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I have a problem with a circuit that would delay me a signal received from an external source. From the beginning - I need to build a circuit that will give the driver a HIGH signal. This signal is supposed to give when it receives a HIGH signal from another controller. The problem is that the second controller does not respond when the first controller is HIGH. The HIGH signal only lasts 1.5 seconds, but that's a big problem. I have no idea how to delay the signal to the second driver so that everything works as it should. In my project I have constant 12V and GND so I have something to use in case of an accident. Image showing what I mean: enter image description here

enter image description here I know how to do it on a microcontroller - I would just use the delay function, but in this project the microcontroller doesn't make sense. Is there any option to do this without any timing circuits like NE555.

Regards.

Update:

The NC555 is cheaper than the MCU so I can go with it if needed. I designed such a circuit on NC555, but I'm not sure if it will work. enter image description here After receiving a signal from module 1, transistor turns on NC timer and on output uses the RC circuit to open NPN transistor that is supposed to reset NC, and also sends a LOW signal to module 2. Switching from low to high will not be a problem, however however, I don't know if such a circuit is correct.

Siemekk
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    I'm not sure that what you consider an MCU and what I consider an MCU are the same. I can get, in ones, an appropriate MCU for about 50 cents. They are tiny and have just 6 pins in SOT23-6 form. Are you thinking about an arduino or something like that? (Because that would never cross my mind as an MCU.) I'm old. So my hind-brain just says "go get 74121 or 74123." (Perhaps there's a more available replacement part these days.) And yes, 555s can be rigged up too. There's not enough info about the signalling for a specific design, my opinion. So that's all from me for now. – periblepsis Jun 19 '23 at 02:02
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    Without more information this is really difficult to answer. How accurate must the 2s delay be? How long is the input pulse? How long is the output pulse? What happens if an input pulse comes along before the 2s delay has elapsed? What happens when the input pulse duration is longer than the 2s delay? Some example waveforms of possible input scenarios and corresponding acceptable responses would help. – Simon Fitch Jun 19 '23 at 07:40
  • This problem just begs for an MCU which will be the cheaper option when you consider your full BOM and production issues. Also, what happens if the driver 1 signal goes low during the 2 s period? Do you cancel it or execute anyway? What if there are several turn on pulses in the same 2 s period? – winny Jun 19 '23 at 08:40
  • In your timing diagram, does the signal to module 2 have to be the same width as the signal from module 1? If the signal from module 1 varies in width, does the signal to module 2 have to track this variation? – AnalogKid Jun 20 '23 at 00:08

3 Answers3

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I wouldn't stray away from timing solutions like the 555 timer when you are trying to solve a timing problem. If you want a really low parts count solution consider the TimerBlox from ADI. The LTC6994-2 can delay your pulse from 1us to 33.6 seconds. The only additional consideration is the maximum voltage of 5.5V, so you would need to divide the input down to a sensible logic level, and then level shift the output logic level back up to 12V.

Another option would be to stack two monostable multivibrators (one-shot pulse generator) in series. The first pulse generator creates a pulse on the rising edge of your input signal, and lasts for your delay time. The second triggers off the falling edge of the first pulse generator, set to the desired pulse width. Unlike the initial solution, this does not retain the input pulse width.

I also wouldn't rule out using an MCU, as it might actually work out as the cheapest option overall, even when including the potential other circuitry needed.

Graham
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Is there some problem with simple RC delay?
The diode just quick discharges the 47u cap during input Low to be ready for next High pulse.
Otherwise if the pulses are coming "rarely" you don't have to place them there (D1 and R4).

schematic

simulate this circuit – Schematic created using CircuitLab

Kuba hasn't forgotten Monica
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Michal Podmanický
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  • I tried such a circuit with a 10k resistor and a 220uf capacitor. I got the impression that the capacitor is charging too fast, so that the signal LOW goes to module 2 while module 1 is still HIGH. There must be a LOW signal on module 1 before I send the converted HIGH signal to module 2. – Siemekk Jun 19 '23 at 10:24
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If you want to use a simple 12 V system logic, a long CMOS shift register may meet your requirements.

The signal level at PulseIn is sampled and shifted in with a frequency of 32 Hz and moves along all 64 flipflops.
About two seconds later it arrives at PulseOut. The delay and the width of the pulse have a tolerance of around +/- 31.25 ms, the clock period of the sample frequency.
If more precision is required, the sample frequency can e.g. be doubled while using 2 cascaded shift registers.

This solution does not cover any questions about maximum pulse width or double pulses, the input signal is just delayed "as is", but the minimum input pulse width is just a bit above the clock period.

After power on the shift register may contain random values and so for the first 2 seconds PulseOut can be anything. Another 64 bit shift register chip with a separate reset input is MC14557B. An R/C combination can produce the reset.

schematic

simulate this circuit – Schematic created using CircuitLab

Jens
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