voltage controlled switch
Printed From: the12volt.com
Forum Name: Car Security and Convenience
Forum Discription: Car Alarms, Keyless Entries, Remote Starters, Immobilizer Bypasses, Sensors, Door Locks, Window Modules, Heated Mirrors, Heated Seats, etc.
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=104035
Printed Date: May 11, 2025 at 1:39 AM
Topic: voltage controlled switch
Posted By: adhuot01
Subject: voltage controlled switch
Date Posted: April 16, 2008 at 12:02 PM
I am wiring an LED to 2 12v sources, there may be an chance that both sources power the LED simultaneously, causing a net voltage of 24v. I believe the solution may be a 3-pin device, maybe a transistor, that has the ability to open a circuit if there is an input voltage (the voltage from the other power source). What specific device might this be and what would be the configuration of the terminals?
------------- Andy Huot
Replies:
Posted By: offroadzj
Date Posted: April 16, 2008 at 7:49 PM
as long as you are wiring in series (+ to +, - to -) you should only stay at 12v. However, if one source is activated, it will very likely backfeed into the second source which may cause issues. Use diodes to keep the 2 sources from backfeeding into each other. that should be all that you have to do. If you really wanted to have the circuit opened, then just install a relay as follows:
85: source 2 12v+
86: chassis ground
30: input side of source 1
87a: ouput side of source 1 (to LED)
I believe that is correct, but if not someone will correct me.
if you dont mind me asking, why are you powering it from 2 sources? why not just wire it to one source?
------------- Kenny
Owner / Technician
KKD Garage LLC
Albany, NY 12205
Posted By: KPierson
Date Posted: April 17, 2008 at 7:36 AM
Yeah, all you need is to diode isolate the two inputs and you can wire them together after the diodes. You can then install the current limiting resistor (if you are using one) in the ground side, or even in the + side after the diode so that you don't need a resistor in each leg.
------------- Kevin Pierson
Posted By: adhuot01
Date Posted: April 19, 2008 at 2:08 PM
I just wrote up a wiring diagram and I feel like I jumped the gun on writing this topic because I now realize that the two voltage sources are in parallel, and 12v in parallel with another 12v source is still 12v, series is when it goes (+)to(-) and (-)to(+). So I should have no problem with a voltage increase. I plan on having a 1N4001 diode on each source to isolate the power to this circuit. The reason I am having 2 voltage sources to 1 LED is because I am mounting the LED inside of the top of a shift knob in a chamber that I bored out and it has a frosted lens over it for lighting effect. I want the LED to come on with the dash lights (fuse box option dashlight), and I also want the same LED to be connected to my keyless entry-the terminal that outputs the blinking armed/unarmed signal that alternates between 0v and 12v when I press the lock button. This light also prompts me during programming the keyless entry. By wiring both sources to the LED there will be a situation when I lock or unlock the doors, the LED will receive power from both sources simultaneously for about 1 second, because when lock/unlock is engaged the dash lights flash once for lock and twice for unlock mean while the blinking signal is in the process of stopping or starting, so there will be a slight overlap on outputs from these 2 sources does this make sense? Nonetheless I don't have to worry about it since I just realized that they are in parallel. Thank you for your comment and I hope I have answered your question.
------------- Andy Huot
Posted By: KPierson
Date Posted: April 19, 2008 at 6:07 PM
Are you sure the LED output works that way? What kind of keyless system are you using? Most of the LED outputs I have seen have been (-) current limited outputs.
------------- Kevin Pierson
Posted By: adhuot01
Date Posted: April 21, 2008 at 2:37 AM
The output for the LED is a 2 prong harness. I have simulated an output by sitting in the car with doors closed, and locked the door with the transmitter, I then used my multi-meter to check the voltage across the 2 prongs. the voltage oscillated between 0v and 12v, untill I unlocked the door. I don't see a problem wireing the 2 prongs to a 1.7v (20mA) LED with a 560 ohm, .25 W resistor at the cathode side which gives total current of 18.4 mA. The system is an Automat model 3. Let me if you think this is a problem and why.
------------- Andy Huot
Posted By: KPierson
Date Posted: April 21, 2008 at 6:32 AM
Probing across the two outputs doesn't tell you much. You need to probe across one output to ground and the other to 12vdc. You should find that one pin is either constantly grounded or constantly tied to 12vdc. The other pin would then be the "switch". Most 12vdc electronics switch the (-) side and have a (+) side that rests at voltage.
------------- Kevin Pierson
Posted By: adhuot01
Date Posted: April 22, 2008 at 1:45 AM
I will check what you said, and if it is in fact a (-) switch, can I use a PNP transistor such as a 3906 with 12vdc to C, (-) switch to B and E to the LED cathode?
------------- Andy Huot
Posted By: KPierson
Date Posted: April 22, 2008 at 6:00 AM
Yeah, that might work.
------------- Kevin Pierson
Posted By: dualsport
Date Posted: April 22, 2008 at 8:30 AM
If you use the transistor to invert the signal, don't forget to add a resistor in series with the base pin to limit the current. Probably 10k would do-
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