How to produce a negative ohms reading on a meter?

[Adrian]

I am trying to construct a digital pressure gauge by using a pressure sender
(10-180ohms over 0-150psi) and a volt/ohmmeter.

I already have a commercial unit and have taken psi-to-ohms measurements.
Performing a regression on the data I achieve an approximately linear
function with the equation:

PSI = 0.8289 x ohms - 13.0786

In other words, if the pressure sender reads 100 ohms, the actual pressure
is 0.8289(100)-13.0786 = 70psi which is the value I would like to display on
a volt/ohm meter.

How would I implement this function with a resistor network? I think the
most difficult part is to get the negative value.

 

[Robert Baer]

I am trying to construct a digital pressure gauge by using a pressure sender
(10-180ohms over 0-150psi) and a volt/ohmmeter.

I already have a commercial unit and have taken psi-to-ohms measurements.
Performing a regression on the data I achieve an approximately linear
function with the equation:

PSI = 0.8289 x ohms - 13.0786

In other words, if the pressure sender reads 100 ohms, the actual pressure
is 0.8289(100)-13.0786 = 70psi which is the value I would like to display on
a volt/ohm meter.

How would I implement this function with a resistor network? I think the
most difficult part is to get the negative value.

1) Are you saying that the transducer is a 2-terminal resistive device?
2) Are you saying the transducer minimum PSI indication is -13?
In any event, one must force some kind of a current to create a
voltage across the transducer output.
If it is a 2-terminal resistive device, then put a resistor in series,
value approximately 13.0786/0.8289 or a 15.8 ohm 1% resistor.
Then force a current of 828uA to produce a voltage equal to PSI; do
this with an adjustable current source.
Add a second "arm" having a 13.078mV source (set by an adjustable
voltage regulator) to subtract that value from the total voltage from
the resistor-plus-transducer network.
Then use a 10 meg DVM to read the voltage across this composite
network; the loading will have no significant effect on the actual VS
measured value.

Why did i choose 828uA to "generate" PSI in mV instead of 828mA to
"generate" PSI in V?
That amount of current would destroy the transducer.
Now if you do not mind a 10X scale factor, then use 8.28mA and
130.78mV subtractive reference.
I would guess that 8mA would not cause heating problems in the
transducer.

 

[Adrian]

Thanks for the reply Robert,

1) Are you saying that the transducer is a 2-terminal resistive device?

Yes. From measurements and datasheets it seems to be a linear resistive
device.

2) Are you saying the transducer minimum PSI indication is -13?

No. The minimum transducer resistance is approx. 10 ohms, but this
corresponds to a 0 psi condition.

Other than that, your calculations seem to work, except that I am trying to
run the whole thing off a 12V supply (the display I have now runs off 12VDC)
so how would I accomplish the equivalent of a constant current source and
the negative voltage?

Thanks in advance.

 

[Ben Pope]

Other than that, your calculations seem to work, except that I am
trying to run the whole thing off a 12V supply (the display I have
now runs off 12VDC) so how would I accomplish the equivalent of a
constant current source and the negative voltage?

Current regulating diode and a dc-dc convertor?

Ben

 

[Fred Bloggs]

I am trying to construct a digital pressure gauge by using a pressure sender
(10-180ohms over 0-150psi) and a volt/ohmmeter.

I already have a commercial unit and have taken psi-to-ohms measurements.
Performing a regression on the data I achieve an approximately linear
function with the equation:

PSI = 0.8289 x ohms - 13.0786

In other words, if the pressure sender reads 100 ohms, the actual pressure
is 0.8289(100)-13.0786 = 70psi which is the value I would like to display on
a volt/ohm meter.

How would I implement this function with a resistor network? I think the
most difficult part is to get the negative value.

Please view in a fixed-width font such as Courier.


R1 R2
+---/\/\-----+-----/\/\--+
| | |
| | |
| 12V / | | ---
| | / -|-+
RL | |/ |
+---/\/\-----+---- < |
| |\ |
| | \ +|-+
| --- \ | |
| |
| |
| R1 | R2
+-------------/\/\--------+---/\/\--+-o 12V
| |
| | 12 R1 |
| | I= -- x -- |
|\|/ RL R2 R4 |
| +--/\/\--+------+
| | |
| R3 | | \ |
+--------/\/\-------|---|+ \ |
psi | | | \|
+-----+ | | >---+--> Vout
| / | | | /| |
| Rs | +---|- / | |
| / | | | / --- |
| \ | | |
+-----+ | |
| R6 | R5 |
+-----------/\/\----+-----/\/\---+
|
-+- R5 R5
Vout=(1+ ------)xIxRs - 12 x --
R4||R6 R4


Vout range 0 - > 2V


R5
12 x -- = 0.13086V
R6

R5
(1+ ------)xI = 8.289mV
R4||R6


sample solution:

RL= 1.21K R1= 49.9k R2=100k I=5mA

R3= 680 R4=100k R5=1.1k R6= 1.69k

OA= LM358 Voltage Reading x 100 = PSI

 

[Tom Bruhns]

Make the resistance one arm of a Wheatstone bridge.

That might not be quite enough info for the OP to make it all work...

OP, the idea is that with a bridge, you can introduce whatever offset
you want by unbalancing the bridge. You do need to take the output as
the voltage between two points across the bridge, so if your indicator
requires a ground-referenced input voltage, you'll want to add a
differential amplifier to the bridge. You can build one with a
rail-to-rail output op amp. The bridge output won't be quite linear
in volts vs ohms, but it just might help linearize your sensor,
actually. To get close to linear, use a resistor in series with the
sensor whose value is large compared with the sensor's max resistance
(and be prepared to deal with relatively low output voltage). You can
play with different values for the three fixed legs using a
spreadsheet, for example...though really, one side will just be a
constant voltage, for a constant drive to the "top" of the bridge. To
maintain stable sensitivity, you should regulate the voltage going
into the top of the bridge, too.

(Simplest hardware would likely be a microcontroller with internal ADC
and a display module...and the sensor with a series resistor.)

Cheers,
Tom

 

[Fred Bloggs]

R5 R5
Vout=(1+ ------)xIxRs - 12 x --
R4||R6 R4


Vout range 0 - > 2V


R5
12 x -- = 0.13086V
R6

R5
(1+ ------)xI = 8.289mV
R4||R6


sample solution:

RL= 1.21K R1= 49.9k R2=100k I=5mA

R3= 680 R4=100k R5=1.1k R6= 1.69k

OA= LM358 Voltage Reading x 100 = PSI

There is no need for a rail-to-rail OA in this application, many can't
even handle a 12V supply anyway. You can use a single-supply workhorse
of the industry like the LM358 and give it a little assist by adding a
diode to the circuit like shown. The current setting 1.21K RL should be
replaced with 1.0K in series with a 500 ohm pot for gain adjust, the R4
should be a fixed 100K in series with a 100K pot for zero adjust. Then
calibrate by substituting 10 ohm for the psi element, adjust the 100K
for zero out, substitute 180 ohm for the psi element, and adjust the 500
ohm pot for 1.50V on the meter etc...

Please view in a fixed-width font such as Courier.


R1 R2
+---/\/\-----+-----/\/\--+
| | |
| | |
| 12V / | | ---
| | / -|-+
RL | |/ |
+---/\/\-----+---- < |
| |\ |
| | \ +|-+
| --- \ | |
| |
| |
| R1 | R2
+-------------/\/\--------+-----/\/\--o 12V
|
| | 12 R1
| | I= -- x -- 12V
|\|/ RL R2 R4 |
| +--/\/\--+
| | |
| R3 | | \ |
+-----------/\/\----|---|+ \ |
psi | | | \|
+-----+ | | >---+--> >------+
| / | | | /| | +|
| Rs | +---|- / | | +-------------+
| / | | | / --- | | _ _ _ |
| \ | | | || | | | | ||
+-----+ | | | - - - |
| R6 | R5 | ||_| o |_| |_||
+-----------/\/\----+-----/\/\---+ +-------------+
| -|
+-----------------------------------> >------+
|

 

[Adrian]

Wow, you guys are awesome! Thanks for all the replies, you definitely got
me thinking in the right direction!

Thanks a lot for the circuit Fred! I had to dig out my old op-amp books to
go over the calculations (because I was confused...) but they seem right and
it seems like it should work!

From what I gather, the top op-amp circuit is to generate a 5mA constant
current source which drives the transducer. The second opamp circuit
performs the multiplication function (through the R4, R5, R6 combination)
and the subtractive function (through R5 and R4). Right?

Does it matter what the value of R3 is since no current flows through that
terminal anyways? Also, what is the diode for in the second circuit?

Thanks Again!

Adrian

 

[Fred Bloggs]

From what I gather, the top op-amp circuit is to generate a 5mA constant
current source which drives the transducer. The second opamp circuit
performs the multiplication function (through the R4, R5, R6 combination)
and the subtractive function (through R5 and R4). Right?

Yes- exactly.

Does it matter what the value of R3 is since no current flows through that
terminal anyways?

No- it is was originally included out of habit- to cancel offset current
effects- but in this case and with this OA , it can be removed.

Also, what is the diode for in the second circuit?

When the transducer hits 10 ohms, the OA will have 5mA x 10 ohm= 50mV at
each input. This means the OA output is sinking 50mV/R5~50uA of current
to sustain the output at 0V. According to the LM358 datasheet, the
output can go no lower than 500-600mV with that loading. The diode
creates a new reference of Vf~700mV @5mA for the OA inputs. The OA
output relative to gnd is now shifted upward by Vf of the diode, the
difference equation remains the same except for a small addition
+Vf*R5/R4~7mV, a constant eliminated with the zero adjust procedure. Now
when the sensor is at 10 ohms, the OA output sinks 50uA as before, but
at a voltage of Vf- a full 700mV above its gnd rail. By making the meter
measurement between OA out and the diode as shown, you measure only that
portion of the OA output due to the difference equation, and you get the
0V reading as required. Also, a shorted sensor should produce a UL
display, and an open sensor definitely should produce an OL display.