# Understanding Responsivity of an IR sensor paired with a black body IR emitter

I'm building an air quality measurement system with the following components:

Reading the sensor datasheet I've seen there's the Responsivity expressed as 150.000 V/W at 10 Hz. Now, I was trying to figure it out how this value is translated on my specific case.

According to Wikipedia

Responsivity measures the input-output gain of a detector system. In the specific case of a photodetector, responsivity measures the electrical output per optical input.

The datasheet of the emitter says that the Optical output power between 4 μm and 5 μm is 4.2 mW. Just for the sake of this example, let's say that the same power is seen by the sensor (of course it is not).

Do I need to check now on the graph of the Responsivity the value for my Frequency, which is, if the wavelength is 4.5 μm, 66.6 MHz? And, with this value, to know the output voltage do I need to multiply the optical power seen by the sensor (that we have supposed to be the same of the emitter, 4.2 mW) to find out the final output voltage of the sensor?

Does it make sense and How the input voltage affects the output as soon as on the datasheet is written: "Output voltage normalized around mid-rail"?

Thanks

• I didn’t see anything in the data sheet about optical power output. Maybe you need to link to something else? – Andy aka Jul 24 at 16:29
• This part of the electromagnetic spectrum is normally described using wavelengths. If you must know the frequency, it is: f = c/λ , where c is the speed of light. 4.5 μm is 6.662E13 Hz – Mattman944 Jul 24 at 16:41
• You said that the optical power is 4.2 mW. Over what solid angle is that specified? If you know this, it becomes a 3D geometry problem. The distance to the sensor and the sensor size must be included in the calculation. – Mattman944 Jul 24 at 16:53
• @Andyaka yeah sorry, wrong link. Now fixed – NicoCaldo Jul 25 at 6:48
• @Mattman944 according to this leister.azureedge.net/-/media/downloads/axetris-docs/… datasheet is Power in < 20° angle – NicoCaldo Jul 25 at 6:49

Do I need to check now on the graph of the Responsivity the value for my Frequency, which is, if the wavelength is 4.5 μm, 222 kHz?

No, whatever power enters through the window of your detector, the detector converts that to a voltage in the ratio of 150 volts per watt.

So, if all 4.2 mW entered the window, you would get a voltage change at the output of 0.15 volts per milliwatt or about 600 mV.

Looking at the data sheet, I believe you meant to write 150,000 volts per watt for the detector’s responsivity.

150000V/W at 10Hz means, that if you modulate your source/the incoming light with 10Hz, you get the 150000V/W sensitivity (pay attention to note 1 in the datasheet).

You usually modulate your light source, and demodulate the measured value, to eliminate any offset errors, which can come from

• temperature changes - sensing element can be very sensitive
• analog frontend

You still need to compensate the demodulated amplitude for temperature changes, or regulate the housing temperature, because the sensitivity of your pyroelectric element is temperature dependent.