Classic Car Puzzle - modern electronics a simple solution?

[rockrockmcrock]

Hi guys,

This seems to be a very widespread issue in the classic car world, but no one has found an easy solution that I can see - I thought perhaps a guru here might be able to help - the classic car community would be in your debt!

The problem:
I want to retain some old bi-metallic gauges in one of my classic cars (got to keep it original on the dash right?). These older gauges are "bi-metallic electronic" (most Smiths, Jaeger gauges pre-79 were). However modern gauges (including 'classic' modern Smiths) work in a different way (i.e. potentiometer based). This means that modern 'sensors' for gauges are mostly 'reversed' in the way they work (more on this later) and that the original sensors for the old bi-metallic gauges are out of production, hard to find and very expensive. Of course without the right signal from the sensor the gauge will not work. Is there a simple circuit to make modern sensors work with old bi-metallic gauges?

How bi-metallic gauges and sensors work:
All gauges are 10v. The gauges work by a 'sensor' which acts as a variable resistor and limits the current on the gauge circuit in line with pressure. The current heats a coil wrapped around a bi-metallic strip in the gauge with responds by moving the needle on the gauge. Low pressure = high resistance (240 ohms) => no heating of coil & gauge needle reads low, high pressure = low resistance (33 ohms) => coil is at it's hottest & gauge needle reads high. The coil is inherently damped (thermo-damping) and the gauge has a linear(-ish) response to current.

How modern sensors work:
Most sensors are 10v. Modern sensors work the opposite way round but over roughly the same range. In other words, Low pressure = low resistance (33 ohms) => gauge needle reads low, high pressure = high resistance (240 ohms) => gauge needle reads high. If you put this modern sensor on the old bi-metallic gauge setup the reading's would be reversed (i.e. the needle would read high when at low pressure and low when at high pressure.

What the circuit would need to do:
Reverse the modern sensors output - i.e. when the sensor gives high resistance, allow more current to flow (i=v/r => 10/33 amps max?) on gauge circuit and visa versa. Not sure if this is a simple thing or a very complex ask!

Sorry for what might be a stupid question but I'm not an electronics wizard - I mess about with RT programming and classic cars so I just don't have that foundation knowledge of electronics needed to get me off the starting blocks on this one...

Thanks guys for any help - do a quick google and you'll see quite how many of us classic and vintage car guys have this issue, so you really would be doing us a favour!
RRMR

 

[Harald Kapp]

It is not the resistance that is important, it is the current through the gauge. At 10V you have:

• 33Ω -> 300mA
• 240Ω -> 42mA (figures rounded)

You can build a current source as shown here. The equation can be changed from I/Vin/Rsense to Vin=I*Rsense.

Your sensors are "inverted", so you want to have:

• 33Ω -> 42mA
• 240Ω -> 300mA

Assuming Rsense=10Ω (an arbitrary value that you may change), using Vin=I*Rsense you end with:

• 33Ω -> Vin = 0.42V
• 240Ω -> Vin = 3V

A simple voltage divider can create this voltage from e.g. the 12V battery:

Using a trim-potentiometer instead of the 933Ω resistor will allow you to adjust the sensitivity of the circuit. Note that the currrent source willl need:

• a rail-to rail opamp for operation at low sense resistances
• a power transistor at the output to carry the current, possibly using a heat sink to carry away excess heat.

 

[rockrockmcrock]

Wow - thanks!

I've taken a little time to get back as I needed to learn a little bit about the concepts you mention (voltage divider, op amp etc.) - I'm afraid they're still a bit of a mystery to me, but at least I know what they do even if I don't yet understand the "why".

So you mention a current source, and give a link to a circuit that I sort of get - but I already have a regulated 10v source on the car (as do most) for all the gauges. It's a very old school sort of regulation that uses a bi-metallic strip to oscillate contacts open and closed to approximate 10v DC.

Can I just use this as a regulated current source? I've been tying myself in knots trying to work out (V=IR) the voltage divider resistances if I did (how do you know the 'trim pot' R value if you don't know the output V?)! I'm guessing the below circuit would not work (sorry for being dumb here - its a steep learning curve for me!)

 

[Harald Kapp]

No, that is not what I meant and will definitely not work. I can understand your desire to get away with a very simple solution, but alas what I proposed is the simplest solution I can think of (short of using a microcontroller to do the conversion). Do you have a friend who could elaborate my sketch to a final circuit and help you integrate and test it on the real gauge?

Have a look at the curent source circuit I linked in post #2.
Replace RL in that circuit by your gauge.
Replave the voltage sources in that circuit by the resistive divider I sketched. The divider will produce a voltage proportional to the resistance of the sensor. This voltage will drive the current source to create a current proportional to the
sensor's resistance.
As I showed the resistive dvider should be powered by 12V. If you power it from regulated 10V (which is a good idea), you'll have to increase Rsense to 12Ω.
But: your 10V regulator is absolutely unsuitable for an electronic device like the one I proposed, The oscillations from the bi-metallic strip will cause al kinds of unpredictable behavior. Use for example an 7810 electronic stabilizer.

You may have to exeriment with the values for Rsense and the upper resistor in the divider to get the readings right.

 

[rockrockmcrock]

Thanks Harald - sadly I'm on my own with this. My mates are losely split into three camps - those that have no technical interests, those that understand everything there is to know about cutting-edge information systems & security and those that are magic at welding, testing oily coils & dynamos and rebuilding old engines....

I am however a fast learner (usually) - so please don't give up on me!

Usually, I need to understand the 'how' of things before I get them right. So I did some self-learning this morning online and hopefully now I am on the right track (see below) - but can I check my understanding of the 'how' it works with you? Also there are some specs - particularly around selecting the right Op Amp that seem confusing to me at present, so any help here would be appreciated!

So here's my understanding of the above circuit and how it works (I hope I'm on the right track):

• Voltage regulator produces 10v & up to 1A (easily enough) at 2 & 6 for 10v rated gauge and RSensor
• Voltage divider (2,3,4) effectively inverts RSensor output, by producing a voltage at 3 which is proportional to RSensor resistance (i.e. increasing RSensor resistance, increases voltage at 3). RTrimmer can be used to vary the sensitivity for the op amp (ie "high-low" voltage change delta - is that correct?)
• No current is carried across the Op Amp (3,5), this is the need for the second circuit 6,7,8,9 - the voltage at 3 is simply used to proportionally vary the current across 6,7,8,9.
• The Op Amp triggers the NPN transistor to control the current and the feedback loop 8 to the Op Amp provides the reference input to the Op Amp (comparison of 8,3).
• To ensure the right current in 6,7,8,9, RSense is used, such that current across RSense (8,9) must equal current across Gauge (6,7). Does this mean I need to know the resistance of the Gauge (Load) though to be accurate?

OK, assuming I'm not completely wrong about the above - any advice on the following hugely appreciated:

1. NPN Transistor needs to carry the current and be triggered by a low current - hence a high gain, 400mA+ specification is ok?
2. The Op Amp - there seems to be loads of variables to consider here....totally confused at present, but slogging on! All help appreciated!

Thanks again
RRMR

 

[Harald Kapp]

You have a very good understanding of the circuit. A few details:

Your image shows Rsensor from 330Ω to 240Ω. Shouldn't that read 33Ω to 240Ω? With 330Ω to 240Ω the circuit will not work as expected.

No current is carried across the Op Amp (3,5), this is the need for the second circuit 6,7,8,9 - the voltage at 3 is simply used to proportionally vary the current across 6,7,8,9.

Actually there is current flowing from the output of the OpAmp into the base of the transistor. But insofar as the current through the gauge is concerned you're right. The OpAmp does not drive the current through the gauge directly.

Does this mean I need to know the resistance of the Gauge (Load) though to be accurate?

No, the current source will try to drive the current as required through the gauge regardless of the gauge's resistance. This has its limits, naturally, as the product of current and gauge-resistance cannot exceed 10V (Rather obvious, isn't it? This is called compliance voltage of the current source.)

NPN Transistor needs to carry the current and be triggered by a low current - hence a high gain, 400mA+ specification is ok?

This sounds reasonable, although high gain is not of the importance. A gain at ~100 will suffice.

The Op Amp - there seems to be loads of variables to consider here....totally confused at present, but slogging on! All help appreciated!

In fact for this application not many parameters of the OpAmp are important. Mainly that it is suitable for single supply operation and input as well as output rang near ground are required. An LT1006 may be used, although quite a few other OpAmps would meet these specs, too.

 

[rockrockmcrock]

Thanks Harald - you are correct and I've now altered the circuit diagram accordingly. OK - now I just need to source the electronics and a sender....exciting! Watch this space.....