Inverting resistance using an op-amp circuit?

Thread Starter

Jarven

Joined Feb 1, 2022
10
Hi
Toatally new to this forum so bear with me.. I've tried searching but I could only find a bunch of threads discussing inverting the voltage output. My dilemma is slightly different.
I have an old classic car with analog instruments that I want to keep intact. But I want to use a new type of sensor since it also has a second switching contact output in the same unit. So here it is;
-Oil pressure gauge in car sends out a 5vdc feed and the input reads ohms. 0psi=75ohm and 80psi=10ohm. I haven't measured but with ohms law this should be then 66mA at 0psi/75ohm and 500mA at 80Psi/10ohm.
-New Oil sensor is capable of 0Psi to 150Psi but the output is switched. 0Psi=10ohm and 80Psi=90ohm (150Psi=180ohm)
After reading up some I thought an op-amp could be the way to go but I'm having a hard time figuring out how the circuit should be designed and what resistors to use?
-Should it be a single feed circuit where I use the 5Vdc to power the op-amp or should I add a 12Vdc feed into the circuit? Whatever works for me.
-The new Oil pressure sensor is only a single wire output that drives the circuit to ground, so it's not like a three-wire pot or similar.

Any help is appreciated. Oh and I added a picture of the classic in question for attention.. and below is a link to my car build project if anyone is interrested.

https://www.forbbodiesonly.com/mopa...arger-ray-barton-528-hemi-pro-touring.135133/
 

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BobTPH

Joined Jun 5, 2013
8,812
Did you actually measure the current? 500mA sound excessive fir a sensor.

And, is the sensor just a two wire connection?

Bob
 

Marley

Joined Apr 4, 2016
502
Often the gauges in older cars are super crude. They quite often use a "hot wire" or a "bimetallic strip" or similar mechanism where the input current causes heat. Therefore increasing current = increase in temperature = expansion of the wire or metal strip = raising the gauge indication. Advantages: very few internal parts, robust and slow operating (good - keeps a steady indication).

To interface the new sensor, where the resistance is different - and reversed operation you need an op-amp circuit driving a controlled current source that then feeds the existing gauge. Can be done - with a bit of careful design.

The new oil sensor is a resistance to ground. This is what I would expect. How was the original sensor connected? Did it have 2 wires? Can you draw out the original circuit diagram?

You say that there is a stable 5V supply in the car for the original gauge. Include this in the circuit diagram.
 

Thread Starter

Jarven

Joined Feb 1, 2022
10
Did you actually measure the current? 500mA sound excessive fir a sensor.

And, is the sensor just a two wire connection?

Bob
Hi Bob, thank's for chiming in.
No, didn't measure. I was just thinking it would logically be this current since it's powered with 5vdc and sensor pulls it down to 10ohms to ground. The gauge itself is just a simple inductance coil and it requires some power to move. I remember from bench testing all gauges that they drew some significant power.
Below is a link to a similar sensor and I can see from the scarce info that it draws up to 5w? See link below and data that I could find.
The sensor output is just 1 wire. The base and threaded connection is what grounds it. The second connection you see on the top of the sensor if you follow the link below is the contact output for i.e a oil pressure warning light.
https://www.amazon.se/KUS-Mekanisk-...8ad20c&pd_rd_wg=napZe&pd_rd_i=B075NDS71W&th=1

Driftspänning: 6 ~ 24 V
Ledande effekt: <5 W
Driftstemperatur: -25 ~ 120 ° C (120°C MAX 1H)
Mätområde: 0~5Bar eller 0~10Bar
Alarm: 0,8 Bar
Utgångssignal: Standardmotståndsvärde: 10~184Ω
Gängbeslag: NPT-1/8, M10X1
 

Thread Starter

Jarven

Joined Feb 1, 2022
10
Often the gauges in older cars are super crude. They quite often use a "hot wire" or a "bimetallic strip" or similar mechanism where the input current causes heat. Therefore increasing current = increase in temperature = expansion of the wire or metal strip = raising the gauge indication. Advantages: very few internal parts, robust and slow operating (good - keeps a steady indication).

To interface the new sensor, where the resistance is different - and reversed operation you need an op-amp circuit driving a controlled current source that then feeds the existing gauge. Can be done - with a bit of careful design.

The new oil sensor is a resistance to ground. This is what I would expect. How was the original sensor connected? Did it have 2 wires? Can you draw out the original circuit diagram?

You say that there is a stable 5V supply in the car for the original gauge. Include this in the circuit diagram.
Hi Marley, you're correct, it's crude but it works. In this thread (follow link below) there are pictures of the pcb and what I've also done to improve the gauges. Originally the 5vdc is made from a switching regulator that was broken in my case. I updated this with a common fix, a 7805 solid state regulator with a heatsink. I'll make a crude sketch of the wiring as it is today and post it here.

https://www.forbbodiesonly.com/mopa...ray-barton-528-hemi-pro-touring.135133/page-4
 

Thread Starter

Jarven

Joined Feb 1, 2022
10
That can't be correct.
With the connection shown there is no change in voltage for the gauge to see.

I would expect the gauge to be in series with the sensor.

Can you measure the resistance of the gauge?
Dang, of course it couldn't.. I mixed up the wiring when I looked at an old picture of the pcb. See attached.
Measuring the resistance i the gauge will be difficult, hard to get there. But if essential to design the circuit I'll just have to get in there somehow..
 

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Alec_t

Joined Sep 17, 2013
14,280
If you don't have 10Ω, just use one of the pressure transducers at 0psi as the current-sense resistor., since you know the transducer resistances at 0psi..
 

Thread Starter

Jarven

Joined Feb 1, 2022
10
If you don't have 10Ω, just use one of the pressure transducers at 0psi as the current-sense resistor., since you know the transducer resistances at 0psi..
That's exactly what I did but the results were slightly confusing. The gauge did not manage to go up to full range (80psi)as it should have? It stopped at roughly 40Psi. The voltage over the sensor was measured to 2,3V. So it seems I had a current flow of 230mA(dang, should have measured amps also to verify, I'll do that tomorrow). Ohms law tells me that the internal resistance of the gauge should be 11,74ohms if I'm calculating this correctly.
 

crutschow

Joined Mar 14, 2008
34,281
So it looks like we should assume, for design purposes, the meter reads 40psi @ ≈260mA of current.
Thus the circuit should provide 260ma of meter current when the new sensor is at 50 ohms, and no more than about 60mA when the new sensor is at 10 ohms.
(Subject to change if you measure the actual current.)
That sound correct?
 

Thread Starter

Jarven

Joined Feb 1, 2022
10
So it looks like we should assume, for design purposes, the meter reads 40psi @ ≈260mA of current.
Thus the circuit should provide 260ma of meter current when the new sensor is at 50 ohms, and no more than about 60mA when the new sensor is at 10 ohms.
(Subject to change if you measure the actual current.)
That sound correct?
Yes, sounds correct.
 

crutschow

Joined Mar 14, 2008
34,281
Ok, here's my first crack at a circuit (LTspice simulation below):
It uses a rail-rail op amp in a constant-current configuration to drive the gauge.
The plot shows the gauge current versus the sensor resistance.
The gauge current (red trace) is 261mA at a 50Ω sensor resistance (horizontal axis), and a little less than 60mA at a 10Ω sensor resistance.
Those values can be tweaked by adjusting pot U2 (set here at 400Ω)

Note that the transistor will dissipate about 0.5W maximum, which will make it hot, so a small clip-on heatsink probably should be added to avoid overheating under hot ambient conditions.

1643768350794.png
 
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Thread Starter

Jarven

Joined Feb 1, 2022
10
Ok, here's my first crack at a circuit (LTspice simulation below):
It uses a rail-rail op amp in a constant-current configuration to drive the gauge.
The plot shows the gauge current versus the sensor resistance.
The gauge current (red trace) is 261mA at a 50Ω sensor resistance (horizontal axis), and a little less than 60mA at a 10Ω sensor resistance.
Those values can be tweaked by adjusting pot U2 (set here at 400Ω)

Note that the transistor will dissipate about 0.5W maximum so a small clip-on heatsink probably should be added to avoid overheating under hot ambient conditions.

View attachment 259395
I'm amazed by this forum. Thank you so much for your help with my dilemma. I think this will work out. I'll order the parts straight away and dig into it. I'll post the results here. Thanks for adding the pot for fine tuning the circuit!
Two questions,
-"R1", is that just a 1ohm resistor?
-Op amp feed and R2 feed "V+" Should this be a 12V feed or 5V feed?
 
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