I have one signal of 5v and i need open one electric lock (door access) of 12 V? I can used it for convert my signal of 5v to 12v?
1493391
That depends on the power / current requirements of your door lock. If it is a 1A 12V lock then that means your input current would have to be 2.4A even if the buck/boost converter was 100% energy efficient and had no loss. Since there will be conversion losses your input current (for a 1A 12V load) is probably at the 3A maximum rating for this converter. If your lock requires more then 1A then this is not the correct DC-DC converter to use for your application. Also make sure that your signal source can supply 5V at the required current. If the input voltage drops because of the load then the input current will increase even further. That depends on the power / current requirements of your door lock. If it is a 1A 12V lock then that means your input current would have to be 2.4A even if the buck/boost converter was 100% energy efficient and had no loss. Since there will be conversion losses your input current (for a 1A 12V load) is probably at the 3A maximum rating for this converter. If your lock requires more then 1A then this is not the correct DC-DC converter to use for your application. Also make sure that your signal source can supply 5V at the required current. If the input voltage drops because of the load then... » Read More That depends on the power / current requirements of your door lock. If it is a 1A 12V lock then that means your input current would have to be 2.4A even if the buck/boost converter was 100% energy efficient and had no loss. Since there will be conversion losses your input current (for a 1A 12V load) is probably at the 3A maximum rating for this converter. If your lock requires more then 1A then this is not the correct DC-DC converter to use for your application. Also make sure that your signal source can supply 5V at the required current. If the input voltage drops because of the load then the input current will increase even further. « Show Less
1493392
That depends on the power / current requirements of your door lock. If it is a 1A 12V lock then that means your input current would have to be 2.4A even if the buck/boost converter was 100% energy efficient and had no loss. Since there will be conversion losses your input current (for a 1A 12V load) is probably at the 3A maximum rating for this converter. If your lock requires more then 1A then this is not the correct DC-DC converter to use for your application. Also make sure that your signal source can supply 5V at the required current. If the input voltage drops because of the load then the input current will increase even further.
1493393
Yes, this converter will work for you. Just make sure wherever you mount this, that you don't ground it out. Mount it on plastic or something non conductive.
1850208
you can connect to the 12v door bell power but wait, don't buy it. As soon as I took it outside to check to be sure it was getting the WiFi it stopped working. so I took it to another one of my houses and the same thing. Works great inside the house and that's where it ends.
562069
This would be the right product for turning light on opening a door... I'm using mine with to turn on LED light strips in my gun safe which are low voltage and works great. Your application may have be different if your trying to use 110 volts.
1953963
The only DC Input is one 12VDC. All run from this single Power input. I do not see or was a 5V lead provide.
291323
Probably not, first, you would need a 12 v source for operation and it is not a clock timer, it will time for up to 999 seconds when activated and will stay activated until the signal is removed.
215546
I confirm what it as been said before by the others, no problem unless you use it with a very sensitive equipment. Another way to do it is to use a 14v adapter to get 12v at the end if you have a voltage drop. But normally the equipment have a tolerance and you should be good.
1540281
I looked around some online and it seems that -5 volts was used in the ISA bus. Being that buss is no longer used it would seem that voltage is no longer needed. Doesn't hurt if it's present or not.
1598371
Not practical. You'd have to have an opener that worked by having 12VDC applied across a set of contacts (not likely). You'd have to toggle the ON , then OFF every time you wanted to open or close the door since the RC switch stays on or off depending on the last command it got. You'd also have to have a 12VDC source to power the receiver. Finally, I bought 2 of these switches and the same remote works either one of them, so there is no security. Hard to beat a garage door opener with a remote for opening a garage dooor.
980990
The system (and control box) uses 48 VDC so that's not an easy conversion to make from 12 volts. A small inverter would probably be a better solution since the antenna only needs power when raising and stowing.
211290
It should be possible, but if you have to ask if its possible you probably couldn't / shouldn't do it. It takes 3 AAs which is about 4.5volts, USB is 5 volts which is probably in the safe voltage range for this. So in theory you could plug in a USB adapter to bring the 12v down to 5v. Then cut the other end off the USB cable. Find a plug type connection and solder one end onto that cord. Solder the other side of the plug to another wire. Finally solder the end of the second length of wire to the battery terminals inside the light. However if you aren't 100% sure how to do it safely don't attempt it. Also, of course, this would void the warranty. It should be possible, but if you have to ask if its possible you probably couldn't / shouldn't do it. It takes 3 AAs which is about 4.5volts, USB is 5 volts which is probably in the safe voltage range for this. So in theory you could plug in a USB adapter to bring the 12v down to 5v. Then cut the other end off the USB cable. Find a plug type connection and solder one end onto that cord. Solder the other side of the plug to another wire. Finally solder the end of the second length of wire to the battery terminals inside the light. However if you aren't 100% sure how to do it safely don't... » Read More It should be possible, but if you have to ask if its possible you probably couldn't / shouldn't do it. It takes 3 AAs which is about 4.5volts, USB is 5 volts which is probably in the safe voltage range for this. So in theory you could plug in a USB adapter to bring the 12v down to 5v. Then cut the other end off the USB cable. Find a plug type connection and solder one end onto that cord. Solder the other side of the plug to another wire. Finally solder the end of the second length of wire to the battery terminals inside the light. However if you aren't 100% sure how to do it safely don't attempt it. Also, of course, this would void the warranty. « Show Less
135854
Sure. You just need to be able to access the wire running from the door opener to the button. In my older house, the low voltage wire is just run on the surface of a rafter. I replaced a keyed entry.
36026
To be perfectly honest with you, if you aren't sure about wiring, maybe this isn't the right product for you. It will mean using trim tools and splicing to a found 12v wire. If you want to press forward, the red goes to a switched +12vdc source, and the black goes to a chassis ground point.Good luck.
1856172
Most likely. would not take anyones word for it would be a good idea to look at your trucks electrical schematic and compare it to the viper schematic.Then go to best buy and ask the install technician if he can do it. depends on wether or not your door locks activate with a 12v positive pulse or ground closing. either way there will be an adapter to make it work. This is not much of an answer but hope it helps. The alarm is still working perfect in my truck 5 years later so its worth getting and a good alarm.
1175149
Thank You! Your response was very informative and give me a much better idea of what I will be faced with in trying to convert my existing system to a locking removal able power cord.
1084907
No it is the door strike only. To set up you need a doorbell type switch which is very inexpensive. When power is applied o the strike it momentarily unlatches allowing the person to push the door or gate open without the need to turn the locked door lever. I bought a 12 v power supply to supply power. It is 12 v DC current
1521748
I have a central switch that provides 12 V DC. This is the only it'll work.
947790
You should specify if you have a databus interface module such as an XpressKit XK01 or similar installed. If you do not you should have relays to power your door lock solenoids directly or you have the green and blue wires connected to your vehicles electric door lock button (Highly not recommended but possible with diode isolation on some vehicles.)If you use a relay interface you may have to increase the pulse time or do a dual pulse lock/unlock to fully engage the solenoid.If you use an interface module ensure it is programmed for the correct vehicle type and that there are no shorts or misswirings. You should specify if you have a databus interface module such as an XpressKit XK01 or similar installed. If you do not you should have relays to power your door lock solenoids directly or you have the green and blue wires connected to your vehicles electric door lock button (Highly not recommended but possible with diode isolation on some vehicles.)If you use a relay interface you may have to increase the pulse time or do a dual pulse lock/unlock to fully engage the solenoid.If you use an interface module ensure it is programmed for the correct vehicle type and that there are... » Read More You should specify if you have a databus interface module such as an XpressKit XK01 or similar installed. If you do not you should have relays to power your door lock solenoids directly or you have the green and blue wires connected to your vehicles electric door lock button (Highly not recommended but possible with diode isolation on some vehicles.)If you use a relay interface you may have to increase the pulse time or do a dual pulse lock/unlock to fully engage the solenoid.If you use an interface module ensure it is programmed for the correct vehicle type and that there are no shorts or misswirings. « Show Less