I have an exhaust fan in my bathroom (used for removing the steam generated by a hot bath).
Currently, it can be turned on and off by a regular switch.
My goal: Exhaust fan should automatically turn-off 5 minutes after it is turned on.
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1\$\begingroup\$ Uhm, most domestic exhaust fans already have this feature built-in? \$\endgroup\$– Jeroen3Aug 25, 2020 at 6:12
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\$\begingroup\$ The fan is AC, yes? If you have DC available, you can use either a DC relay or else an AC relay with a MOC30x3 IC. If you only have AC available at the site, then it may be more convenient to get a simple HLK-PM01 or similar so that you have DC and then develop the timer from there. \$\endgroup\$– jonkAug 25, 2020 at 6:16
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\$\begingroup\$ The other, perhaps preferred, way is to add a humidity sensor, either replacing the wall switch or added at the fan. This way, it turns on and off automatically. \$\endgroup\$– DoxyLoverAug 25, 2020 at 6:39
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5\$\begingroup\$ Switches that do exactly what you want are inexpensive and available in hardware stores. \$\endgroup\$– user128351Aug 25, 2020 at 11:36
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2\$\begingroup\$ Use a Timer Switch. \$\endgroup\$– brhansAug 25, 2020 at 11:37
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3 Answers
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Replace the existing switch with a UL-rated timer switch. Anything else will run the risk of invalidating your fire insurance.
answered Aug 25, 2020 at 11:38
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Such a timer can be purchased. The switch starts the fan (AC) and runs for a few minutes after switching off. The fan is switched by a small triac and the timing is done by a CMOS gate. Its supply voltage (12V) is from mains voltage through a capacitor and a Zener.
(Made in Italy)
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The requirement will be readily met by an off-the-shelf 'Toilet Exhaust Fan Timer'.
Nonetheless, the following schematic, using a 555 monostable, a transistor and an electromagnetic relay, is presented for academics' sake.
The circuit is an adaptation of this power off delay circuit published in Seekic.com.
http://www.seekic.com/circuit_diagram/Control_Circuit/Power_OFF_Delay_Circuits.html
It's an example of a true off-delay timer.
After the switch is turned on the output would be energized. Subsequent turning-off of the switch would signal the commencement of the delay time, at the end of which the output would be de-energized.
The values of 'R' and 'C' would decide the off-delay time.
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1\$\begingroup\$ It is a good idea to expand on comments like provide a circuit diagram, care to elaborate? \$\endgroup\$– Voltage Spike ♦Aug 25, 2020 at 7:32
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2\$\begingroup\$ 555 timer and 5 minutes is not normally a viable combination. \$\endgroup\$– user16324Aug 25, 2020 at 12:51
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3\$\begingroup\$ Homemade circuit and permanent building wiring is not normally a viable combination (building code violation). \$\endgroup\$ Aug 25, 2020 at 15:47
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1\$\begingroup\$ The answer has been edited thanks to all the comments. \$\endgroup\$– vu2nanAug 25, 2020 at 17:37