Calibrating sensors to your Traqmate

#1

After 2yrs of fooling with this issue, I’ve had a real breakthru.

Here’s what doesn’t work.
The TM manual encourages you to use the 5V source on the TM. I won’t go into the details, but what ends up happening is that the TM will only see a voltage swing of about a volt and that’s not enough for accurate data. And getting it calibrated so that it’s reasonable accurate is a lot of trial and error.

Here’s the first thing that works. Gauges with analog outs, like Stack. Connect the analog out to the TM, calibrate it with the info that Stack provides, and it’s pretty damned close to spot on. I’d dumped enough time into this that seeing how close this idea came to perfect in the first try was a serious “holy shit”.

Here’s the second thing that works. Get a US Ohm range sensor (240-33 Ohms), not VDO range (0-180 Ohms) and plumb it in. Then buy part #418-EC5BE03 from Mouser. It’s a 15V power supply that can handle a wide range of supply voltages. That is to say…no matter what your car feeds it, it puts out 15V. It’s about the size of a matchbox.

A VDO range sensor doesn’t work. Even trying it is risky because at 0 Ohms, current demands will spike and something could melt, burn, fail, etc.

Then get a big 10W, 10Ohm resister and wire it in series with your sensor. That will ensure that your 15V supply always sees a decent amount of resistance so you won’t run into trouble because current demands spike.

What this will do is to create an honest 5V swing for your sensor, instead of a 1V swing if you had used the TM’s regulated 5V supply. My second holy shit moment, after looking at the data, was realizing that my first attempt at calibrating data with this was damned close to right. In contrast, my first attempt using the TM’s 5V supply, 2yrs ago, were not even in the ballpark.

My set up.
I have coolant and oil temp Stack gauges sending their analog out’s to the TM. But my OP gauge is mechanical and it’s probably going to stay that way. So I have a separate OP sensor, US Ohm range, and it is fed by the 15V supply via the 10W/10 Ohm resister.

#2

I figured out a couple weeks ago why what jlevie said about teeing a resistor into my signal wire and hooking it into my 5v source worked. It is half of a bridge circuit and the other half is inside of my USM. I enter the ohms of my installed resistor and it calibrates itself. If I knew the resistance of the internal resistance on the other side of the bridge I could pick a resistance that would give me a full 5v swing.

#3

I have no idea what you just said. Signal wire for what? What is a USM? Enter ohms into what? Internal resistance?

#4

I have no idea what either of you said. I think I will go with some big ass gauges and mount them in view of my camera like JTower and call it a day.

#5

0-180 ohm sensors should work fine with a Trackmate. You do need a current limiting resistor though. One end of the resistor goes to +5v the other end goes to the junction of the data input and sensor lead. A resistor of 180-220 ohms will yield a reasonable input swing and not draw too much current.

#6

why do i always see “jtower” and “big ass” in the same sentence?

#7

jlevie wrote:

That’s pretty much what the manual says, except recommending twice that much for a resister between Vsource and analog in. The problem is that it gives you a miniscule voltage delta. What I never understood is why it also turned out to be so difficult to calibrate. Calibration shouldn’t take 6 iterations of trial and error. It should take, at most 2. And their should be an obvious relationship between the resistance measured at the TM, and the resultant graphs. Except there wasn’t. So trial and error.

$30 for an independent 15V source, and the result is about 4V of Vdelta at analog in. And then everything starts behaving predictably.

#8

There shouldn’t be any trial & error iterations in calibrating a sensor. for a linear sensor you compute the Vout at two points (usually at zero and full scale) and plug that into the data logger. For a non-linear sensor you need Vout a set of points on the curve.

#9

jlevie wrote:

Hence my surprise that it took successive iterations of comparing session notes with data to ultimately succeed. Each calibration change I made took the values farther away from what they should have been. At endstate I had calibration values that made no sense, but I had good data.

Tenacity always wins in the end.

#10

Ranger wrote:

i’m gonna have that carved on your headstone

#11

My bad. I have to read every new thread because I can’t stand having (1 new) in orange next to the topic. So I assumed that you had at least scanned over my iq3 thread a few weeks ago. The usm is simply the box that I plug my signal wires into and supplies a 5v out. The bridge circuit is what I found while searching one of my books for an answer to your problem of needing another 5v signal for your tm. I’m sure its more relevant to my electrical confusion than yours.

#12

Ranger wrote:

[quote]jlevie wrote:

Hence my surprise that it took successive iterations of comparing session notes with data to ultimately succeed. Each calibration change I made took the values farther away from what they should have been. At endstate I had calibration values that made no sense, but I had good data.[/quote]
I don’t know what you did, but things to consider are that a computed zero (or max depending on the slope of the sensor) may not be zero volts at the logger input. You have to take zero bias into account for the other data points. And understand that when using a current limiting resistor the Vout is:

Vout = Rs*Vref/(Rs+Rr)

where:
Vout is the input to the logger
Rs is sensor resistance at that cal point
Rr is current limit resistor
Vref is the reference voltage

#13

jlevie wrote:

[quote]Ranger wrote:

[quote]jlevie wrote:

Hence my surprise that it took successive iterations of comparing session notes with data to ultimately succeed. Each calibration change I made took the values farther away from what they should have been. At endstate I had calibration values that made no sense, but I had good data.[/quote]
I don’t know what you did, but things to consider are that a computed zero (or max depending on the slope of the sensor) may not be zero volts at the logger input. You have to take zero bias into account for the other data points. And understand that when using a current limiting resistor the Vout is:

Vout = Rs*Vref/(Rs+Rr)

where:
Vout is the input to the logger
Rs is sensor resistance at that cal point
Rr is current limit resistor
Vref is the reference voltage[/quote]
No offense, but I’m an engineer, not a poet. Maybe you’re explaining the details for those non-technical types that might stumble across this thread in years hence?

#14

[quote]Ranger wrote:
No offense, but I’m an engineer, not a poet. Maybe you’re explaining the details for those non-technical types that might stumble across this thread in years hence?[/quote]
No offense, but if I had a dollar for every engineering mistake I’ve encountered I could retire and investigate the affect of enforced idleness on the human condition.

What you need to straighten things out is a physicist/mathematician.

#15

I don’t like bumping old threads but I figure it’s better to keep my post grouped with the other relevant posts.

So I think I’m going to follow Ranger’s suggestion of buying a DC-DC converter. I’m thinking of going with the 12 VDC just because 15 VDC is too much for my DAQ (0-5V).

I did some maths and decided to use a 500 ohm resistor for the voltage divider. Doing some more maths shows me that when the sensor is at 240 ohms the signal voltage will be 3.89V and when the sensor is at 33 ohms the signal voltage will be 0.74V. This is a reasonable range since my DAQ is 0-5V with an ADC resolution of 1024 divisions which results in a resolution of around 0.12 psi which is very reasonable for measuring things like oil pressure where a resolution of around 0.25-0.50 psi would be sufficient.

My next problem and this is where I need help. Calculating V^2/R for the resistor at both extremes (240 and 33 ohms for the sensor) and I find that the 500 ohm resistor is drawing 0.13 W and 0.25 W but calculating it from the DC-DC converter datasheet which shows an output current of 1250 mA I*R results in 62 W being absorbed which is quite a lot.

Which one is correct? Also, I can’t send a lot of current into my DAQ. I’m not sure what the limit is but it’s pretty low.

#16

Your an engineer. Why did you get a traqmate? Traqmate is for the nontechnical people. I thought you were going to build an arduino logger dash and I was very interested in your progress.

#17

I am. I’m trying to figure out how to wire the sensors up to it without destroying it. The wiring of the sensors is pretty much universal regardless of whether it’s a Motec datalogger, a Traqmate, or my homebrewed setup.

#18

[quote=“MrDomino” post=64429]I don’t like bumping old threads but I figure it’s better to keep my post grouped with the other relevant posts.

So I think I’m going to follow Ranger’s suggestion of buying a DC-DC converter. I’m thinking of going with the 12 VDC just because 15 VDC is too much for my DAQ (0-5V).

I did some maths and decided to use a 500 ohm resistor for the voltage divider. Doing some more maths shows me that when the sensor is at 240 ohms the signal voltage will be 3.89V and when the sensor is at 33 ohms the signal voltage will be 0.74V. This is a reasonable range since my DAQ is 0-5V with an ADC resolution of 1024 divisions which results in a resolution of around 0.12 psi which is very reasonable for measuring things like oil pressure where a resolution of around 0.25-0.50 psi would be sufficient.

My next problem and this is where I need help. Calculating V^2/R for the resistor at both extremes (240 and 33 ohms for the sensor) and I find that the 500 ohm resistor is drawing 0.13 W and 0.25 W but calculating it from the DC-DC converter datasheet which shows an output current of 1250 mA I*R results in 62 W being absorbed which is quite a lot.

Which one is correct? Also, I can’t send a lot of current into my DAQ. I’m not sure what the limit is but it’s pretty low.[/quote]

The idea of “sending current into the DAQ” is wrong. There’s a misunderstanding of electricity there. The DAQ senses voltage, that’s it.

The most current your bridge will pass is 15V/(500+33) so ~ 30mA. 30mA thru 500 Ohms is 15W so I’d get a 25W resister and fasten it to a piece of AL as a heatsink.

Get a chip that is rated at 500mA@15V so it will be nice and stable. Then fasten it to a piece of AL also.

#19

I’m just learning about this sh%% this semester. I’m a 27yo born again freshman. Your the only one who can figure it out. If your still in college ask your professor. If not just PM me and I’ll ask one of mine. I’m at Milwaukee School of Engineering right now. I think I can find someone who knows the answer.

#20

As to the power through the current limit resistor, Scott has the right answer.

I’d sacrifice resolution and use a larger limiting resistance. I can’t think of any reason that you need better than 1-2psi resolution for oil or fuel pressure. The allows for the use of a lower power resistor.

Since the DAC input is limited to 5v, I would drive the measurements with a 5v regulated source. That would eliminate any risk to the DAC should the resistor get shorted. That is how the analog inputs to my IQ3 are set up.