Doubts about temperature measurement simulator AD590[Copy Link]
Recently, when I was measuring high temperature, I used to choose the DS18B20 of the digital converter. Because of its low accuracy, I wanted to use the AD590 But the problem arises----- R2 in the circuit converts the current output by the ad590 into voltage, but I use a 1K resistor to adjust it in zero degree water, and the output current seems to be non-zero. Why?
You ask: The output current does not seem to be zero when adjusted in 0 ℃ water. Why?Why did you ask this question?It can be seen from this question that you have not read (at least not carefully read) the AD950 materials.Note the following statement:For supply voltages between+4 V and+30 V the device acts as a high efficiency, constant current regulator passing 1 μ A/K. Laser trimming of the chip's thin film resistors is used to calibrate the device to 298.2 μ A output at 298.2K (+25 ° C)This passage means:The power supply voltage is in the range of+4 V to+30 V, which can be regarded as a high impedance constant current source with a conversion coefficient (i.e. conversion sensitivity) of 1 μ A/K.(Note K).The product is calibrated at 25 ℃ (=298.2 ℃ k), and the output at zero ℃ should be 273.15 μ A. Of course, there will be more than 270 mV.If you try to use the curve of y=ax+b, it will be clear at a glance.If expressed in degrees Celsius, the relationship is: I=a * T+273.15;A=1 mA/° COf course, these are "theoretical values", or design values.It must work within the recommended temperature measurement range.If you need to output 0 at 0 ° C, you need to do further circuit design or use software to solve it..[This post was last edited by xiaoxif at 23:12, July 8, 2009]detailsreplyPublished on July 8, 2009 23:04
You ask: The output current does not seem to be zero when adjusted in 0 ℃ water. Why?
Why did you ask this question?It can be seen from this question that you have not read (at least not carefully read) the AD950 materials.Note the following statement:
For supply voltages between+4 V and+30 V the device acts as a high efficiency, constant current regulator passing 1 μ A/K. Laser trimming of the chip's thin film resistors is used to calibrate the device to 298.2 μ A output at 298.2K (+25 ° C) This passage means: The power supply voltage is in the range of+4 V to+30 V, which can be regarded as a high impedance constant current source with a conversion coefficient (i.e. conversion sensitivity) of 1 μ A/K.(Note K). Product factory fine adjustment calibrationIt is carried out at 25 ℃ (=298.2 ℃ k), and the output should be 273.15 μ A at zero ℃. Of course, more than 270 millivolts will appear.If you try to use the curve of y=ax+b, it will be clear at a glance. If expressed in degrees Celsius, the relationship is: I=a * T+273.15;A=1 mA/° C
Of course, these are "theoretical values", or design values.It must work within the recommended temperature measurement range.If you need to output 0 at 0 ° C, you need to do further circuit design or use software to solve it. .
[This post was last edited by xiaoxif at 23:12 on July 8, 2009]
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Building owner|Published on July 9, 2009 at 19:05Only the author
Thank you upstairs. I'm dizzy. Go down and think about it------------------------ However, the problem now is that the ad590 outputs 270uA at zero degrees, OK!Put it in 50 ° water, its current is only 284. The circuit is directly connected with a 10K resistance. What's more strange is that if you reduce the 10K resistance, the current will rise. Why?? Is there something wrong with the device?? The answers given by ls are very detailed. I am very benefited,
I want to use AD590 to design a temperature measurement module.Which prawn can help me.Requirements: design the most accurate temperature with the least components and the least price.Thank you, little brother~
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