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Ohm’s Law


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captainzab 
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Posted: October 10, 2006 at 11:16 PM / IP Logged  

There was talk about using Ohm's Law to calculate resistance for negative pulse unlock wire for car keyless entry that requires resistors.

[b]My question[/b] how can u use ohm's law (I = V/ R, i = amperes, v =  volts, r =  resistance) be used to get  the amperage drop across a neg pulse wire?

You know what R is because thats the resistor you are placing on it.

V=? its not 12v is it? if so, why would it be 12v.

The senario is having that [b]neg 200ma through a resistor than having the amperage drop. [/b]

Someone said that you can use ohm's law to figure out your amperage for that scenario above

Havent used ohm's law since highschool physics class

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master5 
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Posted: October 11, 2006 at 12:27 AM / IP Logged  
Ah, I remember you. Hope all is well. Anyhow see if I can help,perhaps you are misunderstanding something. I was not using ohms law to caculate the resistance needed for the doorlocks. I was using it as an example prove that there will be no issues with excessive current draw, which I believe you were initially implying. The reason E or (V) if you prefer is 12 is because we are installing into cars which at least for now and the last 50 years or so, are 12 volt. In the examples I was using it is a constant. since the voltage will always be 12 in this case, the only variables are the resistance, which in turn effects the current. The formula is  I (current) = E or V (voltage)/ R (resistance). I don't really use ohms law much myself other then to prove points but when I was teaching electronics I had it coming out of my ears. If you read back to the original postings it will explain what I need to do to determine the proper R value to trigger lock/unlock so no need to re state that here. I used 100ohms and 1000ohms as an example, the exact resistance needed to operate the doorlocks vary from year to year, vehicle to vehicle, and can be + or - as well. However, the values are almost always in that range, one is 3 digit, the other is 4. The directions that come with the alarm, remote start or relay module come with a list that tells you the resistance needed, most of the time accurate but not 100%. The reason they recommend a relay is also explained in a previous post but in my experience it is not mandatory. If someone wishes to only use relays, or they require a diagram for everything they do, so be it, no sweat. The original post asked about if it can be done without relays and I explained how. It works and has no problems and I have been doing vehicles this way for years. Most of the time I do use relays but if i can get the R value in one or 2 tries, and the other one works 80% of the time without a resistor (the internal resistance is by chance correct to lock or unlock in the databus, then it is in my experience, quicker, more reliable (no moving parts) and less cost for parts. diodes are mandatory when doing it this way as you have 2 wires feeding into one and in addition the doorlock outputs usually switch +/-  If you have any more questions on ohms law i'll be glad to help as I will never forget it after teaching for 6 years. Good evening.
Mad Scientists 
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Posted: October 11, 2006 at 4:55 PM / IP Logged  

There is no such thing as amperage drop.. there's a law that states current is equal in all parts of the circuit..

Correct me if I'm wrong, but the door lock circuit you're talking about is a single wire and when a resistor is added to the wire it will either lock or unlock the door, depending on the value of the resistor?..

What I suspect they are doing is using a voltage divider circuit to determine whether to lock or unlock the door. Imagine that there is a 500 ohm resistor inside the control box. When you hook up a 100 ohm resistor to the wire there will be a specific voltage between the two resistors.. to calculate it, divide system voltage (12v) by total circuit resistance (500 ohms + 100 ohms = 600 ohms). The current in this circuit will be 20 mA.

Now, multiply the current with the 500 ohm resistor.. (0.02 * 500).. it's 10 volts. This means that the 500 ohm resistor will drop 10 volts, so the voltage at the point between the two resistors is 2 volts.

Using the 1000 ohm resistor, the total circuit resistance would be 1500 ohms, divided by 12 volts gives us a current of 8 mA. 8 mA times 500 ohms gives us a voltage drop of 4 volts, which means that the voltage at the point between the two resistors would be 8 volts.

So, if you write the software that says 'if voltage measured is lower than 4 volts, unlock the door, and if voltage measured is over 6 volts, lock the door. you can figure out what range of resistors would work.

Hope this helps.. ask if you have more questions..

Jim

captainzab 
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Posted: October 11, 2006 at 9:00 PM / IP Logged  
Does everyone thinks about ohm's law and other variation of it when yall guys are doing alarm/audio install?
Cuz i just do it without thinking, sorta unconciously,
unless it doesnt work the first time, then i start using reasoning to figure out the problem (dont run into much problem, and when i do it a stupid problem)
Note: You Always Dont Get What You Pay For.
Mad Scientists 
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Posted: October 13, 2006 at 8:25 AM / IP Logged  

That is correct no-use.. it's referring to a 97 Sebring Convert.. we bought one a couple years ago as a toy.. the remote start RKE alarm wasn't working properly so I took it apart and fixed it right.. Scotch blocks etc up under the dash.

I don't do car audio as a profession.. but I do work with electronics. I could put a bunch of letters in my sig. line, but I don't see a need to; I prefer to let my advice speak for itself. What do you call a med student that just barely graduates?.. 'Doctor'. The letters don't mean nothing.

Jim


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