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Time delay for 12V motor using 555 IC

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Now I see the problem. Your switch is a simple ON/OFF toggle (like a light switch in your house). NOT the momentary push button type that I assumed.

This is how your original monostable 555 timer's graph should look using a momentary push button to initiate the 29 second output:
upload_2017-7-20_10-46-0.png

Note the "Trigger" pulse is short (2 seconds). It could be much shorter (milliseconds) and the output timing would remain the same.

It could also be much longer:
upload_2017-7-20_9-37-4.png

And the output is still 29 seconds.

But if the trigger exceeds the 29 second delay dictated by the caps/resistor circuit (47uF + 470uF + 51k resistor), then this happens to the output duration:
upload_2017-7-20_9-41-16.png

In other words, if the trigger pulse duration is greater than the timer's delay circuit duration, the output duration will equal the trigger duration.

So, with your original circuit, when you press the switch you're using to START the timer, if you turn the switch OFF before the end off the timer before the timer's delay duration, the output will stay ON for the normal time.

BUT, if you leave it ON for longer than 29 seconds (in your case, 26 seconds), the output will stay ON until you turn the switch OFF.

So, for your circuit to operate as it should, your switch must be toggled ON and then OFF in less that 26 seconds.

You might also note that while the output is ON, toggling your switch ON and OFF will have no effect for the duration of the normal timing of the output.

Thus, for you to get the timer results you've stated (from your post #10):
... So, best would be having the same switch to start and stop the motor whenever the user wants and also if the user forgets to off the switch it will have a timeout after 26secs which saves electricity and power. ...
The new circuit will achieve that end with this change to the switch (that you are currently using) wiring arrangement:
upload_2017-7-20_10-46-30.png

This change to the new circuit will give you, when you turn the switch ON, a fixed trigger duration, a consistent output duration and the option to stop the output pulse at any time by simply turning the timer OFF.

<EDIT> Grammatical error fixed (while changed to will).
 
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Now I see the problem. Your switch is a simple ON/OFF toggle (like a light switch in your house). NOT the momentary push button type that I assumed.

This is how your original monostable 555 timer's graph should look using a momentary push button to initiate the 29 second output:
View attachment 107186
Note the "Trigger" pulse is short (2 seconds). It could be much shorter (milliseconds) and the output timing would remain the same.

It could also be much longer:
View attachment 107182
And the output is still 29 seconds.

But if the trigger exceeds the 29 second delay dictated by the caps/resistor circuit (47uF + 470uF + 51k resistor), then this happens to the output duration:
View attachment 107183
In other words, if the trigger pulse duration is greater than the timer's delay circuit duration, the output duration will equal the trigger duration.

So, with your original circuit, when you press the switch you're using to START the timer, if you turn the switch OFF before the end off the timer before the timer's delay duration, the output will stay ON for the normal time.

BUT, if you leave it ON for longer than 29 seconds (in your case, 26 seconds), the output will stay ON until you turn the switch OFF.

So, for your circuit to operate as it should, your switch must be toggled ON and then OFF in less that 26 seconds.

You might also note that while the output is ON, toggling your switch ON and OFF while have no effect for the duration of the normal timing of the output.

Thus, for you to get the timer results you've stated (from your post #10):
(From your post #10)
The new circuit will achieve that end with this change to the switch (that you are currently using) wiring arrangement:
View attachment 107187
This change to the new circuit will give you, when you turn the switch ON, a fixed trigger duration, a consistent output duration and the option to stop the output pulse at any time by simply turning the timer OFF.


Omg thank you so much for your explanation!
Yes thats exactly whats happening to my circuit right now!
Does this means i can circuit my board without the use of the Diode? or does the diode have to be part of the circuit. What effects will it bring to my circuit whether having the diode or not?
Also for the 1uf capacitor that i had to include do i use the 35V or 50V ?
Thank you so much!! Its night time here in my country so ill try out the circuit u have shown me tomorrow when i reach school! And ill get back to you to see whether it works or not! Thank you
 
You're welcome, edmondcjj.

Yes, the diode is necessary. Here's why. In those events where you turn off the timer BEFORE it has timed out, there is a good chance that the 47 and 470uF caps might retain a portion of the electrical charge that they acquire during the course of the timing event. That means that if you were to immediately turn the circuit back ON (after having prematurely stopped the timer by turning it OFF) , the new timing event might be incorrect since the caps might not start their charging sequence at zero volts. The diode is there to "bleed off" any leftover DC charge on the caps (it does it very quickly).

I should add that if you allow the circuit to "time out" normally (no premature OFF), the 555 internal circuit will immediately "ground" (basically short out) the caps, returning their charge to zero volts and ready for the next trigger to start fully timed output duration

You can try it without the diode, but I think you'll occasionally see inaccurate output timing without it.

Since your circuit operates at 9VDC, a 35VDC cap is fine.

Also, don't forget that even if the motor is no longer running, your circuit will continue to draw power from the battery. Not much, but not zero...
 
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You're welcome, edmondcjj.

Yes, the diode is necessary. Here's why. In those events where you turn off the timer BEFORE it has timed out, there is a good chance that the 47 and 470uF caps might retain a portion of the electrical charge that they acquire during the course of the timing event. That means that if you were to immediately turn the circuit back ON (after having prematurely stopped the timer by turning it OFF) , the new timing event might be incorrect since the caps might not start their charging sequence at zero volts. The diode is there to "bleed off" any leftover DC charge on the caps (it does it very quickly).

I should add that if you allow the circuit to "time out" normally (no premature OFF), the 555 internal circuit will immediately "ground" (basically short out) the caps, returning their charge to zero volts and ready for the next trigger to start fully timed output duration

You can try it without the diode, but I think you'll occasionally see inaccurate output timing without it.

Since your circuit operates at 9VDC, a 35VDC cap is fine.

Also, don't forget that even if the motor is no longer running, your circuit will continue to draw power from the battery. Not much, but not zero...

Oh so u meant like if i didnt use the diode and i stopped the motor after like 10secs thn the remaining time left will bring over to the next time i start the motor? Which means like the next timeout will be like around 40secs?

Sorry what do u mean by "Also, don't forget that even if the motor is no longer running, your circuit will continue to draw power from the battery. Not much, but not zero..."
do u mean like even if i dont operate the motor or on the switch, the battery will continue to lose power even if i just leave it there?
 
I just tried the circuit and yes finally the switch works now! It stops whenever i off it.! Thank you so much!
However, the time delay doesnt work it doesnt stop unless i off the switch. Well i did not try it with the diode yet but since the diode is meant for accuracy , the motor should at least stop i tested it for like 1min+ it still didnt stop . I wonder why though
 
This is currently how my circuit looks like without the Diode. which i will try to get it ASAP. Is there any like parts should be connected to so that the timer will work? because for now the time delay does not work but the switch thing works already. Thank you
 

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    IMG_2246[1].JPG
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Now i am getting a time delay of around 19 seconds, what should i change to allow the time to be longer? like around 26seconds?
 
However, the time delay doesnt work it doesnt stop unless i off the switch. Well i did not try it with the diode yet but since the diode is meant for accuracy , the motor should at least stop i tested it for like 1min+ it still didnt stop . I wonder why though
Wiring error. See below:
upload_2017-7-21_8-9-15.png

Move the indicated cap lead from row 14 to either row 13 or row 12 (either one will be fine).

Oh so u meant like if i didnt use the diode and i stopped the motor after like 10secs thn the remaining time left will bring over to the next time i start the motor? Which means like the next timeout will be like around 40secs?
Nothing that precise or predictable. But for the time being, don't worry about the diode.

If, down the road, you experience timing errors, we'll revisit the issue.
 
Wiring error. See below:
View attachment 107191
Move the indicated cap lead from row 14 to either row 13 or row 12 (either one will be fine).


Nothing that precise or predictable. But for the time being, don't worry about the diode.

If, down the road, you experience timing errors, we'll revisit the issue.


Ahh i see thank you so much!!!
I will try and get the diode today and ill test it out later! thank you
 
Single Pole Single Throw switches have only two connectors.

Latching SPST switches also only have two connections, but the button itself, when pressed down (which often makes a "click" sound with each pressed) will remain in the down position (ON) until pressed again, releasing the latch (again clicking) and returning to the up or OFF position.

Note differences between old:
View attachment 107164
and NEW
View attachment 107165


sorry but for the diode arrangement u drawn, which pin is the diode on? i cant really see clearly isit on row 13 or 14 haha thank you
 
help after i added the diode my whole circuit doesnt work now. even after i remove the diode. the timing doesnt work again omg!!! it doesnt stop and continues to run atw unless i off the switch
 
this is the circuit with the diode. the diode is quite short so i did not pin exactly how u showed on the picture but i still pin to a negative terminal so is it still correct? also im not sure whether the other side u showed is row 13 or 14 . i used row 14 as row 13 is already used up for the capacitor u showed at the top that the wiring is wrong
 

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    IMG_2247[1].JPG
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help after i added the diode my whole circuit doesnt work now. even after i remove the diode. the timing doesnt work again omg!!! it doesnt stop and continues to run atw unless i off the switch
Forget about the diode for now. Remove it.

Then re-check ALL your connections to make sure they are inserted in the board correctly and that there are no shorts between leads, especially in this area of the board:
upload_2017-7-22_9-59-58.png
 
You appear to have 2 wires stuffed into 1 hole in row 13. That's a no-no.
 
Wiring error. See below:
View attachment 107191
Move the indicated cap lead from row 14 to either row 13 or row 12 (either one will be fine).


Nothing that precise or predictable. But for the time being, don't worry about the diode.

If, down the road, you experience timing errors, we'll revisit the issue.


However now i moved the capacitor lead back to the original position from row 13 to row 11. it is working again. however the time out timing is around 19 secs.
but if i want to try it with the diode i dont kniw how to connect it as the diode i hve is v short and i cant connect the way like how u shown in the other picture. what should i do
 
there isnt any stuffed in the same hole tho
That's ok then. The photo is unclear as to what is inserted in the hole next to pin 6 of the 555.
 
Row 11 is "-", or ground. Moving the cap lead there shorts it and, basically, removes it from the circuit. The cap's lead needs to be in either row 12 or 13.

Also, might be time for a new battery.

And, again, forget the diode for now. You're worrying this death.
 
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