# Power off delay for door electric bolt

Discussion in 'General Electronics Chat' started by Fawaz Al-Rushaid, Oct 25, 2015.

1. ### Fawaz Al-Rushaid Thread Starter New Member

Oct 25, 2015
4
0
Hi,
I have made a simple circuit for a 12V door electric bolt from the below components:
1- 12V power supply
2- Door Electric Bolt
3- Magnetic Contact Sensor

When the sensor on close state the door bolt unlocks, when the sensor open state the door bolt locks, pretty straightforward. I want the door bolt to stay unlocked for few seconds (4 to 8 seconds) when the sensor goes into open state. How can I achieve that? I read in the forum something similar can be achieve with 555 timer. Is it possible to design a circuit using only simple components such as capacitor, transistor, diode? if possible can someone point me out to the right direction or sharing schematic would be highly appreciated.

Regards,
Fawaz

2. ### ScottWang Moderator

Aug 23, 2012
5,285
818
Will you attach the Magnetic Contact Sensor and spec?
Maybe you can drawing a block diagram to show what you want?
If the 555 can do the job, why don't you want to use it, does the 555 not easy to buy?
Using bjt, resistor, diode, capacitor could do some delay work, but you have to show what you want to do.

3. ### Fawaz Al-Rushaid Thread Starter New Member

Oct 25, 2015
4
0
Thanks for the reply, The contact sensor is a simple switch (on/off). I can't use it because simply I don't have it. I was kinda sure this could be done with a simple components that I already own. Anyways, I already solved my problem by using diode, capacitor, transistor, and resistor to introduce a 5 seconds power off delay. I have attached a schematic if anyone interested.

4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,070
I see a delay of only about 3s.

D1 does nothing, and can be eliminated.

You have a serious reliability flaw. It is only a matter of time before you weld the contacts of S1 together, or burn the contacts so that they no longer work or blow-up the diode. There is nothing to limit the in-rush current into C1 as the switch contact closes (except the forward resistance of the diode, which is quite low..., or the output impedance of your 12V power supply).

Replace D1 with a 100Ω resistor to cure the problem.

5. ### Fawaz Al-Rushaid Thread Starter New Member

Oct 25, 2015
4
0
I see your point, I guess because of the switch there is no need for the diode. Also thank you for pointing out the flaw. This was my first circuit design I am just learning about electronics. Could you please show me how you calculated 3 seconds delay? Thanks again.

Jul 18, 2013
11,939
2,984
It sounds like you really need a magnetic latch solenoid, these have a permanent latch/unlatch condition.
They just require a pulse in either direction.
Max.

Oct 2, 2009
5,451
1,070
8. ### dl324 Distinguished Member

Mar 30, 2015
4,267
844
Using the formula to calculate the discharge of a capacitor:
$\small V_f=V_ie^{-t/RC}$
Solving for t you get:
$\small t = -RCln(V_f/V_i)$

9. ### Fawaz Al-Rushaid Thread Starter New Member

Oct 25, 2015
4
0
Thank you all for your support and feed back

Fawaz