# Turn on switches sequentially - ONLY ONCE - when triggered

I have some circuits in which I have to give two signals (I mean signals as positive) through push buttons to do some work like latch, and it is the only work of these buttons.

I feel that this should have been simpler, pressing just one button to provide two positive currents, one by one to the circuit

Fig:1 Currently I need to press switch 1, then switch 2 sequentially, The violet and pink wire is connected to my circuit (which is top secret),the circuit is not closed as it's just to explain

Fig:2 What I really need, is pressing just one button, that produces a signal through the green wire, which goes through some chip/gates* which will produce two signals one by one, these two signals will go to my actual circuit. (top secret)

*this chip/gate refers to the solution/answer I need.
So how to do this with a non-programmable IC/circuits/gates (answer needs to be with simple IC's like shift register)

Note: This is just like shift register with 555 timer, but timer needs to turn on only when there is input from the user (button),and only one cycle/repeat, not forever loop

# Timeline/Time graph

Note: Gap is required because my circuit cant understand so fast,Even 1s and 3s are needed to be variable according to resistors and capacitors I use

• Just wire the button to as many inputs as you need? You want to denounce it though. May 15, 2022 at 7:43
• Are you saying that one press of the button produced a pulse on one wire, followed by a pulse on another wire? May 15, 2022 at 8:10
• @HandyHowie I need exactly the same. May 15, 2022 at 11:10
• @winny No I do not need that(in fact I have tried it before). what you say will turn on them at the same time(or even at times turn on the second then first button) what I need is to turn on the first then after a gap then the second May 15, 2022 at 11:11
• Then your question is unclear. Draw a truth table or timing diagram of what you need. May 15, 2022 at 15:43

I have a nice idea for you. Use two 555 timers, each set up as a one shot and then you can adjust the R and C to get the pulse width you want. The one shots will get triggered and then deactivate and will not respond till the next trigger. One of them should be connected to an input through a NOT gate at the TRIG so that the rising edge will trigger it. Usually, falling edges trigger one-shots made of 555 timers.

Another thing you should do is make an edge detector. This is basically 3 NOT gates connected together to one leg of an AND gate and the input is passed to the NOT gates and the AND gate to give a very short pulse due to the propagation delay of the NOT gates. This gives a rising edge detector. The falling edge detector can be made by replacing the AND gate with a NOR.

• I have only known that 555 timers trigger infinitely + then -...,how do you make them one shot, can you show the circuit? May 20, 2022 at 11:31
• I will upload it soon. May 20, 2022 at 11:31
• Inside the 555 timer lies an SR latch. At first, the R input is 1, so that the Q1 is on, making the capacitor discharge. Then S & R are both 0, so the latch maintains the same state When a falling edge is applied, R=0 and S=1, so the Q1 is off and the capacitor charges. When the falling edge finishes S=0 and R=0, so the 555 timer stays like that. When the capacitor voltage reaches two-thirds of Vcc, R=1 ans S=0, so it discharges . Then S=0 and R=0, so that the latch stays like that till the next trigger. The SR latch is responsible for the one-shot behaviour. May 20, 2022 at 11:37
• The two 555 timers should be used , but they should not be connected together. One 555 timer is on one input, the other is on the other input at the TRIG terminal. May 20, 2022 at 11:41
• You'll need a third 555 to get a 1s gap between both pulses, and perhaps another 4th to avoid having to press the button for 3+ seconds yourself. Unnecessarily large circuit in my opinion, but at least it uses identical blocks so that's something. May 21, 2022 at 9:08

Timing Diagram & Schematic credit: All About Circuits

1. Timing Diagram

2. Schematic

• To be clear, what are the red, blue, green highlighted parts in this image, I guess red:-ve, blue: 4input NAND gate, green: 2input NAND gate, Am I right? May 20, 2022 at 11:20
• Is this a Schmitt trigger? May 20, 2022 at 11:47
• @Neptotech-vishnu You'll find the answers in the CD4093 datasheet. May 20, 2022 at 12:45
• @Seir but i understand nothing(I am a Arduino guy,don't know lot on EE) May 21, 2022 at 4:16
• Hi Neptotech -vishnu, CD 4093 is a Quad 2-Input NAND Schmitt Trigger IC i.e. four 2-input NAND Schmitt Triggers in one package. You have highlighted one of them in blue and the other three in green. The four of them are identical. Pin 14 and pin 7 are + 5 V and common (highlighted in red). May 21, 2022 at 6:54

I just tried a solution using a Johnson decimal counter 4017 as sequencer and a couple of 4093 Schmitt-Trigger gates for clock, signal conditioning and locking. It works:

And this is the schematic of that artwork:

The first gate inverts the input and C01/R02 extracts the falling edge and feeds this into the second gate. The voltage will soon rise again via R02, this makes the circuit independent of the input signal duration.

This second gate can forward the edge only if Q0 of the counter is high (AND gate). This is the case after reset of the counter and all the time this circuit is in standby. The egde (inverted again) enters the third gate and finally feeds the counter's clock input (/CP1).

The counter advances one count, Q0 will be low and lock the input trigger section during the coming sequence. The third gate forms a square wave generator using C02/R03 as long as Q0 stays low (=counting).

This Johnson counter works like a shift register, that moves a single "high" signal along it's outputs. On the way it creates a pulse at Q1, Q2(not used here), Q3 and so on. The pulse at Q2 is the gap between the two used outputs. Q6 finally activates Master Reset (MR) and the game can start again.

C03 has the job to force MR high at power on and R04 will drop down this signal because Q6 is low after counter reset.

Current consumption at 10V standby = 11.5uA, counting = 1.3mA

Have a nice day!

• looks promising,will try it soon May 21, 2022 at 4:19
• Make C2 bigger for slower timing, this 4u7 is faster than 1s per pulse
– Jens
May 21, 2022 at 5:09
• 4.7uF and 1.0uF
– Jens
May 21, 2022 at 5:16
• There are different families of these logic chips, names startig with CD operate up to 15V (as used in my test), the SN74HCxxx work with the same logic but only up to 5.5V supply. The circuit is the same. Don't use HCT versions here.
– Jens
May 21, 2022 at 5:20
• And there is C01 with 4n7 = 4.7nF
– Jens
May 21, 2022 at 5:29