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I have a question regarding the usage of the phototransistor SFH 203 P, which I want to use for a project, where I need a small fall- and rise-time. The data sheet (I think) implies that if I use the photodiode as follows (in reverse bias) I should get a photocurrent of about 800µA (see page 3 under "Rise and fall time of the photocurrent").

schematic

simulate this circuit – Schematic created using CircuitLab

But unfortunately I only measure 2.3µA with bright room lighting. Did I misinterpret the datasheet, use the wrong circuit or is my photodiode maybe broken?

planespotter5000
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    800uA is nearly 1mA. That's a butt load of current for a photodiode unless your photodiode is huge and/or you're shining a laser directly into it. – DKNguyen Jul 30 '20 at 22:26
  • Are you re-inventing the opto-coupler? – Mast Jul 31 '20 at 12:46
  • Not really. I want to use the mentioned photodiode and an IR LED to measure the speed of light (over a distance of a few meters). – planespotter5000 Jul 31 '20 at 12:56

2 Answers2

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Actually, that sounds about right. I don't know where you got the 800 µA figure from — both the table and the graph indicate much less.

"Bright office lighting" is about 500 lux. The sensitivity of the SFH 203 P is specified as minimum 5 µA, typical 9.5 µA @ 1000 lux.

The graph in the middle of page 4 agrees with this.

The rise and fall times were measured with a much brighter pulse of light than room lighting -- probably a laser or flashtube. Interestingly, none of the graphs go up that far, implying that that would not be a "normal" regime of operation.

Dave Tweed
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  • I used the parameters on page 3 under "Rise and fall time of the photocurrent". There a current of 800µA is mentioned. – planespotter5000 Jul 30 '20 at 22:29
  • @David Tweed a photodiode is not a linear component. Doubling lux won't double current. – Helena Wells Jul 30 '20 at 23:13
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    @HelenaWells: Actually, it pretty much is, at least over a reasonable range. That's the whole idea behind "quantum efficiency" -- a certain number of photons gives you a certain number of photoelectrons. – Dave Tweed Jul 30 '20 at 23:15
  • Not all photoelectron diffuse. – Helena Wells Jul 31 '20 at 00:15
  • @HelenaWells: I'm not sure where you're going with this. Photoelectrons are created in the depletion region of the reverse-biased diode junction, and they are strongly driven out by the E-field that exists in that region. So you're right in a way -- they don't "diffuse". – Dave Tweed Jul 31 '20 at 00:22
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    I am talking about the photovoltaic mode. – Helena Wells Jul 31 '20 at 00:54
  • @Helena Wells: Perhaps provide some data or an example? – Peter Mortensen Jul 31 '20 at 12:46
  • @PeterMortensen: PV mode is completely off the topic of this question, and in particular, my answer. If you want to discuss it, take it to chat. – Dave Tweed Jul 31 '20 at 14:31
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I should get a photocurrent of about 800µA (see page 3 under "Rise and fall time of the photocurrent").

From the datasheet:

Photosensitive area: 1 mm^2

Sensitivity: 0.86 A/W (at 850nm), about 0.5 in the visible.

To get 800uA in the visible, you would need 1.6mW. From a 1x1 mm area, that is 1600 w per square meter, or about 2.5 times as bright as noon-time sunlight. No way you're going to get that kind of current indoors without an arc lamp or a laser beam.

If you need more current, get an avalanche photodiode (about 100x more current) or a silicon photomultiplier (about 1,000,000x more current). They're actually not that much more expensive than the diode you have, although bias voltage will be 30-40v.

user1850479
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