Electronics gurus please help!!!

[jimbote]

OK...In my never ending quest to fully integrate my tacoma TDI swap I have come across yet another issue that requires the input of a electronics guru.... on my swap I'm using ecm Pin 24 of 121 for the CEL signal straight to the toyota's CEL....one side of the CEL gets 12v from the truck harness the other side relies on a ground signal from the VW ecm... what I have discovered is that using this pin causes a reverse operation of the CEL ...ie: when I have a DTC the bulb is NOT lit ...when I have no DTC's the bulb is lit!! ...funny huh? ...so what I need is an electronic circuit or a component that would act much as a two position or five terminal n/o, n/c relay to reverse the signal and give me a CEL at the appropriate time ...any input would be greatly appreciated and acknowledged in the swap forums....tia!!

 

[where2]

I'm torn between suggesting a reed relay to reverse the actuation you are seeing now, and a diode to stop potential back flow causing the bulb to illuminate when you don't expect it.

If you have a digital multi-meter and can do some testing, this may help sort out the best route to take.

With the bulb disconnected, and the engine running, do you see +12V on the Tacoma wire?

With the bulb disconnected, do you see any DC voltage on the ECM pin #24 wire?

 

[jimbote]

I'm torn between suggesting a reed relay to reverse the actuation you are seeing now, and a diode to stop potential back flow causing the bulb to illuminate when you don't expect it.

If you have a digital multi-meter and can do some testing, this may help sort out the best route to take.

With the bulb disconnected, and the engine running, do you see +12V on the Tacoma wire?

With the bulb disconnected, do you see any DC voltage on the ECM pin #24 wire?

thanks for the quick response... ok with engine running and bulb disconnected I get 13.4v on the truck side (as expected).... on the ecm side I get about 10mv..... I read up on the reed switches earlier and there was some mention of them becoming magnetized and remaining closed...would this be a consideration in my application or is this after many thousands of cycles?

 

[greengeeker]

Personally I would just use relay.

 

[jimbote]

Personally I would just use relay.

I've thought of using a relay but two things come to mind...the current required by the windings may be too much for the ecm's driver and the voltage spike from the relay (think ignition coil voltage) when the relay shuts off...I guess the spike can be handled with a diode..... i'm thinking there is a way to make a transistor do the same thing but again I'm woefully ignorant when it comes to eelectronical deevices

 

[jimbote]

just found this.. http://www.allaboutcircuits.com/vol_3/chpt_5/2.html ... but I'm having a hard time visualizing the actual component assembly.... JFET transistor?...

 

[Alchemist]

A JFET would work just fine, but for ease and simplicity, I would use an NPN bipolar. Do you know how much current the MIL needs, or the lamp type number? The circuit is very simple, just one transistor and two resistors. If I knew how much current the lamp required, I could calculate the resistor values and select the transistor.

Paul

 

[jimbote]

A JFET would work just fine, but for ease and simplicity, I would use an NPN bipolar. Do you know how much current the MIL needs, or the lamp type number? The circuit is very simple, just one transistor and two resistors. If I knew how much current the lamp required, I could calculate the resistor values and select the transistor.

Paul

thanks Paul...the bulbs are tiny #74 and around 1w ...so the NPN transistor will cut current to the bulb when current is applied to the NPN? ....EDIT: excuse my ignorance but did you mean a PNP? which is normally closed and opens when current is applied?

 

[Alchemist]

No, NPN is the one to use. All transistors are off unless biased on by Base current. Apply +12 V to lamp, connect other side of lamp to Collector, ground Emitter. Add resistor from Collector to Base and connect Base to ECU MIL pin. The Collector to Base resistor will bias the transistor on and light the lamp, unless the ECU has grounded the Base. When the Base and Emitter are connected together, the transistor is cut off and the lamp is out. I would start with a 2N2222 transistor and about a 6K8 1/2 watt resistor. The same circuit will work with a PNP transistor, but on a + Ground vehicle.

Paul

 

[jimbote]

No, NPN is the one to use. All transistors are off unless biased on by Base current. Apply +12 V to lamp, connect other side of lamp to Collector, ground Emitter. Add resistor from Collector to Base and connect Base to ECU MIL pin. The Collector to Base resistor will bias the transistor on and light the lamp, unless the ECU has grounded the Base. When the Base and Emitter are connected together, the transistor is cut off and the lamp is out. I would start with a 2N2222 transistor and about a 6K8 1/2 watt resistor. The same circuit will work with a PNP transistor, but on a + Ground vehicle.

Paul

Headspin ....OK, I think I have it....another trip to radio shack is in order ... so let me reiterate :
12v to one side of lamp
collector to other side of lamp
emitter to ground
6k ohm from collector to base
Base to MIL pin 24 of ecu.... let me know if I have it straight.... thank you so much the TDI swap world will be forever grateful!!

 

[Alchemist]

Pretty close. Just one small point 6K8 means 6.8 K ohms, which is a standard value. It's a short form to use the multiplier prefix as a decimal point. So 5M6 would be 5.6 Megohms etc.

Paul

 

[jimbote]

Pretty close. Just one small point 6K8 means 6.8 K ohms, which is a standard value. It's a short form to use the multiplier prefix as a decimal point. So 5M6 would be 5.6 Megohms etc.

Paul

...now I feel like I'm in high school all over again....thanks for the correction! ...learning alot today!! ...EDIT...I just drew this circuit out and now it makes perfect sense and also helps in my understanding of how transistors work in a circuit....going to breadboard it manana and test it out!

 

[jimbote]

OK...I've given this some more thought and bear with me...so when the ecu grounds the base it's turning "off" the transistor so no more current flows to ground through the emitter, but now there is a path for 12v through the resistor to the ecm @ pin 24...I'm assuming that 6.8k ohms is sufficient to keep current low and the ecm from short circuiting ? EDIT!!...I guess this was kind of a dumb question considering the bulb itself has much less resistance and can pass much more current than a 6.8k ohm resistor....

 

[DPM]

does the CEL circuit go straight to the bulb, or is it possible there's an inverting circuit in the cluster already? Airbag and ABS lights tend to be configured this way.

In this case, it might be easier to just bypass that circuit...

 

[greengeeker]

...now I feel like I'm in high school all over again....thanks for the correction! ...learning alot today!! ...EDIT...I just drew this circuit out and now it makes perfect sense and also helps in my understanding of how transistors work in a circuit....going to breadboard it manana and test it out!

Jim, can you share your drawing?

 

[jimbote]

does the CEL circuit go straight to the bulb, or is it possible there's an inverting circuit in the cluster already? Airbag and ABS lights tend to be configured this way.

In this case, it might be easier to just bypass that circuit...

the 1996 tacoma cluster is a dumb cluster so the CEL output goes directly to the lamp ....I kinda figured that on some VW's in other markets or for industrial app's this is how these outputs are utilized, of course the glow lamp and tach function perfectly normally

Jim, can you share your drawing?

sure, let me redraw it a little bit better my initial is somewhat "sloppy"

 

[Alchemist]

This is what I would do. If you want detailed theory of how it works, just say so and a dissertation will follow.

Paul

 

[jimbote]

photobucket was three minutes too slow!! ^^
EDIT!!!! This is a corrected diagram I've deleted the previous sorry for any confusion !

 

[jimbote]

This is what I would do. If you want detailed theory of how it works, just say so and a dissertation will follow.

Paul

dissertation would be appreciated

 

[greengeeker]

dissertation would be appreciated

X2....

 

[Alchemist]

This is what I would do. If you want detailed theory of how it works, just say so and a dissertation will follow.

Paul

After a bit more thought I realize this circuit may not work as it is drawn. It may be necessary to connect the resistor directly to a switched 12 V on the source side of the lamp instead of the the switching side.

An NPN transistor is normally in a non-conductive state between the Emitter and Collector. When current flows between Base and Emitter, the transistor allows current to flow between Collector and Emitter. The Collector current can be much larger than the Base current which allows the transistor to amplify. In the case of the 2N2222 the ratio of Base to Collector current is 100:1. This means 2 ma of Base current will permit 200 ma of Collector current. The 6.8K resistor will allow a Base current of 2.1 ma @ 14.5 V, causing the transistor to conduct and light the lamp. When the ECU grounds the Base, no current can flow in the Base-Emitter circuit and therefore no Collector current can flow, extinguishing the lamp.

 

[jimbote]

thanks for the clarification and circuit edit.... now does the resistor need to be connected directly @ the bulbs 12v source or will any 12v source @ the cluster be sufficient? and does the resistor need to be connected at the source because the resistance the bulb adds cuts down on the base current, thus not "opening" the transistor gate?

 

[whitedog]

Paul, I'm an electronics dummy, so I don't understand what the resistor part of the circuit is for. In a typical relay, pin 30 is constant power, 87 and 87a are switched power out, 85 and 86 are signal and ground for the solenoid. (I think I have the right from memory)

So with your drawing, 30 is the wire from the light, 87 is the wire to ground, and 85 would be the wire from the ECU. So where does the resistor come in? Maybe that is part of your dissertation?

 

[jimbote]

also when the ecm "grounds" the base current (providing an alternate path) is it essentially bleeding off the amperage required by the base to open the junction between C and E?

 

[jimbote]

Paul, I'm an electronics dummy, so I don't understand what the resistor part of the circuit is for. In a typical relay, pin 30 is constant power, 87 and 87a are switched power out, 85 and 86 are signal and ground for the solenoid. (I think I have the right from memory)

So with your drawing, 30 is the wire from the light, 87 is the wire to ground, and 85 would be the wire from the ECU. So where does the resistor come in? Maybe that is part of your dissertation?

I "think" the resistor provides adequate resistance to keep the lamp from lighting when the circuit @ pin 24 is closed, yet allows enough current for the base to activate or "open" the transistor allowing current to freely flow from bulb to ground (light on) when the "switch" @ pin 24 is open.... please correct if I am wrong...

 

[whitedog]

So the resistor part is like the pin 85 on a regular relay providing power to the "solenoid" and the ECU wire is pin 86, providing ground?

 

[Alchemist]

Jim

The 12 V source can be any one that is switched with ignition. When the ECU causes the MIL pin to go to ground, it shorts the Base-Emitter so the current path is from 12 V source through the resistor then to ground preventing any Base current.

Whitedog

Think of the Collector as terminal 30 of a relay, Emitter as 87, and Base as 85 with 86 grounded. The resistor limits current to about 2 ma and the voltage drops to about 0.6 (This is a characteristic of Silicon semiconductors.) at the Base. I don't have the exact value, but if the Base current exceeds about 15 ma there will be a smoke escape! Because the resistor limits current, it is possible to apply power and ground to the Base without letting the smoke out. A conventional relay could do the same job if 30 connects to the lamp, 87a is grounded, 85 connects to 12 V and 86 connects to the ECU.

 

[jimbote]

thanks for the clarification...I'm starting to understand much better now.... a conventional relay would probably be too much load for this circuit correct?...coil loads and coil collapse spikes may kill the ecm? ....if this circuit works I'll build another for my oil pressure lamp as I'm using a relay @ the moment to reverse the signal between the VW sender and the toyota cluster...

 

[whitedog]

OK, it makes sense on how to hook it up.

When the ECU causes the MIL pin to go to ground, it shorts the Base-Emitter so the current path is from 12 V source through the resistor then to ground preventing any Base current.

That's the part I'm confused about.

 

[jimbote]

OK, it makes sense on how to hook it up.



That's the part I'm confused about.

the base current required to open the the collector to emitter is tiny so when the ecm interrupts or diverts the current by "grounding" not enough flows from the resistor to to "B" base to keep the valve between "C" and "E" open.... yeah I guess you can think of "B" as a valve that opens the main channel of current through the transistor ...the main channel (collector/emitter) is what carries the CEL current from 12v to bulb to ground