Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
It has an output voltage that is proportional to the Celsius temperature.
The scale factor is .01V/C
The LM35 does not require any external calibration or trimming and maintains an accuracy of +/-0.4C at room temperature and +/- 0.8C over a range of 0C to +100C.
Another important characteristic of the LM35DZ is that it draws only 60 micro amps from its supply and possesses low self-heating capability. The sensor self-heating causes less than 0.1C temperature rise in still air.
You will need to use a voltmeter to sense Vout.
The output voltage is converted to temperature by a simple conversion factor.
The sensor has a sensitivity of 10mV / C.
Use a conversion factor that is the reciprocal, that is 100C/V.
The general equation used to convert output voltage to temperature is:
Temperature (C) = Vout * (100C/V)
So if Vout is 1V , then, Temperature = 100C
The output voltage varies linearly with temperature.
Let's say your temperature range is 0'C to 60'C. So, output voltage varies between 0V and 0.600V. So, if you have an operational amplifier circuit configured for 5X gain, the output voltage will be amplified to a range of 0V to 3.00V. So, each 'C is now (0.01V*5)=0.05V.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
