# A bit tricky..

Discussion in 'General Electronics Discussion' started by wcs61, Mar 29, 2013.

1. ### wcs61

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Mar 29, 2013
..for me that is but should be a simple fix for some of you. I will try and explain what I am trying to do here without the vast knowledge of circuits and dialect.
What I what to fabricated is a 12v switching relay that is activiated by exactly 160ohms. This 160ohms will also continue to control another device at the same resistance since I will be splicing into the existing circuit for my coil feed.
So I believe what I will need is some sort of relay in which the coil is activated by 160ohms. Once activated the coil will close a set of contacts which in turn will allow 12volts to flow through the closed contacts to lets say a 12volt lamp @ 10amps+-.. Also I will need to maintain the resistance of the coil feed circuit @ 160ohms.
Any ideas? I'm sure this is a simple fix but again I am not that experienced with these types of circuits. Any help will be greatly appreciated!

2. ### Harald KappModeratorModerator

10,820
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Nov 17, 2011
Ohm is a unit for resistance. A relay isn't activated by a resistor, but by a current (or voltage).
Volt is a unit for voltage. Volts don't "flow". It is current that flows. Current is measured in Ampere (A).
Obviously it doesn't help not to have an understanding of the "dialect".
Instead of mixing units and twisting our brains, tell us what you want to do. It looks like your main idea is to light a 12V/10A light. Where does the signal come from that turns the relay on or off? What has all this to do with 160 Ohm?

Last edited: Mar 29, 2013
3. ### shrtrnd

3,783
499
Jan 15, 2010
Yeah, you're going to need to specify what you're working with.
It sounds like you're using an AC relay to run an inductive circuit (though the circuit
is given to you in ohms, I would assume they mean impedance in an inductive circuit).
And you're using that to run a DC lamp circuit.
There's just too many unknowns with the information you've provided so far, to give you
Can you be more specific? A circuit, or make/model of the devices you're working with?

4. ### wcs61

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Mar 29, 2013
Guys I am working on Automotive circuits therefore no AC current is involved.
What I have is a transmission gear position sensor which provides certain resistance I guess for each gear the transmission is in. I say guess because the wiring diagram I located for this shows resistance at certain levels per gear. If I read correctly,the digital readout or instrument cluster works off resistance and for specific values will post a different indicator letter per gear on the cluster. Perhaps I can locate a link to the wiring diagram and post here.

5. ### BobK

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Jan 5, 2010
If it truly is a pure resistance, AND, it is not connected to anything else, you can check for 160 Ohms by simply running a known current through it and testing the voltage with a comparator. If the sensor is already connected to something else, like the instrument cluster or the engine computer, you will have to determine what the other circuit is doing with it, you can no longer just treat it as a resistance.

Bob

6. ### wcs61

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Mar 29, 2013
Exactly Bob, I do not want to damage the cluster by just simply connecting wires and grounds to the switch!
Have a image and switch schematic I will post onto photobucket then share.

http://i413.photobucket.com/albums/pp215/rolltide777/ATV/internet/switch_zps54f666a9.jpg

Last edited: Mar 29, 2013
7. ### Harald KappModeratorModerator

10,820
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Nov 17, 2011
You're on the wrong track.
You can't use a specific resistor to set a relay in action - at least not if the different resistors are that close in value.
You need to sense the resistance by an electronic circuit and activate a relay.
As Bob suggested:
Run a known current through the resistor box, e.g. 10mA. If the switch is in the 160Ohm position, the volatge across the box will be V=R?I=160Ohm*10mA=1.6V.
Use a window comparator that is active if the voltage is e.g. between 1.5V...1.7V. The output of the windows comparator can drive a relay.

You should take note of Bob's typ about the circuit being connected to other circuitry. Sending your own current through the sensor will interfere with the other components and possibly endanger your safety when driving the car.

8. ### wcs61

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Mar 29, 2013
With the cluster being involved and no schematic or manual to tell me what is happening here I would only be guessing. On the switch you can see that the B ( BN brown) wire is going to a ground terminal with multiple other grounds and the A (White wire) going back to apparently the cluster. (The actual full diagram does not show where the termination is for this White wire).
I would think the white is simply a ground wire from the cluster. If assumed correctly, wouldn't the switch only be a multi position switch providing different levels of resistance to ground from the cluster?
As for sending other current into this circuit... no I will not be adding current into this circuit. I only want the resistance value to be noticed and somehow through a totally separate circuit activate a 12V relay which will have it own power source.
Is it possible to wire a separate circuit for the relay, and also have it set up to activate at 160ohms resistance?
Tapping into the switch lead will provide the 160 ohm resistance when in the correct gear?

Last edited: Mar 30, 2013
9. ### BobK

7,682
1,686
Jan 5, 2010
You cannot measiure a resistance without sending current thorugh it.

If the resistance it already connected to the instrument cluster, it is probably sending current through the resistor. In this case, most likely one of the wires will be at ground and the other will show a different voltage depending on what gear is selected. You can use this voltage in a window comparator to determine the gear.

Try measuring each of the wires with a voltmeter and switching gears.

Bob

10. ### wcs61

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Mar 29, 2013
So if there is voltage changes I will be able to fabricate some type of switching system and hopefully not change the current properties of the original circuit. Very well I will take some readings on the two wires and get back later. Later being in about 4 weeks due to work related delays. Thanks Bob.

Last edited: Mar 31, 2013