How to trigger a 5 sec delay without using 555 ic

Thread Starter

tpny

Joined May 6, 2012
220
I want a pushbutton when pressed and released to trigger a 5 second (or x seconds whatever) voltage level change (high to low or low to high doesn't matter) without using a timer ic. Any guidelines? Thanks!
 

Thread Starter

tpny

Joined May 6, 2012
220
Yes, I knew I had to have a cap in there, just didn't know how to put it in a circuit to serve this purpose. Do you have a drawing? Thanks!!
 

ScottWang

Joined Aug 23, 2012
7,504
You can use two NPN transistors connect them similar as darlington, but the each C of two NPN connecting a 10K to +V.

The circuit as this:
+V → Start Key → 1K → Cap([-] toGND) [+] → 1M Ω → two BJT → C output of second NPN.

The similar as darlington driver on the top of right side of page 2.
Take off the Rbe,Re resistors from the first NPN Tr1.
http://panasonic-denko.co.jp/ac/e_d.../common/catalog/mech_eng_cau_appli.pdf?via=ok
 

crutschow

Joined Mar 14, 2008
38,601
You could use a single MOSFET which has the advantage of a very high input impedance, allowing a larger value of resistance and a smaller value of capacitance then a BJT.

The circuit would be +V\(\rightarrow\)Start Key\(\rightarrow\)capacitor\(\rightarrow\)MOSFET gate & resistor with resistor\(\rightarrow\)ground. Connect the MOSFET source to ground and the drain through a resistor to +V (resistor value determined by load). This will cause the MOSFET drain output to go momentarily low and then back high when the Start Key is pressed. Try R = 10MΩ and C = 0.5μF to start.

To reset the capacitor you will need to add a resistor, 1MΩ or less, from the switch output to ground.
 

Dodgydave

Joined Jun 22, 2012
11,395
The relay will operate immediately the push switch is pressed, and stay operated until the capacitor discharges,time depends on coil resistance and capacitor value, bigger values = longer time delay.
 

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