output voltage of LM35 ...........

The voltage output of LM35 IC is very low. How can we amplify this voltage? would anyone plz give any suggestion in this regard..... ?

Use op-amp to amplify the signal. How much amplification you need?

Give more info for better results

you can use either an op-amp or a transistor to amplify the voltage

Examples :

**broken link removed**






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.


http://www.ladyada.net/learn/sensors/tmp36.html

thank you... thanx to all of you....:-D
all this material is really helpfull for me......

Here you can see example with uC with code explanation :

Interfacing LM35 Temperature Sensor with PIC Microcontroller
https://extremeelectronics.co.in/mi...-temperature-sensor-with-pic-microcontroller/

https://www.ti.com/lit/ds/symlink/lm35.pdf

I need Maximum output voltage after amplification should be less than 5 volts but greater than 1 volts.

you can use a zener diode or 7805 ic to regulte the output voltage

maria khan said:

I need Maximum output voltage after amplification should be less than 5 volts but greater than 1 volts.

Click to expand...

What is your required temperature range?

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.

Hope this helps.
Tahmid.

lm35 provide output 10mv per degree total 1.25v

regards
Fragrance