Hi I'm a part time farmer and I have a quick question. Hopefully someone here can help me out. I have a tractor with an oil pressure gauge for the engine. Normal operating pressure should be 40 psi or more. I'm wanting to make an alarm for when the pressure drops below 40 psi. I'll need to use an oil pressure sending unit which is basically a sensor that changes resistance as oil pressure increases or decreases. At 0 psi it's 10 ohms. At 40 psi it's 60 ohms. I'm needing a circuit that will turn on an alarm or buzzer when the pressure is below 40 psi which will be in the range of 10 to 60 ohms. Could someone point me in a general direction that would help me find a circuit or schematic for this kind of device? Thanks

 

What is the voltage drop across the sensor at 0 and at 40#?

 

A quick search came up with this: https://www.pegasusautoracing.com/productdetails.asp?RecID=2943

 

I will start by making the assumption that you have a 12 volt power supply negative ground. I will also assume that the sensor only has one connector and the other end of the sensing element is grounded as it will probably be screwed into the engine block. ( Most oil pressure sensors I have seen on cars are like this.) If we choose to have a voltage drop of about 2 volts across the sensor when it's resistance is 60 ohms (40 PSI) (Current through sensor 33 mA) then a 300 ohm resistor up to the +12 volt supply will give this current. (10 volts across 300 ohms = 10/300 = 0.033 Amps = 33 mA) If we then have another potential divider with the same ratio of resistor values then that will also have 2 volts at the junction of the resistors. The actual value of these two resistors is not critical. If we choose a 1K resistor at the ground end the the other resistor will need to be 5K. As 5 K is not a standard value we can use two 10 K resistors in parallel. If we then use a comparator IC (Such as an LM393) with the non inverting input to the junction of the sensor and the 300 ohm resistor and it's inverting input to the junction of the 1K and 5K resistors then it's output will be low when the pressure is less than 40 PSI. You can connect it's output to a PNP transistors base via a resistor with the emitter of the transistor connected to +12 volts and the buzzer between it's collector and ground. You will need to check that sensor is rated for a current of at least 47 mA as this will be the current whenthe battery voltage is 14.5 volts and the sensor resistance is 10 ohms.

Les.

 

.... and if the buzzer is an electromagnetic type then a reverse-biased diode should be connected in parallel with it to protect the transistor from voltage spikes.

 

1) Do you need to display the pressure ?
2) When alarm is tripped do you need to power something off, like with a relay ?
3) Can you write software, do embedded micro work ?
4) Need to monitor and display other stuff, like battery voltage, water jacket T ?

Regards, Dana.

 

I would just go with a switch to ground as KISS suggested. A 20-40 psi Adjustable Oil Pressure Warning Switch, 1/8 NPT. A potential problem with using a voltage divider network combined with a comparator might be that a 12 volt automotive system, which I assume you have, is actually not 12 volts. A fully charged lead acid battery is about 12.6 volts and with the engine running anywhere between 13 and 14.5 volts depending on the charging system. Rather than worry about regulation I just see a mechanical pressure switch as the easy approach. Then add a low oil light, audible sound device or whatever works for you.

Entire kit about $32 USD with a wide adjustment range.

Ron

 

I will start by making the assumption that you have a 12 volt power supply negative ground. I will also assume that the sensor only has one connector and the other end of the sensing element is grounded as it will probably be screwed into the engine block. ( Most oil pressure sensors I have seen on cars are like this.) If we choose to have a voltage drop of about 2 volts across the sensor when it's resistance is 60 ohms (40 PSI) (Current through sensor 33 mA) then a 300 ohm resistor up to the +12 volt supply will give this current. (10 volts across 300 ohms = 10/300 = 0.033 Amps = 33 mA) If we then have another potential divider with the same ratio of resistor values then that will also have 2 volts at the junction of the resistors. The actual value of these two resistors is not critical. If we choose a 1K resistor at the ground end the the other resistor will need to be 5K. As 5 K is not a standard value we can use two 10 K resistors in parallel. If we then use a comparator IC (Such as an LM393) with the non inverting input to the junction of the sensor and the 300 ohm resistor and it's inverting input to the junction of the 1K and 5K resistors then it's output will be low when the pressure is less than 40 PSI. You can connect it's output to a PNP transistors base via a resistor with the emitter of the transistor connected to +12 volts and the buzzer between it's collector and ground. You will need to check that sensor is rated for a current of at least 47 mA as this will be the current whenthe battery voltage is 14.5 volts and the sensor resistance is 10 ohms.

Les.

Had a quick look on Google, sensors range from 10 to 180 ohms (10ohms =0psi,, 180 ohms =100psi)

I would say 40 psi is around 78 ohms.

 

Hi Ron,
The variation in battery voltage would not matter as both the reference level and input level are both derived with a potential divider from the same voltage. When the sensor has a resistance of 60 ohms both inputs to the comparator will be one sixth of the supply voltage. I agree that just using a switch type sensor is a simpler solution than using an analogue sensor.

Les.

 

Hi Ron,
The variation in battery voltage would not matter as both the reference level and input level are both derived with a potential divider from the same voltage. When the sensor has a resistance of 60 ohms both inputs to the comparator will be one sixth of the supply voltage. I agree that just using a switch type sensor is a simpler solution than using an analogue sensor.

Les.

Now that you mention it, that's right, as voltage changed so would go the reference. Duh, I never thought that through. Anyway, I still like the switch thought unless the thread starter wants to get fancy and have a readout with an alarm function or along those lines.

Ron

 

Thanks for the info. I'm going to go ahead and buy a pressure switch and wire it to a light and buzzer. Seems like the most simple solution.

 

I would just go with a switch to ground as KISS suggested. A 20-40 psi Adjustable Oil Pressure Warning Switch, 1/8 NPT. A potential problem with using a voltage divider network combined with a comparator might be that a 12 volt automotive system, which I assume you have, is actually not 12 volts. A fully charged lead acid battery is about 12.6 volts and with the engine running anywhere between 13 and 14.5 volts depending on the charging system. Rather than worry about regulation I just see a mechanical pressure switch as the easy approach. Then add a low oil light, audible sound device or whatever works for you.

Entire kit about $32 USD with a wide adjustment range.

ReloadRon is right, that would have been my suggestion as well, because it is both far more reliable and also safer. GM used to use exactly such a device, both a switch and a sender in one three terminal package. So you may be able to find it in an auto parts store as a new replacement part. Then you would know that it was good.