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I'm using a 555 timer to turn off an LED with delay. Here is the circuit. enter image description here All I want to implement is, to get the input voltage changes in output with a constant delay.

When I close the switch, the capacitor in RC network being charged and the OUTPUT will be LOW after n fraction of time. But, after opening the switch, it takes time for capacitor to being fully discharged.

A. Is there a way to discharge it quickly? B. And is there a more simple implementation without using 555 timer? C. How can you revert the output from LOW to HIGH instead of HIGH to LOW?

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    \$\begingroup\$ a microcontroller will be a simpler implementation \$\endgroup\$
    – MCG
    Commented Sep 26, 2019 at 9:11
  • \$\begingroup\$ @MCG it's also a more expensive for a simple operation \$\endgroup\$
    – M. Pi
    Commented Sep 26, 2019 at 9:22
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    \$\begingroup\$ You asked is there a more simple implementation. There is no mention of cost. \$\endgroup\$
    – MCG
    Commented Sep 26, 2019 at 9:29
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    \$\begingroup\$ @M.Pi PADAUK PMS150C-U6 microcontroller is less than a nickel from LCSC, under 3 cents each if you buy $9 worth \$\endgroup\$
    – Jasen Слава Україні
    Commented Sep 26, 2019 at 9:59

2 Answers 2

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A: To get it to discharge faster add a 330 ohm between the switch output and ground, to get it even faster also add a diode parallel with the 2.2K resistor. (pointing upwards) pretty much any diode will do eg: 1n4148 or 1n4001 etc.

B: more simple, the 555 behaves more predictably than most simpler circuits and is pretty simple itself, you could maybe do something with a SCR if the voltage drop is not a problem

C: swap the 2.2K resistor (including the diode from part A) with the 100uF capacitor.

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  • \$\begingroup\$ Thanks! Isn't sinking 36 mA current constantly a bad idea? Isn't there a way to implement the discharge path, dependent on switch's open state? \$\endgroup\$
    – M. Pi
    Commented Sep 26, 2019 at 15:10
  • \$\begingroup\$ true it's a bad idea if your electricity supply is expensive, or the small amount of heat is a problem. you could switch the 2.2K to 220K and the 100uF to 1uF and get much the same timing and not need the 330 ohms (because the 555 has internal resistors across the supply it's about 24K iirc.) \$\endgroup\$
    – Jasen Слава Україні
    Commented Sep 27, 2019 at 11:36
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A. Basically, greater current will make faster discharge. to do this, change the R to smaller value (increase current) B. IDK yed hahah C. You can use transistor, just read the datasheet to determine is using NPN or PNP.

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  • \$\begingroup\$ I don't wanna decrease the delay time, so I cannot change the R to a smaller value. \$\endgroup\$
    – M. Pi
    Commented Sep 26, 2019 at 9:25

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