AD590 with arduino uno connection and calculation

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1

hi,

I have a ad590jh 3 ping round head sensor , which i have no clue how to connect and retrieve
temperature data . I am an hobbyist, so just trying to play with it and got some results,
so far i have connected sensor vcc to 5v , sensor vdd to ground and the third pin to A0 which is also connected to ground via 5k resistor.
i am getting a result which i assumes in Kelvin so i subtracted the absolute kelvin with the received value to obtain result in celcius. the result data is

Kelvin :306
Celcius :32.85
Kelvin :304
Celcius :30.85
Kelvin :305
Celcius :31.85
Kelvin :305
Celcius :31.85
Kelvin :307
Celcius :33.85
Kelvin :306
Celcius :32.85
Kelvin :306
Celcius :32.85
Kelvin :307
Celcius :33.85
Kelvin :309
Celcius :35.85
Kelvin :304
Celcius :30.85
Kelvin :307
Celcius :33.85
Kelvin :307
Celcius :33.85
Kelvin :308
Celcius :34.85
Kelvin :309
Celcius :35.85
Kelvin :306
Celcius :32.85
Kelvin :307
Celcius :33.85
Kelvin :305
Celcius :31.85
Kelvin :311
Celcius :37.85

this code i have written is a very simple analog read, and subtraction . the output spits once per second. And i am 100% sure the values are jumping very much here and there. I am having a pre built temperature sensor which i am using as control. it was showing consistantly at 33 deg Celcius when this output is coming.

I know this is not correct, so any help and pointers how to connect ad590jh with arduino uno to get reliable value.

thanks in advance.

2

Which arduino do you have ?
What is power source for arduino ?
Have you measured power source for arduino with meter ?
Havd you measured arduino 5V pin with meter ?
Post your schematic and code.

3

So how have you wired it up, schematic please and what is this code, please use the <> icon when posting it, post between the tags that appear when this icon is pressed.

4

Hi,

, which i have no clue how to connect and retrieve
temperature data

That is what the data sheet you linked us to is about, have you read it?

Although your device is the three pin case, only two are still used, connecting + directly to 5V and the - to gnd may have damaged the sensor.
Do you have the 580 or 590, the 580 is the three pin operation.

You need to connect +ve to +5V and a current limit resistor from -ve to gnd and measure the voltage between gnd and -ve terminal.

The 590 is a current output device.

Tom.... :slight_smile:

5

Nominal current at 25 degrees Celsius is 298.2 uA, so with a 10 k 1% resistor in SERIES , the voltage drop across the resistor is 2.982 Volts.
I the resistor is connected to ground, with thd sensor in series, connectrd to +5V , the analog input can measure the voltage drop across the resistor ,divide by 10000 to obtain how many uA At 1 uA / degree K , 298.2 degees K = 25 degees Celsius = 2.982 V .

6

raschemmel:
Which arduino do you have ?
What is power source for arduino ?
Have you measured power source for arduino with meter ?
Havd you measured arduino 5V pin with meter ?
Post your schematic and code.

  1. Arduino UNO
  2. Desktop windows 7 intel i3
  3. No
  4. Yes

the schematic is attached and the code is well i am lets just say pretty ordinary one though,
don't scold me

float tempvalue=0;
void setup() {
  // initialize serial communication at 9600 bits per second:
  Serial.begin(9600);
}

// the loop routine runs over and over again forever:
void loop() {
  // read the input on analog pin 0:
  
  // print out the value you read:


  int sensorValue = analogRead(A0);
  Serial.print ("Kelvin :");
  Serial.println(sensorValue);
  tempvalue=sensorValue-273.15;
  Serial.print("Celcius :");
  Serial.println(tempvalue);
  delay(1000);        // delay in between reads for stability
}

AD590 schematic.jpg
AD590 schematic.jpg1179×756 75.5 KB

7

When you said you measured the 5V pin you didn't post the voltage you measured so I'll take that to mean you simply verified it was in fact 5V.

First of all, I'd like to say that it appears you have not done any homework on this.
This is one of the first links that pops up on Google with the search "AD590JH SCHEMATIC"

The datasheet shows the third pin unmarked , as it should be since it is the case.

AD590 WIRING SCHEMATIC.jpg
AD590 WIRING SCHEMATIC.jpg1920×1155 515 KB

The following circuit should have an output of 2.982 V @ 25 deg C

AD590 SCHEMATIC.jpg
AD590 SCHEMATIC.jpg3060×2310 151 KB

AD590 AP NOTES

AP NOTE AN-273

NOTE: AN-273 SHOWS AD590 as CURRENT SOURCE driving 1 k /0.1% precision resistor create voltage source (1 uA/deg K)

R = 1000 ohm

I = 298.2 uA @ 25 deg C

V(25 deg C) = I * R = 0.0002982 A * 1000 ohms = 0.2982 V

Vamplified(A = 10) = 0.2982 * 10 = 2.982 V @ 25 deg C

Regardless of how many pins are on the device , the datasheet clearly states it is a TWO terminal device , so the pin you labeled "ADJ" is science fiction.

The AD590 is a 2-terminal integrated circuit temperature transducer
that produces an output current proportional to absolute
temperature. For supply voltages between 4 V and 30 V, the device
acts as a high impedance, constant current regulator passing
1 μA/K. Laser trimming of the chip’s thin-film resistors is used
to calibrate the device to 298.2 μA output at 298.2 K (25°C

I found this schematic which shows the device connected in series with a fixed resistor and a pot with the object of calibrating the device by adjusting the series resistance as close to 10000 ohms as possible.

EVERY SINGLE SCHEMATIC I can find online using Google Images shows the device connected in series with 10 k ohms.
NOT ONE shows the device having a THIRD pin labeled ADJ.
Did you just make that up ?

The schematic you posted is nonsense, since you are trying to read the case.

The question now , is have you damaged the device by connecting it directly to the power supply with no series resistor. The circuit schematic appears to be quite complex so I would not rule out the possibility of short circuit protection. If the current through the device is 1 uA per degree K and 298.2uA @ 25 degrees celsius, and it has some protection, the device would look like an OPEN circuit to the 5 V power supply .

Try connecting it with a 10 k resistor in series first with the resistor end connected to GND and see if the voltage across the resistor is 2.982 V dc. If not , then you probably fried it when you connected it directly across the supply, but somehow I doubt it is damaged, I think you have yet to connect it correctly. We'll know after you try that. Next time try doing some homework before you start messing around with electronics you are not familiar with. Normally , I would be nicer about this but in your case you didn't seem have made an effort to research the subject. You could not have read the datasheet or looked at any schematics online. Where you got your schematic is a mystery. Where did you get it ?

8

NOT ONE shows the device having a THIRD pin labeled ADJ.
Did you just make that up

Well I will bet that there was not a symbol for his sensor so he just used another one from Fritzing. This is why we hate Fritzing here as it leads to a lot of time wasting like this.

9

Well I will bet that there was not a symbol for his sensor so he just used another one from Fritzing. This is why we hate Fritzing here as it leads to a lot of time wasting like this.

Fritzing isn't to blame here. It took me 10 seconds to find schematics for that part and they all show it as a two-port device , with the three pin version having the third pin tied to case. The datasheet said it was a two terminal device. I couldn't find any parameter suggesting there was a maximum current which leads me to believe the device is actually short circuit protected so if it is connected without a current limiting resistor it will simply not burn up.

10

Hi,

raschemmel:
Fritzing isn't to blame here. It took me 10 seconds to find schematics for that part and they all show it as a two-port device , with the three pin version having the third pin tied to case. The datasheet said it was a two terminal device. I couldn't find any parameter suggesting there was a maximum current which leads me to believe the device is actually short circuit protected so if it is connected without a current limiting resistor it will simply not burn up.

Ditto.....I didn't provide as much info a raschemmel as I was trying to get the OP to read the data sheet.
Tom..... :slight_smile:

11

The more you read about the AD590JH , the more you want one.

12

After reading the first page.. This is a most accurate and valuable/usable part...
A 10 K .1% resistor, better a .01% resistor makes a Most accurate temperature sensor.
Designed for long twisted cable with virtually no errors.. IE Cat 5 cable..
Preferable to use a 10 K Very low Temperature Coefficient resistor.
10 PPM is the best, But very expensive....
1 % is a 1% error added to the .5%/c stated error is ? OK but not the best... that the part is capable of.
Just my .02 worth...

Doc

13

Hi,
Thanks for all the details provided, yes I think i bricked the sensor, And I am sure i will take time read the docs first then play with them next. Any way as i am not having a second sensor, i have to purchase it, one more question

LM35DZ is much more cheap and probably gives (+)(-) .5 celcius difference than AD590JH which by
datasheet says 1.5 deg celcius difference and much more costly .
which one is more reliable.

note@raschemmel :

Special Thanks for taking such a huge time to teach.

14

Reliability will be the same, the devices are all the same, the more accurate ones are selected at the factory.

Thank raschemmel by clicking on his Kama icon.

15

Grumpy_Mike:
Reliability will be the same, the devices are all the same, the more accurate ones are selected at the factory.

Thank raschemmel by clicking on his Kama icon.

Thanks for karma , I cannot understand what you are saying, "the more accurate ones are selected at the factory" means?

16

note@raschemmel :

Special Thanks for taking such a huge time to teach.

Your welcome. I have to do something to offset all the OP's who don't appreciate my methods, ha ha.

Are you sure you don't have the AD580 ?

Can you take a CLOSE UP photo of the top of the part showing the part number and post it ?

You understand now that the sensor needs a 0.1 % (or 0.01%) 1 k resistor in SERIES with the sensor and that the "+" pin of sensor goes to ARDUINO 5V and the other pin goes to the resistor and the OTHER end of the resistor goes to GND (arduino GND). This converts part from current source to voltage source and the analog pin connects to the end of the resistor that connects to the sensor ?

Please confirm you understand that now .