Turn on switches sequentially - ONLY ONCE - when triggered

Question

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

Answer 1

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.

Answer 2

You may adapt the following for your purpose.

Timing Diagram & Schematic credit: All About Circuits

https://forum.allaboutcircuits.com/threads/two-seperate-pulses-from-rising-and-falling-edge.91080/#post-662807

1. Timing Diagram

2. Schematic

Answer 3

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!