!!!Working fuel gauge, FOR REAL this time!!!

[RagingMain]

Would this be the same thing.
I drew it like I have it wired up with the wires going in out through the light
The only thing I can see with this set up is when the thermresistor starts passing enough current to light the light it wont light up the LED just go around it through the lower resistance resistor???
Obviously this will allow for the circuit to work as normal but may not light the light.

 

[Bon-86]

Yup, that's the circuit. but, no, the current won't bypass the LED because of the parallel resistor. Rather, the current will flow through both circuit paths without regard for the one another. This is because each of the parallel paths are connected to the supply voltage and ground. They don't actually interact at all.

You will still have a very small current for your LED and a much greater current (in parallel) that simulates the original bulb.

 

[RagingMain]

Okie Dokie. Will give it a try
Thanks for your help and
WELCOME to the forum
If it works I will be very very happy :th_party33::th_peace:arty:

 

[Bon-86]

Cool. Radio shack has a 50-ohm 10-watt wire wound power resistor for $2. Good luck; Interesting project!

 

[RagingMain]

Ok, why dont I just wire, solder, heatshrink in the orginal bulb in parallel???
That should do the same thing and not have to guess on the resistance, power yada yada yada

 

[Bon-86]

That'll work. The bulb will eventually burn out though (100 years?) and will be hotter than the resistor. But it'll work!

 

[satariel666]

Ok, why dont I just wire, solder, heatshrink in the orginal bulb in parallel???
That should do the same thing and not have to guess on the resistance, power yada yada yada

Some guys here use to make that trick when instaling the LED's indicators.
Its not proffesional but it works! unk:

I second Bon, do as he said.

 

[RagingMain]

Yup, that's the circuit. but, no, the current won't bypass the LED because of the parallel resistor. Rather, the current will flow through both circuit paths without regard for the one another. This is because each of the parallel paths are connected to the supply voltage and ground. They don't actually interact at all.

You will still have a very small current for your LED and a much greater current (in parallel) that simulates the original bulb.

Since I am not an expert at this by any means I still have some questions.
Sorry if I seem exasperating.

The total resistance of a circuit with resistors in parallel is equal to.. see image inserted.
Here is a link as well
http://www.wikihow.com/Calculate-Series-and-Parallel-Resistance

The total resistance of a 470 ohm resistor in parallel with a 50 ohm resistor now becomes 45.192 ohms.

So they do actually interact to form a different resistance in the circuit.
when resistors are in parallel, there are many different means to an end, so the total resistance will be smaller than each pathway. The total resistance is always smaller than the smallest contributor for a parallel circuit, which when we do the math you can see that 45<50

Now this will allow the fuel pump relay to work properly but it will also cause too much current for the LED and cause it to pop. Unless of course I am completely way out in left field.

 

[Bon-86]

No you're not exasperating but sometimes electronics can be. I'd forget applying the parallel resistor formula in this circuit because the LED changes things a bit.

Roughly speaking the current (I) for the Led and its resistor is:
I = (Vs - Vled) / Rled where Vs is around 12 volts, Vled is the voltage drop of the LED when it is flowing current and Rled is the current limiting resistor you're using to avoid frying the LED.

The current flowing through the 50 ohm resistor is:
I = Vs / R where Vs is around 12 volts and R is the 50 ohm resistor.

You need to consider the two circuits described above as two separate circuits. The additional current drawn by the added 50 ohm resistor doesn't flow through the LED but rather it flows "around it".

The sum of the above equations is your total current in the new circuit. The current in the LED is about 0.012 Amps and the current through the 50 ohm resistor is .24 Amps. The current due to the 50 ohm resistor is 20X greater than the contribution of the LED. This is why your fuel circuit doesn't work right without the contribution of the 50 ohm resistor or bulb.

I hope this helps and doesn't just muddle things for you.

 

[satariel666]

No you're not exasperating but sometimes electronics can be. I'd forget applying the parallel resistor formula in this circuit because the LED changes things a bit.

Roughly speaking the current (I) for the Led and its resistor is:
I = (Vs - Vled) / Rled where Vs is around 12 volts, Vled is the voltage drop of the LED when it is flowing current and Rled is the current limiting resistor you're using to avoid frying the LED.

The current flowing through the 50 ohm resistor is:
I = Vs / R where Vs is around 12 volts and R is the 50 ohm resistor.

You need to consider the two circuits described above as two separate circuits. The additional current drawn by the added 50 ohm resistor doesn't flow through the LED but rather it flows "around it".

The sum of the above equations is your total current in the new circuit. The current in the LED is about 0.012 Amps and the current through the 50 ohm resistor is .24 Amps. The current due to the 50 ohm resistor is 20X greater than the contribution of the LED. This is why your fuel circuit doesn't work right without the contribution of the 50 ohm resistor or bulb.

I hope this helps and doesn't just muddle things for you.

That would be Kirchhoff rules.

 

[RagingMain]

That would be Kirchhoff rules.

Yeah I understand Kirchoff.
My only question still is, when the level sender finally allow enough current to activate the relay to shut down the pump so we have to go to reserve will there be enough flow through the LED to light it up if there is a lower resistance resistor in parallel with the LED???

 

[Bon-86]

Yes. The 50 ohm resistor makes the reserve function work. The LED is only for you to see when this happens.

 

[satariel666]

Yeah I understand Kirchoff.
My only question still is, when the level sender finally allow enough current to activate the relay to shut down the pump so we have to go to reserve will there be enough flow through the LED to light it up if there is a lower resistance resistor in parallel with the LED???

Yes.
LED itself as an semiconductor takes very little current to make semiconductor structure to conduct. And You need to limit that "little" current so by the same time theres nothig you can suply behind the led in the same circut.

By bypassing the led with resistor you can suply something behind the led.
So exactly what Bon said will work fine.


I will have the same problem as i plan to utilize the stock fuel sender.
Thes is a very simple way to make so called "ajustable relay".
4 resistors and one operational amplifer working as an comparator.
If you want i can create one system.

 

[RagingMain]

Yes.
LED itself as an semiconductor takes very little current to make semiconductor structure to conduct. And You need to limit that "little" current so by the same time theres nothig you can suply behind the led in the same circut.

By bypassing the led with resistor you can suply something behind the led.
So exactly what Bon said will work fine.


I will have the same problem as i plan to utilize the stock fuel sender.
Thes is a very simple way to make so called "ajustable relay".
4 resistors and one operational amplifer working as an comparator.
If you want i can create one system.

I was looking at that route as well but I wanted to use off the shelf components. I like the idea but for accuracy I would use more than 4 resistors. Just my .02

Thanks for the advice on the LED. Now I just have to go pull it all apart again :bang head:

 

[satariel666]

I was looking at that route as well but I wanted to use off the shelf components. I like the idea but for accuracy I would use more than 4 resistors. Just my .02

Thanks for the advice on the LED. Now I just have to go pull it all apart again :bang head:

Good luck and keep us posted.

 

[RagingMain]

Update
Ok if you guys are going to use a 3 terminal sender, Pos, Neg, Send you are going to have to wire your power supply and negatives for the gauge and the sender to the same spots.

In other words
I used an add a fuse and spliced in 2 wires, one going to the sender and one going to the power for the gauges, so they are seeing the exact same voltage.

Also the ground for the sender and the gauges needs to be grounded to the same spot.

The range for the gauge is only 1.6 vdc to 3.95 vdc. Even if there is a .5 vdc difference between either the supply or the ground it will give false readings on the gauge and or can scramble the program in your sender.

dont ask me how I know this :bang head::bang head::bang head:

 

[satariel666]

Update
Ok if you guys are going to use a 3 terminal sender, Pos, Neg, Send you are going to have to wire your power supply and negatives for the gauge and the sender to the same spots.

In other words
I used an add a fuse and spliced in 2 wires, one going to the sender and one going to the power for the gauges, so they are seeing the exact same voltage.

Also the ground for the sender and the gauges needs to be grounded to the same spot.

The range for the gauge is only 1.6 vdc to 3.95 vdc. Even if there is a .5 vdc difference between either the supply or the ground it will give false readings on the gauge and or can scramble the program in your sender.

dont ask me how I know this :bang head::bang head::bang head:

Not that simple as it was supouse to be, huh?

 

[RagingMain]

Not that simple as it was supouse to be, huh?

Yeah, R&D is never simple. Then again I am a rookie at electronics so I am making some rookie mistakes.

At least I am learning all the tricks so if anyone else does this they can learn from my mistakes :clapping:

 

[RagingMain]

Well another set back
The 50 ohm resistor did not do the trick to allow enough current for the relay to keep the fuel pump on.

Will have to try a lower one.
Think I will try an 8 ohm resistor just like the original bulb

 

[satariel666]

Yeah, R&D is never simple. Then again I am a rookie at electronics so I am making some rookie mistakes.

At least I am learning all the tricks so if anyone else does this they can learn from my mistakes :clapping:

Thatn is the essence of electronics. It never goes as it was planned...

Well another set back
The 50 ohm resistor did not do the trick to allow enough current for the relay to keep the fuel pump on.

Will have to try a lower one.
Think I will try an 8 ohm resistor just like the original bulb

I wonder why?
What wires size did you use?