am i doing this right?

[philip k]

i'm attempting a build.  

I will be using 5 pressure sensors.  here are the specs:

Pressure Range: 0-200 psi
Output: 0.5V – 4.5V linear voltage output. 0 psi outputs 0.5V, 100 psi outputs 2.5V, 200 psi outputs 4.5V
Working Temperature: -40—+125ºC;.
Accuracy: ±1%FS;
Thread: 1/8"-27 NPT.
Overload Capacity: 2 times;
Pressure Medium: The gas and liquid which is compatible with 316L stainless steel;
Load Resistance: ≤(supply power-6.5V/0.02A)Ω;
Long-term stability: ± 0.2%FS/year;
Temperature Effect on Zero: Typical: ±0.02%FS/ºC; Maximum: ±0.05%FS/ºC;
Temperature Effect on Sensitivity: Typical: ±0.02%FS/ºC; Maximum: ±0.05%FS/ºC;
Shock Resistance: 100g;
Anti-Shock: ≤±0.01%FS(X,Y,Z axes, 200Hz/g);
Response Time: ≤1ms;
Insulation Resistance: >100mΩ  500VDC;
Explosion-proof Class: ExiaTTCT6;
Electromagnetism Compatibility: EN50081-1; EN50082-2; IEC6 1000-4-3 ;
Wiring connector: water sealed quick disconnect.  Mating connector and wire harness (pigtail) is included.

Wiring: Red for IN+.  Black for GND.  Green for OUT.

would i be correct by running a 600Ω resistor parallel to each sensor to bring the voltage down to ≤1.8v and go from pin 7 or 8 to the five sensors, then run sensor1 signal and a 600ohm resistor in parallel to pins P9 35, sensor2 signal and a 600ohm resistor in parallel to P9 36, and so on?

thanks for any input.  

[Dennis Lee Bieber]

On Tue, 30 May 2017 10:50:55 -0700 (PDT), philip k
<kel...@gmail.com> declaimed the following:

>i'm attempting a build.
>
>I will be using 5 pressure sensors. here are the specs:
>
>Pressure Range: 0-200 psi
>Output: 0.5V – 4.5V linear voltage output. 0 psi outputs 0.5V, 100 psi
>outputs 2.5V, 200 psi outputs 4.5V

I'm presuming this is the critical specification...

>Working Temperature: -40—+125ºC;.
>Accuracy: ±1%FS;
>Thread: 1/8"-27 NPT.
>Overload Capacity: 2 times;
>Pressure Medium: The gas and liquid which is compatible with 316L stainless
>steel;

Don't think knowing what size pipe thread it uses means much here...

>Load Resistance: ?(supply power-6.5V/0.02A)?;

This might be important...

>Wiring: Red for IN+. Black for GND. Green for OUT.
>

>would i be correct by running a 600? resistor parallel to each sensor to
>bring the voltage down to ?1.8v and go from pin 7 or 8 to the five sensors,

>then run sensor1 signal and a 600ohm resistor in parallel to pins P9 35,
>sensor2 signal and a 600ohm resistor in parallel to P9 36, and so on?
>

I doubt it.

What you most likely need is a voltage divider system (use fixed width
font):

4.5V ------+
|
~ ?1
|
+ 1.8V ----->
|
~ ?2
|
GND ------+----------->

where ~ is resistor, values to be determined (do a Google on how to
determine a voltage divider -- I'm not going to do all the work <G>)

{Unstudied: 4.5 / 1.8 = 2.5; say a 240Ohm for ?2, total needs to be 240*2.5
=> 600, so ?1 would be 600 - 240, or 360Ohm...}

--
Wulfraed Dennis Lee Bieber AF6VN
wlf...@ix.netcom.com HTTP://wlfraed.home.netcom.com/

[philip k]

ah cool, cool.  i had read and done some math, but i was totally wrong in my understanding of how the circuit was set up.  thanks for steering me in the right direction!

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[Dennis Lee Bieber]

On Tue, 30 May 2017 19:03:14 -0500, philip k
<kel...@gmail.com> declaimed the following:

>ah cool, cool. i had read and done some math, but i was totally wrong in
>my understanding of how the circuit was set up. thanks for steering me in
>the right direction!
>

Might be able to use a potentiometer to work out the resistances...

Outer legs on the GND-SensorOut, feed max voltage (4.5?), then adjust
wiper until a DVM reads 1.8V between wiper and GND. Then, without changing
potentiometer setting, use measure resistance between GND leg and wiper,
and between wiper and sensor leg.

[John Syne]

You need to pay attention to data acquisition circuits that use a input mux and a sample and hold circuit like that used by the BBB analog circuits. When the ADC samples the analog input, it charges a small capacitor (sample and hold) and then the ADC converts the voltage stored in the capacitor. When the capacitor is connected to the input circuit, it represents a short circuit and then the capacitor charges based on the time constand t = rc, where r is the source impedance. This is apparent if the channel before is fully 0 or full scale as the capacitor in the sample an hold could be 0 or 1v8 or somewhere in between and that can affect the accuracy of your measurements.

A better approach is to use an opamp circuit as a voltage divider as it will present as a low impedance to the sample and hold circuit and eliminate bleed through from adjacent circuits.

Regards,
John

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