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Pressure Sensor Amplification and Power Questions

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vne147

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Hello everyone. I am beginning a project and before I make too many design decisions, I’d like to get some input. Here is what I’m doing:

I want to display a measured pressure on an LCD screen. I plan to accomplish this by using a differential pressure sensor to measure pressure, an op amp to amplify the output of the sensor, and a PIC for the onboard A/D and to drive the LCD. I’m fairly comfortable with PIC programming, using the onboard A/D converter and LCD displays so most of the advice I need has to do with the pressure sensor, output amplification and design of the power supply to power everything.

First off, I would like to power the whole thing off of 2 AA or AAA batteries so I think I should probably use a low voltage PIC maybe 3.3V. I think I have seen PICs in the past that can run off of 2 or 2.5V but I haven’t gone hunting yet so I’m not certain. I am open to suggestions for other PICs or voltage levels that might be more appropriate.

Next, I will need a power supply stage for the circuit. If I go with 3.3V or even 5V I will need some sort of DC to DC boost converter. Or if I go with a voltage less than the battery voltage, I can probably just use an LDO voltage regulator. Any thoughts, ideas, or suggestions on this to save cost and make the circuit more efficient are welcome?

The pressure sensor I will most likely be using is from Freescale Semiconductor model # MPXV2010DP (link to datasheet below). It is a temperature compensated differential pressure sensor with a ratiometric output of 0 - 25 mV. The data sheet lists the input voltage specs as 10V typical and 16V max. I haven’t gotten the sensor yet and played around with it to see if it will function at all off of a lower voltage but if it does, I imagine the output range will be smaller than 0 - 25mV. If the sensor will not work with less than 10V, I will most likely need to modify the power supply to supply both 10V for the sensor and a lesser voltage for the PIC and everything else.

After the pressure sensor, I think I will need to amplify the output before it goes into the PIC’s A/D converter so I was just thinking of using an op amp with the gain set appropriately. Any thoughts on this? Is there a cheaper, better, less expensive or more efficient way to do this?

I think those are all the questions I have for now. I also included a block diagram of my concept for the circuit in case the written explanation was unclear. Thank you to everyone in advance for the help.

Datasheet for pressure sensor:

http://www.electro-tech-online.com/custompdfs/2009/06/MPX2010.pdf

Block Diagram:
 

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MikeMl

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Reading the spec sheet, no minimum excitation voltage is listed, so I think you could use 2 or 3V; however, since the device is ratio-metric, at 2V excitation you will only get 20% of the output compared to exciting it with 10V.

You need a DIFFERENIAL amplifier between it and your ADC. An IC instrumentation amp is ideal. Home-brewing a diff amp out of an opamp is difficult because you need to match resistors to better than 0.1%.

Also, dont forget that you will need an anti-aliasing filter between the diffamp and the ADC input. What will your device be measuring?

Try to avoid using a switcher to provide the excitation for the sensor. The excitation must be super clean because the sensor is ratio-metric. Any noise on the excitation appears superimposed on the pressure signal.
 
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vne147

Member
Mike,

Thanks for the advice. This sensor will be used to measure differential pressure (one wet and one dry) and that measured pressure will be used to calculate a liquid level inside of a closed container.

I understand your comments about the output span being less with a lower excitation voltage. In theory that shouldn't matter as long as the amplification is done properly. Would you say that is an accurate statement? Also, I was unsure from the data sheet if the output was the voltage measured between the two outputs (+Vout and -Vout) or if it was measured between +Vout and ground. If it is the first case then I understand why I must use a differential amp but if that's not the case then I don't understand. I will begin looking into using an IC instrumentation amp instead of an op amp. Can you suggest an appropriate IC or family of ICs to get me started in that area? I have no knowledge or experience with ant-aliasing filters but I'll start looking into that as well.

From your last statement, it would appear that the better option would be for me to use a lower power PIC and an LDO voltage regulator vs. a boost converter. Or if I did end up requiring a higher voltage for the sensor, I should at least place a voltage regulator downstream of the switcher. Would that be a suitable work around?

Let me know if I've interpreted all your statements correctly and if anyone else has any thought or ideas please chime in. Once again thanks for your help.
 

vne147

Member
Eric,

I'll start studying the info you provided. At first glance it looks like it's exactly what I need and will be very helpful. Thank you.
 

MikeMl

Well-Known Member
Most Helpful Member
...
I understand your comments about the output span being less with a lower excitation voltage. In theory that shouldn't matter as long as the amplification is done properly. Would you say that is an accurate statement? Also, I was unsure from the data sheet if the output was the voltage measured between the two outputs (+Vout and -Vout) or if it was measured between +Vout and ground. ...
Yes, you should be able to make up for the lost signal by having more gain in the instrumentation amp. However, the SNR will be worse compared to exciting the sensor with 10V.

Yes, the output from the sensor is the difference between +Vout and -Vout.

The anti aliasing filter issue is related to your ADC sampling rate and the high frequencies that may exist in your fluids being monitored. Look up "Sampling Theorem" in the context of sampled data. If you use a switching step-up power supply to excite the sensor, then an aggressive low-pass anti-aliasing filter with a cuttoff freq below the switching rate might solve the problem of the switching frequency feed through.
 
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vne147

Member
OK, I'm pretty sure I understand all your comments. I'm going to start working on it and I'll post back with the results or if I have any questions. Thanks.
 

MikeMl

Well-Known Member
Most Helpful Member
More reading for you:

1

2

3
 
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vne147

Member
I have a follow up question not directly related to the earlier conversation. I'm trying to decide on an appropriate LDO voltage regulator so I kind of need to decide on a voltage for the circuit. As I stated earlier, I want to power the whole thing off of 2 AA or AAA batteries. I know that those are 1.5V each and 1.5V X 2 = 3V. But I also know that the batteries won't output a constant 1.5 V for their entire lives. Basically, I'm curious about how low the output voltage drops to before the battery dies. Does it maintain 1.5V until the battery decreases to 20% capacity remaining or is it more like 80%? I don't want my circuit to stop working with 80% life left in the batteries becasue they aren't ouputing a high enough voltage for the voltage regulator to work. If it is going to go down to 1.1 V for example then 1.1V X 2 = 2.2V. I would therefore want the output of my voltage regulator plus the dropout voltage to not exceed that 2.2V (i.e. Vout + Vdropout < 2.2V). The 2.2V is just an example. I just thought it would help to make my question clearer. I also remember reading somewhere that the voltage drop and life of a battery is also affected by the current the battery must supply. I'm not certain yet what that current will be but an initial rough guess is somewhere in the neighborhood of 100 - 200 mA. In the circuit there would be a small pressure sensor, and IC instrumentation amplifier, an anti-aliasing filter, and low voltage PIC MCU, a small LCD screen with an occasionally lit LED back light, some intermittent indicator LEDs, maybe a constantly lit power LED, and some NO pushbuttons for user interaction. Please let me know if I need to give anymore details. Thanks.
 

MikeMl

Well-Known Member
Most Helpful Member
I would design the circuit so that the batteries can get down to 1.2V. It sounds like to me that at your proposed 200mA you should be looking at C or even D cells :(
 

vne147

Member
Hopefully, it won't be quite up to 200mA. That was just a really rough guess. I'm sure I could work on decreasing the current draw significantly. I'll be back for constructive criticism once the circuit is designed. The pressure sensor is on it's way. I should be able to start playing with it soon. I saw a voltage regulator at mouser with 2.2V output and 100 mV dropout. It sounds like that might work for me. Thanks for the help.
 
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vne147

Member
Mike,

Can you give me a sanity check here please? A link to the data sheet for the LDO voltage regulator I'm going with is below. Right now for testing purposes I'm planning to turn the voltage regulator on by connecting the enable pin to the battery through a switch. Note 2 from the first table on page 5 says "Applied voltage must be current limited to specified range." I'm interpreting that to mean that I must place a resistor somewhere between the enable pin and the battery. The specified range I think is 0 - 1µA and I got that from the first table on page 6. So if I interpreted everything correctly and my battery voltage is 3V that means I'm going to need a 3 MΩ resistor, correct? That just seemed a little high to me (not like I have a basis for comparison anyway). Thanks.

http://www.electro-tech-online.com/custompdfs/2009/06/FAN2502.pdf
 

MikeMl

Well-Known Member
Most Helpful Member
I read the data sheet and here is my take: I have attached the relevent part of it.

The Enable input is just a CMOS logic input. It is a logic 1 if it is higher than 2V. The leakage into the pin is 1uA or less. It cannot be left floating. To turn on the regulator, the pin can be tied to Vin or any voltage > 2V. No series resistor is asked for or needed. If the regulator is turned off by driving EN low, the regulator will leak 1uA from the power source (battery).

For your application, I would tie EN to VIN, and then put your battery switch between VIN and Battery+. This way, the switch is switching >100mA. The reliability of switches is suspect when switching only 1uA.
 

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vne147

Member
I am progressing with the circuit design and surprise, I have another question. Here it is.

I want the finished device to turn on and off with a single pushbutton (i.e. one push turns everything on and a subsequent push will turn everything off). I know this sort of thing can be done a number of ways using logic IC's like flip flops for example but I'm using a PIC in the circuit. I'm thinking it would be easier to perform this function with the PIC instead. For a slight increase in PIC program complexity I could reduce the component count and cost associated with having a portion of the circuit dedicated to turning everything on and off. I attached a simplified schematic showing how I'm thinking this can be accomplished as well as some pseudo PIC code below showing how I expect the software to control this function. Does anyone have any thoughts on this? Is there any reason that this is a bad idea or wouldn't work? Is there a better way to go about accomplishing what I'm after? Thanks.

Here are some definitions of terms that I used in the pseudo-code below:
POF = power off flag. The purpose of this flag is to not allow a high logic signal at RB1 to satisfy condition 2 until the pushbutton SW1 has been released following the initial push to turn on the PIC.
LO = low logic signal at an I/O pin
HI = high logic signal at an I/O pin

pseudo code begins here:

PIC powers up
Registers and status flags are set
SET RB0 = HI
SET POF = 0

MAIN LOOP {

#Comment - This will become true once the pushbutton SW1 is released.
#condition 1

IF (RB1 = LO){
POF = 1
}

#Comment - This will be true for a subsequent press of the pushbutton SW1.
#condition 2

IF (POF = 1 and RB1 = HI){
RB0 = LO
#Comment - Once SW1 is released after RB0 is set to LO, RB0 and RB1 will both be LO so the EN pin on IC1 will be LO and IC1 will shutdown.
}

...Other program stuff here

} #End Main
 

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MikeMl

Well-Known Member
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Doesn't this leave the PIC powered up forever? Why don't you just get a miniature mechanically latching push-button on-off switch which powers everything, and then just boot the PIC when the switch is first closed. I used to have some very small toggling push switches which even had a passive reflective On indication (like a backlighted pushbutton, only it uses no power)
 

vne147

Member
It won't leave the PIC powered on forever because the supply voltage for the PIC is the output from IC1. Once the EN pin on IC1 goes LO, the output goes to OV and the PIC becomes unpowered. I had considered the latching ON/OFF pushbutton but my goal here is to make this thing as cheap as possible. That probably would add some cost to the project. Whether or not that cost would be significant, I can't say. Additionaly, all the buttons in this project, including the on off pushbutton will be part of a custom made membrne swich and overlay. Do they make latching ON/OFF membrane switches? Thanks.
 
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