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I am attempting to use a retriggerable monostable multivibrator datasheet to tell me if an engine is on or off. The input to the IC is a square wave usually between 100-200Hz 0-3.3V. The signal is coming off of the engine's ignition coil, but is being taken to 0-3.3V using an optocoupler circuit. The multivibrator will be continuously retriggered producing a steady 3.3V logic high signal when the engine is on.

The part where I am confused about is what should tw (pulse width) be such that the IC is always retriggering? According to the datasheet tw can be found using: tw = KRC where R is in KOhms and C is in pF, and tw is in ns. K is approximately 0.45. Assume I select 200kOhms and 1uF my tw will be 90ms. Because my signal is 200Hz I should see a peak every 5ms. Therefore for every multivibrator I have 90/5 or 18 chances to retrigger the IC.

If I wanted to increase the number of retrigger possibilities per multivibrator pulse I could just keep increasing the capacitance for example. However there appears to be a "minimum retrigger time" which grows as the tw increases. Computing this for the 200kOhm 1uF case yields @Vcc 3V: 1.5s. Am I calculating this wrong? If this were true than I could not produce a continuous logic high output under these conditions.

What RC values would be best to assure retriggering and at the same time having some noise immunity such that if the input pulse stopped for a few ms the output wouldn't indicate the engine was off?

retrigger equations

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  • \$\begingroup\$ I'd like to know a lot more about the signal coming off of the engine's ignition coil. But given that you have an opto involved, I think it can be arranged to work with this circuit, which is retriggerable and for which the computation of values is quite easy. (The collector/drain of the opto can be arranged to replace the indicated switch.) \$\endgroup\$
    – jonk
    Commented Nov 2, 2022 at 18:47
  • \$\begingroup\$ I took a look at this, it seems rather complex. I would prefer to use an IC if possible due to the footprint and ease of understanding. Also I think with proper selection of resistors and capacitors the IC I selected should work \$\endgroup\$
    – Feynman137
    Commented Nov 2, 2022 at 19:19
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    \$\begingroup\$ It's a nice dual and I've buckets of them from decades back (not that particular version, though.) I've used them for similar purposes. I was just tossing out a more boutique (less sensitive to IC availability issues -- plus it is very low-power) approach that also works well. You have every right to use what is more comfortable for you. I only wanted to see if you were open to another idea. It's enough that you aren't. If I get a moment, I'll help. \$\endgroup\$
    – jonk
    Commented Nov 2, 2022 at 19:23

1 Answer 1

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You're doing the calculation correctly, but I think the equation is wrong.

If you change the * symbol to a +, it gives a result that makes more sense.

With the 200 kΩ resistor and 1 μF capacitor, you get 9.5 ms with the 5 V equation. Changing that one operator in the 3 V equation makes the calculation 7.5 ms as shown in MATLAB.

enter image description here

There's no way to know for sure what the correct equation is unless there's enough info in the datasheet to derive it, but if you have a signal generator and a scope, you could test how it functions with different input frequencies.

If that is the correct equation, you could change the resistance to 265 kΩ or greater to get a retrigger time of 10 ms (100 Hz)

500 kΩ would be a better option to ensure the system will always retrigger by stretching the retrigger time to around 20 ms. Here are the calculations for such:

rt1_fix = retrigger time in seconds for 3.0 V (possibly fixed)

rt2 = retrigger time in seconds for 5.0 V

enter image description here

Here is the equation for solving for an R value based on time in seconds. For 20 ms => t = 0.020:

$$\frac{2\cdot 10^{10}\cdot t-820}{3\cdot C^{0.9}+20}$$

You can also replace C with any value you want.

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  • \$\begingroup\$ You may be correct about the equation issues, I hope this is the case. I requested some tech support from nexperia, and I can confirm this. Usually they reply within 24hrs or so. \$\endgroup\$
    – Feynman137
    Commented Nov 2, 2022 at 21:50
  • \$\begingroup\$ I see why you are questioning the 3V equation, because the 5V equation has a + not a *. Based on the current documentation the 3V and 5V supply would give radically different retrigger times. \$\endgroup\$
    – Feynman137
    Commented Nov 2, 2022 at 22:02

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