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how can I generate Output 2 to give a fixed delay pulse (about half a second or less) on a rising AND falling edge of Input?

                     ________________________
Input    ___________|                        |___________________
                     ________________________
Output 1 ___________|                        |___________________
                     __                       __
Output 2 ___________|  |_____________________|  |________________`

Specification: Input / output: 2.5 - 3.3 V DC at < 40 mA.

I need a board or PCB or something that makes Output 2 possible. this board can be powered by anything, from an wall outlet to a micro usb power supply. does not matter that much.

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closed as unclear what you're asking by Chris Stratton, Passerby, Andy aka, pjc50, RoyC Feb 19 '18 at 11:32

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ "Low voltage like a power wire" 120-240V (AC) isn't really "low voltage" unless you are talking about switchboard equipment. \$\endgroup\$ – Ron Beyer Feb 19 '18 at 2:49
  • \$\begingroup\$ Probably best to look for timer relays. On Delay relay. OFF Delay relay. \$\endgroup\$ – Marla Feb 19 '18 at 2:53
  • \$\begingroup\$ i think its like 4-7 volts ac. ill look into the delay relays and off delay relays now thanks! ill be back. \$\endgroup\$ – Acrolance Feb 19 '18 at 2:59
  • \$\begingroup\$ thanks mr grammar person... well i looked up OFF Delay Relays i don't think that's what i need. that i think is a constant flow though the relay. i need it to pulse once when it 1st gets power, then again when it looses power. \$\endgroup\$ – Acrolance Feb 19 '18 at 3:11
  • \$\begingroup\$ probably it will take some combination of several relays to do what you want. the more details you give in the question the better the answers will be. specifying voltages and currents for the inputs and outputs would be a good start, \$\endgroup\$ – Jasen Feb 19 '18 at 3:49
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Interpretation of OP's question.

                     ________________________
Input    ___________|                        |___________________
                     ________________________
Output 1 ___________|                        |___________________
                     __                       __
Output 2 ___________|  |_____________________|  |________________

Figure 1. System timing diagram.

Specification:

  • Input / output: 2.5 - 3.3 V DC at < 40 mA.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A possible solution.

How it works:

  • NAND1 to 4 are CMOS Schmitt trigger NAND gates. When both inputs are high the output goes low. Three of them are wired with the inputs tied together so these work as inverters.
  • On power up NAND3's inputs are pulled high by R1 and R2. When both inputs are high the output, OUT2, goes low.
  • NAND4 is a double-inverted version of the input signal. This satisfies the requirement for OUT1.
  • NAND1 and NAND2 provide an inverted and a double-inverted version of the input signal. We will use one to detect rising edges on the input and the other for falling.
  • When NAND1 output goes low (input going high) the negative going output causes the right-hand side of C1 to go to 0 V by capacitive coupling. This turns off one of the inputs of NAND3 so it's output goes high.
  • After a delay set by C1 x R1 (about 0.5 s) R1 charges C1 back up to 3.3 V and OUT2 goes low again. D1 protects NAND4 from overvoltage.
  • The circuit around NAND2 works the same way.

Cautions:

  1. Your requirement for inputs as low as 2.5 V may be on the edge of the CD4093 working correctly. Also the minimum operating voltage of the TI CD4093B is 3 V which is very close to your 3.3 V supply. When run on a 4 V supply anything above 2.2 V is read as a high and below 0.9 V is a low. This means it should work for your application. Try it and see.
  2. The chip can only source or sink about 0.5 mA at that operating voltage. You will need to buffer the output. You haven't given the information on the source signal or the load so it isn't possible to offer advice on this.
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  • \$\begingroup\$ so in your diagram the top line is powered and the bottom line is unpowered? @Transistor if yes thats EXACTLY what i want! \$\endgroup\$ – Acrolance Feb 19 '18 at 21:21
  • \$\begingroup\$ What I have shown is a timing diagram with logic levels low (off) and high (on). They're much better than words for this sort of thing. (1) Hit the edit link on my question and copy the relevant contents from there into your question and delete what's irrelevant. (2) Make sure that you specify the voltages correctly and maybe say what they're powering. (3) What is going to power the pulse timers? Is there available power and, if so, at what voltage? e.g., 12 V? If not are you hoping to power it from the 4 - 7 V supply? (4) Check that you've addressed 1, 2 and 3. \$\endgroup\$ – Transistor Feb 19 '18 at 21:50
  • \$\begingroup\$ kinda figured out how to remake what you posted... \$\endgroup\$ – Acrolance Feb 20 '18 at 2:41
  • \$\begingroup\$ OK. That's a lot better. You should un-accept my answer until I or someone else actually solves the problem. I've requested that the question be reopened. \$\endgroup\$ – Transistor Feb 20 '18 at 6:57

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