Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Latched Relay Car Immobilzer

Status
Not open for further replies.

ic07722

Newbie level 4
Joined
Oct 13, 2019
Messages
6
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
44
Hi, I'm new here, please be gentle. I'm also a complete novice, so apologies if I'm a bit thick!

Some years ago I bought a commercially available car immobiliser which used a latched relay. This proved so successful that even when I exchanged my car at the main dealer (and explained the immobiliser) I had to return to start the car for them. I can no longer find this device in the obvious places (Amazon & eBay) so I'm trying to design my own. I have drawn up a circuit diagram and would be most grateful if one of you experts out there could look over the circuit and advise me of its viability or any improvements I need to make. It is very simple circuit, but alas not to me. The circuit diagram is attached to this post.

Basically it functions as follows: When the car ignition is switched on the relay is open and hence the fuel pump/coil does not function. Whilst the engine will turn over, it will not start. Momentarily pressing the button causes the relay to close two circuits: The first is the power to the relay to keep it closed (latched); the second is to close the circuit to the fuel pump/coil to enable the car to start. When the ignition is switched off, the power to the relay is cut and it returns to the open state (hence I can't forget to set it).

Thanks in advance for any help, I really appreciate it.

Ian.
 

Attachments

  • Image4.png
    Image4.png
    32.7 KB · Views: 85

You almost have it right. the relay contacts on the left go across the momentary switch instead of chassis. Basically, it makes it think the momentary switch is being held closed. Wired as shown the left contacts will close and short out the 12V feed and blow the fuse.

Brian.
 

Thank you so much Brian, for helping me out. Do I have it right now?

Cheers

Ian.
 

Attachments

  • Image5.png
    Image5.png
    33.9 KB · Views: 87
Last edited:

That is correct.
With no power applied, there is no path to power the relay coil. When the switch is pressed, power reaches the coil and the contacts close. From then on, until the power is removed, the contacts keep the power connected to the coil.

There is a simple modification you could consider, it isn't essential but it will increase the life of the relay. You connect a diode (example 1N4006 which costs pennies) directly across the relay coil with the cathode (banded end) to the positive side. Assuming your car has a negative chassis, the cathode would be at the top in your schematic. The purpose is to reduce the high voltage spike you get when the relay turns off.

I'll try to explain without getting too technical: when a current flows through the coil it produces a magnetic field, it is this field that pulls the contact mechanism to close them. When the power is removed, the magnetic field collapses and the process reverses, a voltage is produced by the coil. How much voltage depends on how fast the field collapses but it can be several hundred volts and importantly, the polarity is also reversed. So for example, if you connect +12V to one end of the relay (as you do by pressing the switch) then disconnect it, the +12V end of the coil might go to more than -100V. The spike is very short, maybe only one thousandth of a second but it happens at the time the contacts have only just started to open so the gap will be very small and spark will jump across it. The spark eventually erodes the contact surface and reduces the relay life expectancy. A diode only conducts in one direction so if you wire it in the normally non-conductive direction it does nothing but when the polarity reverses, it does conduct and safely 'shorts out' the spark voltage.

Brian.
 

Now that you mention it, I remember something about a diode. However, I can't claim to understand what it is or why it was included in the circuit. Now that you've explained, I think I understand. Is this correct...?

Image6.png
 

Almost, the diode is in the right place but the arrow should point upwards. As shown it will conduct when the switch is pressed and divert the current away from the coil, the intention is it DOESN'T conduct until the power is removed, when that happens the polarity across the coil (and nowhere else) reverses so the diode does conduct and prevent a high voltage being produced. The arrow points to the cathode end which physically is the end with a distinctive marking, usually a colored band around the body.

Brian.
 

OK, so I believe this circuit is now correct...

ImmobiliserCircuit.png

But, and please excuse my ignorance, I understood that cathode = negative and positive = anode. As the vehicle is negative (cathode) earth, surely the cathode is at the bottom and the top +12v supply from the battery is the anode?
 

You understood correctly, but in this application you DON'T want the diode to conduct so it is wired the other way around. It will still conduct when the power is removed because at that instant both the push switch and relay contacts will open and a negative (remember what I said about the polarity reversing when the magnetic field collapses) will appear at the top of the coil. It is only the reverse voltage you want the diode to catch, if wired with the cathode to ground it would conduct when the switch was closed and divert all the current that should go to the coil through itself.

Brian.
 

Ahhhh, got it! When you said "the cathode would be at the top in your schematic", you were referring to the top of the diode - I thought you were talking about the top of the diagram! Now I understand.

Thank you so much for taking the time and trouble to help me. I assume this is pretty basic stuff to you, so I'm all the more grateful for you taking this on and helping a novice such as myself.

I'm one happy bunny.

Thanks Brian.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top