skiier
Electrical
- Nov 12, 2003
- 140
Unusual question here.
I want to sense when a car lighter is on. A car lighter is a self contained simple unit. When you press the lighter "in" you "load" a spring while engaging the contact points of the lighter resistance element to 12VDC and in turn this causes the lighter element to heat up. This heat causes expansion and the unit moves to a point where the loaded spring releases and pops the lighter out. Easy enough right?
There is no way to sense a voltage change at the lighter as the lighter is self contained. The +12V and -12V leads to the lighter never change state.
I also think that if you placed anything in series with the lighter - a resistance of some sort say to create a voltage divider - you might reduce the voltage to the lighter to such a magnitude that the lighter will not function properly mechanically. Power developed in the lighter element being the inverse square of voltage. Hence a 30% or 3.6V reduction in voltage to the lighter element means 1/2 the original power to the lighter element.
Also keep in mind the lighter resistance is about 1.0 ohm. Any sensing element placed in series would have to be much smaller than 1.0 ohm for the lighter to function properly.
So basically I don't want to mess with the lighter properties I just want to sense current flow.
So what I was thinking was using a power transistor as a switch. The transistor will reduce the voltage at the lighter slightly but I am hoping to use the base of the transistor as a switch while offering the lighter a low resistance path thru the collector in this application. If you think of the collector driving the base instead of the base driving the collector you will understand what it is I am doing.
I connect the 2 leads of my lighter to the emitter of my transistor cct. and to GND (-12V).I connect the collector directly to +12V. I know the original current draw of the lighter is 8A@12V so I can calculate the lighter resistance R=12/8= 1.5ohms.
Now I assume a certain votage drop at the lighter that I hope will allow the lighter to work properly. I need this voltage to overcome the base voltage drop and allow me some level of voltage in the base cct that I can use to drive some sort of relay. Let's say my lighter will work properly at 10V. So my emitter current is now 10/1.5=6.66A. If my transistor had a gain of 100 then my base current would be about 6.66/100=6.6mA. A resistance in the base cct would have a voltage drop of 12-0.7-10=1.3V.
So what I need is a relay that operates at or around 1.3V and draws close to 6mA with an impendance of 1.3/0.006=216ohms. Any thing commercially available?
Yea right! This is where I become stumped. Is there an easier way to do this. I want simple thank you. Just simple electronic components. No complex ccts needed or required thanks. You get the idea.
Any way to make the base drive something else?
Any help would be appreciated. I do not want to modify the lighter in ANY WAY. I mean any way. The lighter has to operate as normal. A replacement lighter must also work in the same receptacle.
I want to sense when a car lighter is on. A car lighter is a self contained simple unit. When you press the lighter "in" you "load" a spring while engaging the contact points of the lighter resistance element to 12VDC and in turn this causes the lighter element to heat up. This heat causes expansion and the unit moves to a point where the loaded spring releases and pops the lighter out. Easy enough right?
There is no way to sense a voltage change at the lighter as the lighter is self contained. The +12V and -12V leads to the lighter never change state.
I also think that if you placed anything in series with the lighter - a resistance of some sort say to create a voltage divider - you might reduce the voltage to the lighter to such a magnitude that the lighter will not function properly mechanically. Power developed in the lighter element being the inverse square of voltage. Hence a 30% or 3.6V reduction in voltage to the lighter element means 1/2 the original power to the lighter element.
Also keep in mind the lighter resistance is about 1.0 ohm. Any sensing element placed in series would have to be much smaller than 1.0 ohm for the lighter to function properly.
So basically I don't want to mess with the lighter properties I just want to sense current flow.
So what I was thinking was using a power transistor as a switch. The transistor will reduce the voltage at the lighter slightly but I am hoping to use the base of the transistor as a switch while offering the lighter a low resistance path thru the collector in this application. If you think of the collector driving the base instead of the base driving the collector you will understand what it is I am doing.
I connect the 2 leads of my lighter to the emitter of my transistor cct. and to GND (-12V).I connect the collector directly to +12V. I know the original current draw of the lighter is 8A@12V so I can calculate the lighter resistance R=12/8= 1.5ohms.
Now I assume a certain votage drop at the lighter that I hope will allow the lighter to work properly. I need this voltage to overcome the base voltage drop and allow me some level of voltage in the base cct that I can use to drive some sort of relay. Let's say my lighter will work properly at 10V. So my emitter current is now 10/1.5=6.66A. If my transistor had a gain of 100 then my base current would be about 6.66/100=6.6mA. A resistance in the base cct would have a voltage drop of 12-0.7-10=1.3V.
So what I need is a relay that operates at or around 1.3V and draws close to 6mA with an impendance of 1.3/0.006=216ohms. Any thing commercially available?
Yea right! This is where I become stumped. Is there an easier way to do this. I want simple thank you. Just simple electronic components. No complex ccts needed or required thanks. You get the idea.
Any way to make the base drive something else?
Any help would be appreciated. I do not want to modify the lighter in ANY WAY. I mean any way. The lighter has to operate as normal. A replacement lighter must also work in the same receptacle.