rmcdaniels

In an effort to make the hatch less stealable, I wanted to install a kill switch. I had some cheap RFID modules that I got off of ebay (I see them there for $10-$20 all the time), so I rigged it up as a kill switch. That way I don't have to have a switch that is visible or even accessible from the interior of the car. As long as the RFID tag is within a few inches of the antenna, it activates the switch. You could put it behind the dash or an interior panel, maybe by the ash tray, just toss the tag in the ashtray to drive the car. I won't go in to what to kill with the switch, just don't make it the fuel pump, kill multiple items necessary to start the car (you will want a manual with a wiring diagram of the car), run your wires inside the OEM wiring harness, physically separate the RFID controller/antenna/relay, and back it up with a paging system. My goal isn't to make the car impossible to steal, just to notify me if someone is trying to steal it and slow them down long enough for me to arm myself and go make their acquaintance. The units that I got didn't have an output section, aside from a TTL output, so I rigged up a transistor to act as a switch for a relay. It appears to be working well, although mine is overheating a bit because my transistor is too small (trying to run a relay that takes 2.5 watts with a transistor that can handle 1.8 watts), but a bigger transistor costs about a dollar, so I'll fix it later. I may also try a smaller relay, the large/heavy-duty automotive relays draw much more than a regular electronics relay, and none of the stuff that I am killing requires anywhere near the 30 amp relay that I am currently using.

For anyone not familiar with RFID, there is a small antenna in the car that constantly looks for a tag, and I have a small tag that looks like a plastic key fob. The tag gets it's power from electromagnetic energy radiated from the antenna (so no batteries for the tag), so when it gets in range of the antenna (about four inches), the tag transmits a code to the RFID controller. If it transmits the correct code, then the RFID controller deactivates my kill switch. Each RFID tag has a unique code, and I can program my controller for which tags I want it to recognize.

I'd include pictures of my antenna mounting and details of how I wired the kill switch, but that would be stupid.

wrx-killer-Sti-eater

iTrader: (1)

good lookin

nsxxtreme

automotive relay should be about 800mA. How are you calculating wattage? You should not need a very large transistor.

You should be using the RDSon of the transistor to calculate power through it.

1fastVX

Nice idea, though all the RFID tags on Ebay seem to be quite large. Would you post up a pic of yours?

Thanks,

John

rmcdaniels

Yeah, the transistor is rated for 800mA, but it still gets really hot powering a 180mA (66 ohms at 12v) relay and a .5W resistor on the base also gets really hot during operation. I did notice that it would not operate with a 1K resistor on the base, I had to put a 150 ohm resistor on it before I could get it to turn on, so maybe I just have too much base-emitter current, although this is supposed to be a medium power switching transistor, so I don't know why it needs that much current, the Hfe is 200, so it really shouldn't need that much base-emitter current. It's been years since I biult any basic circuits, so I'm not ruling out something stupid on my part at this point. Thyere is an LED in parallel with the transistor, although taking it out of the circuit had no effect on the operation of the transistor. I should put it in series with the load though, just to make everything correct. I need to to some more work on the details, but the concept seems to work at this point. I probably should put the LED in series with a very small relay inside the RFID controller case, then use that output to provide a 12v output. That would be the cleanest solution.

My RFID tags are 1.25"x1.25"x.25"

02 accord

Post up a schematic of your transistor circuit and I'll take a look at it.

rmcdaniels

I noticed that with a larger resistor on the base I can get it to work if I tap the relay to activate it, then it will stay on and not heat up. I wonder if the initial current to activate the relay is higher than the nominal current to keep it activated, and I'm having to put too low a resistor on the base to get it to turn on. If that's the case then I could use a very small board-mounted relay and just run the output of that to the larger 30A automotive relay.

Here's what the circuit looks like, currently the resistor is 135 ohm, .5w, the transistor is a 2N2222:

nsxxtreme

A BJT transistors collector to emmiter current is based on the Base to emitter current. Thats why when you change the base resistor it changes the current flow through the collector to emmiter. you also muct drop the entire 12V across the transistor in order to activate that relay. which = heat.

I would suggest using a pmos mosfet transistor or Nmos.


Modified by nsxxtreme at 3:53 PM 12/23/2007

02 accord

Yes, the relay needs more current to turn on than it does to stay on. The transistor/relay should turn on with a 10k base resistor. Measure the signal, base, and collector voltages: signal should stay at 12v, base should sit at ~0.7v, collector at ~0.2v if the circuit is working correctly.

nsxxtreme, the only time the transistor is dropping the full 12v is when it's off. At that point, the current through the relay is 0, and since P=VI, I is 0, therefore the transistor doesn't dissipate any power. When the transistor is on, the drop across it should be ~0.2v, so P = .2v * .2A = .04 watts.

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 02 accord &raquo;</TD></TR><TR><TD CLASS="quote">nsxxtreme, the only time the transistor is dropping the full 12v is when it's off. At that point, the current through the relay is 0, and since P=VI, I is 0, therefore the transistor doesn't dissipate any power. When the transistor is on, the drop across it should be ~0.2v, so P = .2v * .2A = .04 watts.</TD></TR></TABLE>

Actually it's droping it when its "on" it doesn't drop anything when its off. So the only time it drops the entire 12V is when it first turns "on" .The relay pulls the voltage up to 12V. When the transistor turns "on" it must drop 12V across the transistor in order to pull the collector to .2-.4v and thus activating the relay.

To determine the heat that is disapated into the transistor you need to know the RDSon resistance. This is the resistance internal to the transistor when its fully turned on. So the equation we are looking for to determine the power is I^2 *RDSon.

We could use the final voltage to calculate power but its somewhat of a guessing game. Its better just to use the RDSon given in the data sheet. Many ways to accomplish it, nmos,pmos,BJT. Mosfets function better as switches which is why I recommended using one. The mosfet also wont represent much of a load to the driving device.


Modified by nsxxtreme at 3:58 PM 12/23/2007

02 accord

The low side of the relay sits at 12v when not activated. The voltage across the transistor is 12v, because the collector is at 12v and the emitter is grounded. When the transistor is on, the collector drops to 0.2v, while the emitter is still grounded. The only voltage across the BJT is the 0.2v. The relay drops the rest of the 11.8v.

What you're saying with the transistor dropping the full 12v is only during the turn on transition. The collector voltage drops from 12 to 0.2, while the current increases, but this is insignificant in power calculations.

He's using a BJT, not a MOSFET. To determine the heat dissipated, you need to know the voltage drop across (~0.2v) and the current through (~200mA) the transistor.

Speaking of loading the source, 1mA drawn by the BJT should not have any significant effect on the output of the source.

rmcdaniels

So my problem is probably that I'm lowering the resistance on the base too much to get it to switch on, causing excessive current on the base-emitter junction. My intention is to saturate it, so emitter-collector current shouldn't be an issue. I think that I'll try a smaller relay and see how it behaves with a bigger resistor on the base.

BTW, the diagram is a little off, the anode of the protection diode is actually connected to the collector of the transistor, vice the emitter.

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 02 accord &raquo;</TD></TR><TR><TD CLASS="quote">What you're saying with the transistor dropping the full 12v is only during the turn on transition. The collector voltage drops from 12 to 0.2, while the current increases, but this is insignificant in power calculations.</TD></TR></TABLE>

Correct, on the voltage drop. Power will drop exponentially.

The power calculation is not insignificant if you want the transistor to last.

lets not also forget (14.4-.7)*.015=.20W across the resistor even though a 1/4 resistor is rated for .25watts try running one that way for a while. All calculation should be done at 14.4V down to 10V if you want this thing to work in worst case (weak battery while cranking) conditions. So if you decide to go the BJT route ensure that you have the proper base to emitter current at both voltages to properly turn on the transistor.

I'll make this really simple for you go to radioshack buy an IRF510 transistor. Put anywhere from 4-14.4 volts on the gate and be done (it wont event get warm to the touch). Also put about a 47k or &gt; resistor from the gate to source. Nothing will get warm. Its way more transistor then you need but sucky radio shack doesn't sell a wide selection of mosfets. Mosfets work better as digital switches and this is what you are trying to do. You also wont need to worry about the change in voltage. Good luck with your project.


Modified by nsxxtreme at 9:44 PM 12/23/2007

rmcdaniels

Thanks, I'll give one of those a try.

Any thoughts on the LED being in parallel with the base-emitter junction?

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by rmcdaniels &raquo;</TD></TR><TR><TD CLASS="quote">Thanks, I'll give one of those a try.

Any thoughts on the LED being in parallel with the base-emitter junction?</TD></TR></TABLE>

I dont see any current limiting resistor in your diagram for the LED. Which means you will either burn out the LED or the device driving the LED. An LED is just a diode. Which in all practicality its a short from VCC to ground. You use a resistor to limit the current. (14.4-.7)/r=(current rating of LED).

If it were me I would put the LED in parallel with the relay coil. With a resistor in series with the LED. Let the transistor do all the work.

I might add one other benifit to using the transistor I recomended you can ditch your extra diode going from the collector to emmiter. The transistor I recomended was designed to drive inductive loads it already has the diode built into it.

Let me try explaining your heated transistor in another way. When you don't supply enough base-emitter current the transistor partially turns on. Meaning most of the voltage is droped across the transistor. The only way to prevent this is to supply enough current to immediatly saturate your transistor so it's fully turned on.
Your BE current requirements are related to your collector-emitter load. Meaning the larger load you use the more BE current you will need to turn the transistor on. If its getting hot and you have not exceeded the transistors rating (your dumping the voltage across the transistor) then you are probably not dumping enough current into the Base-emitter. Lower the resistance of your base resistor. Not knowing your load I would gestimate a 2k resistor should do the job. You can use a BJT if that's what you really want to do. I dont want to mislead you, there are many ways to accomplish the same task. I would choose the mosfet for various reasons.

I would still go the mosfet route. Mainly if the battery is week and this thing try to turn on its likely to overheat and burn out.


Modified by nsxxtreme at 5:53 PM 12/24/2007

rmcdaniels

There is a resistor in line -w- the LED, I just didn't include it in the drawing.

I finally broke out a 12v power supply and breadboarded the circuit. I got it to work (without heating up) with a smaller relay and the BJT with a 1K resistor on the base, but it was unstable when the voltage dropped from cranking the starter.

The MOSFET fixed everything, rock solid with any voltage, pretty much anything on the gate works, no discernable heat, good stuff.

Thanks for all the help, I think this thing should work now. I'll update if it gets flaky or anything.

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by rmcdaniels &raquo;</TD></TR><TR><TD CLASS="quote">The MOSFET fixed everything, rock solid with any voltage, pretty much anything on the gate works, no discernable heat, good stuff.

Thanks for all the help, I think this thing should work now. I'll update if it gets flaky or anything.</TD></TR></TABLE>
That was my original concern as voltage varies a BJT would not be the best choice. It's why I recomended the mosfet to begin with.

Glad you listened.
Make sure you put a 47k or greater resister between the gate and ground. This will drain any charge on the gate. If not you risk leaving this device on because the charge has no where to go.

And Happy New Year I am in New York. Crowd here is nuts!!!!!!!!!!!!!!!!!!!

blundar

if you have any dead honda ECUs laying around, rob a SK5050, SK5151 or SK5154 from one of them. They're designed to be driven by TTL logic and can switch a pretty good deal of power at 12v - 8/10/15 Amps if I remember right. Been awhile since I looked at the datasheet. They'll be in any automatic or vtec ECU from 92-00/01

and they're made by Allegro / Sanken if you need to track down the datasheets btw.

ShaunRR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by blundar &raquo;</TD></TR><TR><TD CLASS="quote">if you have any dead honda ECUs laying around, rob a SK5050, SK5151 or SK5154 from one of them. They're designed to be driven by TTL logic and can switch a pretty good deal of power at 12v - 8/10/15 Amps if I remember right. Been awhile since I looked at the datasheet. They'll be in any automatic or vtec ECU from 92-00/01

and they're made by Allegro / Sanken if you need to track down the datasheets btw.</TD></TR></TABLE>

Hmm, interesting.

rmcdaniels

I gave one of my extra (I bought ten of them) RFID setups to a local guy who got it to recognize an OEM Honda RFID key.

It would be interesting if someone could program code recognition into a Honda ECU, as this is really made to read RFID tags and output the code, I'm just jury-rigging it to drive kill relays.

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by blundar &raquo;</TD></TR><TR><TD CLASS="quote">if you have any dead honda ECUs laying around, rob a SK5050, SK5151 or SK5154 from one of them. They're designed to be driven by TTL logic and can switch a pretty good deal of power at 12v - 8/10/15 Amps if I remember right.</TD></TR></TABLE>

Why hunt for and destroy an ecu when you can go buy an off the shelf transistor like the one I mentioned above and drive the same loads? Every transistor will burn out if not used correctly.

People run into problems when using a transistor because its not a relay and does not function like one. So if you don't think about the application it will be used in you will run into problems like the original poster. There are many things that need to be thought about.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by rmcdaniels &raquo;</TD></TR><TR><TD CLASS="quote">It would be interesting if someone could program code recognition into a Honda ECU, as this is really made to read RFID tags and output the code, I'm just jury-rigging it to drive kill relays.</TD></TR></TABLE>

So why dont you integrate this into an ECU? Use your newly aquried transistor to switch on something inside the the ECU? You dont need a relay. Relays should only be used for 30 amps or above, and even them you dont need them. Relays are for weenies

rmcdaniels

I'm running an AEM EMS, so it's even easier, but RFID key recognition would make it interesting vice using a switch. Essentially I'm using a transistor as a kill switch, and activating it via a RFID key. I'm thinking something more along the lines of a factory RFID system where the ECU is useless without the right key.

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by rmcdaniels &raquo;</TD></TR><TR><TD CLASS="quote">I'm running an AEM EMS, so it's even easier, but RFID key recognition would make it interesting vice using a switch. Essentially I'm using a transistor as a kill switch, and activating it via a RFID key. I'm thinking something more along the lines of a factory RFID system where the ECU is useless without the right key. </TD></TR></TABLE>

Then integrate this device into the AEM EMS. A circuit is a circuit no matter where its located. Just physically move this circuit inside the EMS. Place the acuall RFID sensor outside the ECU.