Honeywell Pressure Sensor

[James_sgp]

I am using a Honeywell Pressure Sensor to measure water pressure, my
question is that the terminals are +, Grd, Output +, Output -. I need
a voltage on a sigle line to go into my micro controller; can i connect
the Output- to Grd? And still be able to get the voltage from Output+?

 

[Chris]

I am using a Honeywell Pressure Sensor to measure water pressure, my
question is that the terminals are +, Grd, Output +, Output -. I need
a voltage on a sigle line to go into my micro controller; can i connect
the Output- to Grd? And still be able to get the voltage from Output+?

Hi, James. I kind of wish you'd mentioned the Honeywell part number --
that would have increased the chance of getting a correct answer.

Many pressure sensors have a bridge-type output, where you measure the
differential voltage across the bridge. Honeywell makes some of these,
and it sounds likely that's what you've got. A bridge sensor works
like this (view in fixed font or M$ Notepad):
|
| .--------o-----------.
| | + | |
| | .-. .-.
| | R1| | R2| |
| | | | | |
| | '-' _ '-'
| | |Out-/ \Out+|
| +| o---( V )---o
| ---V | \_/ |
| - .-. .-.
| | R3| | R4| |
| | | | | |
| | '-' '-'
| | GND| |
| '--------o-----------'
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The pressure sensor itself in this case is one or two of the resistors
R. The other resistors in the bridge are set so, when there's no
differential pressure, the sensor resistor balances the resistor on the
other side of the bridge, and the voltmeter reads zero. But when a
presure is applied, the sensor resistance changes, so there's a reading
across the voltmeter. Simple, sensitive to small changes in pressure,
safe.

Connecting the Out- to GND should be OK if you're running the sensor
off an isolated battery. But if you're using the uC power supply to
power the sensor, it's not a good idea. Not that you''l break
anything, but your resistive divider doesn't work. You're shorting
across one of the resistors -- no balance, no measurement.

If you want to amplify the difference between the two outputs without
loading them down, you'll need to use an op amp configuration called an
instrumentation amp. It's basically a difference amp preceeded by two
voltage followers. A schematic of the circuit is in National
Application Note 31 Op Amp Circuit Collection :

http://www.national.com/an/AN/AN-31.pdf#page=8

Look at the "Differential-Input Instrumentation Amplifier" on the top
of p.14. You might be able to make something serviceable with 3/4 of
an LM324 and four resistors (forget the balance pot in the diagram --
that's not for an LM324).

Since you didn't mention which sensor you were using, I guess you'll
have to post back with the Honeywell part number to get an idea of the
voltage gain you want & which resistors to choose for the IA, or we'll
have to leave that as an exercise for the student.

Good luck
Chris

 

[Chris]

Hi, James. I kind of wish you'd mentioned the Honeywell part number --
that would have increased the chance of getting a correct answer.

Many pressure sensors have a bridge-type output, where you measure the
differential voltage across the bridge. Honeywell makes some of these,
and it sounds likely that's what you've got. A bridge sensor works
like this (view in fixed font or M$ Notepad):
|
| .--------o-----------.
| | + | |
| | .-. .-.
| | R1| | R2| |
| | | | | |
| | '-' _ '-'
| | |Out-/ \Out+|
| +| o---( V )---o
| ---V | \_/ |
| - .-. .-.
| | R3| | R4| |
| | | | | |
| | '-' '-'
| | GND| |
| '--------o-----------'
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The pressure sensor itself in this case is one or two of the resistors
R. The other resistors in the bridge are set so, when there's no
differential pressure, the sensor resistor balances the resistor on the
other side of the bridge, and the voltmeter reads zero. But when a
presure is applied, the sensor resistance changes, so there's a reading
across the voltmeter. Simple, sensitive to small changes in pressure,
safe.

Connecting the Out- to GND should be OK if you're running the sensor
off an isolated battery. But if you're using the uC power supply to
power the sensor, it's not a good idea. Not that you''l break
anything, but your resistive divider doesn't work. You're shorting
across one of the resistors -- no balance, no measurement.

If you want to amplify the difference between the two outputs without
loading them down, you'll need to use an op amp configuration called an
instrumentation amp. It's basically a difference amp preceeded by two
voltage followers. A schematic of the circuit is in National
Application Note 31 Op Amp Circuit Collection :

http://www.national.com/an/AN/AN-31.pdf#page=8

Look at the "Differential-Input Instrumentation Amplifier" on the top
of p.14. You might be able to make something serviceable with 3/4 of
an LM324 and four resistors (forget the balance pot in the diagram --
that's not for an LM324).

Since you didn't mention which sensor you were using, I guess you'll
have to post back with the Honeywell part number to get an idea of the
voltage gain you want & which resistors to choose for the IA, or we'll
have to leave that as an exercise for the student.

Good luck
Chris

While you're at it, James, you might also want to protect the inputs of
the op amp and your uC by adding a few resistors and diodes (1N4148 for
D) like this:
|
| VCC
| + .------.
| | | |
| D - | |\ |
| ^ '--|-\ | ___ ___
| ___ | ___ | >o-|___|-o-|___|--.
| .----------------------|___|-o-|___|--|+/ | |
| | VCC 100 | 1K |/ | VCC |
| | + D - | + |
| | | ^ | |\| |
| | o-----------. | '--|-\ | Vout
| | + | | VCC | | >--o--o---o
| | .-. .-. + === .------. .--|+/| |
| | | | | | | GND | | | |/=== .-.
| | | | | | D - | |\ | | GND | |10K
| | '-' _ '-' ^ '-|-\ | ___ | ___ | |
| | |Out-/ \Out+| ___ | ___ | >-o----|___|-o-|___|--. '-'
| '----o---( V )---o-|___|- -|___|--|+/ | |
| | \_/ | 100 | 1K |/ | ===
| .-. .-. D - ===GND
| | | | | ^ GND
| | | | | |
| '-' '-' |
| | | ===
| o-----------' GND
| |
| ===
GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

This should get you where you want to go. Please post again with the
Honeywell sensor model number, and any other info if you have it (like
a link to the Honeywell data sheet) for advice on choosing resistor
values.

Good luck
Chris

 

[James_sgp]

Chris,

Sorry for not including the part number, here it is: 24PCFFM1G
Its a 0-100 PSI Gauge measuring PCB mount pressure sensor from
Honeywell.
I understand what you`ve mentioned. SO my question would be is how to
configure it so I get one line with a voltage output going to my micro
porcessor?

 

[Chris]

Chris,

Sorry for not including the part number, here it is: 24PCFFM1G
Its a 0-100 PSI Gauge measuring PCB mount pressure sensor from
Honeywell.
I understand what you`ve mentioned. SO my question would be is how to
configure it so I get one line with a voltage output going to my micro
porcessor?

Hi, James. First off, as a Google Grouper (I'm one, too), you might
not know it's considered good form to bottom post. On Google, click
"show options", then "reply". Put your response to the post below the
previous post -- that gives greater continuity, and makes it easier to
answer the previous post with minimum scolling. Also, it helps for
continuity -- most newsreaders don't act like Google Groups. Please
check out Google Groups Help Topic "What's good 'netiquette' when
posting to Usenet?"

http://groups.google.com/support/bin/answer.py?answer=12348&topic=250

Now I've scouted around the Honeywell website, and found something that
may be pretty close -- the 24PCFFM6G:

https://sensing.honeywell.com/index.cfm/ci_id/140301/la_id/1/pr_id/80994.htm

Now assuming your sensor is pretty close, the docs suggest that you
should provide a 2mA costant current source at a nominal voltage of 10V
(but up to 12V will do) to help with temperature compensation. Now for
the sake of simplicity we can just use a resistive voltage divider to
get to 10V, by putting a 1K ohm resistor in series with the nominal 5K
resistance of the bridge, like this:

|
| .--------.
| | |
| | .-.
| | | |1K
| | | |
| | '-'
| | |
| | o-----------.
| | + | |
| | .-. .-.
| | R1| | R2| |
| | | | | |
| | '-' _ '-'
| | |Out-/ \Out+|
| +| o---( V )---o
| ---+12V | \_/ |
| - .-. .-.
| | R3| | R4| |
| | | | | |
| | '-' '-'
| | GND| |
| '--------o-----------'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Now the data sheet suggests that the differential voltage at the output
of the bridge will be 2.25mV per psi with a 100psi max rating. Since
you want that to correspond to a max 5V input, you have to find a way
to amplify that 225mV to5V. That means you need a gain of about 22.
Again, referring to the "Differential Input Instrumentation Amplifier"
schematic on p. 14 of National's AN-31, that means you should use 19K
resistors for R2 and R3, and 220K resistors for R4 and R5, giving you a
gain of 22. Make sure you match the two 10K resistors and the two 220K
resistors with an ohmmeter as closely as possible. Again, the pot R1
is not used with an LM324.

All of this is assuming your sensor has the same specs as the one
mentioned above. You should definitely try to obtain the data sheet
(if necessary, call or email Honeywell for a link) and verify yourself.
Also, if you're interested in temperature compensation, it gets more
complicated -- you should post again if you need this capability.

In the event of overpressure, it's possible the differential voltage
could exceed the maximum input voltage of your uC. It might be good to
put a 5.1V zener from the output of the third op amp of the IA to GND
to limit the maximum output to 5V.

I hope this is enough to get you started. If you need more help,
please include in your post
some information on what this project is about, as well as if this is
just hacking around, a real industrial control project, or a class
project. If the need requires, there are a number of additional
features you can include which will enhance the project. And if this
is a senior project, the above suggestions should be modified quite a
bit for academic purposes.

Good luck
Chris