Adjusting the gain in a differential pressure sensor circuit

Thread Starter

projix

Joined Nov 17, 2021
14
Hello,

I got myself into a bit of a pickle. Long story short, I have a few 0-1000 Pa differential sensors, plus a circuit that turns this into a 0-10VDC signal. The supply voltage is 24 VDC.
I would actually need 10V to be 250 Pa or so instead. I could order the same sensors again, which would mean binning the current ones and waiting for a few months for them to arrive, but because all the sensors seem to actually use the same 0-1 kPa sensor instead, I am suspecting that it should be possible to control the gain by changing a resistor on the board.

Here are the pictures of both sides of the circuit board:
front.jpg
rear.jpgYou can also find the the datasheet for the pressure sensor in the attachments.
The 16 pin IC is a mystery to me. It is marked as 3094AS, but I can not find any information about it. Usually a differential pressure sensor circuit would have some sort of op-amp, so I would hazard a guess it is some sort of an instrumentation opamp. Those usually have external gain resistors... which is what I am hoping for here.

The wire colors that come into the board from the sensor are as follows:
S+ - yellow
S- - white
I+ - red
I- blue.

Now, S+ is connected through a 49.9k resistor to Pin 5 of the IC and S- is connected through a 49.9k resistor to Pin 6. Measuring the resistance between S+ or S- and ground I get 3.2k (or 2.2k to I-, as there's a 1k resistor from I- to GND), as expected the S+ resistance goes down when pressure is applied to the higher pressure port of the sensor and S- resistance goes up by the same amount.

The output is a lot less clear to me. I thought initially it was fairly simple, as it goes through R10 to Pin 13 on the IC, and then it would mean that the IC is outputting the 0-10V signal, with R10 (2.2k) being a current limiter and R11 (2k) being a pulldown resistor, but then I noticed that T3 (BFS20 NPN) is also connected to the output directly, and this is where my knowledge is insufficient to make heads or tails of anything anymore.

I thought about just changing the I- 1k resistor to ground for a bigger one, but a quick simulation with LTSpice seems to be telling me, that this is probably a stupid idea.

So I figured before I bin a few hundred euros worth of sensors, I'd ask for advice. Most likely one of the resistors (or a combination) are gain resistors on the op-amp and by changing them it will be simple enough to alter the output gain.

I'll also note that the absolute reading of the sensor is actually completely irrelevant. The sensor is meant to control the pressure in my ventilation system, and the target pressures are found empirically anyway by equalizing intake and exhaust using a flow meter, then noting down the pressures. After that there is a closed loop PI controller in the control unit. The problem is that the signal is too low for the control, and I have no control over the software in the control unit.

If anyone feels like helping out, it'd be most appreciated.
 

Attachments

Thread Starter

projix

Joined Nov 17, 2021
14
I started slowly reversing the circuit, here is what I got for the sensor side:

1637276042854.png

The sensor is represented by 2.2k resistors because that's essentially what it is when the pressure between the two ports is equal, as the pressure changes so does the resistance of the sensor.

The sensor side does not look at all like what is in the sample schematic. My knowledge is limited, but could it be that the unknown IC is a uC, the sensor is being fed with a voltage instead of a current source and then the output sensed with an ADC?

Something like this rather than the constant current source example in the datasheet:
1637276297783.png
 

Thread Starter

projix

Joined Nov 17, 2021
14
Here is the resistor/transistor setup at the output, chances are I drew something wrong, and the one marked "Q1" has zero labeling on the chip, so no idea if that even is a transistor.

1637313977185.png

Could it be a uC that is using PWM to drive the output and pin 12 as feedback? The resistor divider between R9 and R8 comes out as 0.47, so if the output is 0-10V, that'd put it at 4.7V.
If this is the case perhaps changing the R9 (the numbers in LTSpice do not correspond to the graph) then changing R9 from 2.2k to 6.4k might double the output?
 

jeffl_2

Joined Sep 17, 2013
63
It's not really possible to make complete sense of the schematic, you've left some critical pins like power off and some of the pins and components don't make sense exactly where they are and how they're connected. I would be inclined to "hazard a guess" that this is nothing more complex than an LT6372 instrumentation amplifier (especially since you're showing the Linear Technology logo), but you "can't really get there from here" with the information you've provided. It COULD be that some of this is some arcane analog thermal compensation or linearization scheme but it's just not crystallizing.
 

Thread Starter

projix

Joined Nov 17, 2021
14
It's not really possible to make complete sense of the schematic, you've left some critical pins like power off and some of the pins and components don't make sense exactly where they are and how they're connected. I would be inclined to "hazard a guess" that this is nothing more complex than an LT6372 instrumentation amplifier (especially since you're showing the Linear Technology logo), but you "can't really get there from here" with the information you've provided. It COULD be that some of this is some arcane analog thermal compensation or linearization scheme but it's just not crystallizing.
I am slowly filling it out, as I can. The power supply voltage is 24VDC, it is not connected directly to any pin (obviously). The board is actually quite well visible in the photos. I have not added any of the capacitors yet either.

The analog devices logo has nothing to do with anything, it's just the default symbol for an unknown component in LTSpice.

If it's just an op-amp then what's the transistor logic on the output? Wouldn't the transistors point to PWM control and then smoothing the signal out with caps?
 

jeffl_2

Joined Sep 17, 2013
63
"Transistor logic"? I don't see any gates, logic symbols or even a popular logic-family-compatible power supply (TTL, CMOS or anything else). The presence of discrete "transistors, diodes and resistors" doesn't imply a logic circuit (I guess it might have in the 1970s), in fact nowadays it's much more likely an indication of NON-logic circuitry. The only way you could "make an argument" that it's some kind of MCU would be if there was evidence that it was providing a digital output in some common serial format like I2C or SPI, but that would imply either the output used either 2 or 3 pins, there's really no reason for an MCU if it's just providing an analog output (most 16 pin devices wouldn't be compatible with high-precision analog anyway). And if that logo doesn't mean anything, then there are 4 or 5 extra vendors of analog instrumentation amplifiers that you need to look through their data sheets to see if you find anything with compatible pinouts, you might try going through the meticulous listings for such products on Digikey or Mouser.
 
Last edited:

Thread Starter

projix

Joined Nov 17, 2021
14
Makes sense, I am just not very familiar with usage of transistors in analog circuits. I am no electronics engineer.

Good call on looking through datasheets of instrumentation amplifiers.

I do not really understand the need for the Vout -> Divider -> Pin 12 though.
 

Thread Starter

projix

Joined Nov 17, 2021
14
Why not just add an amp on the output with a gain of 4. That will get you 10 volts at 250 Pa.
Yes, this could be done as well. But not sure how much sense it makes.

Traced another resistor, that now should account for all resistors on the board:
1637356456397.png
 

Thread Starter

projix

Joined Nov 17, 2021
14
I think getting the parts/making said amp makes it a little pointless compared to just buying new sensors. Was hoping that one of the resistors might be a gain resistor and changing it would be all that is needed.

But after mapping out the circuit there is no obvious candidate for such a thing, I don't see any resistor connected to two pins.
 

sghioto

Joined Dec 31, 2017
2,702
I think getting the parts/making said amp makes it a little pointless compared to just buying new sensors.
That might be true if the sensors were readily available.
The parts for the amp are readily available and that might eliminate "the downside is the waiting time and the waste." plus the cost of new sensors. Just my 2 cents.
 

michael8

Joined Jan 11, 2015
243
What sensor is it feeding the circuit board? What is S+ S- I+ I+ Where does the +24V supply connect in that latest diagram? (is that S+ and S-)?
 

Thread Starter

projix

Joined Nov 17, 2021
14
I do not have the +24V supply mapped out. I think +24V is not fed in directly.

The sensor is in the diagram drawn out. I made a red box around the component.
1637358839092.png

You can see the datasheet of the sensor as a PDF in the first post.
 

Thread Starter

projix

Joined Nov 17, 2021
14
Added +24VDC, it goes through a protection diode and then to pin 2 it seems. And it's connected to some of the transistors.
I am really unsure about Q1 being a transistor in the first place, there is no marking, might just as well be anything else.

1637361002836.png
 

Thread Starter

projix

Joined Nov 17, 2021
14
Another thing of notice - the output actually clips exactly at 10V for me. Even though the pressure sensor can actually read a little more. I wonder if the resistors on the outputs can be altered to change the gain (R10, R11, R13, R12 on schematic). Another odd one is 24k resistor R14.

Unfortunately I do not know enough about analog circuits and transistors.
 

RPLaJeunesse

Joined Jul 29, 2018
195
I suspect Q3 is a PNP part, not NPN as shown. That makes for a Darlington output stage with R9 and R8 the feedback resistors that set the amplifier gain. If you want 4x the gain reduce R8 to 500 ohms. For a quick test just try a 665 ohm 1% resistor in parallel with R8 (or use 680 if that's all you have).
 
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