http://www.sparkfun.com/datasheets/Sensors/Temperature/MLX90614_rev001.pdf
http://www.ge-mcs.com/download/temperature/920_159a.pdf
http://www.ge-mcs.com/download/temperature/920_159a.pdf
http://www.atmel.com/dyn/products/product_card.asp?part_id=4720
I don't understand, how capacitors would reduce the gain of a thermopile amplifier, which should be basically DC capable for a thermometer.The output from the ztp135 is around 1.8 mV then 1000 gain with capacitors involved I got out only .288 V
I don't understand, how capacitors would reduce the gain of a thermopile amplifier, which should be basically DC capable for a thermometer.
No. The amplifier circuit needs a virtual ground, like previously suggested in post #8.So I need to have a third resistor that equal R1 // R2 on positive terminal on the thermopile ampifier?
As said it's a resistor. It has no "output" on it's own. When designing a voltage divider, the fixed resistor value should be in the same order of magnitude as the thermistor resistance, about 100K, if I remember the datasheet right. The circuit can be according to post #8 as well.The thermistor itself output is really small
In a slightly simplified view, the thermopile measures a temperature difference between an object and the thermopile temperature. To get the object temperature, you have to add both measurements. I have describe in post #6, how the thermistor measurement can be converted to a temperature value.Could you please explain it to me on the ambient compensation part in this design ?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?