3

enter image description here

Taken from: http://www.geni-uv.com/download/products/GUVV-T20GD-U.pdf

I'm trying to do some estimates here and I've run into something that seemingly doesn't make sense. In the table above, it states that the photocurrent is 14.7uA at 1mW/cm^2. To estimate the photocurrent at 10uW/cm^2, I scaled down that photocurrent to get 147nA. But 147nA is getting close to the dark 90nA current that is listed.

Yet, the minimum optical power source range that this photodiode is recommended to be used at is 1nW/cm^2 which is way, way below the 10uW/cm^2 that was being used in the estimate above.

Also, note that this is a huge 6.894mm^2 photodiode, yet 10uW/cm^2 doesn't seem like it should be an unreasonably low optical power to use such a photodiode at, but you can see from the estimate above it is already approaching the dark current.

Am I misunderstanding something?

DKNguyen
  • 56,670
  • 5
  • 69
  • 160
  • 1
    What type of read circuit are you using? Some type of TIA? Something else? With a capactive TIA, appropriate integration time, and a bit of calibration this can be made to not be an issue. – Matt Mar 22 '21 at 12:10
  • @Matt I would be using a TIA for these. I've not heard of capacitive TIA. I would have to investigate the response time of a capacitive TIA since speed is somewhat of a factor if I were to use these (think a UV-green horizon sensor mounted on a monocopter such that it scans as it spins at 7.5Hz). – DKNguyen Mar 22 '21 at 18:03
  • Given how expensive those photodiodes are, I wonder if using UV filter glass like UG5 or UG11 and a cheap silicon diode would be cheaper. You can get filter glass for a few dollars, although you need to be careful about NIR transmission for some of these glasses. – user1850479 Mar 22 '21 at 19:31

3 Answers3

2

The concern when your dark current approaches your photocurrent is SNR. The three big noise components you are likely to encounter are dark current shot noise, photocurrent shot noise, and noise from your readout circuit. The first two can be easily calculated as:

\$S=2qI\$

Where \$S\$ is the input referred noise power spectral density in A2/Hz. The third is something you may have to measure, unless you have good noise models in your simulator. Sum all of these components, multiply by your bandwidth, and take the square root to get a measure of your expected noise in A.

If you wanted to measure photocurrents below the dark current, you can do so using a capacitive TIA, which is just a regular TIA with a capacitor in feedback, also known as an op-amp integrator. Adjusting the integration time allows you to increase the signal strength proportional to integration time, but the noise only increases as the square root of integration time. Increase integration time to obtain whatever SNR you need. Be sure to size your integration capacitor to account for the dark current. You will additionally need to calibrate your particular detector as dark current varies between detectors.

Matt
  • 1,699
  • 11
  • 16
2

Yet, the minimum optical power source range that this photodiode is recommended to be used at is 1nW/cm^2 which is way, way below the 10uW/cm^2 that was being used in the estimate above.

Note that the photodiode is advertised for use in photovoltaic mode, but the dark current is speced with a reverse bias (so in photoconductive mode where dark current is a bigger problem). For very low light they expect you to use photovoltaic mode.

Because there is a version of this photodiode with much smaller area and the dark current is lower but so too is photocurrent and they just pretty much scale with each other. If so, what's the point of choosing a larger sensor (for applications where you're not blasting a laser at it)?

This is a great optical engineering question. The maximum radiance of a photodiode goes up with the active area, but the number you are looking at (10uW/cm^2) has units irradiance, which does not have the same relationship. The difference is that radiance cannot be increased by optics (or anything), but irradiance can be using a simple lens. If you really care about irradiance, then you would get a lens and use it to focus down onto a tiny photodiode. Since dark current scales with area, this would allow you to divide your dark current by a factor of hundreds or thousands at no loss of irradiance. The smaller diode would also be cheaper, faster and introduce less electronic noise into your opamp.

The reason larger photodiodes exist is due to radiance. A lens trades (via the Fourier transform) angle for position, so focusing only works if the source has a narrow range of angles (which you can Fourier transform into a narrow range of positions). Try to focus sunlight on a sunny day works well. Try it on a cloudy day (when the light comes from clouds at all angles) and the lens does nothing. In your case, if your source of UVA comes from a single, compact source (like an LED) and that source is stationary, you can point a 2 cm lens at it with a 100 micron photodiode and get the same photocurrent as using a 2 cm detector. This will make your dark current insignificant.

user1850479
  • 16,883
  • 1
  • 21
  • 46
  • Oh, so even 0.1V is considered reverse bias? They just didn't list the current for dark current photovoltaic mode then? I'll have to simmer on the radiance vs irradiance. It sounds like it's the difference between just receiving what's there versus collecting everything and focusing it down. It always felt like there was a difference to me but I could never pin it down. – DKNguyen Mar 22 '21 at 18:14
  • @DKNguyen I'm not certain for InGaN diodes, but for silicon the dark current keeps decreasing until you hit 0V. I think they picked 0.1V as an example of the lowest possible bias voltage where the diode is effectively reverse biased (so lowest possible dark current in photoconductive mode). Dark current in photovoltaic mode will be extremely small, probably negligible. – user1850479 Mar 22 '21 at 19:06
  • @DKNguyen A lens just Fourier transforms the incoming light distribution. If your light was narrow in angle but wide in area (so a small source far away), it gets focused down. The width of your detector will therefore become a spatial low pass filter, with the pass band range of angles extending to +/- arctan(diode_radius/focal_length). A large detector is therefore a high (angular) bandwidth system. Set the bandwidth to match your signal and you get the best possible SNR. Usually you can do some simple trig to see what range of angles you need to collect. – user1850479 Mar 22 '21 at 19:19
  • Well, I don't think a lens is going to work in my application anyways since it will be used as a horizon sensor so it's job will be to observe the brightness of the sky or the ground as a reference, and to observe the mixed brightness at the horizon between the sky and ground to determine attitude. So that's definitely a wide area and diffuse. – DKNguyen Mar 22 '21 at 19:30
  • @DKNguyen Are you sure about that? If your goal is to get light from the horizon (and not everywhere else in front and above you), then a lens is something I would want. Otherwise you'll be be detecting direct UV emission from the sun, specular from the ground, etc. – user1850479 Mar 22 '21 at 19:36
  • Yes, it is everywhere and above that because it is trying to observe the average brightness between the two. It then compares the difference in average brightness on opposing sides against the difference in brightness between ground and sky to determine the attitude. The green channel is used to normalize the UV channel to handle the sun. – DKNguyen Mar 22 '21 at 19:43
1

Typically for precision work, you have to calibrate out the dark current (along with other offsets). This may happen implicitly if you're measuring the AC component of the optical signal. If you need to work with DC, then you're in the world of periodic recalibration, temperature control to reduce drift, etc.

pericynthion
  • 6,174
  • 20
  • 34
  • Does that mean that this power level is just unmeasurable though without calibration? Because there is a version of this photodiode with much smaller area and the dark current is lower but so too is photocurrent and they just pretty much scale with each other. If so, what's the point of choosing a larger sensor (for applications where you're not blasting a laser at it)? – DKNguyen Mar 22 '21 at 07:06
  • Dark current can be filtered (AC) or calibrated (DC) out, but for weak signals, the photocurrent of a small diode might vanish in the noise. – CL. Mar 22 '21 at 07:56
  • @CL. Damn. This needs to be DC and cannot be re-calibrated in use. The only other alternative to me is SiC photodiodes which cost much much more for the same price and the spectrum isn't quite what I want either. But very low dark current. Not that these InGaAs photodiodes are cheap to begin with. But for the same price you can get like 14x the active area. – DKNguyen Mar 22 '21 at 08:02
  • If the photodiodes are being used in an antagonistic fashion, can dark currents of two opposing photodiodes be relied on to track each other? Because that I can handle. – DKNguyen Mar 22 '21 at 08:04
  • 3
    @DKNguyen You can subtract off the average component of dark noise (as in a balanced photodetector), but not the shot noise of the dark current (which is random and so will add in quadrature). The real issue with dark current is keeping the dark shot noise low enough that it doesn't obscure your signal. The average dark current is just a DC offset and indistinguishable from the opamp offset. – user1850479 Mar 22 '21 at 13:23