Houston.. I got problem!

[Garrinn]

Hi there

(I did move this post to this area)

I'm new here and just as good as average beginner in electronics. I'm a old programmer and done a lot of heavy stuff in that department. I know simple terms and understand somehow some basic concepts as transistors, relays, resistors, diode, capacitors and can I have made some simple logic gate's when I was younger.

The problem I'm facing is I think a simple one for you guys. I have a a circuit for motorcycle to show what gear it is in, i.e. Gear Indicator. Ok. I bought this very cheap from China. Every thing works as it should on the bench.. that is, until I test it in the bike.

The neutral wire was supposed to return 0v when in neutral (no gear), otherwise it returns 12v (for every other gear setting's) which is the main voltage on the bike. The problem is, the neutral sensor is is not returning 0v when the bike is in neutral. It's returning 0.5 something volts which is not low enough for the gear indicator circuit to trigger neutral. I haven't bother to contact the manufacturer but I did some simple resistors to ground check and with I think with 10ohm resistor I got 0.22V (if I remember right) and that did trigger the gear indicator properly.. but of cause that kind of resistor allows to much power drain and not usable when the bike is running in gears and the signal is 12V.

So.. I must add some extremely simple way to kind of "mask" all low voltage to almost zero without affecting 12v. That is, at least all voltage below 1V should be masked down to as close to 0V (0.1-0.2V seems to be enough)

Any idea how to do that without drawing current when signal is 12V?

[danadak]

Have you got a rough schematic, even hand drawn, you can post ?

A wire that is grounded, and still presents a voltage, eg. the .5 you are
seeing, tells you either ground contact ohmic, not low enough resistance,
or that there is a lot of current flowing in the ground wire and that X parasitic
resistance develops the .5 V.



Regards, Dana.

[PA0PBZ]

A diode drops 0.5 - 0.7 volts when conducting...

[MrOmnos]

Post schematic.

[Garrinn]

Have you got a rough schematic, even hand drawn, you can post ?

A wire that is grounded, and still presents a voltage, eg. the .5 you are
seeing, tells you either ground contact ohmic, not low enough resistance,
or that there is a lot of current flowing in the ground wire and that X parasitic
resistance develops the .5 V.



Regards, Dana.

Thanks Dana

I think this sensor doesn't ground, it only drops voltage as low as it can (0.5v)

Here is a picture of this sensor:
 

I do however recall some kind of voltage drop circuit. Perhaps like this?

[Garrinn]

Hi there

I did some google and found this side about "Universal Gear Indicator"

Found this under section "Possible Improvement"

In the current design, when the neutral switch is open (there is a gear on), there appears to be a very small current (< 0.5 mA) sinking through R3, due to the voltage difference between the neutral switch connection (TO_POWER-4) and the microcontroller. If the neutral indicator is of LED type (not a resistor bulb), there is a possibility that it stays dimmed, instead of being completely off. In that case, a small switching diode (1N4148) can nicely replace R3 (on the same PCB) in order to block this small incoming current when the neutral switch is open, as shown in the figure below :

[PA0PBZ]

That is exactly what I suggested an hour ago... 

[Garrinn]

That is exactly what I suggested an hour ago... 

I think you are misunderstanding something.. this quote I used is of the page I linked to is not what I'm using and I'm not designing this circuit, I bought this item from China in sealed structure or box as can been seen in this image.

[PA0PBZ]

Ok, let me be more specific: Add a diode to the neutral wire like this   ------wire-------->|-- , it will 'subtract' 0.5V from the voltage on the wire.

[Garrinn]

Ok, let me be more specific: Add a diode to the neutral wire like this   ------wire-------->|-- , it will 'subtract' 0.5V from the voltage on the wire.

And what will happen when the wire transfer 12v ?

[PA0PBZ]

Ok, let me be more specific: Add a diode to the neutral wire like this   ------wire-------->|-- , it will 'subtract' 0.5V from the voltage on the wire.

And what will happen when the wire transfer 12v ?

Then 11.5v will end up at the other end, enough to do the job.

[PA0PBZ]

Ok, let me be more specific: Add a diode to the neutral wire like this   ------wire-------->|-- , it will 'subtract' 0.5V from the voltage on the wire.

Never mind, I just realised that the neutral wire is pulled to ground, it is not supplying any voltage. Forget what I wrote. 

[Garrinn]

Ok, let me be more specific: Add a diode to the neutral wire like this   ------wire-------->|-- , it will 'subtract' 0.5V from the voltage on the wire.

And what will happen when the wire transfer 12v ?

Then 11.5v will end up at the other end, enough to do the job.

Ok.. I'm not sure I understand this to full.
 
Ok, let's say I will buy 12v diode 0.5A wich will withdraw 6W when the voltage is 12w and drop voltage to 11.5V. The real question is, will that voltage drop still be 0.5V when the wire has total 0.5V also?

I have always thought that diode was kind of resistor. And voltage drop therefor would be linear with voltage.. but as I said, I'm a beginner!

[PA0PBZ]

I have always thought that diode was kind of resistor. And voltage drop therefor would be linear with voltage.. but as I said, I'm a beginner!

See above, I just realised that the diode will not work. Look at the diode as a one way device, current will only flow in one direction and it needs 0.5v to open the port If you reverse the positive and negative it will stay closed.

[Garrinn]

I have always thought that diode was kind of resistor. And voltage drop therefor would be linear with voltage.. but as I said, I'm a beginner!

See above, I just realised that the diode will not work. Look at the diode as a one way device, current will only flow in one direction and it needs 0.5v to open the port If you reverse the positive and negative it will stay closed.

Hmmm... if it needs 0.5V to open and (withdraw power) how on earth would that be usable to drop 0.5v to close to zero which is my main goal?

This sensor is loaded 12v regular by this one wire. When it detects neutral, it voltage drop to 0.5v That drop isn't enough for this gear indicator, if it would drop the voltage to 0.2v it would be enough, as I explained in my first post.

(Modified). And reverse current isn't any part of my problem, so resistor would be enough.. but as I described in the orginal post, it would withdraw to much power because it have to be so "small" to lower 0.5V to 0.2V

[PA0PBZ]

Yes, like I already said I realised a bit late that the wire was not supplying any voltage so the diode trick will not work.
You probably have to look at the circuit that detects the voltage and make a modification there, or you could try the diode trick in the wire going from the indicator to ground. That way you will raise the point where the indicator thinks is ground by 0.5v

So, ground wire from indicator box ------------------>|----------------- ground.

[jm_araujo]

As I understand the neutral light is pulling current from the switch and increasing it's voltage to 0.5V.

Raising the ground of the gear indicator can create a lot of other problems.

The simplest without drawing too much current I think is using a relay. Coil to +12V and the neutral switch. Common to GND,  NO to gear indicator. Use a generic small automotive relay.

[PA0PBZ]

Raising the ground of the gear indicator can create a lot of other problems.

Since this is made for a motorbike I don't think it would create a problem stealing 0.5V from the supply voltage, it's not that it is a regulated 12V anyway.
It's certainly the easiest way and it will not hurt to try.

[Garrinn]

Thanks guys for the effort.

Just to clarify.. here is simple schema I did draw.

As you can see it's only one wire to the sensor. I reckon it's power is controlled from the ECU as it is only one wire. How, I don't know.

The problem is I need almost zero volt from this sensor so my gear indicator can show me neutral (zero sign)

The sensor is always 12v when gear is 1..6 but 0.5v when gear is in neutral.

[PA0PBZ]

And this is what you will get adding the diode:

[tpowell1830]

From the picture and the description, the "sensor" is a switch that takes the circuit to ground only in neutral.
Make sure your grounding to your motor/transmission is adequate by tightening the ground lug.  This can cause a problem such as yours.

PEACE===>T

[Garrinn]

From the picture and the description, the "sensor" is a switch that takes the circuit to ground only in neutral.
Make sure your grounding to your motor/transmission is adequate by tightening the ground lug.  This can cause a problem such as yours.

PEACE===>T

Thanks!

I did check that indeed.

Why?

You see.. when I was wondering where I could get the ground for the indicator I saw two black wires (ground's) fasten to bolt on top on engine close to the neutral sensor.. the bolt was so stuck I could'nt move it easily.. so I did fetch the ground for the indicator from other position close by (from the engine also).

In fact I do not understand how single wire sensor works at all or as this one does. It has 12V when bike is in gear. Those 12V must come from some power source. When the sensor sense neutral all the sudden it got almost zero volts.. so my guess is ECU is controlling the voltage and as you can see from this schema it is rather complicated how this information is used in this start system.

[jm_araujo]

You're over-complicating things.

The motor body is all grounded, that's the second "wire" to the sensor. The switch shorts the wire to ground when in neutral.
According to that schematic,  when in gear the switch is open and will see the 12V from the connection 2-3-4-5-6-7. When in neutral it will energize the relay 6 allowing you to start the motorcycle.

That diagram is missing the connection to the neutral indicator light, that also energizes the wire (fully annotated wiring diagram in here: http://circuitswiring.com/circuit-diagrams/2/2011/06/complete-electrical-wiring-diagram-of-yamaha-yzf-r1.jpg ,neutral switch located middle-up)

[Garrinn]

You're over-complicating things.

The motor body is all grounded, that's the second "wire" to the sensor. The switch shorts the wire to ground when in neutral.
According to that schematic,  when in gear the switch is open and will see the 12V from the connection 2-3-4-5-6-7. When in neutral it will energize the relay 6 allowing you to start the motorcycle.

That diagram is missing the connection to the neutral indicator light, that also energizes the wire (fully annotated wiring diagram in here: http://circuitswiring.com/circuit-diagrams/2/2011/06/complete-electrical-wiring-diagram-of-yamaha-yzf-r1.jpg ,neutral switch located middle-up)

Thanks

Great diagram.. I own the service manual but this (vidal) remarks are missing.

And.. yes, I did suspect the sensor did shortcut the power to ground.. and by that, lower it's voltage and "drain" it's source so to speak. But had some tough thoughts about the draining part..  But that explain why I do not get absolute zero value.

So.. the problem still stands. I will try the diode method, that is, drain this wire to ground through diode.. any recommendation about size of diode? And will it lower my 0.5V to 0 as PA0PBZ suggested?

[tpowell1830]

Can you check the diode marked #7 in your circuit? This could be bad. Check all of the diodes at bottom of your drawing to see if they are shorted or closed.

PEACE===>T