k..I have designed the circuit..obviously this will not exact one ....please suggest what are the mistakes have i done....
for 30 deg it will produce 0.001018V .so the gain should be 982 to produce 1V
for 60 deg it will produce 0.00677V .so the gain should be 738.5 to produce 5V...here i have designed the instrumentation amplifier

 

k..I have designed the circuit..obviously this will not exact one ....please suggest what are the mistakes have i done....
for 30 deg it will produce 0.001018V .so the gain should be 982 to produce 1V
for 60 deg it will produce 0.00677V .so the gain should be 738.5 to produce 5V...here i have designed the instrumentation amplifier

 

This circuit takes a signal of .156 mV to 6.7766 mV and outputs approx. 58 mV to 4.94 V It is just a simulation and should not be followed exactly but represents something easy to assemble and test.

It does not deal with temp. issues or supply drift, it is simple and performs the required function. After building something like it you can test it in real world conditions and determine if performs acceptably or needs to be improved.

 

d

one input can be ground.

sorry for my inconvinience
Due to some technical issues i can t open this page...so only i give late reply
according to this circuit i designed such that Rgain=1k(if we adjust this as 909 ohm & 738.5 ohm we can get 1 to 5v)at the output
2. and R=10k

is it correct circuit to produce my required output voltage

 

you do not need 5 volts exactly. you simply need an amplified signal. you do not need more than one amplifier. you do not need more than one gain value.
Pick a single gain value that gets you close to 5 volts at 60 deg. C
keep that gain value.
do the math to calculate temp with the high and low voltages in software. it does not matter what the high and low are exactly. you must record the values, calculate the span covered and do math to figure the voltage increase pet degree. this will be done in software.
if you get .675 volts at 25 C and 4.875 volts at 60 C, you write a program that has the voltage change per degree centigrade included as a constant. the software csn then use that value and mathematically calculate your temp from the voltage value it is given by your analog to digital converter

 

Hi Kermit2 thanks for your explanation

Here i have enclosed simulation results
0.00108V i get 0.755V (for 25 deg)
0.0067766V i get 4.955V(for 60 deg)
is it right?
if is it right what will be the further step?
thanks in advance...