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I am using an IRROMETER sensor for soil moisture measurement. This sensor requires AC excitation, so I found an interfacing module for it from online. This is the circuit recommended by them:

enter image description here

The circuit was easy to simulate and these were the results:

enter image description here

Currently I am trying to understand the exact working of the circuit, since this is similar to the astable vibrator I am trying to relate them.

enter image description here

From the general circuit diagram, we can conclude that R1, R2 & C form the wave shaping network.

The following are my questions:

  1. Which of the resistances form R1 & R2 in the case of my soil moisture interfacing circuit? Basically I need those values to theoretically calculate the duty cycle & ON time and then compare them with my LTspice simulation.

  2. What exactly is the difference between the OUTPUT pin & DISCHARGE pin, since I am seeing the same waveforms from both pins?

    1. Is the direct feedback mode same as the Astable mode? why is it called direct feedback?
    2. Why is the OUTPUT pin called totem pole? 5.Why is the duty cycle different in my simulations and the one I have observed on the CRO, when I have used the exact same values of R & C components?
    3. Also why is it said that the current through the circuit is of AC type when we are providing a DC supply?

The brief working of the circuit is given here by the manufacturers. Ignore the voltage regulator part and refer the circuit of the simulation. Till now I have figured out that the 390 Ω, 150 kΩ & 0.1 uF and the sensor resistance form the wave shaping network, but I am still not able to understand which ones are R1 or R2?

enter image description here

enter image description here The same circuit I built in my lab, and observed it's waveforms on the CRO, sorry the pics are not very clear. I am also adding the data sheet of this 555 timer module. https://www.emesystems.com/smx/documents/SMX_2018.pdf

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    \$\begingroup\$ You're using the NE555 with a 3.3V supply. That can't work in reality, the minimum operable supply voltage of the NE555 is around 4.5 V. $$\,$$ Also, the NE555 in a battery-powered long-term observation device is a terrible idea – far, far, far too high power usage. The PDF at the bottom (crop your screenshot and link to the source!) mentions the LMC555, which is a similar, but different device, and does indeed work at lower voltages; so maybe fix the top circuit! \$\endgroup\$
    – Marcus Müller
    May 21, 2020 at 7:57
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    \$\begingroup\$ Then, you're asking about basic functionality of the xx555 chips – have you read the LMC555 datasheet? It comes with pin descriptions and a lot of functional descriptions that look a lot better than what you show in your whatsapp screenshot (again, when posting such pictures, crop to the relevant content and link sources). \$\endgroup\$
    – Marcus Müller
    May 21, 2020 at 7:59
  • \$\begingroup\$ Actually I am using the LMC 555 timer , couldn't find it in LTspice so used NE555 for simulation. \$\endgroup\$
    – Rohit Kumar
    May 22, 2020 at 6:29
  • \$\begingroup\$ That makes no sense. I didn't find sugar, so I simulated my cake with salt will not make a cake simulation that has anything to do with your real cake. \$\endgroup\$
    – Marcus Müller
    May 22, 2020 at 7:53
  • \$\begingroup\$ Also, I've given you 24h to crop your screenshots, -1 for lack of diligence. \$\endgroup\$
    – Marcus Müller
    May 22, 2020 at 7:53

1 Answer 1

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Your circuit is a little different to the "normal" 555 circuit. The normal 555 circuit gives a duty cycle of < 50%. your circuit gives exactly 50% duty cycle.

Take a look at the 555 datasheet, you can see how TRIG and THRS work to turn the OUT pin high and low. The basic idea of this circuit is that it's charging and discharging C1 through the combination of R1 with R3-C3-R4-C2. As the value of R4 changes, the charge and discharge rate of C1 changes, so the period of the waveform changes.

The DIS pin is almost the same as the OUT pin, but it is open-drain (Pull down only). OUT pin will drive both high and low so it will source current to charge C1, and sink current to discharge C1. It looks like the junction between R1 and R5 is the place to read the output from the circuit.

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  • \$\begingroup\$ How did u find out that the duty cycle is exactly 50%? Also can i theoretically calculate the timer periods? Is it possible that the duty cycle won't be 50% bcoz when I built the circuit and replaced R4 by the actual sensor and observed the waveform on the CRO the duty cycle wasn't 50% \$\endgroup\$
    – Rohit Kumar
    May 22, 2020 at 18:46
  • \$\begingroup\$ 50% because high time is the time to charge c1 From trigger voltage to threshold voltage when OUT is high, and low time is the time to discharge c1 from threshold voltage back down to trigger voltage. Trigger and threshold are 1/3 and 2/3 voltage so the charge and discharge times will be the same. Which circuit did you build? The first circuit should be 50%. The second circuit is not the same. It has a shorter discharge time than the charge time. \$\endgroup\$
    – harry courtice
    May 29, 2020 at 9:07
  • \$\begingroup\$ I have built the first circuit which consists of 5 resistors, I have only changed R4, then got the following output on the oscilloscope, my confusion is which components decide the ON time and which of them decide the OFF time? Is there any way I can theoretically calculate the ON time and OFF time? \$\endgroup\$
    – Rohit Kumar
    May 29, 2020 at 17:06
  • \$\begingroup\$ Also all 555 timers in astable mode have trigger and threshold between 1/3 and 2/3 so shouldn't all of them always have 50% duty cycle. \$\endgroup\$
    – Rohit Kumar
    May 29, 2020 at 17:07

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