I’m worn out had to work late. I will get this done soon, I promise.

 

OK, so can you:

For each gauge:

1. With the gauge disconnected from sender, measure the voltage, using your DM20, at the gauge connection for the sender to ground.
2. While still measuring the voltage at the gauge, use the analog meter to measure the current from gauge to sender. Voltage should drop, record voltage & current.

Repeat for BOTH gauges connected together.

View attachment 244627

Finally got some accurate numbers to both gauges. G is oem gauge and G2 is the add on gauge.
G Disconnected from sender = 11.85V
Current across sender lead and G = 42MA
current testing = 2.27V (big voltage drop)
in same order for G2

G2 disconnected = 5.03V
Current G2 = 51MA
G2 Voltage while current testing = 2.66V

I had to cut the sender lead to oem gauge. There was no way I could get that lead pinned out. But that allowed me to get accurate readings long as the old Simpson is accurate. What next? Can we work with this? I know voltage is much different but voltage and current aren’t too far apart? Thanks much fellas.

 

Looks like they are using different ways of measuring:

11.85/42 mA = 52 ohms
5.03/ 51 mA = 54 ohms
and getting the the same result.

I would not be surprised that the 52 ohms is ratiometric to the supply voltage.

This is "thinking out loud" where you don't necessarily come up with an answer. You just brainstorm.

It makes it real messy, I think. i did find this https://www.hicomponent.com/ohm-to-...eter-resistance-ohm-to-voltage-converter.html, a 0-500 ohms to 0-10V converter. Not sure if it's of any use.

I think you can use it with a voltage to current converter and get the 5V gauge to work.

Can you use a 4-quadrant multiplier, https://www.analog.com/en/products/ad734.html#product-documentation and get the other one. Feed it V and I (from an I-V converter) in a feedback loop via a FET to get R.

The circuits get really messy and you need a bipolar supply.

You may just be able to measure current in the 5V system. V is fixed. feed that into the 4-quadrant multiplier in a feedback loop.

Ignore the resistance to current converter.

Then there is the idea of putting a microprocessor in there like a www.picaxe.com, but you have no floating point arithmetic.

Cubloc http://comfilewiki.co.kr/en/doku.php?id=cubloc:index has floating point. You can always use an I2C D/A converter.

So, put the sensor in the 5V system and use a feedback ammeter to measure I. For the motorcycle measure both I and V. I with an I-V converter (no voltage drop). So, you effectively measure R using 5/I and then make the product of V and I of the second system to equal R.

Good news, A sinngle supply except for a possible slightly negative bias voltage generator from TI. https://www.ti.com/product/LM7705?keyMatch=LM7705

Vos and Ib are the biggest issues with I-V converters. You can always center around Vcc/2
See MCP3551 http://www.t-es-t.hu/download/microchip/ds21841c.pdf#page=8

These are only some starting points. Hopefully there are much better ideas.

let's define two types of A/D converters.
1. Uses a fixed reference like 5V. 0-5V maps to 0 to 1 Bar.
2. One uses the supply voltage/2. i.e whatever it happens to be. This is how car sensors work. it's not 5V = 1 Bar. it might be 100% is defined as 90% of the supply voltage and 1 bar is 90% of the supply voltage. It is .9*5.00 IF the supply voltage is 5V. If it happenes to be 5.25V at the time of the measurement then 0.9*5.25 is 100% or 1 Bar.

car sensors cannot reach zero and cannot reach 5V without a complex power supply.


With the 5V sensor, V is constant, so it's easy to measure k*R.

The other sensor likely depends on both V and I, so they both need to be measured and V/I divided and made equal to k*R above.

Itt can be done in real time with an 4-quadrant multiplier or with a microprossor possibly with floating point.

52.156862745098​

54.0476190476​

 

One simple way out of the predicament, I think is;

Measure the voltage across the sensor using the 5V gauge, Below a certain voltage you have low fuel.
> that number, you keep the digital gauge in place, Below that number let the engine computer AND your guage see see one out of 5 resistors.

e.g. It would supply a resistor to your guage and the same resistor or just a low fuel value to the bikes computer. You would supply 5V to the sensor and use a comparitor to select what resistor to substitute to your guage.

So, basically from low-fuel to full, you will use your gauge and tell the bike the tank is full.
When the fuel is low, you will measure the sensor with your own 5V supply.
Then window comparitor and select 1 of 5 resistors to the digital guage and a low fuel resistor to the bike.

the OEM guage, you don;t care about anyway, so make it read full or low-fuel.
Between low and full, use the digital guage and set the OEM gauge to1/2 full with a resistor.
At low fuel, you add your own 5V supply and measure the voltage. You tell the OEM computer that your low on fuel. with a resistor.

To keep things indicatingon the digital meter, use the voltage that you measured and supply a resistance to the digital guage for whatever graduations you have.

maybe all you have to do is switch in a low fuel resistor to the OEM sensor. So, just switch in 1/2 full and full to the OEM sensor based on the voltage of low-fuel.

You don;t have to solve the problem of two gauges. OPTOMOS or PHOTOMOS relays could be used. They are opto FETs.

I "think out loud" so you know by thought process.

it's always good to do this and notice how a complex problem gets really simplified. I think the desired outcome is not two guages reading the same. It;s to keep the low fuel indicator functioning.

There could be reasons why it won;t work like RANGE calculations or something that does soemthing with fuel vapors at low fuel level.

if the bike ECM can deal with 1/2 tank and low fuel your done. Your digital guage is independent of the bike's ECM.

 

One simple way out of the predicament, I think is;

Measure the voltage across the sensor using the 5V gauge, Below a certain voltage you have low fuel.
> that number, you keep the digital gauge in place, Below that number let the engine computer AND your guage see see one out of 5 resistors.

e.g. It would supply a resistor to your guage and the same resistor or just a low fuel value to the bikes computer. You would supply 5V to the sensor and use a comparitor to select what resistor to substitute to your guage.

So, basically from low-fuel to full, you will use your gauge and tell the bike the tank is full.
When the fuel is low, you will measure the sensor with your own 5V supply.
Then window comparitor and select 1 of 5 resistors to the digital guage and a low fuel resistor to the bike.

the OEM guage, you don;t care about anyway, so make it read full or low-fuel.
Between low and full, use the digital guage and set the OEM gauge to1/2 full with a resistor.
At low fuel, you add your own 5V supply and measure the voltage. You tell the OEM computer that your low on fuel. with a resistor.

To keep things indicatingon the digital meter, use the voltage that you measured and supply a resistance to the digital guage for whatever graduations you have.

maybe all you have to do is switch in a low fuel resistor to the OEM sensor. So, just switch in 1/2 full and full to the OEM sensor based on the voltage of low-fuel.

You don;t have to solve the problem of two gauges. OPTOMOS or PHOTOMOS relays could be used. They are opto FETs.

I "think out loud" so you know by thought process.

it's always good to do this and notice how a complex problem gets really simplified. I think the desired outcome is not two guages reading the same. It;s to keep the low fuel indicator functioning.

There could be reasons why it won;t work like RANGE calculations or something that does soemthing with fuel vapors at low fuel level.

if the bike ECM can deal with 1/2 tank and low fuel your done. Your digital guage is independent of the bike's ECM.

Looks like I could run the oem gauge lead to a resistor so it would just read full all the time. That way it doesn’t flash low fuel. Strange how same fuel sensor can work both gauges independently but one gauge has 5v on its sender input and the other has 12v on its sender input yet they read the same. Why is it when both gauges are connected to sender at same time, both gauges drop down to just above 3/4 tank from full? Would a diode in each gauge sender input lead stop this?

 

What might be happening is that your paralleling a voltage source with a current source. That would make the resistance appear lower.

Can you get another current reading (You can use a fixed resistor) at a different fill level?

IF the current is the same for all sensor positions, we can possibly subtract it off. See https://www.ti.com/product/LM334

i.e. subtract 42mA. Put the two terminal device from the OEM out to ground. It could go anywhere, but that may be the best place.

Anywhere::= across sensor, across OEM output to ground and 3 across 5V sensor and ground. They are all in the same place.
If the current sink is on the OEM, then if you disconnect the OEM, the 5V gauge would read OK and the resistance would be in range.
If you put it on the 5V side, then your like returning it to stock when the digital gauges are removed.

Analog Devices makes this: https://www.digikey.com/en/product-highlight/l/linear-tech/lt3092-current-source part.

 

You didn't do a measurement with them paralleled?

 

You didn't do a measurement with them paralleled?

It appears not, it would be useful.

Looks like they are using different ways of measuring:

11.85/42 mA = 52 ohms
5.03/ 51 mA = 54 ohms
and getting the the same result.

Not sure you meant that, but:

G: 2.27/0.042 = 54.05ohm
G2: 2.66/0.051 = 52.15ohm

The apparent difference in sender resistance from the two gauges is 'interesting', as is the dfference in voltages. The first gauge is taking directly off the 12v feed, through a 220ohm resistor (thus empty would be roughly 6v assuming the sender reads ~210-240ohm and is probably comparing the voltage ratiometrically with a voltage divider of 1/2 supply volts. This is roughly analogous to the split coil approach.

G2 is clearly using a stabilized 5v supply via a sense resistor (maybe 47ohm?) giving ~20mA at empty, or it could be a constant current feed of 50mA (giving 11.65v at empty).

 

I won't quibble over 2 ohms. OK ~4%.

 

You didn't do a measurement with them paralleled?

I did what was requested in that last message Irving posted. I didn’t realize the parallel reading was necessary. What are we going to measure with both gauges hooked up? Voltage for each gauge to sender and current from both gauge leads through Simpson? Something is definitely dropping when in parallel. If we could just somehow amplify the drop back to original whether it be voltage, current, or whatever is dropping, then we could just keep it wired in parallel. For example: if voltage drops on both gauges when hooked in parallel, is there a device or could an op amp put boost both voltages just enough to make it back up to original voltage you would get when only one gauge was connected? Thoughts?

 

I did what was requested in that last message Irving posted. I didn’t realize the parallel reading was necessary. What are we going to measure with both gauges hooked up?

We don't really know. Your measurements at one sensor point are not really conclusive either.
Just hoping both together would verify that the OEM system is a current source which we hypothesize it is.
We don;t have two points, and that would help too. Using two resistors would be fine.

Injecting current, makes the guage read lower. i.e. an effective lower resistance.

I'm proposing based on the evidence so far, that if you siphon off 42mA with either suggested part and a resistor, it might work.

open circuit voltage (open sensor) and short circuit current(shorted sensor) also gives us information. We didn't ask for that either. We're assuming a shorted sensor would not damage anything.

There is an "output impedance" associated with the OEM gauge. It can also be measured "indirectly: by measuring the "open circuit voltage" and connecting a potentiometer with R much greater than the sensor and adjusting for 1/2 what you measured open circuit. The resistance of the removed potentiometer is the input Z of the OEM guage.

The fuel gauge might act differently if the battery voltage was 8V. If the OEM used a ratiometric technique, the battery voltage would not matter, you would still read the right tank level. With 5V, that voltage is much lower than expected.
8V might occur when cranking the car.

The current sink might not work if the OEM sytem uses a "ratiometric technique", but it would probably be close enough.

You have about 4 proposals and we're trying to analyze a "black box".

We don;t know if the OEM gauge will be used "at the same time" or always visible and you would like it to be correct?

The "current sink" could work, but not if it had to vary with battery voltage.

Not knowing much of anything, I said it's likely you can get the resistance knowing the current through the sensor using the digital gauge, You need a low voltage drop to do that and an I-V converter does that with <1mV drop.

Voltage controlled resistors https://en.wikipedia.org/wiki/Voltage-controlled_resistor are not easy to come by.
By measuring I and V and manipulating the gate drive of a JFET, you can close loop simulate a resistor.
That's another proposal, but the design is "messy".

 

You didn't do a measurement with them paralleled?

Not angry that it wasn't done. It's like 20/20 hindsight. There could be useful information in that test.

 

Not angry that it wasn't done. It's like 20/20 hindsight. There could be useful information in that test.

Of course. I appreciate everything second you guys spend trying to help me figure this out. There is an online manual for this bike and it has a small section on testing fuel sender and maybe gauge. I’m gonna try to find the link and post it. It may help you guys by giving some insight as to how it operates. My thought is that if both gauges work independently, but each drops equally when in parallel, then it seems logical if we could simply raise the sensor lead resistance by a fixed amount to compensate whatever the loss is when both connected, then it should work. We know the sensor is changing the gauges by resistance. The manual says that and gives the value range it’s supposed be. So, keeping it simple, let’s get a reading while both connected and do the calculations to determine how much resistance change we have. If the gauges read higher with more resistance then it’s obviously lowering the resistance when both are connected, maybe not the sensor itself, but both gauges are ultimately getting lower resistance at the gauge lead. Maybe due to adding the second gauge 5v creating the resistance drop. If so, would the diode idea prevent the voltage from traveling to the sender by only allowing the gauge to receive and not send any power to the circuit? If that doesn’t work, how could we manipulate the sensor resistance? Say we are at 240ohms full, we drop 40ohms (for example only) when 2nd gauge connected, so how can we raise sensor a fixed 40 ohms but sensor still operate through out the range, we just changed the range from say 233 to 30 or whatever to 273 to 70. When both gauges connected it’s now back to 233 to 30 ohms at gauges. Thinking out loud. That will require figuring out how much R changes when both gauges are connected. Thoughts?

 

Here: https://www.tanksinc.com/pdf/MeterMatchInstructionManualTanksInc.pdf is something I'm worried aboutL

Voltage Correction Most gauges are connected to a voltage regulator in the instrument wiring so that they get a constant voltage whether the engine is running or not. However, if this is not the case, when you calibrate your gauge at one voltage (for example, engine not running, battery at +12.5V), then start the vehicle and the voltage rises (for example to +13.8V), the gauge reading might change significantly. If you run into this, MeterMatch has an automatic voltage correction feature that corrects the reading for different voltages. To enable this feature: (1) remove power from the MeterMatch, (2) set the rotary switchѲ to position "0" (normal operating), then (3) press and hold the Up button while applying power to the MeterMatch. Hold it until the LED blinks after power is applied. MeterMatch will remember this setting in future use. If you wish to remove the voltage correction feature, repeat the same process, but hold the Dn button rather than the Up button. Note that if you change this setting, the gauge may need to be recalibrated with the new setting.

This is the "ratiometric technique" disguised. We don;t really know how the OEM gauge operates.
1) A current source would have the same current at most battery voltages.
2) Something doing a ratio-metric technique would not.

 

I apologize in the delay at times for me testing the gauges. I’m a health care provider for the last 4 years and covid crap is keeping me working 13 to 14 hours almost every day. Didn’t go to college until age 37. I used to do HVAC when young, then a mechanic running my own garage for years. I can understand most of the basics, but dang, you guys are flipping brilliant. No joke. It amazes me how much you guys know. Way smarter than me.

 

There are two pages here. Check out both. One is sender, other is oem gauge

 

Guessing:
You have a 5V voltage source and a resistor for one. The aftermarket measures current.
Old school.

You have an OEM guage that when paralleled made the gauge resistor smaller and lowered the "tank level"
It's injecting current and reading voltage. By "sinking" that 42 mA away, it MIGHT be all you need, BUT if the current varies with battery voltage the game gets played differently.

I was probably about 10 YO and we had problems with a 1968 chrysler and it did blink. There was a ammeter type movementum a box in the dash and I put a meter on it so we could use it on a 250 mile trip. It definately blinked. I think the meter had to be replaced, but the tank got dropped.

I has an 82 Celica and the gauge was very viscous, so You could not use it to estimate when filling the tank There was a self-heating thermister for low fuel. It was an independent system. I had no problems with the system,

With electronics, you can remove the gas tank irregularities and even get estimated range information. The sensor is linearized for the gas tank and thus the car computer has a better estimate of MPG and gallons left in the tank and it can estimate range.

Before that I was fixing lawn mowers. Dad taught me. I was fixing TV's and radios too. I didn't exactly know what I was doing and didn't have the money for a full restoraton.

There is a 1957 Motorola console AM/SW/Phono I want to restore. I managed to axquire the FM section without tubes or shields.

My cousin has a T-bird and the sensor is removable from the trunk which is cool. The fuel pump in a 2000 Impala is behin the back seat.

I paid $25.00 for my first car, a 1965 Ford Galaxy 500. It was wrecked and I wasn't driving at 15 YO. I learned to replace body panels, do body work (I even etched the metal), I replaced body panels, replaced the brake system, took the steering colum out to fix the horn.

Then fun stuff like remove the heads and had them refinished, Upgraded the ignition to electronic. Rebuilt my second carb.
The 1982 was the worst carb to rebuild. 120 parts and had to drill out the mixing screw caps and caarb body.

Work was a solar energy research lab with lots of toys to play with and fix. The Scanning Electron Microscope broke, can you fix it? The e-gun (Electron Beam evaporator power supply) is broke, can you fix it? Certain things I couldn't fix. Certain things I could and certain things I made better. We worked with a spare/swap mentality where I might fix or might not,.
Power supplies had the same problems (capacitors replaced). The e-gun always had power supply issues until I rebuilt it. There were vacuum guages that had similar problems - negative supply filtering and bypass caps. I was able to keep some spare parts.

It's nice when a company creates a "spare kit". Most of the IC's needed for an Amray SEM were boxed. Adding EBIC (Electron Beam Induced Current) measuremnts was fun, Designing a Faraday cup and interfacing to the SEM. This was a time when stuff was for the most part repairable and parts were available from HP and Tektronix. Now, if something breaks you have to buy 1/2 an instrument for the repair cost.

Another company, Digital Equipment Company, I learned the hard way to not touch their boards, but they had a program where if you could determine what board was faulty, you could send it for repair. You would not get the same one back.
That was a nice system. There were two of us that could troubleshoot to the module level. I learned on my owm.
A system consisted of:
A backplane and power supply.
A CPU card
Memory Cards
Bus grant cards for unused slots
Disk/Floppy controllers
Boot ROM
Terminator

You needed multiple books, schematics, diagnostics but it was easy if you had a computer you could rob from.
replacing fans, checking the power supplies. You could have ab ABAB backplane or an ABCD backplane. Cards functionality got combined. This was a Q-bus system for PDP 11/2, 11/23 or 11/23+ systems.
I'd briefly helped work on a PDP 11/50. It had a maintenece contract. One problem was particularly irritating - it dropped a bit in the processor status word

 

Here's a possible solution.
Do we have a list of the voltage on each gauge at various different fuel levels?
e.g What is the voltage across the sender with gauge 1 connected for a full tank, and what is the voltage with gauge 2 connected instead.
Same again for empty and a few more points.
Then we can plot gauge 2 vs. gauge 1, and apply linear regression.
Then we can go back to @Irving 's circuit of post #8, and use the values obtained from the linear regression to set the gain and offset of the op-amp, so that the output translates the gauge 1 voltages to gauge 2, and thus give a reading which is as close as we can manage within the limits of a straight-line approximation.

 

Imagine taking two ohmmeters to read a resistor.
OLD SCHOOL: You had a 1.5V battery for the ohms scale and a selected resistor. You had to calibrate full scale and zero everytime you used it. My meter was huge and had three 9V batteries in parallel.

New school: A single 9V battery and little or no calibrations. One had an ohms adjust to remove up to 20 ohms of lead resistance. That was rare.

The technologies were different.

 

One is a constant current source and the other is a series resistor. So in one case the voltage has a linear relationship to sender resistance, and in the other the voltage follows part of reciprocal curve, but the gauges are not renowned for their accuracy, and linear regression will give a least-squares fit from one to the other, which, considering all such things as the shape of the tank will probably be accurate enough.