# How do determine characteristics of oil pressure gauge so I can buy matching transmitter

#### Circuits123

Joined Dec 7, 2012
93
I have an electric, Smiths oil pressure gauge from an antique car (100 PSI, 12v, model PE2300/00). I need to replace the transmitter/sender but the original is no longer available (I think it was model PT 1804/10). Is there any way I can determine the characteristics of the gauge so I can shop for an appropriate replacement sender? The old sender was bimetal.

#### Jerry-Hat-Trick

Joined Aug 31, 2022
450
Typically, the oil pressure sensor has a resistance which falls with increasing pressure so that, connected in series with the gauge across 12V, the current increases thereby heating the resistive coil around the bimetal in the gauge.

First test the resistance of the gauge with a multimeter. Then, if you have a bench top power supply connect it across the gauge and start with a low voltage, increasing slowly and watching the needle move to indicate the expected pressure of your engine. At zero volts the needle will probably sit to the left of zero so the voltage required to move it to zero will give you information about the expected resistance of the sensor when it’s not under pressure, probably in the region of 200 ohms. The voltage needed to indicate the expected working pressure will give you information about the expected resistance of the sensor under normal working pressure, possibly in the region of 100 ohms.

If you then find a sender which fits your car and measure its resistance with the engine off and the engine running, it should be possible with series and parallel resistors to adapt the sensor and gauge to work together.

#### Circuits123

Joined Dec 7, 2012
93
The gauge has a resistance of 25 ohms. The needle sits at 0 PSI at about 1.3V, 50 PSI at 3.3V, and 100 (max) at 4.9V. How would I use that to calculate the sensor's resistance?

As a separate question: I tried to measure the current but the gauge and the DMM read zero - why would that happen?

My setup was like this:
[+V]------------[gauge]--------[DMM]-------[neg. V]

The gauge worked properly if I removed the DMM.

#### Ian0

Joined Aug 7, 2020
8,947
The gauge has a resistance of 25 ohms. The needle sits at 0 PSI at about 1.3V, 50 PSI at 3.3V, and 100 (max) at 4.9V. How would I use that to calculate the sensor's resistance?

As a separate question: I tried to measure the current but the gauge and the DMM read zero - why would that happen?

My setup was like this:
[+V]------------[gauge]--------[DMM]-------[neg. V]

The gauge worked properly if I removed the DMM.
That means that you have blown the fuse in the current range of your DMM and not realised it.

#### Jerry-Hat-Trick

Joined Aug 31, 2022
450
The gauge has a resistance of 25 ohms. The needle sits at 0 PSI at about 1.3V, 50 PSI at 3.3V, and 100 (max) at 4.9V. How would I use that to calculate the sensor's resistance?
Okay. In the days when gauges were bi-metallic the circuit was usually powered by a (rudimentary) 10V voltage regulator since the voltage across the battery could vary significantly when charged by a dynamo. I'm going to calculate with a 10V supply but using 12V or even 14.5 Volts from an alternator would be possible with the necessary adjustments.

Suppose your sender has a resistance of 200 ohms when the engine is not running. Connected in series with the gauge, the voltage across the gauge would be 10 x 25/(25 + 200) = 1.11V. This would work fine for the zero point. 12V would be 1.33 across the gauge which would nudge it above zero. From your readings, 65 PSI would need about 3.8V - not exactly, but close enough to make calculation easy. With a 10V supply the sensor resistance Rs would need to be calculated by 10 x 25/(Rs + 25) = 3.8 which simplifies to Rs = 10 x 25/3.8 - 25 = 40 ohms.

You'll need a sensor with the correct thread that has a resistance which goes down with increasing pressure and has a characteristic curve similar to 200 ohms with no pressure and 40 ohms at 65 PSI. If these exact numbers are not available you can calibrate with series and parallel resistors. For example, if the 'no pressure' resistance is higher, put a resistor in parallel with the sensor, or in series if the sensor resistance is lower. And you can bleed away some current from the gauge by putting a resistor in parallel with it. Note that the resistances are quite small so the current is fairly large - which is maybe why you may have blown your multimeter. Series and/or parallel resistances if needed will have to have sufficient wattage resistors calculated.

My view is that the precision and accuracy of automotive gauges is not so important as repeatability. It's more important that they stay the same. So if your gauge shows 50 PSI when the motor is running it is showing lower than it should, but the time to worry is when it drops below where it usually sits.

#### MisterBill2

Joined Jan 23, 2018
16,582
The gauge has a resistance of 25 ohms. The needle sits at 0 PSI at about 1.3V, 50 PSI at 3.3V, and 100 (max) at 4.9V. How would I use that to calculate the sensor's resistance?

As a separate question: I tried to measure the current but the gauge and the DMM read zero - why would that happen?

My setup was like this:
[+V]------------[gauge]--------[DMM]-------[neg. V]

The gauge worked properly if I removed the DMM.
To determine the current from this data use the formula I= V/R, with "V" being the applied voltage across the gage and "R" being the measured 25 ohm resistance.
So the conclusion is that you need a sender that decreases resistance as the pressure rises.
If you want to make the reading more stable and accurate then yiu can add a simle voltage regulator to supply the voltage supply.

#### Circuits123

Joined Dec 7, 2012
93
Thanks. I had to scrape the rust off my calculator to understand this but I think I've got what I need now.

• Jerry-Hat-Trick

#### MisterBill2

Joined Jan 23, 2018
16,582
What has not entered the discussion yet is accuracy of the reading. It seems that the gage response is not as linear as would be hoped. But if the TS is able to compare the gage indication with an actual accurate pressure gage and know where the minimum and normal pressures show, that should be good enough.
"precision" is an advertising word, Accuracy and resolution are the useful descriptors.
"Accuracy" describes the uncertainty of the correctness of the indication, and "Resolution" describes the size of steps between adjacent reported values.

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