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Time Delay to Unlatch Relay

Thread starter #1
Greetings,

The attached circuits refer.

Being a mechanically minded person, I know very little about electricity and even less about electronics.

However, on a "paint-by-number" basis, I have managed to successfully build the 2 circuits shown. On their own, each circuit works perfectly.

Now, I would like to combine these circuits in order to end up with relay that latches with the push of the momentary button, but remains latched for about 2 to 3 seconds after the power is switched off.

I've tried to work out how to do this, but all my attempts resulted in failure.

Can these 2 circuits be combined, or is it wishfull thinking on my part?

If it is at all possible, how could this be achieved?

Any guidance will really be greatly appreciated.
 

Attachments

rjenkinsgb

Active Member
#2
A relay can be made to latch (and unlatch) just by using an extra contact on the relay to bypass the "on" button, so the circuit remains when the button is released.

eg. https://encrypted-tbn0.gstatic.com/...3HGURx4p7yC9MznJRMC6FBhK7CsFgJv962vxMg6mKtuhA
You can link out the push-to-break "off" button if manual switch off is not needed.

Keeping the relay on without any power is a different thing.

You need a capacitor across the relay coil to store power so it can hold in for a while.
That needs a low value resistor in series with it, so it does not take excess current at the instant the circuit turns on, while the cap is initially discharged. Something like 100 ohms should be reasonable.

The value of the cap depends on the particular relay; the higher the coil resistance, the longer the relay will hold in for a given capacitor value.

I'd try eg. 470 or 1000uF to start with.
 
Thread starter #3
rjenkinsgb

Many thanks for your response.

It'll probably take a while for me to understand what you have suggested - but I'll do my best.

I will provide feedback.
 

MikeMl

Well-Known Member
Most Helpful Member
#4
Do you want to use a single momentary push button for both actions: first push to latch the relay on; second push to unlatch the relay after a two second delay?

Or can you use two different push buttons: left button latches the relay; right button unlatches the relay after a two second delay?

Can the relay have two sets of poles (DPDT)?
 

dr pepper

Well-Known Member
Most Helpful Member
#5
Not sure what exactly you want, in some ways fig1 does what you want.
When you say power switched off do you want this thing to work without power, that could be done if you were to use something like a pneumatic timer, or a very large value capacitor across the coil.
 

AnalogKid

Well-Known Member
#6
remains latched for about 2 to 3 seconds after the power is switched off.
If by this you mean that all power is removed from the circuit, then you need 2-3 seconds of energy storage to power the relay coil after power is gone. Typically, this is a relatively large capacitor; the size depends on the relay coil nominal voltage and current.

ak
 

MikeMl

Well-Known Member
Most Helpful Member
#7
Both of these circuits use power from the battery to provide the hold-in current for the relay after the switch is opened; thereafter, the battery current is < 10uA of leakage. Is this acceptable to Kasie52?
npn.png nf.png

It would be easy to power these circuits through a NO contact on the relay so that the leakage goes to zero after the relay times out...
 
Thread starter #8
Firstly, thank you all for the responses I've received.

I'll try to explain - should probably have done so in the first place. Apologies.

I'm building an immobilizer for my 30 year old run-around light pick-up. The vehicle is very basic in that there is no fuel injection and no engine management unit. It does have electronic ignition though. The immobilizer interrupts the power supply to the ignition controller.

The immobilizer is powered from switched power. It is installed, connected and works great - ie, switch on ignition, push the momentary button, the relay latches, turn the key further and the engine starts. Being powered from a switched power supply, the relay unlatches as soon as the ignition is turned off.

The idea for adding the 2 second time delay is that, if the vehicle does not start first time around, or stalls, I don't have to arm the immobilizer again after turning the ignition off and on again (if I'm quick enough).

I hope this makes sense.
 

MikeMl

Well-Known Member
Most Helpful Member
#9
So you have an "always available, even if the key is off" power source.
It also seems that you only need one "hidden" momentary pushbutton to start the vehicle. Turning the key to off begins the shut down sequence, which immediately stops the engine if running, but you want the immobilizer relay to stay on for a few seconds.

Here is a circuit that does what you requested using a DPST relay:
im.png

The immobilizer relay can only be latched by the momentary push button if the ignition switch is on. Once latched, the relay stays on for a time constant defined by R1 and C1 after the ignition switch is turned off. D2 prevents the timing circuit from discharging into the other vehicle loads. You only have to build the stuff inside the dashed box. Note that it has four connections to the outside world...
 
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Thread starter #10
Thank you so much for the circuit MikeMI. I really appreciate your assistance.

Would it be okay to use the following components?
Diodes: 1N4007
NPN: 2N2222

I assume that I could play around with the values of capacitor C1 and resistor R1 to get the optimum time delay.
For starters, would the following work?
C1: 470μF
R1: 1KΩ

Thanks again MikeMI.
 

MikeMl

Well-Known Member
Most Helpful Member
#11
Here is a more detailed analysis of just the timer part. For the purpose of the simulation, I am omitting the various other switches and relay contacts required to latch the relay, and I am simulating only the timer details.

I am assuming that the relay has a 12Vdc coil with a resistance of 85 Ohms or higher. The diodes are 1N4007 Si rectifier types. I am adjusting the R1-C1 to get a relay drop-out time of about 10sec.

V2 is used to simulate the V(key) signal (ignition switch). In the sim, it is initially 0V (ign switch off), it becomes 12.8V at 2sec (ignition switch on), and back to 0V (ign sw off). Diode D1 conducts to charge C1 to the battery voltage (12.8V) minus one diode voltage drop. Note that V(RC) follows V(key) as it goes from 0V to 12.8V, but decays slowly after V(key) goes back low. To limit that charging current in-rush, it is necessary to add R2 (add this to the actual schematic you build).

After the ignition is turned off, the charge stored in C1 flows through R1 into the base of Q1. The base current is amplified by Q1 such that its collector current is sufficient to pull-in the armature of L1. For most relays, they drop out when their coil current becomes about 25% of their normal rated pull-in current. I plot I(L1), the current through the coil, and show where I think the relay drops out.

With the values shown for C1 and R1, it looks like that will produce a delay of about 9sec. To increase the delay, you can increase the value of C1. To shorten the time, you could make R1 less than 10K, but do not increase R1 above 10K to get longer delays. Obviously, if the ign sw is re-closed before the delay elapses, the relay never drops.

tsn.png
 
Thread starter #12
Thanks again for all the help, patience and time spent MikeMI.

I can understand it the way you explained everything, including the time delay circuit.

I've tried a couple of local stores today, but have found it quite difficult in sourcing a DPST relay - even a DPDT relay for that matter.. Could 2 SPST relays in parallel be used instead? I've drawn the circuit with the 2 relays, and is attached for your comment. I've also included R2 as you recommended.

In the event that 2 relays can be used, I've measured the coil resistance on some I found laying around. One is 40 Ohms and the other 68 Ohms. Unfortunately, they are not from the same manufacturer, presumably the reason for the difference in resistance.

Purely for my "education", would it be advisable to replace the left hand relay with a suitable SSR?

12v Latching Circuit.jpg
 

MikeMl

Well-Known Member
Most Helpful Member
#13
I have some reservations about paralleling the two relays you have. It will take ~0.5 A to power them and the 2N2222 is not up to the job.

I will work out a circuit later tonight.
 
Thread starter #14
No need for a revised circuit. I've managed to get hold of a 12v DC DPDT relay - EUREKA!

The spec sheet and image of the relay is attached. I hope that the relay will meet requirements.

12v DC DPDT Relay.jpg
 

MikeMl

Well-Known Member
Most Helpful Member
#15
That relay is ideal for the job...

I did come up with a circuit that would have worked with the two relays:
78-2.png

Use the relay with the higher coil resistance for L1.
 
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Thread starter #16
Thank you so much MikeMI, I am truly grateful for all your help and guidance.

I'm definitely saving both circuits. Chances are that I'll need to build one or two more immobilizers once my brother finds out what I've built. Now I even have the option of a "one relay" or a "two relay" circuit. This is so great.

Kind regards.
 
Thread starter #17
Apologies for bugging you again MikeMI.

Your post #11 above (paragraph 3, last sentence) and its attached schematic refers.

Could you advise on the value of R2 please? I've breadborded the circuit, but could not get the relay to latch. In desperation I replaced R2 (100k Ohm) with one of 21,6k Ohm. Tried again - this time it worked (another EUREKA moment!).

Is the 21,6k Ohm resistor sufficient to limit charging current in-rush to the capacitor? If not, what value resistor would be ideal for this application?

Looking forward to your reply.

Thanks MikeMI.
 

MikeMl

Well-Known Member
Most Helpful Member
#18
In the schematic, R2 is 100 OHMs (not 100,000 Ohms)
 

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