Mistery circuit winner.

So, who won the mistery circuit? and what was it?

Luftek, you made the best analysis of it, so the circuit goes to you. You only had a bit left wich would have given away the purpose for it: “Q1”
That transistor sits in parallel with 555’s main capacitor…so…shorts it out whenever it’s told to. And if you short the 555’s capacitor, what happens? Yep, it starts charging again, efectively retriggering it.

Laser watchdog

So, what does the circuit actually do?

The circuit is an EDGE detector plus retriggerable 555, so, it can detect the change in state (not the state itself) of an input, and, given enough frequency (0,1 Hz or less) it mantains the output ON. If the water sensor (or other) fails to give life signals for some seconds, it will shut down the whole laser machine. (provided you wire it that way, of course)

Laser watchdog 001
Capacitor voltage vs switch/sensor input.

Laser watchdog 002
The edge detector gives away a short 2ms pulse. (switch press was 150ms)

Edge detection circuit:

Laser watchdog 003

Internet says that just a NOT + AND, will work, however the 555 required an inverted pulse for it’s activation, hence, the NAND. R3 as pulldown so input doesn’t float, and C3, because otherwise, the propagation delay from U2 as so short, the circuit didn’t work. (yep, 20uF is a LOT, I know)

But, why bother to make the edge detection rather than just retrigger the 555 from the sensor? Safety (heh, XD) reasons. Imagine the worst case scenario:

Water pump stops and the simple flow sensor also stops in the ON position, you have no way to know that the pump has stopped if you retriger by level (+5V, for example). The edge detector waits for the rising level shift, so, pump stopped = laser shutdown.

I hope the winner has some use for the circuit, and that someone can use the edge detection info.

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5 responses to “Mistery circuit winner.

  1. Nice. I went with a MSP430 to monitor temperature and water flow for my laser. I considered monitoring the pulses to the steppers so the laser would only fire when moving. The idea would be to remove that dot at the start of a cut and ensure the laser couldn’t be left burning away in one spot.

  2. If you placed a low value resistor (eg. a strong resistor) between the inverter and the NAND gate, your inverter wouldn’t have a hard time changing the transition.

    Before you pull out the 20uF capacitor, try adding a very small resistor and see a huge change. For instance a 10 Ohm resistor should make a huge difference; even a 1 Ohm resistor should make a significant change.

    You would then be able to lower the value of the capacitor significantly. You could start by using a 100nF capacitor and then find a resistor value that would give you the right timing.

    Note: it’s common to make a glitch-filter on the button.

    That sums up to … A capacitor could be placed on the button itself, then a resistor between the inverter and the NAND gate. Place a small capacitor after the resistor (eg. closest to the NAND gate) to adjust the timing.

    • The inverter doesn’t have any problem with the states, the problem lied in the fact that it was SO FAST, the NAND didn’t had enough time to detect the diference and generate a spike of incongruence detection. The capacitor after it turns the otherwise “vertical” falling edge from the invertor, into a smooth slope…(realtime thinking) but now that you say it, I think I understand what you say. Yes, YES! I SEE IT NOW! /madman voice/. XD
      Just kidding, but yes, the problem lied with the inverter draining the capacitor, If I used a resistor between the capacitor and the NOT, that should prevent the fast discharge. I know I did some tests with resistors, but I don’t think I tried this one.
      THANKS!
      Will try and let you know.
      /Still surprised someone read the blog, XD/

      • If you write something interesting, there *will* be readers. ;)
        Hmm, I think I might have been too tired when I wrote the above reply.
        I’m not sure why I wanted to put the capacitor after the resistor; it should really be before the resistor, so the inverter would change the state of the signal quickly and the NAND gate will discharge the capacitor slowly.
        (you can also see it this way: the NAND-gate has a built-in resistance, which can be extended by adding an external resistor).
        I’m still sleepy, but I do believe that this approach should work as intended.
        Please let me know if I’ve messed up again. ;)

      • (Waking up now) …. Argh, no, the first approach should be right.
        (I hear myself mumbling something about testing what I’m saying id right before posting, but it’s good to make some noise, heh).
        The inverter is the one to discharge the capacitor when the input to the inverter changes.
        The NAND gate would discharge only slowly, as it’s only “reading” the state.
        Normally, the circuit is made by using 3 inverters in series, because one IC often holds 6 inverters, and two of the inverters are there to delay the signal and the last one is to make the edge right.

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