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Simple Delay Timer Circuits Explained

Here we discuss how we can make simple delay timers using very ordinary components like transistors, capacitors and diodes.

Importance of Delay Timers


In many electronic circuit applications a delay of a few seconds or minutes becomes a crucial requirement for ensuring correct operation of the circuit. Without the specified delay the circuit could malfunction or even get damaged.

Let's analyze the various configurations in details.

Using a Single Transistor and Push Button


The first circuit diagram shows how a transistors and a few other passive components may be connected for acquiring the intended delay timing outputs.

The transistor has been provided with the usual base resistor for the current limiting functions.

A LED which is used here just indication purposes behaves like the collector load of the circuit.

A capacitor, which is the crucial part of the circuit gets the specific position in the circuit, we can see that it's been placed at the other end of the base resistor and not directly to the base of the transistor.

A push button is used to initiate the circuit.

On depressing the button momentarily, a positive voltage from the supply line enters the base resistor and switches ON the transistor and subsequently the LED.

However in the course of the above action, the capacitor also gets charged fully.

On releasing the push button, though the power to the base gets disconnected, the transistor continues to conduct with the aid of the stored energy in the capacitor which now starts discharging its stored charge via the transistor.

The LED also stays switched ON until the capacitor gets fully discharged.

Te value of the capacitor determines the time delay or for how long the transistor stays in the conducting mode.

Along with the capacitor, the value of the base resistor also plays an important role in determining the timing for which the transistor remains switched ON after the push button is released.

However the circuit using just one transistor will be able to produce time delays which may range only for a few seconds.

By adding one more transistor stage (next figure) the above time delay range can be increased significantly.

The addition of another transistor stage increases the sensitivity of the circuit, which enables the use of larger values of the timing resistor thereby enhancing the time delay range of the circuit.





PCB Design





Video Demonstration


 Using a Triac:


The following image shows how the above delay timer circuit may be integrated with a triac and used for toggling a mains AC operated load


Without a Push-Button


If the above design is intended to be used without a push button, the same may e implemented as indicated in the following diagram:



The following circuit shows how the associated push button may be rendered inactive as soon as it's pressed and while the delay timer is in the activated state.

During this time any further pressing of the push button has no impact on the timer as long as the output is active or until the timer has finished its delay operation.



Two Step Sequential Timer


The above circuit can be modified to produce a two step sequential delay generator. This circuit was requested by one of the avid readers of this blog, Mr.Marco.



A simple delay OFF alarm circuit is shown in the following diagram.

The circuit was requested by Dmats.



The following circuit was requested by Fastshack3

Delay Timer with Relay


"I am looking to build a circuit that would control an output relay. This would be done in 12V and the sequence will be initiated by a manual switch.

I will need an adjustable time delay (possibly displayed time) after the switch is released, then the output would go on for an adjustable time (also possibly displayed) before shutting off.

The sequence would not restart until the button was pressed and released again.

The time after the button release would be from 250 milliseconds to 5 seconds. The "on" time for the output to turn on the relay would be from 500 milliseconds to 30 seconds. Let me know if you can offer any insight. Thanks!"



So far we have learned how to make simple delay OFF timers now let us see how we can build a simple delay ON timer 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.

Delay ON Timer Circuit Working Details


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

PCB Design


Application Note


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

Circuit Problem:


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.


Solving the Circuit Problem


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.

Feedback from Mr. Bill


Hi Swagatam,

I stumbled across your page doing research on the web to make my delay more consistent.Some back ground information first.

I am a bracket drag racer and launch the car on first sight of the 3rd amber bulb as the christmas tree is coming down.

I use a transbrake switch that is depressed to lock the automatic transmission in forward and reverse at the same time.

This allows you to rev up the engine to build power for launch. When the button is released the transmission comes out of reverse and moves the car  forward under high rpm.

This is like popping the clutch on a manual transmission car, anyway my car reacts to quickly and the result is a redlight, leaving to early,  and you lose the race.

In dragracing your reaction time on the launch is everything and it is a game of hundreths-thousanths with the big boys, so I have put the transbrake switch on a relay and put a  1100uf  cap combo across the relay to delay its release.

Because of the car electronics I don't believe there is a precise voltage charging this cap every time I activate this circuit and precision is key so I bought a power stabilizer off of Ebay that takes 8-15 volts in and gives a consistent 12volts out.

This turned my season around but i believe this circuit could be made to be more precise and to vary the delay time in an easier way rather than swap cap combos.

Also should I run a diode in front of the relay, not currently because all that is there is the on off switch- where will the current go? I am not an electrical engineer by any means but do have some knowledge from trouble shooting high end audio for many years.

Would love your thoughts-  thankyou

Bill Korecky

Analyzing and Solving the Circuit


Hi Bill,

I have attached the schematic of an adjustable delay circuit, please check it out. You can use it for the mentioned purpose.

The 100K preset can used and adjusted for acquiring precise short delay periods as per your specifications.

However, please note that, the supply voltage will need to be minimum 11V, for the 12V relay to operate correctly, if this is not fulfilled then the circuit might malfunction.

Regards.


Simple 5 to 20 Minute Delay Timer


The following section discusses a simple 5 to 20 minute delay timer circuit for a specific industrial application.

The idea was requested by Mr. Jonathan.

Technical Requirements


While trying to figure out a solution to my problem on google, I came across your posting at 

https://homemade-circuits.com/2012/05/simple-delay-timer-circuits-explained.html

Because you posted your email address there, and have been responsive to requests, I thought I’d take a shot at getting your help with a circuit design.

I'm trying to figure out how to build a better Sous Vide controller. The main problem is that my water bath has a very high hysteresis, and when heating from colder temperatures will overshoot about 7 degrees from the temperature at which power is terminated.

It is also very well insulated, with a gap between the inner and outer vessel which makes it act like a thermos jar, because of this it takes a very long time to decline from any excess temperature. My PID controller has an SSR control output and a relay alarm output.

The alarm can be programmed as a below limit alarm with an offset from the set-point. I can use a five volt supply I already have for my circulation motor to run through the alarm relay and drive the same SSR the control output is driving.

To be on the safe side and protect the PID controller I'll add a diode to both the alarm voltage and the control voltage to prevent one output from feeding back into the other.

I'll then set the alarm to stay on until the temperature rises above the set-point minus 7 degrees. This will allow the PID tuning to be adjusted without having to account for the initial temperature ramp-up.

Because I know that last few degrees will be achieved without any power input, I'd really like a way to delay any recognition of the control signal for about five minutes after the alarm shuts off, as it will still be calling for heat.

This is the part I've yet to figure out the circuitry for. I’m thinking of a normally closed relay in series with the control output, which is held open by the alarm signal.

When the alarm signal is terminated, I need a delay on the order of five minutes before the relay returns to its ‘off’ normally closed state.

I would appreciate help with the delayed off portion of the relay circuit. I like the simplicity of the initial designs on the page, but I get the impression they wouldn’t handle anywhere near five minutes.

Thank you,

Jonathan Lundquist

The Circuit Design


The following circuit design of a simple 5 to 20 minute delay timer circuit can be suitably applied for the above specified application.

The circuit employs the IC4049 for the required NOT gates which are configured as voltage comparators.

The 5 gates in parallel form the sensing section and provides the required time delay trigger to the subsequent buffer and the relay driver stages.

The control input is acquired from the alarm output as indicated in the above description. This input becomes the switching voltage for the proposed timer circuit.

On receiving this trigger, the input of the 5 NOT gates are initially held at logic zero because the capacitor grounds the initial trigger via the 2m2 pot.

Depending upon the 2m2 setting, the capacitor starts charging up and the moment the voltage across the capacitor reaches a recognizable value, the NOT gates revert their output to logic low, which is translated as a logic high at the output of the right single NOT gate.

This instantly triggers the connected transistor and the relay for the required delay output across the relay contacts.

The 2M2 pot may be adjusted for determining the required delays.

Circuit Diagram





Need Help? Please leave a comment, I'll get back soon with a reply!




Comments

  1. I guess your need matches the one explained in the following article:

    https://homemade-circuits.com/2013/06/programmable-bidirectional-motor-timer.html

    Though it's quite complicated, probably it's the only way of doing it through discrete components.

    ReplyDelete
  2. The sequential timer can be surely designed, in fact I have already posted one such design in this blog, however I could not understand the connection between the motor rotation and the timing, if you could clarify that part then I could possibly suggest you an appropriate design.

    ReplyDelete
  3. Replace the push-switch in the second circuit with the tilt switch terminals with a series 1uF/25V capacitor.

    ReplyDelete
  4. I can provide a 4060 IC circuit, not a transistorized one.

    ReplyDelete
  5. Good day,
    I have a 24vdc door activation circuit that I need a 5 second timer for. When the push paddles is depressed, I need to activate the latch relay for 3-5 seconds (to hold the latch open long enough for the operator to activate). Can one of your 12v circuits be modified to work with the 24vdc power supply?

    ReplyDelete
  6. Good day!
    you can try the second circuit presented in the above article, replace the LED/resistor with the latch relay with protection diode.
    The push button may be replaced with the paddle switch.

    ReplyDelete
  7. ...the 1000uF cap should be tried with different values ranging from 10uF to 100uF for acquiring the required delay.

    ReplyDelete
  8. The 12V light should illuminate and timing initiated when the tilt switch is tilted the first time or after its tilted back to its original position??? please clarify this point.

    ReplyDelete
  9. I want to create a small, variable delay. Changing the resister on the transistor base will allow the delay to be adjusted?
    Using a 12v supply, can you recommend what resistance and capacitance values I should experiment with to induce a delay between 0.1 and 0.5 seconds?

    ReplyDelete
  10. you can try the second circuit given in the above article, use a 1M pot in series with a 1K resistor at the base of the transistor and a 10uF capacitor

    ReplyDelete
  11. Hey everyone,
    I am looking for a time delay circuit that I can use for some drop photography. I am looking to conenct a light gate to it so that when a drop breaks the gate it starts the time delay circuit then triggers my flash gun. (flash gun has a 2 wire cable to connect to circuit). Does anyone know of a circuit which can help me? If you could include the light gate into the circuit would be a bonus.

    Thanks :)

    ReplyDelete
  12. Swagatam... i need circuit, that after power on the circuit , output should be 5 or 10 sec, delay...Please help.

    ReplyDelete
  13. you can try this:

    https://homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html

    ReplyDelete
  14. Good morning sir,
    I am totally confused with so many timer/delay ckts. Please refer me a ckt.
    as per my requirements.
    1) I want to start a 12v/500mA (max)
    d.c gear motor after 40-60sec of switching on and will run untill it is switched off.
    2) Another previously related problem.please refer any 12-0-12 a.c to 12v d.c rectifier ckt.with component
    list and diagram......thanks.....
    With regards,
    K. Kausik

    ReplyDelete
  15. Good morning kaushik,

    you can try the following circuit, the values of C2, R2, D1 will decide the length of the ON time delay.

    https://homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html

    please repeat your previous request, I am not able to remember it.

    ReplyDelete
  16. hi can you suggest to me a simple circuit with 15 components for me as an electronics student ............pls i know u can help me...

    ReplyDelete
  17. hi, you can try the following design:

    https://homemade-circuits.com/2012/04/semi-automatic-water-level.html

    ReplyDelete
  18. i want to make ten LED light using time delay feature having fed one by one after specific time day.

    ReplyDelete
  19. Hi Swagatam,
    I wanted to make a timer circuit which cuts off the supply to a circuit after 5-6 seconds after being triggered , even when the trigger is still on

    ReplyDelete
  20. Hi Shyam,

    you can try the following design:

    3.bp.blogspot.com/-FHAn8wSZwQ0/UrFIJJpYjgI/AAAAAAAAF7I/M8L5IqLxCwA/s1600/automatic+choke+solenoid+circuit.png

    ignore the relay contact capacitors and wiring set up, it was drawn for a different application.

    ReplyDelete
  21. Hi Swagatam,
    I Tried the circuit which you gave , but the relay never turns off... I have changed the capacitor values ,but also no change.... can you please tell me what could be the problem.

    ReplyDelete
  22. Hi Shyam, I think I misread your requirement, the given circuit is good but is not applicable as per your need.

    I'll give you a suitable one but let me know whether it should be a relay operated or without a relay, meaning the current requirement of the circuit is relatively higher or nominal??

    ReplyDelete
  23. Hi Swagatam, I wish to operate a 12v relay with this circuit. the current requirement is nominal.

    ReplyDelete
  24. Hi Shyam, I think the previous 555 based circuit which I recommended is perfect for your application and it'll surely work. Connect the relay between pin3 and ground, and make sure the line connecting pin4/8 of the IC gets the (+) 12V supply....remove the entire relay connections that's shown in the original diagram....your relay must be between pin3 and ground....let the 1N4007 diode be there with the coil, its anode will go to ground and cathode to pin3.

    I thought that you wanted the entire circuit to get cut off after 5/6 sec but later realized that you wanted to execute the cut for another circuit via this timer circuit, therefore the suggested 555 circuit is right for the application.

    ReplyDelete
  25. Dear Sir ,it still doesn't work because, as I had mentioned earlier the switch remains on . Only when I turn off the switch the relay gets off immediately,there is no 5 sec delay even then.

    Sorry Sir if I am bothering you too much..

    ReplyDelete
  26. Dear Shyam,

    It seems there's something seriously wrong in the connections or the IC may be faulty, because the circuit is extremely standard and straightforward and should start working immediately.
    pressing the push button should trigger the relay, and switch it off after the predetermined delay decided by the values of the 47K res and 10uF cap, irrespective of the switch position.

    3.bp.blogspot.com/-FHAn8wSZwQ0/UrFIJJpYjgI/AAAAAAAAF7I/M8L5IqLxCwA/s1600/automatic+choke+solenoid+circuit.png

    ReplyDelete
  27. Hi Swagatam

    Hoping you will be able to suggest a suitable circuit for a problem we are having. We need a circuit that can be fed via an engine ignition switch which will output a 12v feed to a solenoid for a short time, about a second, then power down once the feed from the ignition switch is turned off. Basically we have a piece of equipment that no longer turns off with the ignition switch, our current best option is to directly wire in a switch to the solenoid but if the circuit I describe is feasible then that would be a preferred option.
    Thanks in advance.

    ReplyDelete
  28. Hi Hex,

    There are a few confusions:

    Do you want the solenoid to stay ON for 1 second after the ignition is turned OFF or should it be in response with the ignition switching, or is it regardless of the ignition ON time? Meaning once the ignition is switch ON, the solenoid will switch OFF after a second regardless of the ignition ON/OFF situation??

    What kind of solenoid are you using, a spring loaded one or the one which requires a push-pull opposite supplies for locking and unlocking.

    ReplyDelete
  29. hello sir . i need a similar circuit but with 1 hour delay. What changes should i do?
    Thankyou

    ReplyDelete
  30. Priyanka, you can try the following design:

    https://homemade-circuits.com/2014/11/long-duration-timer-circuit-using.html

    the 1000uF/2m2 components will need to selected by some trial and error for achieving the desired delay response.

    ReplyDelete
  31. how would i do a circuit like this that turn off 5 seconds after being turned on with a 12v source?

    ReplyDelete
  32. use the first circuit, connect the load or the circuit in place of the LED/resistor

    ReplyDelete
  33. Hi Swagatam

    I have a circuit what is exactly like the one in "simple delay timer circuit". The only minor difference is +5V, and the major one is that it has to be approximately 1 sec "Delay ON" and immediately off again.

    How would I do that?
    Thank you

    ReplyDelete
  34. Hi Detlef,

    In the first circuit, you can remove the 1000uF capacitor and replace the push switch points with a 100uF capacitor, now it'll behave in the way you have suggested, you may need to tweak the value of the 33k resistor in order to achieve the required 1 sec timing.

    ReplyDelete
  35. Hai, I need to place three LEDs L1, L2, L3 in a single circuit. So when I power ON the circuit means, the LED L1 should glow after 30 mins and the LED L2 should glow after 45 mins and the LED L3 should glow after 60 mins. Could u pls provide a circuit for that???

    ReplyDelete
  36. Hai, you can use a IC 555 and IC 4017 "chaser" circuit, set the 555 as a 15 minute timer astable and then use the outputs of the 4017 IC for getting the required sequences.

    for 30 minutes you can just skip one output of the 4017.

    make sure to connect pin15 with pin10 of the IC 4017

    ReplyDelete
  37. Thanx. I tried an 555 astable calculator to obtain 1 sec time high and 15 mins time low. But not able to obtain a correct value for R1, R2, C. Could u pls help me by giving the value for R1, R2 and C for the 555 astable circuit for 1 sec time high and 15 mins time low? Thanx...

    ReplyDelete
  38. 15 minutes high, and 1 sec low will also do.

    R1 = 1.3 Meg

    R2 = 1.5K

    C = 1000uF

    ReplyDelete
  39. Hi, Swagatam,

    i need a delayed off timer working on 3v dc power supply , kindly share your idea if you have.

    Best Regards,

    Obaid

    ReplyDelete
  40. Hi Obaidullah, you can try the second or the third design from the above article

    ReplyDelete
  41. Dear Sir Please help me ....

    I Want A Circuit Wich can Relay On 6v For 20 Second from get A Puch....


    After 20 Second It Will Automatically Goes off

    Again push then get on for 20 second

    Thanx Sir

    ReplyDelete
  42. Dear Sharoj,

    You can make the first circuit from the above article, just replace the LED assembly with your relay

    ReplyDelete
  43. Hi, for the shown circuits you can use any voltage between 3V and 24V

    ReplyDelete
  44. Hi, i need a circuit that has 1 push to make switch but can control 2 things (in this case a water pump and fuel injector) i want the water pump to be going before the injector does its thing to build up enough pressure in the hose and all this needs to work on 12V. Can you help?

    ReplyDelete
  45. Hi you can use the second circuit from top and do the following modifications:

    replace the LED/resistor with a relay coil (connect a diode across the coil).

    Now connect the pole of the relay with the 12v positive, N/O with (+) of the water pump and the N/C with the (+) of the injector...the (-) wires of the two units may be joined together and connected with the circuit negative.

    The above mods will allow you to implement the actions as specified by you.....reduce the 1000uF value for setting the required time delay between the pump and the injector activation.

    ReplyDelete
  46. hi there, i need a 15v circuit with a push-button that when i press and hold for 3 seconds then release the circuit LED will remain on for 10-15 seconds without dimming effects. Is it possible? Thanks for your help.

    ReplyDelete
  47. hi,

    the first circuit in the above article will fulfill your requirement, but you won't need 3 seconds to initiate it, just a fraction of a second press will be enough to trigger the LED for the desired length of time...

    ReplyDelete
  48. hi mr swagatam,

    is it possible i add one more LED to the circuit but it will delay on around 2seconds after the 1st LED light up and both LED will off at the same time.
    Thanks for your help.

    ReplyDelete
  49. Hi Babusan,

    I will have to figure it out, can't able to simulate it quickly...if it's possible I'll produce it in the above article soon...

    ReplyDelete
  50. hi swagatam,

    sure, hope to hear from you soon. Thank you in advance.

    ReplyDelete
  51. Hi Swagatam,

    I am impressed with the simple design you manage to came out to create a time delay without using the common 555 timer IC which most users would have used.

    I am interested in the formulas on how did you manage to determine the values of the resistors and capacitors and I believe one does not simply just random pick the values and try it out to determine it. It would be great if you can provide me a rough explanation on the formulas part and also how does the circuitry works

    For instance, in the section where you mentioned about adding an additional transistor to the circuit would increase the time delay but why and how does it affect the time delay? Does it have anything to do with the transistor switching on/off and meanwhile does it mean the more transistors the circuit has, the longer the delay it can provide?

    Hope to hear from you soon. Thanks:)

    ReplyDelete
  52. Thanks John,

    Definitely there should be a universal formula for determining the RC timing values, however determining the same using a practical trial and error method is much simpler than solving the formula.

    If you are interested to know regarding the formulas, you can refer to the following wikipedia article:

    https://en.wikipedia.org/wiki/RC_time_constant

    By adding more transistors we are making the circuit more sensitive to the voltage rise across the charging capacitor...so if one transistor requires 0.6V to trigger, with two transistors we could reduce it to 0.3V with three transistors it can become more sensitive and trigger at 0.1V..this improvement allows smaller capacitors to be used and generate longer time delays...however making the circuit more sensitive can make it prone to false triggering from external spurious signals, that's the drawback, although this can be tackled by adding caps across the base/emitters of the transistors.

    ReplyDelete
  53. Hi,
    I need a circuit which has two out on and off.
    I mean one on and the other off and reverse with transistor or mosfet.

    Thanks

    ReplyDelete
  54. Hi,
    I need a circuit which has two out on and off.
    I mean one on and the other off and reverse with transistor or mosfet.
    So i can pulse it with 555.
    Thanks

    ReplyDelete
  55. you can use a 4047 IC...a 555 will not be needed as it has it's own oscillator.

    ReplyDelete
  56. Hi Swagatam,

    Thanks for the link and explaination on how the side effects it will cause for having more transistors and it really give me a better idea of how the circuit work out. Currently I have another quiry which is why when do you use a pnp transistor and connect in that format instead of using the NPN transistor and connect back the similar connection like in the first circuit?

    Does it have any differences? (like lesser power consumption? neater? lesser voltage required?)

    Hope to hear from you soon:)

    ReplyDelete
  57. Hi John, the PNP transistors activates in response to a negative (-) pulse at its base...the in the discussed design the NPN is supposed to produce a negative pulse when activated which is intended to be responded by the next stage, and therefore a PNP stage is used in that position.

    together they make the circuit more sensitive than a single NPN

    ReplyDelete
  58. Hey, I need a circuit that can delay a signal output to a electric motor. I would be using a 9v battery to
    power the circuit and a contact switch to turn it of and on. I need the motor to turn on about 0.5s after the contact switch is pressed and to stay on until the contact switch is no longer pressed.

    ReplyDelete
  59. you can use the following circuit:

    https://homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html

    adjust C2/R2 appropriately for the intended delay period....you can eliminate the emitter zener diode D1, since it's required only for longer delays.

    ReplyDelete
  60. Hi!
    I need a circuit with auto stop when the dc motor stops.

    ReplyDelete
  61. Thank you so much for the article! I am working on a analog circuit that controls the glowing time period of a bulb.
    I'm using a relay,2 resistors and 2 capacitors,one is electrolytic and the other one is ceramic. I understood the role played by electrolytic capacitor i;e it stores enerygy but I'm unable to figure out why on earth is the other capacitor used for. PLEASE check the link for circuit

    ReplyDelete
  62. Hi Swagatam Majumdar

    I find that circuit requested by Fastshack3 is useful to my case, well.... almost, since it require 12V signal to activate. Is there any way to activate the circuit with ground or 0V signal?. Thanks in advance

    ReplyDelete
  63. Hi balakutak,

    you can do it by eliminating the BC547 stage and replacing the collector/emitter position with the push button and the capacitor.

    the base resistor of the BC557 could require an increase to may be upto 100k

    ReplyDelete
  64. Hello. I am trying to build this circuit -
    s768.photobucket.com/user/dpbayly1/media/BMW%20X5%20Folder%20%202/autofoldmirrors_zps53079bfd.png.html - but the problem is, the time delay relays that I am using, does not reset, when the SPDT relay changes its state from NO to NC, as mechanical switch of the central pin is too fast. Is there a way to "slow" down the flow of electricity, when the pin switches? I was thinking about two coils, one at pin 87A and the other at pin 87. what do you think?

    ReplyDelete
  65. you can reduce the current to the relay coils for reducing the electromagnetic pull, but still that would create a difference in milliseconds only....

    ReplyDelete
  66. Good day

    i would like a circuit that counts up in hours maybe to a maximum of 5 hours continuously when every time it is switched ON.i have been having a problem with this for a long time now.

    Thank you

    ReplyDelete
  67. you can use a 4060 and 4017 timer circuit as given here:

    https://homemade-circuits.com/2012/04/how-to-make-long-duration-timer-circuit.html

    ReplyDelete
  68. Is is possible to make a circuit that lights up a bulb but then a momentary switch can cut the power to turn the led light off? The tough part would be to keep the light off for longer than when you release the switch. I want the bulb to stay off for 5 seconds before it getting power again.

    ReplyDelete
  69. you can use the second circuit from the above article, and replace the BC557 with another BC547 transistor..after this you can tweak values of the 1000uF and the 2M2 for getting the desired off time delay

    ReplyDelete
  70. Hi, good job on these. Extremely useful, so thanks! Is it possible that when switch is pressed it only charges the capacitor but does not power on the light? The LED is only required on for 5 seconds after the switch is released.

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  71. you can try connecting a 220uF capacitor in series with the LED with a 1M resistor in parallel with this capacitor, this will probably take care of the mentioned situation.

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  72. I see this is a rather old article, but I have a small project that I need help with. I'm building a "Groundhog day" clock that will play "I got you babe" when the alarm goes off. So I need a circuit that will activate a relay after about 5 seconds and only last about 500 milliseconds. I have both SSD and mechanical relays of different voltages, so I'm pretty flexible with the design.
    The reason for the delays is that the MP3 player takes a few seconds to turn on, then I have to bridge the play button to get it to play the song. I have tapped into the alarm clocks pre-amp, so the volume can be controlled by the original knob.
    I will post a link to my project, once I have it completed.
    Thanks for any assistance.

    ReplyDelete
  73. Instead of the above circuits the following would suit your application better:

    https://homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html

    just make sure to add a 0.22uF (arbitrarily selected) capacitor between the base of T2 and the junction of R3/R4, this will take care of the 500ms activation of the relay

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  74. I saw that to design a latch you told to use a 100K resistor across the collector of BC557 and the base of BC547 removing the capacitor entirely. Could you please suggest a modification to this design that would de-latch the power to the same led upon click of another button.

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  75. you can add a push button across the base of BC547 and ground for the required de-latching feature

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  76. Thank you! That was perfect... but I have another question... how do I de-latch by pulling a GPIO pin from low to high? Could you please help me with this..

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  77. you can do it by adding another BC547 to the design.

    collector to the existing BC547 base, emitter to ground, and the base to the high logic via a 10K resistor

    ReplyDelete
  78. Thanks for this ...very helpful..:) With the above latching/delatching circuit, I am driving a microcontroller board instead of the LED. Before connecting the board, the supplied voltage comes across the points. But just after connecting the board, the voltage at the same points goes down by nearly 30% and is not sufficient to turn the board on. Is there a way we can reduce the resistor values in the circuit you provided to avoid the voltage drop? Or any other recommendations?

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  79. what could be written while i submit synopsis for this project ??

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  80. please try increasing the 1K resistors to 10K and see the response :)

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  81. Hi Mate,

    I was looking for something like this. Can this be modified such that the delay time is less than a second. like 0.1 to 0.9.


    Thanks

    ReplyDelete
  82. hi Mate,

    Sorry if this is a resubmission. i want to delay the timer to less than a second. I am guessing it can be done by reducing the farad of the capacitor. Could you please advise.

    Seen several modules in ebay but did not find anything that could meet my requirement.

    ReplyDelete
  83. you are right mate, just reduce the 1000uF capacitor to something like 1uF, or even less to get the intended small delay effect

    ReplyDelete
  84. Thank you mate. I don't see the 1000uf capacitor that your referring to on the last circuit. Well that is the circuit am after. What if i need to delay the energizing once i hit the switch to about a second or two. i am guessing i need to add a resistor in series to the positive end of the capacitor.

    I've also tested a circuit where i just add a 10v 2200uf capacitor in parallel with a 5v Relay. I seem to get what i want, the relay stays energized for less than a second. Would that be ok to operate an ac transformer for a DIY spot welder.

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  85. I was referring to the first circuit which looks more appropriate for your requirement....simply replace the LED/resistor with a relay/diode, and adjust the 1000uF capacitor value to some lower level, and you will get exactly what you are looking for.

    Connecting a capacitor in parallel to the relay coil will delay the switch OFF as well as the switch ON of the relay which probably we don't want...

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  86. Thanks A lot buddy. Your the best.

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  87. Hi,

    I'd like to ask if some of these circuits would fit for alarm system? A few months ago my basement was robbed. Now I made very simple alarm system (opened door - ON, Closed door - OFF) and additionally installed "jail" doors. I would like to have delayed-OFF alarm (for one or few minutes), Can anyone suggest some proper and functioning circuit? [Siren properties: 12V; 150mA]

    Thank You in advance

    ReplyDelete
  88. Hi, yes definitely you can use any of the first two circuits and use it for the mentioned purpose in conjunction with a reed switch and magnet assembly...I hope you know how to configure the proposed idea.

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  89. Thank you. Well I hope too that I know how to carry out. The problem is delay time, if it will be too short what should I change in the circuit to prolong alarm?

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  90. the timing can be varied by altering the input capacitor and resistor at the BC547 base individually or together. If you have problems let me know about it, I'll create the design for you.

    ReplyDelete
  91. Hello,

    I have a similar requirement, but I need to reset a 5V touch alarm (by cutting power to it just momentarily) after it trips and starts buzzing a piezo buzzer for about five seconds. Thanks very much!

    ReplyDelete
  92. ...the solution is also the same as explained by me in the above comment.

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  93. Hi, So I've done the circuit according the second scheme, but the delay OFF time was very short (~0.5 sec). Then I changed capacitor from 1000 uF to 2200uF and resistor from 2m2 to 3m3, but delay off time extended maybe up to one second. Do you have any suggestion what should I do to prolong alarm? Thank you.

    ReplyDelete
  94. Hi, In the second diagram, using a 1000uF and a 2M2 resistor at the base of the NPN should produce a delay of over 2 minutes....if it's giving 0.5 seconds then there could be something seriously wrong with your circuit...
    similarly with a 2200uF and 3M3 the delay should be well over 5 minutes.

    I think you might have connected the transistors wrongly or something could be faulty in the circuit.

    ReplyDelete
  95. Hi again, might be I made mistakes, I'll check it later today. Thank you again.

    ReplyDelete
  96. you mean, delay on power switch ON?

    for that you will need to replace the switch points with the emitter/collector of a BC557 transistor, then connect a 0.22uF capacitor across its base and ground line, and make sure to connect two 10k resistors on either side of the 0.22uF capacitor with their free ends connecting with the positive line.

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  97. if you can interpret my advise through a drawing then I could check it and suggest you the needful.

    you could send it to my email ID which is provided in my "contact" page.

    ReplyDelete
  98. Hi Swagatam. I have built the single transistor version of this for an application to hold a 3V relay on for å period after a pushbutton is pressed and the timing function works well - I can see the LED growing slowly dimmer. I have connected the relay's negative side (with a diode across to the positive) to the 1K resistor, but the relay does not latch and hold (it clicks though). I also tried reducing and then removing the 1K resistor and even the LED. Still no latch. What am I doing wrong?

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  99. Hi Glen, a single transistor will not be enough to sustain the relay operation, you may either try the second circuit or configure another BC547 with the existing one in the "Darlington mode"....this will do the job for you.

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  100. A-ha. Thanks, I'll try that.

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  101. Thanks Swagatam. I have built a Darlington bridge with two 2N2222 transistors (that's what I had available) and the relay now latches. But I cannot get a delay of more than about 5 seconds (I need 10-15.) I have increased the caps to 3 x 6800uF but when I increase the timing resistor beyond 40K it stops working - the transistor will not turn on. Any suggestions?

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  102. that's very strange, because 6800uF is huge and should easily produce above 5 to 10 minutes of delay...I think may be your relay has a low coil resistance which is causing quicker discharge of the cap.

    in that case you may need to build the second circuit, and still if it does not provide the required delay then you could probably go for an IC 555 monostable circuit

    ReplyDelete
  103. Since I am now using two transistors as a Darlington pair the circuit is almost the same as your circuit two. See the circuit diagram I sent via Google Drive. I just added your cap and timing resistor and used two 2N2222s.

    I agree that the coil resistance may be the problem since when I remove the relay and just use the LED the delay is much longer. The specs for the 3V relay say 60 ohms at .15W to 20 ohms at .45W.

    When I replace the relay with the LED I can also increase the timing resistor beyond 40K and the transistors still turn on - why does the relay affect how high the timing resistor can be?

    Does the choice of transistors have any effect?

    And finally, once we solve this problem, I need to get the final circuit as small as possible. I have read that there are Darlington pair transistors in a combined package. Would these work and can you recommend a particular model?

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  104. Sorry Swagatam I didn't understand your answer to my question above about the 5V alarm circuit. I have a signal coming from an alarm to a buzzer. I want the buzzer to go for 5-10 seconds, then momentarily cut off power to the alarm unit so it resets. Since there is continuous power to the buzzer I don't know how to cut power to the alarm after the delay.

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  105. Yes I saw the diagram in my email, it's good.
    However now I understand the real cause of the problem....it's because of the 3V supply which is much nearer to the holding threshold 0.6V compared to 12V...and that's why the capacitor charge is able to reach this limit pretty quickly.

    there's one option that you can try to counter this issue....connect 4 nos of 1N4148 diodes in series with the base resistor of the transistor, this will help to increase the delay significantly even with smaller capacitor.

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  106. 2N2222 is OK, the problem is not with the transistor but the voltage level as explained above.

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  107. Hi Glen, you can use an IC 555 monostable between the signal and the buzzer. This can be set for triggering for the desired length of time....but I could not understand why the alarm needs to be reset? It should have its own resetting system right?...what kind of alarm circuit are you using by the way?

    ReplyDelete
  108. The "alarm" is actually a capacitance touch board from Adafruit and I have attached it to a metal clock to be used in a sculpture. It goes off when anyone touches the clock, but at the moment it is locking on after the touch.
    I could combine your timer circuit with your transistor latch circuit and use the NC relay contacts to cut off power for a fraction of a second. But I wondered if there is a way to use a PNP transistor as an NC contact and when it is turned on after the delay it could reset the alarm? (The alarm takes very little power.)

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  109. ...sorry correction: the 4 diode in series will not increase the delay in fact it will do the opposite....

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  110. Yes, I wondered how that would help. :-) But I still don't understand why with the relay connected I cannot increase the timing resistor beyond 40k ohms? (The transistor stops turning on.)

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  111. The 4 diode idea would effectively work for a delay ON timer, but since here we are discussing a delay OFF timer, the idea would be disadvantageous.

    the base resistor directly influences the collector/emitter current specs. as it increases the collector/emitter current delivering capability proportionately decreases.

    you can refer to the following article for the formula:

    https://homemade-circuits.com/2012/01/how-to-make-relay-driver-stage-in.html

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  112. OK, you mean to say the system requires a positive pulse for resetting the alarm...which could be probably achieved through a suitable BJT stage as soon as the delay timer is triggered...

    yes that's possible.

    ReplyDelete
  113. OK, so it may not be possible to build a 10 second delay timer circuit for a 3V relay because of their low resistance (without using an IC.) Thanks anyway for all the advice.

    ReplyDelete
  114. Your reply: "OK, you mean to say the system requires a positive pulse for resetting the alarm...which could be probably achieved through a suitable BJT stage as soon as the delay timer is triggered...yes that's possible."
    The alarm just needs to have it's 5v power cut - either + or -. Could you explain what a suitable "BJT stage" would look like? Thanks!

    ReplyDelete
  115. Thanks for the formula Swagatam. I think I found a solution - I also have access to 5V in this project and when I raised the voltage to 5V the delay time increased considerably. I still need a big cap (6800 uF) but I can live with that. Maybe by experimenting with the value of the delay resistor I can reduce the cap a bit. Thanks again!

    ReplyDelete
  116. That's correct, it's because of the higher load current and also the 3V value which is very relatively closer to the holding threshold of the transistor's 0.6V. If supply could be increased then the timing could also be increased

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  117. yes 5V is higher than3V so it will proportionately produce higher delay, in that case if you have access to 12V, you can apply it only to the RC network of the circuit and feed 3V to the relay...this will allow you to get much higher delays without the need of high value caps....or alternatively you can employ a voltage multiplier circuit to raise the 3V to 12V for the RC network.

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  118. you can introduce an NPN transistor in series with the negative of the alarm circuit, the supply negative now gets connected with the NPN emitter and the collector to the alarm circuit negative line, meaning the NPN now acts like a switch for the circuit.

    the base is connected with the delay timer NPN transistor's collector via a 10uF capacitor, the base also has a 10k resistor connected with positive.

    that's all..this would do the trick.

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  119. Thanks. But I need the transistor to provide power to the alarm UNTIL it is turned on, then it needs to CUT power to the alarm for a fraction of a second to reset it. Your solution sounds like the alarm circuit negative line would only be connected to negative power when the transistor is turned on. Or am I missing something?

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  120. the negative line is also a part of the power line, if this line is cut off, the supply to the alarm circuit also gets cut of

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  121. HI, Do you have a variation on this circuit so that after you release the switch, the circuit remains dead for about 5 minutes and then allows the switch to be pushed again?

    ReplyDelete
  122. Hi, It can done but not with few components, it might require the involvement of an IC 4017 for a sequential delay effect, and this could make the above simple design much complex.

    ReplyDelete
  123. Hello, Swagatam, and thank-you for your work on this site. It is very impressive. I wish to make a "firefly" circuit that will light an LED for 10 seconds, turn it off for two minutes, then turn it on again for 10 seconds and so continue. I want to power this by battery and hope that this "firefly" can run for several months before draining the battery. I would like to make many of these and hang them above a street. Can you tell me whether this is possible, and, if so, how would I go about doing it? Thank-you, David.

    ReplyDelete
  124. Thanks David, a 20 second/2minute timing sequence looks too slow for a fire fly simulation, it should be kind of 1/2 second ON and 1 second OFF...to actually imitate a firefly light.

    anyway you can try the first circuit configuration from this link for your application, the R1, R2, adn C can be appropriately calculated through the attached software

    the LED may connected across pin#3 and ground via a 1K resistor.

    ReplyDelete
  125. sorry here's the link:

    https://homemade-circuits.com/p/ic-555-calculator.html

    ReplyDelete
  126. I'm hoping you can help me with this .... I want to add a timer circuit to a solar powered garden light. When the photo cell goes dark and turns the LED on, I would like a timer circuit to turn the LED off after an adjustable period of 4 to 5 hours. The lights are powered by a single AA battery. Thanks.

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  127. I'll try to design and post it soon....you can check back some days later

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  128. Thanks for considering this.

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  129. It would be so helpful if you could tell how to make the time 30 s

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  130. Hi Swagatam,

    Good day...
    I hope you can help me with this, can you make me a circuit where-in when the door open,it will give an alarm so that i can hear it so that i know that what i am doing is already finish. And also,can you put a reset switch to clear the alarm once i attended to it...Thank you very much.

    ReplyDelete
  131. Hi, you can use a reed relay and magnet mechanism for sensing the opening or closing of the door.

    the alarm can made by using 555 IC timer, I'll try to update the diagram if possible..

    ReplyDelete
  132. Hi,in the first circuit, once I press the button and release it, is the LED working only because of the potential provided by the capacitor, or does the battery have some effect on LED even when the button is not pressed?

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  133. Hi, the capacitor is responsible for the delay only, the LED is getting its power from the supply

    ReplyDelete
  134. Can you tell me how to calculate the time constant for the first circuit? I used the formula T=RC, and used the value of base resistor as R, and got a time constant of 33 seconds, and the time to fully discharge a capacitor is 5T, that is 165 seconds, however practically it took a lot more time to discharge the capacitor. Can you please explain why this happened?

    ReplyDelete
  135. you can find the formula in wikipedia easily...

    for discharging the capacitor quickly and effectively you can try the following design instead

    https://homemade-circuits.com/2014/11/long-duration-timer-circuit-using.html

    ReplyDelete
  136. Sir, I want a simple continuous time delay buzzer ringing every 2 or 3 minutes with volume control, buzzing time 5 seconds. Please provide me Circuit Diagram.
    Prachi
    praneelpune@gmail.com

    ReplyDelete
  137. Prachi, you can try the second circuit from this link

    https://homemade-circuits.com/p/ic-555-calculator.html

    by some trial and error identify the values of R1, R2, C such that T1, T2 are as per your timing specs...that is T1 could be 150 seconds and T2 could be 5 seconds or vice versa.

    once these are fixed you can add a buzzer across pin#3 and ground for the required results

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  138. Is there someone here that would be able to give some tips about how design this circuit?
    I am building a combat robot for my kids, I am a machinist; this circuit is over my head.

    The purpose of the circuit is to activate an air powered flipper on a combat robot.

    I need a list of components and circuit diagram.
    I can do the soldering.

    Here is how it needs to work.

    The circuit has 2 inputs:
    Input 1 is 24 volts connected to LiPo batteries
    Input 2 is a trigger also 24 volts

    Output 1 is energized when the input 2 (trigger) is energized. It supplies current to activate an air solenoid that has a 24 volt 24 watt coil. After it is energized, it will stay on for 1 second and then turn off. (this supplies air pressure to raise the flipper)

    Output 2 is energized 0.5 seconds after output 1 is turned off. It will remain energized for 2 seconds, then turn off. It also supplies current to activate a second air solenoid that has a 24 volt 24 watt coil. (this releases air pressure to allow a spring to lower the flipper)

    Should I be able to do this with a circuit directly or should I expect to make the circuit low amps and use external relays?

    ReplyDelete
  139. you will need two of the following circuits

    https://homemade-circuits.com/2014/11/long-duration-timer-circuit-using.html

    and integrate the relay contact with the second circuit for executing the mentioned procedures.

    please make a single module..... test it if you succeed then we can proceed for the final integration.

    the 2m2, and the 1000uF will need to be drastically reduced for achieving the 1 second relay ON/OFF time....

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  140. Have you had a chance to look into this yet ?

    ReplyDelete
  141. actually I already had a similar design in this website....I have modified it as per your request and posted at the bottom of the same article, please check out the last diagram posted in this article

    https://homemade-circuits.com/2014/09/solar-garden-light-with-programmable.html

    ReplyDelete
  142. Hello there, I am student in electronics, and I was given this circuit (Two Step Sequential Timer) to rewrite it on SPICEnet, and do allkinds of simulations on it. I would like to know if any of you have any xperience in SPICE and how to simulate on it. If not I would apreciate if anyone could atleast explain me how the circuit works, and wich are the cruical elements in it.

    Thanks!

    ReplyDelete
  143. Why i can't use dc motor with first circuit? It work fine with led but it didn't work when i replace a dc motor for led.my dc motor even work with 1.5v voltage.what can i do? Thank u and sorry for my bad english

    ReplyDelete
  144. because motor require much larger current than LED....just replace the BC547 with TIP122 and it will start working

    ReplyDelete
  145. HI, the circuit I am looking for is a latch which will toggle states as a flip flop will, but will not change states unless the button is held down for say 4 seconds. This is to emulate the digital life we have become accustomed to. State does not change unless we long push. I'm sure many of us wanted this circuit without knowing we did. Thanks

    ReplyDelete
  146. you can make the following circuit

    https://homemade-circuits.com/2013/08/single-push-10-step-selector-switch.html

    connect a series resistor with the push button and dimension this resistor value for getting the 4 second delay

    and replace pin#10 with pin#7

    ignore the relay stages, you can replace them as per your preference.

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  147. helo sir i made circuit very last one 4.bp.blogspot.com/-TFvftGeJn-4/UUVoW5r9OvI/AAAAAAAADmo/JxjM1N5U5Vk/s1600/sequential+delay+relay+circuit.png
    and found an error that relay continously turning on and off i mean it is triping again and again also please tell me capacitor and resistance values for veriable timing range 10 sec max thanks in advance.

    ReplyDelete
  148. one more thing that i want to turn manually on the circuit and after pushing the button it turns the relay on for 3 to 10 sec varible and then turn the relay off and when again i push the button it again run relay for required time and then set the relay in off state... can this ciruit will work( very last one)

    ReplyDelete
  149. the relay cannot oscillate since there's no feedback loop in the circuit...I am not sure what may be causing this. The concept is very simple and is explained in the article.

    you can connect a 1n4148 diode in series with the BC547 base and see if that helps to rectify the issue.

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  150. Hi,mr.Swagatam,I'm looking for a substitute for a NipponDenso starter relay, which is used on a Kubota truck to shut down the diesel engine after switching the contact key to "0" position. In order to do so, after turning off thr contact key, the original timer relay activates a valve in the high-pressure fuel pump for few seconds, which leaves it without fuel and the engine shuts down. The sequence can be repeated only if the key is switched back to "ON" position, and then to "OFF".
    With other words the circuit I need should activate the output relay (and hence the shutdown valve) for adjusted time only by passing the control input (the contact key) condition from "ON" to "OFF" and not the opposite otherwise the engine will not start.
    As long as I can see, the last shown circuit (requested by Fastshack3) does the right sequence of actions for me. Correct me if I'm wrong. Thanks in advance!

    ReplyDelete
  151. Hi Ivan, sorry I am finding it difficult to compare your existing relay sequence with your required circuit sequence.

    Do you mean to say that you want the delay timer relay to be controlled through an external push button and not directly by the contact key?

    or is it the entire sequence that you want to control through a push button by completely replacing the NipponDenso starter relay

    please clarify so that I am able to help you further.

    if possible please provide a step by step operational details of the circuit

    ReplyDelete
  152. Sorry for the missunderdtanding! I'll try to explain better.
    1) The Denso relay is burned out and is out of production (nowhere to find it), so I need another circuit to repeat it's functions.The Denso company names it "starter relay", but in fact here it's used as a shutdown one.
    2) The external signal of 12V that controls the relay comes only from the contact key, no buttons of any kind.
    3) The diesel engine has a 12V electromagnetic shutdown valve which purpose is to stop the fuel to the injection pump, so the engine stops working when the user desides to shut it down.
    4) The relay should close the circuit of this shutdown valve for few seconds only when the contact key is switched from "ON" to "OFF"(i.e. the engine is working and it needs to be shutdown). It (the relay) should not do anything when starting the engine (i.e. when the contact key is switched from "OFF" to "ON"), because if it closes again the shutdown valve for few seconds in this case, the engine will not start.

    ReplyDelete
  153. OK got it! for an accurate performance you may have to employ an IC 555 based circuit as shown in the first diagram of this article:

    https://homemade-circuits.com/2016/11/intruder-position-indicator-security.html

    Here please remove the piezo which is connected with R2, and feed the (+) input from the contact to this R2.

    The LEd and its resistor will also need to be removed and replaced with a relay, as shown in this example article:

    https://homemade-circuits.com/2014/06/input-trigger-synchronized-monostable.html

    please see only how pin#3 of the IC needs to be connected with the relay, ignore the rest.

    that's all, this will take of your requirement.

    In the diagram from the first link, R3/R2 and C2 can be tweaked for setting up the desired delay time.

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  154. Hi Swagatam, congratz on the good work. I also look for a same solution, in the reverse of the above.

    I have simple $5 dollar alarm which is activated when a cord is pulled.

    I do not want the alarm to ring for more than 2 minutes (100dB)..

    When we replace the push button for the capacitor, will the capacitor create a currentleakage and drain the battery in a few days time?

    Thanks, Ronald

    ReplyDelete
  155. Thanks Ronald, no it won't drain the battery, because as soon as it gets charged it will behave like an open circuit across its terminals.

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  156. Hi Swagatam,

    I build the simple delay timers. They work just fine with the push button. Nice work!

    However, as you have explained it is possible to remove the capacitor from its position and put it in place of the push button. This way the delay on-switch will result in the reverse, that is, in a circuit that will switch off in 2 minutes after giving power.

    I removed the switch button and replaced it with the capacitor. However, it does not seem to work; the LED keeps on lighting infintely. How is this possible when the capacitor is fully loaded after a few seconds and should not feed the base? (also, after a few seconds the LED looses a little bit of intensity)

    What I am looking for seems simple, my 9V boat alarm to shut down automatically after 2 minutes, so when I am not at home, the neighbourhood does not go completely haywire. The alarm gives 145 dB and will scream for hours! I only need to scare away the burglers.

    Can I build this with a capacitor-transistor solution, without the use of a relay?

    It looks the solution might be simple.

    Thanks again.

    ReplyDelete
  157. Hi Ronald,

    Both the designs will behave in the same manner, that is they will switch ON initially then switch OFF after the predetermined delay.

    I think your capacitor could be leaky, because normally any capacitor once charged fully will not conduct any further...to prevent this leakage you can try adding a diode in series with the emitter of the NPN transistor....I'll try to update a more appropriate circuit soon in the above article for your reference.

    ReplyDelete
  158. Hello Sir,pls What Is The Value Of The Unknown Transistor?

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  159. which resistor are you referring to?

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  160. Hi,
    I have this device which runs on 3.7v li-ion battery
    Require a simple button to activate the Same.

    How much power will the first circuit consume during idle mode

    ReplyDelete
  161. Hi, all the circuits shown will not consume any current during idle mode, however these are delay circuits, pressing the relevant buttons will produce a specified delayed output

    ReplyDelete
  162. hello sir, I'm planning to build my breaklight to perform like this clip (F1 style)
    https://www.youtube.com/watch?v=PQh-0bRqx3c

    I did manage to buy a tiny circuit that do the flashing part, but it flash continuously
    So I'm in need to build a dead simple relay circuit to perform the steady lit part, with a delay of 2 sec after connect the power supply
    My led array consist of 16 leds, and there is not much room left behind the board

    Could you give me a hint?

    ReplyDelete
  163. Hello Minh, due to my slow internet I could not watch the video, are you looking for a relay oscillator circuit that will switch your existing flasher after certain time interval? Just like we have in police car strobe light flashers where the lamps flash rapidly between their ON/OFF intervals??
    I hope I understood your requirement.

    ReplyDelete
  164. Hi Swag, I want my brake light to operate like this: [hit and hold the brake] > light flash flash flash (maybe for 1 to 2 seconds) then lit up steadily
    The "flash flash flash" part was solved with a little prebuilt circuit ($1.5), wired serial in between the power source and the light
    Now I need a delay circuit to bypass it after 1 or 2 seconds to get the "steadily lit" part
    Thanks in advance

    ReplyDelete
  165. Ok got it, instead of shutting the lamps during the intervals you want them to stay solid ON, this can be done using any standard IC 555 flasher circuit as shown in the below article

    https://homemade-circuits.com/2016/07/alternate-switching-relay-timer-circuit.html

    ignore the existing wiring of the relay contacts, and replace it as given below:

    connect the N/O and the pole of the relay with separate wires and connect the ends of these wires across the transistor which is controlling the flashing of your lamps, or whatever "switch " that's used for the flashing of the Lamp

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  166. ...sorry I thought you wanted the process to keep repeating...if the lamps are required to stay ON permanently as long as the brakes are applied in that case the above circuit will not work...instead you can try the first circuit as given in this article:

    https://homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html

    wire the relay contacts exactly as explained previously

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  167. so to get a 1 sec delay with that circuit what value of R2 & C2 should be used? base on your experience
    Since I don't have all those parts around to test which suit my need and have to buy them new in bulk quantity

    ReplyDelete
  168. R2, C2 will need to be experimented to get the required delay, and since 1 sec is very less, you can replace D1 with a short link....you can even eliminate the relay if you are able to configure the PNP transistor's collector with your lamps.

    the 12V to this circuit will come from your brakes....

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  169. in the end, I still have to go along the try-redo way
    about the "PNP transistor’s collector" you mentioned, I have almost no knowledge about it since I'm new to this electronic stuff and still learning
    All my electronic knowledge came from the days at elementary school, now it comes from gg :D

    Thanks alot for your help. Good day, sir

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  170. it's always fun to have try-redo in electronics, and for this circuit it could be just a matter of minutes to finalize the delay parts.

    If you can explain how your LEDs are configured with your flasher circuit, I may try to solve the PNP configuration details for you.

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  171. my led array consists of 16 leds, divided in to 4 parallel rows.
    in each row there are 4 leds with 1 330ohm resistor wired in serial.
    the array is running fine with 12v source from my brake switch for 3 years
    hope you can figure it out, thanks

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  172. OK, but how is it connected with the flashing circuit, through a relay or transistor?

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  173. right now it connect directly in serial with the flashing circuit, no thing in between

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  174. use separate diodes for supplying the positive voltage to the LED module, one from the flasher and one from the delay timer, this will do the job.

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  175. meaning, break the existing flasher positive connection and put a diode in the middle.

    next connect a diode from the collector of the delay timer PNP and join it with the LED positive line

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  176. I see, so instead of using a relay to turn off the flasher and switch to continuous power source, you just let them run in parallel. So the led will be solid on after the time is up and the continuous power is on, am I right?

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  177. yes that's right, just make sure the delay timer's PNP is appropriately rated to handle the LED current.

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  178. Hi Swagatam, I am trying to built IOT with esp8266-01 (wifi module) to operate the 5V relay my circuit. Every think is working fine but if I will restart the system wifi get stuck (may be the drawback from load). If I remove the esp GND from relay GND wifi restart normal and after few seconds connect the GND again and working normal.

    I want to add delay for 02 Sec between esp and relay.

    Thanks,

    ReplyDelete
  179. Hi farid, you can try the following delay concept and see if it works for you?

    https://homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html

    ReplyDelete
  180. Sir,

    I would like to make a 10 minute delay circuit to be fitted via a relay of 3v, to power up the water motor pump. It should use either button cells of 3v or 6v. Can you suggest a suitable circuit so that when powered on the water pump will run for 10 minutes and shut off.

    ReplyDelete
  181. Gyaneshwar,

    you can try the following concept:

    https://homemade-circuits.com/long-duration-timer-circuit-using/

    ReplyDelete
  182. make sure to replace the BC557 base 10K with a 1K

    ReplyDelete
  183. Thank you Sir for responding very very fast.
    My circuit design will require to use buton cells to power up the circuit. This circuit will have to switch off the AC motor pump after exactly 10 minutes. So please let me know whether such circuit is available to run a relay of lower battery voltage.

    ReplyDelete
  184. If the supply is rated at minimum 6V, then a 5V/6V relay can be tried, however this will demand relatively high current from the cells and can deplete it quickly.

    you can go for a triac instead, that will work better.

    ReplyDelete
  185. Thank you Sir for the reply. Can you name a triac to be used int he circuit instead of the transistor (i.e. label of the triac). The circuit will remain the same and the TR will be replaced by the TRIAC. Right!

    ReplyDelete
  186. Hello P GNANESWAR , I have updated the required diagram for you, in the article

    ReplyDelete
  187. I seek a simple transistor-based timer circuit for use in a car. There is a light which has two functions: one is to illuminate the rear number (licence) plate and the other is to illuminate the boot when the boot lid is raised and the general car lighting is off. I want the light (drawing about 1 A at 12 volts) to switch on when the circuit is first powered by opening the boot lid, and for the light to stay on for five to 10 minutes then switch off while the circuit remains powered (boot lid open and switch still 'on'). The circuit would be reset when the boot lid is closed and the circuit is open again. In the particular car there is also a small relay which is, I think, normally closed, linking the light to all other lighting in the car. I assume that when the boot lid circuit is powered, it goes open and isolates the light from the rest of the car's lights.
    Thank you.

    ReplyDelete
  188. you can try the following design

    https://homemade-circuits.com/wp-content/uploads/2017/12/monostable.png

    the trigger from the boot lid cab be connected to the base of the BC547, this will switch ON the connected LED at pin#3 of the IC, and will keep it ON for a time period determined by the values of the 100K pot and the 220uF capacitor.

    ReplyDelete
  189. I have an LED light strip that has white and red LEDs. In order to light the colors independently, the manufacturer suggests using an on-off-on rocker switch. The LED strip has three wires.

    However, in my application I can only use a simple on-off switch.

    I’m looking for a circuit that would allow me to achieve this:

    Turn switch on: RED lights
    Turn switch off then back on within 4 seconds: WHITE lights

    If more than 4 seconds go by after the switch is turned off, the next time it turns on: RED lights

    I have some bulbs from a company called Imtra that does this, but they do not make light strips and I would like to achieve the same results.

    Can you help? Thanks!!!

    ReplyDelete
  190. It will require some thinking...it seems it can be accomplished with a 4017 IC circuit and a delay timer...

    ReplyDelete
  191. Hello, for the last diagram, what is the number unknown transistor

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  192. Hello, for the last diagram, what is the number unknown transistor in dalinton pair with Bc557 on the relay coil

    ReplyDelete
  193. Hi,
    could you please suggest me a simple circuit for "Delayed ON IC"? ( The IC should get the supply or turn ON after 3 seconds after switching ON the circuit)

    ReplyDelete
  194. Can you tell me the IC number because different ICs may have a slightly different way of implementing this....a general way can be to connect the positive supply pin of the IC through a 10 k resistor, and add a 100uF capacitor across this supply pin and the negative line

    ReplyDelete
  195. Am using IC 4093. +'ve pin is pin 14. The should get the +'ve supply after 3 seconds. So can I implement the same as u said?

    ReplyDelete

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