[SOLVED] Suggestion on trigger circuit

I plan to do a circuit than will power up a LED for 5s after i press a button without using a microcontroller. Can any one give me a suggestion on how to design such a circuit???

For a first guess, have a look at 555 timer application notes. If you want zero current consumption in off-state, a different approach might be required, e.g. using CMOS logic ICs or transistors.

in my circuit, I plan to press the switch and release it. Then the LED will light up for about 5s then off until the switch is press again

As FvM pointed it out; what is the power (voltage) supply of your circuit? Is it a battery?

As FvM pointed it out; what is the power (voltage) supply of your circuit? Is it a battery

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ya i am using a 9 v battery. also have a 5v voltage regulator

Do you like using CMOS ICs?

CD4011, Quad 2-Input NAND gate
or
CD4093, Quad 2-input Schmitt trigger NAND gate

Notes:
(1) I think you heard of "RS flip flop using two NAND gates".
(2) If your circuit is just about the LED, using a CMOS IC doesn't need a 5V regulator. 9V battery is ok.
(3) If your LED is red, by the same battery current 3 LEDs (or 4) in series could be turned on via a current limiting resistor.
(4) If the LED color is white or green, only 2 LEDs could be connected in series.

CD4011, Quad 2-Input NAND gate
or
CD4093, Quad 2-input Schmitt trigger NAND gate

Notes:
(1) I think you heard of "RS flip flop using two NAND gates".
(2) If your circuit is just about the LED, using a CMOS IC doesn't need a 5V regulator. 9V battery is ok.
(3) If your LED is red, by the same battery current 3 LEDs (or 4) in series could be turned on via a current limiting resistor.
(4) If the LED color is white or green, only 2 LEDs could be connected in series.

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Sir can u provide me a circuit or a reference

Could you search for "RS flip flop NAND gates"?
When you will get it, we can talk about the next move

Could you search for "RS flip flop NAND gates"?
When you will get it, we can talk about the next move

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Sir i read the RS flip flop NAND gates from this site **broken link removed**
and not understand how it can work as I only press the switch let say "S" once and release.

If you search around on the 'net you will find something.

If you need approximately 5s, I'd suggest a simple RC circuit triggering a transistor or FET which drives the LED. Simple, easy, and cheap. There are a little more complex designs which have low power consumption.

http://www.mycircuits9.com/2012/12/simple-delay-timer-for-one-minute.html

or

http://www.bowdenshobbycircuits.info/page2.htm#delay.gif Power-Off Time Delay Relay
(replace the relay with resistor and LED, should work fine with 9V, use a different capacitor say 470uF/16V)

Usually you find power-on delay circuits but these can be modified to do what you want.

Good work "kbtan".

You are right to be confused, so let us analyze, step by step, the circuit at the right side (of NAND):

(1) In a real circuit, the pins, R and S, shouldn't be left open.

(2) In case of a 2-input NAND, R and S should be at a high state normally, so that the second input (coming from the other output) determines the state of the gate output in the steady state.

(3) For this, we connect a resistor of a high value (100K to 1MEG) between Vcc and each of the two pins R and S (two pull-up resistors).

(4) At start up (applying Vcc), the state of the previous RS flip flop is undetermined. So, we add a small capacitor (C1, say 10nF) between S and ground.

(5) In this case, at start up and while C1 is charged by the pull-up resistor, the S voltage is low and this lets the output Q' (pin 6) be high. The output Q (pin 3) becomes low because pin 1 and 2 are high.

(6) Even after C1 is fully charged making S at a high state, the state of the flip flop doesn’t change since the input pin 4 (connected to Q) is now low.

(7) Now, we have Q (pin 3) low and Q’ (pin 6) high. Even if we press the switch (between S and ground), Q and Q’ won’t change.

In the next steps, we will use another gate with an RC circuit at its input. Its ouput can reset the flip flop after a time delay (say 5 sec) if connected to the input R (pin 1).

Note:
If you don't have questions about the above steps, we can go on.

In the next steps, we will use another gate with an RC circuit at its input. Its ouput can reset the flip flop after a time delay (say 5 sec) if connected to the input R (pin 1).

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Sir I don quite understand this meaning

I hope you meant that you understood the first 7 steps in the least.

From here and when I refer to IC pin's numbers, I assume that you have a CD4011 IC and its datasheet
I mean the below pin numbers may differ from the above ones... ok?
Also, please note that the two inputs of each gate are alike (and the 4 2-input NAND gates are alike)
For example, for your real circuit you can exchange pin 1 and 2 (also pin 5 and 6, pin 8 and 9, pin 12 and 13).

(8) Let us recall that there is a pull-up resistor R1 (say 470K) between pin 1 and Vcc (9V) for gate (U1).
(9) There is also another pull-up resistor R2 (say 470K) between pin 6 and Vcc (for gate U2).
(10) We add a capacitor C1 (2.2nF) between pin 6 and ground (for start up) also the push-button (PB1) for triggering.

(11) The output of U1 (pin 3) is connected to the second input of U2 (pin 5).
(12) Also the output of U2 (pin 4) is connected to the second input of U1 (pin 2).

(13) Let us use the gate U3 for the delay. A resistor R3 (330K) is connected between the output of U2 (pin 6) to an input of U3 (say pin 8).
(14) The timing capacitor C2 (22uF) will be connected between pin 8 and ground.
(15) The second input of U3 (pin 9) is connected to Vcc.
(16) The output of U3 (pin 10) is connected back to pin 1.

In the coming steps we will find out how we can use the remaining gate U4 to drive the LED via a PNP transistor, so that when its output (pin 11) is low the LED (or the LEDs in series) are turned on (for about 5 seconds). And there will be some extra notes.

If you can build the circuit so far, you will notice that the output of U3 (pin 10 as pin 1) stays high for about 5 sec every time you briefly press the push-button PB1.

Sir I briefly understand what u had explain. Is the voltage supply must be 9v ???
I try draw out the circuit

You did well once again, kbtan.

Since you use a 9V battery and the CMOS IC can be supplied with 3 to 15V, I thought the battery voltage is the best option for Vcc in your project.
As you know, the 5V regulator draws about 5 mA even for no load at its output.

The current of the circuit, we are working on, is very small (a few micro Amp) when the LED is off.

From your drawing, I noticed I forgot to tell you that pin 7 is for ground and pin 14 is for Vcc.
Please note that the PB1 (switch) and C1 (2.2nF) are in parallel, between pin 6 and ground (not in series as drawn).

For instance, do you know the function of a 2-input NAND gate?

Only if both inputs are high, the output becomes low.
So if one input is low, the output becomes high.

When the battery is connected to the circuit, the 5 sec timer will also be activated because C1 (2.2nF) needs a short time to charge (during which pin 6 is low) and this is equivalent to a press on PB1.

Before further analysis, could you tell me if you can get the CD4011 IC and build the circuit?

Before further analysis, could you tell me if you can get the CD4011 IC and build the circuit?

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Ok Sir, I will try to build the circuit 2moro. Thank You for your guide

Hi kbtan,

I just liked to know if you could build it or not.
It seems you can do it, so I can go on too.

A general IMPORTANT note:
For CMOS ICs, if it happens there is an 'input' pin 'not used', it should be connected to Vcc or Gnd (directly or via a resistor).

If an unused CMOS input pin floats, its state (low or high) may change rather easily due to interference or noise since the CMOS input has a relatively high impedance. One of the consequences is draining unnecessary current from the voltage supply which is better to avoid when using a battery.

On post #13, there is a mistake

WRONG: If you can build the circuit so far, you will notice that the output of U3 (pin 10 as pin 1) stays high for about 5 sec every time you briefly press the push-button PB1.

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As we will see, the previous line should be written as:

RIGHT: If you can build the circuit so far, you will notice that the output of U2 (pin 4 as pin 2) stays high for about 5 sec every time you briefly press the push-button PB1.

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Before completing the design, could you tell me more abour your LED?
What is its color?
Is it rated for 20mA, for example?

Since the LED forward voltage is much less than 9V, the same battery current can lite more than one LED by connecting in series with it one or more LEDs and decreasing the value of the current limiting resistor. Are you interested to use more than one LED since the load current won't change?

Before completing the design, could you tell me more abour your LED?
What is its color?
Is it rated for 20mA, for example?

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I am using a white LED.

In this case, the LED forward voltage is about 3.3 V
If you like we can put 2 LEDs in series... ok?
I assume it is a small (standard) LED hence rated @20 mA... right?

In this case, the LED forward voltage is about 3.3 V
If you like we can put 2 LEDs in series... ok?
I assume it is a small (standard) LED hence rated @20 mA... right?

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Yes. But i only use 1 LED. So I need put a resistor series with it???