to relevant limits and starts charging C2.

Once C2 charges up to a level which develops a potential of 0.3 to 0.6V (+ zener voltage) at the base of T1, T1 is instantly switched ON, toggling T2, and the relay subsequently....finally the load gets switched ON too.

The above process induces the required delay for switching ON the load.

The delay period may be set by appropriately selecting the values of R2 and C2.

R1 ensures that C2 quickly discharges through it so that the circuit attains the stand by position as soon as possible.

D3 blocks the charge from reaching the base of T1.







Parts List

R1 = 610K
R2 = 330K
R3= 10K
R4 = 1K
D1 = 3V zener diode
D2 = 1N4007
D3 = 1N4148
T1 = BC547
T2 = BC557
C2 = 33uF/25V
Relay = SPDT, 12V/400 Ohms


Application Note

Let's learn how the above circuit becomes applicable for solving the following presented issue by one of the keen followers of this blog, Mr. Nishant.


Hello Sir,
I have a 1KVA automatic voltage stabilizer.It has one defect that when it is switched on, very high voltage is outputted for about 1.5s (therefore cfls and bulb got fused frequently) after that the voltage becomes OK. I have opened the stabilizer it consist of an auto-transformer,4 24V relay each relay connected to a separate circuit(each consisting of 10K preset,BC547,zener diode,BDX53BFP npn darlington pair transistor IC,220uF/63v capacitor,100uF/40V capacitor ,4 diodes and some resistors).These circuits are powered by a step down transformer and output of these circuit are taken across corresponding 100uF/40V capacitor and fed to corresponding relay.What to do in order to tackle the problem.please help me.Hand drawn circuit diagram is attached.



Solution:

The problem in the above circuit might be due to two reasons: one of the relays is switching ON momentarily connecting the wrong contacts with the output, or one of the responsible relays is settling down with the correct voltages a little while after power switch ON.

Since there are more than one relay, tracing out the fault and correcting it can be a bit tedious......the circuit of a delay ON timer explained in the above article could be actually very effective for the discussed purpose.

The connections are rather simple.

Using a 7812 IC, the delay timer can be powered from the existing 24V supply of the stabilizer.

Next, the delay relay N/O contacts may be wired in series with the stabilizer output socket wiring.

The above wiring would instantly take care of the issues as now the output would switch after some time during power witch ONs, allowing enough time for the internal relays to settle down with the correct voltages across their output contacts.

37 comments:

  1. Hello Sir,
    1.What is the power rating of each resistor?

    2.Will SPDT 12V relay,7A would work?Please specify the current rating?please check this link is it ok?
    http://evelta.com/index.php?route=product/product&path=158_81&product_id=58

    3.Can I power the above circuit with the same 12 0 12 step down transformer which is already there in the stablizer used to power 4 above hand drawn circuits.

    If yes then how?
    connecting 24v ac to bridge rectifier (using 1N4007) then a filter capacitor(please tell the best value preferred ), also tell the preferred value for filter capacitor(ceramic or electrolytic?) to be connected at input and output of 7812,Is this power supply will comfortably run the delay circuit?

    ReplyDelete
    Replies
    1. Hello Nishant

      1) All resistors are 1/4 watt CFR

      2) Yes the shown relay will work.

      3) you will have to use the 0-12 tap and connect a bridge rectifier/capacitor(1000uF/25V) with it and then the output would become OK for the above circuit.

      Alternatively you may use a 7812 IC and feed its input with the existing 24V DC...the output of the 7812 would then become applicable for the above circuit.

      Delete
  2. hello sir,
    i tried ur above circuit with 2 transistor in simulation software but its not working.......plz help me out....

    ReplyDelete
    Replies
    1. helo rocky, try it practically, it will work.

      Delete
  3. I have tryed it practically and mine doesen't Work.

    ReplyDelete
    Replies
    1. I have tried this circuit and it works just as explained.

      Delete
  4. hi mr swagatam,

    I am graduating engineering student can u help me to have a piece of idea for what i am going to invent as my theses please?
    thank you more power

    ReplyDelete
  5. hi mr swagatam please help me i am going to do a project for my theses i am just run out of time please provide me an idea to have a starting point in doing a ece project as my theses thanks. more power

    ReplyDelete
    Replies
    1. Hi Sandy, It will be difficult for me to suggest any particular project, you can choose any one from this blog, it has plenty neat stuffs of all levels.

      Delete
  6. What is the formula for determining the delay time from R2 and C2? I need about 1 second of delay. Thank you

    ReplyDelete
    Replies
    1. the formula is complicated, Google "RC time delay formula", you'll get an idea

      Delete
    2. ...you can set it with some trial and error.

      Delete
  7. sir! if want to use this circuit with two relays, how to connect it?
    i have an 15v 150mA adapter can i use it to supply this circuit? if can, do i need a regulator or 12v zener to step down it?

    ReplyDelete
    Replies
    1. you can connect the relay coils in parallel in the above circuits.
      you can connect 15V directly, no need to step it down.

      Delete
  8. The circuit works fine, thanks for the article. I have a requirement where I need delay ON and delay OFF. The required delays are very small, about 10 ms. How can I modify the above circuit to achieve ON and OFF delays?

    ReplyDelete
    Replies
    1. connect another capacitor right across the base and ground of T1, let's call it C1.

      Dimension C1/C2 for getting the 10ms delay,

      Delete
  9. hi swagatam.
    i want a delay of 3-8 seconds. will it work? by the way what are is the time delay of this circuit in this configuration?
    i want to run this directly on 220V. but dont want to use transformer or power supply. can you please give me a link or suggest me a capacitor diode based simplest power supply as an addon.
    thanks. waiting for your reply as you have not replied me before.

    ReplyDelete
    Replies
    1. Hi Muhammad, any delay between 1 second and 1 minute can be achieved from the above circuit by varying R2/C2 accordingly.

      you can try the following transformerless power supply:

      http://1.bp.blogspot.com/-9MKa1nQydEY/U6fgPJheSeI/AAAAAAAAHVA/FQbUQOL1qa8/s1600/simple+1+watt+LED+driver+circuit.png

      Remove the shown LEDs, and use the transistor emitter as the positive supply for the delay circuit. You can also remove the 56 ohm resistor, it's not required.

      Delete
  10. sir,
    i want to use above circuit work for 10 min
    is it possible by changing R2 and C2????????

    ReplyDelete
    Replies
    1. yes it's possible by increasing R2/C2 and also by increasing D1 voltage rating

      Delete
  11. Hi Swagatam,

    I am looking for a circuit for the following application:
    My pellet stove has a control board which I am not capable of modifying. It controls the pellet feed rate by activating a 115vac .5amp motor driving a feed auger. The control board allows me to adjust the time that the motor is powered from 3 sec to 8 sec. in a 15 sec cycle. I need to be able change the lower limit to 1 sec. I can do it in a couple ways with a circuit between the control board and the motor:
    either 1. Apply a two second delay each cycle (when the board activates the motor)
    --or-- 2. Only allow one of every three 15 sec cycles to activate the motor

    I think I prefer option one, but could use either.

    Could you provide a circuit to accomplish this?

    Thank you for any help you can provide.

    ReplyDelete
    Replies
    1. Hi Jencliff, from the explanation what I could understand is that you basically want an intermediate stage that would allow you to transform the existing 3 to 8 sec feature into 1 to 8sec, (right?)

      If that's the case I could recommend you a IC 555 monostable timer stage to be inserted in between the motor power device and the control board output.

      Delete
    2. Hi Swagatam,

      Thanks for the quick response. I do have a few questions after looking at various circuits on your blog which use the IC 555.
      1. I assume I will need a relay in the circuit to handle the 120ac. That being the case, how does the IC 555 get triggered, since the only I have is the 120ac and the IC 555 seems only to handle low (5-9dc) voltage?
      2. I assume the circuit when completed could be placed between the motor and control board on the neutral leg from the relay.
      3. Are the above assumptions correct and could you point me to a circuit like those on your blog which shows the whole circuit I need?

      Delete
    3. Hi Jencliff,

      The trigger input will need to be taken from the DC output of the control board which may be responsible for activating its relay.

      If you don't wish to disturb the control unit circuit, then you can think about installing a separate 12V AC/DC power supply for feeding the IC 555 circuit and use another external relay at the output of the IC for controlling the motor.

      The above set up could then be powered from the 120V relay output from the control board.

      You could try out the following design for the monostable:

      http://homemadecircuitsandschematics.blogspot.in/2014/10/power-switch-on-alarm-with-auto-off.html

      Delete
  12. Hi Swagatam,

    Can I connect a push button between R2 to instantly switch the circuit on?

    Thanks,
    Vijay

    ReplyDelete
    Replies
    1. Hi Vijay,
      you mean parallel to R2? Yes you can do it but make sure that the switch has a 1K resistor in series otherwise the BC547 will get damaged instantly on pressing the button.

      Delete
    2. Hi sir can you give me please any electrical project for my theses...thank you

      Delete
    3. Hi Naldz, It'll depend on your preferred difficulty level, there are plenty of circuits from easy to the most difficult

      Delete
  13. hi ,i am a big fan of yours. i want to build a circuit like.. when power is present- relay on(relay got power) then off (relaxed) automatically after few seconds..please reply..thanks

    ReplyDelete
    Replies
    1. thanks skl, you can use the above explained circuit with a slight modification.

      replace T2 with another NPN....remove R3 and let the base of T2 connect directly with the T1 collector....now, T2 emitter will connect with the ground, and the relay assembly will need to be fixed across the collector of T2 and the positive line.

      I hope you got it!!

      Delete
  14. thanks for reply, i redraw the circuit as you said. should i connect R4 with base of T2 or remove that too? i am little curious to know how 'delay on' circuit become 'delay off' by replacing a transistor ? actually i wanted to know the working principle... thanks again.

    ReplyDelete
    Replies
    1. without R4 your T2 (NPN) will never conduct, so it's required.

      By changing T2 into a NPN transistor we are inverting the process (the driver stage) and making delay ON into delay OFF, T2 is now doing the opposite of T1, previously with a PNP it was following T1

      Delete
  15. sorry to disturb you again. actually i want to use this circuit replacing a push button switch. in my circuit relay coil needs a positive spike for a second to start. your circuit providing negative spike for relay...can i use a pnp transistor as T2 ? hope you got my question.

    ReplyDelete
    Replies
    1. there is no push button in the discussed design, the spike is received by R2 from the positive line...or you can disconnect R2 from the positive and use it across any source which generates the spike

      Delete
    2. sorry it won't to a spike...the input must be consistently present for R2.

      A PNP at T2 will make it delay ON, as it's in the existing diagram.

      please explain your requirement in detail once again.

      Delete
  16. I have a battery charger circuit (you post a circuit like this with main ac on-off by relay; but i want to cut the 12 dc supply), its like - a push button switch turns on the whole circuit by giving (12v) positive supply to a relay - relay on - circuit on. now relay stay on by the circuit. when battery got full charge - relay off.
    i just want to replace that push button switch so that battery got automatically connected with charger when i;m not in home. or
    when the electricity come after loadsheding i do'nt want to push that button to give charge to the battery

    ReplyDelete
    Replies
    1. your application has nothing to do with a delay timer....I think you are referring to the following post:

      http://www.homemade-circuits.com/2011/12/high-current-10-to-20-amp-automatic.html

      make the second circuit from the following article it will do exactly what you are looking for:

      http://www.homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html

      Delete

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Readers are advised to proceed with the construction of the presented circuits only after understanding the concepts from the core. Not adhering to this can lead to failures and frustrations.
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