# thermopile circuit design

#### riyashini

Joined Sep 18, 2014
20
hi..
i have collected the following data from ztp135sr datasheet
for 25 degree the output will be 0.156mV(approximately)
for 80 degree the output will be 10.56mV(approximately)
if i want to produce 0 to 5 v respectively for 30 degree to 60 degree what are the steps should i follow?
0 to 5 v should be given in to ADC..
this ZTP135SR consists thermister which will use for temperature compensation.i have to consider that one too..
can any one suggest me the things which i have to follow to build the signal conditioning unit for thermopile

#### wayneh

Joined Sep 9, 2010
17,498
Take a look at op-amp "instrumentation amplifier". You want to be able to have control over offset (zero) and gain (span).

Oh wait, you can do the zero and calibration in software. In that case you just need an op-amp to give a gain of about 5v/10mv =500.

#### Kermit2

Joined Feb 5, 2010
4,162
you have a 55 degree span and 10.404 mv change across that span.
Figure out the mv change per degree. Add that amount to your base value of .156 mv for each degree above 25. adding that amount 5 times will tell you the output in mv at 30, etc. etc.

#### wayneh

Joined Sep 9, 2010
17,498
Just stay aware that the response is not linear and will need to be calibrated.

#### riyashini

Joined Sep 18, 2014
20
thanks for ur reply wayneh and kermit2.actually I m the beginner for electronics..
first i have to calculate the voltage for one degree celcious right?
after that i have to build a signal conditioning circuit using amplifier..here i got some application diagram of ZTP135SR...but it is quit confusing me

#### riyashini

Joined Sep 18, 2014
20
actually 80-25=55 and 10.56-0.156=10.404mV
hence for 1 degree i get=0.189mV..
for 30 degree i get 5.67 mV which i have to convert into 1v right?

#### wayneh

Joined Sep 9, 2010
17,498
first i have to calculate the voltage for one degree celcious right?
Not really. Your main concern is the maximum mV you need to measure, which sounds like it is ~10mV in your case. You want that to map out to 5V fed to your ADC. That defines an amplifier with a gain of 500. A single op-amp can accomplish that. That op-amp will need a DC power supply giving at least 5V (more would be better, say 9V or 12V).

You will also want to be sure that op-amp can sense down to the lowest input voltage you need. If you really need to precisely handle a voltage of <1mV, you'll need a dual power supply, one that supplies ±9V for instance.

#### riyashini

Joined Sep 18, 2014
20
Not really. Your main concern is the maximum mV you need to measure, which sounds like it is ~10mV in your case. You want that to map out to 5V fed to your ADC. That defines an amplifier with a gain of 500. A single op-amp can accomplish that. That op-amp will need a DC power supply giving at least 5V (more would be better, say 9V or 12V).

You will also want to be sure that op-amp can sense down to the lowest input voltage you need. If you really need to precisely handle a voltage of <1mV, you'll need a dual power supply, one that supplies ±9V for instance.
but i want to get 5v for 60 degree ....then what will i do?

#### riyashini

Joined Sep 18, 2014
20
then one basic question
i want to check the sensor in sunlight..will it need a powersupply or it simply produces the voltage corresponding to the sunlight temperature without the powersupply

#### riyashini

Joined Sep 18, 2014
20
you have a 55 degree span and 10.404 mv change across that span.
Figure out the mv change per degree. Add that amount to your base value of .156 mv for each degree above 25. adding that amount 5 times will tell you the output in mv at 30, etc. etc.
sorry.first point i understood...remaining things can u explain breifly

#### Kermit2

Joined Feb 5, 2010
4,162
You must amplify the thermopile output, if you desire a higher voltage level.
Decide where Max. will be on temp scale. You refered to 60 deg. as a max earlier.
Calculate the gain needed to get your desired voltage level by dividing it by the thermopile voltage at your max. temperature.
You said 5 volts. thermopile output at 60 deg. is 6.7766 mv, so 5 div. by .0067766 is 737.8
that would be the amplification factor(gain) which would produce a 5 volt ouput when the thermopile output is 6.7766 mv.

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#### Kermit2

Joined Feb 5, 2010
4,162
.18916 mv per degree
.156 mv output at 25 degree.
.156 + [(temp - 25) x .18916] = new millivolt level at new temp.
Example:
50 degree

.156 + [(50-25) x .18916] = 4.885 mv

#### wayneh

Joined Sep 9, 2010
17,498
then one basic question
i want to check the sensor in sunlight..will it need a powersupply or it simply produces the voltage corresponding to the sunlight temperature without the powersupply
It will produce a voltage. However to condition that small voltage for your ADC, you will need to 1) amplify it, 2) compensate it for the temperature of the sensor itself using the thermistor signal, and 3) filter out noise. These functions require a power supply and circuitry. Step 2 might best be done after the ADC but it will require a channel for the thermistor measurement.

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#### Kermit2

Joined Feb 5, 2010
4,162

one input can be ground.

#### riyashini

Joined Sep 18, 2014
20
You must amplify the thermopile output, if you desire a higher voltage level.
Decide where Max. will be on temp scale. You refered to 60 deg. as a max earlier.
Calculate the gain needed to get your desired voltage level by dividing it by the thermopile voltage at your max. temperature.
You said 5 volts. thermopile output at 60 deg. is 6.7766 mv, so 5 div. by .0067766 is 737.8
that would be the amplification factor(gain) which would produce a 5 volt ouput when the thermopile output is 6.7766 mv.
k...
0.156+[(60-25)*0.189]=6.7766mV at 60 degree.we r going for the instrumentation amplifier for linear response..here i have attached basic circuit for thermopile ZTP135SR.may i follow this?
and how can we compensate the thermopile temperature using thermister for 30 degree to 60 degree?

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#### riyashini

Joined Sep 18, 2014
20
or if i follow the instrumentation amplifier how can i compensate the thermopile temperature?

#### Kermit2

Joined Feb 5, 2010
4,162
take this in small bites.
Build a circuit to amplify the milli-volt output of your thermopile. work on that until you get what you desire. Be aware that often we want more precision in our circuits than they need or than we are capable of building. you can easily work with an amplified range of 1 volt - 4.9 volt. it need not be 0 to 5 to achieve your goal.
browse some suppliers and compare specs on instrumentation op amps. pick 2 or 3 and draw up a simple circuit using app notes or spec sheet and post it back here. We will help all you can stand, but will not "do it" for you. you try. you ask. we suggest or correct.

#### riyashini

Joined Sep 18, 2014
20
take this in small bites.
Build a circuit to amplify the milli-volt output of your thermopile. work on that until you get what you desire. Be aware that often we want more precision in our circuits than they need or than we are capable of building. you can easily work with an amplified range of 1 volt - 4.9 volt. it need not be 0 to 5 to achieve your goal.
browse some suppliers and compare specs on instrumentation op amps. pick 2 or 3 and draw up a simple circuit using app notes or spec sheet and post it back here. We will help all you can stand, but will not "do it" for you. you try. you ask. we suggest or correct.
thank u so much ...i ll be back with a circuit ..

#### wayneh

Joined Sep 9, 2010
17,498
I agree with Kermit but would add one point: Define your goal, with numbers. Think about the accuracy you need, in ±°C, and how you will verify the accuracy of your measurements. What standard will you use? Then think about precision, again in ±°C, the repeatability of your measurements. Circuits and sensors are affected by ambient conditions, so if you need tight precision you will need to consider these.

How much processing of your data can be done in software? For instance it will be far easier to compensate for sensor temperature in software than with a circuit.

#### Kermit2

Joined Feb 5, 2010
4,162
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.