1
\$\begingroup\$

I have been able to use the following photodiode sensor board to detect changes in the output of the sensor when a UV plasma is active. However, changes sensed when the UV plasma is active are very small (0.01 - 0.05 mV). This is when the following board is directly connected to the the 5V supply of the arduino. How would I go about amplyfying this output signal? I've carried out some research on Op-Amps, and thought I might be able to use it in a transimpedance configuration. Would this work?

Photodiode board: https://www.adafruit.com/product/1918

Datasheet: https://cdn-shop.adafruit.com/datasheets/1918guva.pdf

Thank you for all your replies.

This is the schematic of the photodiode break-out board. Other photodiode's I have tested prior to this operated in large wavelength ranges (350nm to 1100nm) and as expected the output contained a lot of noise, probably due to ambient lighting. However, with a wavelength range of 240nm to 380nm this break-out board does not respond to changes in ambient lighting but only to the presence of weak-UV radiation. It is consistently at 0V till plasma is active. The only problem is that this signal is very small (mV magnitude)

enter image description here

\$\endgroup\$
5
  • 1
    \$\begingroup\$ The board already has an amplifier on it per the description, so I suspect you're picking up noise rather than actual UV there. \$\endgroup\$ – pjc50 Dec 20 '17 at 14:49
  • 1
    \$\begingroup\$ UV sterilization uses UVC which is 100-280nm. Your sensor is designed for UVB (280-315nm) and has only a little bit of sensitivity below 280nm according to the datasheet. Your best bet may be to find a UVC sensor. \$\endgroup\$ – Frosty Dec 20 '17 at 15:23
  • \$\begingroup\$ I consider UV measurements to require careful attention to everything in the path, as well as the detector itself. UV photons are short and do not normally penetrate deeply into the detector. They may not even make it to the depletion zone. There are a few common techniques to enhance this: (1) thin layer of gold on the surface, charged to drive photogenerated charges into the depletion zone and (2) laser dyes to Stokes-shift the wavelengths into better quantum efficiency. Also, glass blocks shorter UV completely, so fused silica or sapphire windows/lenses may be indicated. \$\endgroup\$ – jonk Dec 20 '17 at 16:10
  • \$\begingroup\$ Thank you for your reply. I had a search for UVC sensors but they seem rather expensive for this application. I think my first port of call may be to try boosting the gain of the current setup. \$\endgroup\$ – Johan Dec 20 '17 at 16:11
  • 2
    \$\begingroup\$ It doesn't have to be expensive. Self-educate. I have made my own phosphors. Laser dyes aren't terribly expensive. You can diffuse gold by yourself with a bell jar and modest vacuum pump. Sure, it takes knowledge and that takes time. But the out of pocket costs can be low, if need be. The problem you may be experiencing is that far, far too few photons of interest are making it through to the detector and then photogenerating charges that you are actually collecting. No amount of electronics will fix that problem and energy is best spent fixing things before they enter the electronics system. \$\endgroup\$ – jonk Dec 20 '17 at 16:25
3
\$\begingroup\$

Do you know how much UV light there is? Maybe the photo current is just very small (nA instead of uA.) You could use the photodiode spec sheet to estimate how much current you expect. You could try making R1 bigger (say 100 Meg ohm) and make C2 smaller (say 1 nF).

\$\endgroup\$
2
  • \$\begingroup\$ Hi, I'm not able to modify the existing breakout board. Would you suggest purchasing a seperate photodiode and buying the op-amp, capacitors and resistors to make my own configuration? \$\endgroup\$ – Johan Dec 21 '17 at 8:48
  • \$\begingroup\$ @Johan, well first run the numbers and find out how many photons/ micro-watts of power you have. Then why can't you mod the existing board? The R marked 105 is the 1 meg ohm one. \$\endgroup\$ – George Herold Dec 21 '17 at 14:32
1
\$\begingroup\$

That break-out-board has an op-amp built in. Resistor R1 sets the gain. You should be able to get more gain by swapping it. Doubling it to 2M would double it, 5M would quintuple it, for example.

A side effect of increasing R1 is that it will move the cutoff frequency of the low-pass filter formed between R1 and C2. You may need to reduce C2 by the same factor if you find your circuit responds too slowly. In addition, the op-amp has a gain-bandwidth product which will have a similar effect when you increase the gain. All you can do about this remove C2 entirely and get as much bandwidth out of the op-amp as you can (or replace the op-amp).

You may also find that you have alot of noise and the problem may really be that your plasma is a very weak source of UV. Also consider that your plasma may be emitting UV in a wavelength that is off the peak of the responsivity curve for this detector. These may be greater contributors to your lack of a strong signal and the better solution for those may be finding a different detector.

That said, boosting the gain is a very easy and cheap thing to try so it's probably worth a go.

\$\endgroup\$
2
  • \$\begingroup\$ Hello. Thank you for your help. I have added the schematic to the original question. \$\endgroup\$ – Johan Dec 20 '17 at 15:18
  • \$\begingroup\$ OK I updated my answer \$\endgroup\$ – kjgregory Dec 20 '17 at 15:55
0
\$\begingroup\$

What you are looking for is generally known as "signal conditioning". This is a circuit that applies gain and/or offset to a signal in order to make it more suitable for subsequent processing.

You can design your own (yes, using one or more opamps), or you can purchase a commercial module designed specifically for this purpose.

\$\endgroup\$
0
\$\begingroup\$

Here are the instructions from the AdaFruit link:

To use, power the sensor and op-amp by connecting V+ to 2.7-5.5VDC and GND to power ground. Then read the analog signal from the OUT pin. The output voltage is: Vo = 4.3 * Diode-Current-in-uA. So if the photocurrent is 1uA (9 mW/cm^2), the output voltage is 4.3V. You can also convert the voltage to UV Index by dividing the output voltage by 0.1V. So if the output voltage is 0.5V, the UV Index is about 5.

When the device is connected properly and working correctly it does not require any additional amplification. If you double check all of your connections and they are right you may have a defective board.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for your reply. I have doubled checked the connections and am unsure why the output signal is very small. The plasma radiation is from a sterilisation machine and is fairly weak. The sensor only works when placed very closely to the plasma UV source. \$\endgroup\$ – Johan Dec 20 '17 at 15:17
  • 2
    \$\begingroup\$ Do you have a known UV source that you can use for testing/calibration? I recommend getting one if you don't. You're going to want to calibrate it later on especially if you wind up needing to adjust the gain or add additional amplification. \$\endgroup\$ – kjgregory Dec 20 '17 at 15:30
0
\$\begingroup\$

I would remove R3 as it is killing the negative feedback loop gain. Beyond that, I would probably add a couple of 7400 series inverters after that to clean up the square wave if it needs it.

\$\endgroup\$
0
\$\begingroup\$

I would try to add a low pass filter on 5V supply - inductor, capacitor and low ESR cap. Alternatively you can make your own board with better opamp for this purpose as you mention transimpedance amplifier.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.