8/10 mm IR photo diode

Some time ago I downloaded data sheets from digikey - there were loads of large & expensive looking photo diodes among those.

See

or take a bunch of small sized ones, e.g. TSHF5400

- Henry

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"ian field" schrieb im Newsbeitrag news:oOHbh.1651$ snipped-for-privacy@newsfe6-gui.ntli.net...

10

If you find that big photodiodes are too expensive, you can try a solar cell. That is just a very big photodiode, but usually poorer quality than a small one. In particular, solar cells will have a shunt resistance, so it is better to operate them with zero bias, using a low-offset voltage opamp to create a virtual earth on one side of the photodiode, and grounding the other side.

Chris

10

They have a VERY low bandwidth.

- Henry

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I know zip about photodiodes, so here is a question. It seems the 1mm and 15mm diodes have the same sensitivity according to the datasheets. The larger diode is slower (more capacitance), but not more sensitive.

photo diode: The sensitivity by area is constant. If you use a larger one you will have more current but also proportional more capacitance you have to fight against. In general use a smaller one if the frequency is higher. Don't forget to use the same transmitting and receiving optical frequency.

You can use optics (e.g. lense) to amplify optical power (But this will narrow the reception field) or use more transmitter power. Pulsing the transmitter is possible, an APD receiver works more sensitive.

On the data side you can compress the information and the modulation scheme can be adapted. More on, use a good IR filter, synchronize the system with the mains power, etc....

All depends on price and complexity.

- Henry

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Yes they can be slow, and I have no way of knowing whether the OP is happy with that. Some big photodiodes have poor bandwidth also. Chris

You can get about 20 kHz bandwidth from a 1x2 inch solar cell by cascoding it. Verrry useful sometimes.

Normally the best way to get more light is by using a lens to collect from a larger area. To the OP: What's your exact application?

Cheers,

Phil Hobbs

"Chris Jones" schrieb im Newsbeitrag news: snipped-for-privacy@corp.supernews.com...

so

I had no luck with solar cell and Op Amp circuit and because of the bandwidth problem I never investigated further the Op Amp-capacitance problem. Maybe there is a solution.

Use a PIN-structure photo diode and much transmission power. The easy way to go.

- Henry

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But the datasheet specs are in A/W. I don't see any notation that this is per unit area

If you use a larger one you will have

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This is the spectral sensitivity. It says how much current you can expect depending on wavelength.

You must look on the short circuit current. This current is almost proportional to die area. And capacitance is proportional with die area. Lowest for PIN structure.

This is a good one. Cheap, sensitive, large and PIN:

The variant BPW104 is with included IR filter (excluding normal light).

- Henry

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a larger area. To the OP: What's your exact application?

This is a special application. A 10 khz IR data transfer ( 8 meter's with no linses ).

Benny

Look for irDA spec or just for "ir remote control" like for tv used. That should work if you use such a ready-to-go receiver and make the transmitter a little more powerful. I suggest a MOSFET as driver and a couple of LEDs around in different directions.

Avoid to send NRZ-data. Use a reasonable carrier frequency. Otherwise you have a lot of problems with interference coming from sunlight and especially indoor from lamps.

- Henry

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"Benny" schrieb im Newsbeitrag news:4570a301$0$183$ snipped-for-privacy@dread11.news.tele.dk...

no

There is a corresponding graph that is wavelength dependent, but every reference I have found indicates A/W is the current output based on light input. For example:

So I am still at a loss why the particular datasheets show the same A/W for different die sizes.

Thanks for the link. The built in filter looks useful.

Why are you looking discretes for this? Cheap devices for such a task are made by the tens, if not hundreds of millions per annum.

Look at, for example, the Vishay TSOP's - there are v.many other manufacturers of similar.

8 metres outside is quite straightforward

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A/W is the spectral sensitivity and NOT the output short-current to input absolute optical power. Ten years back I had the same (your) problem and read the Siemens/Infineon application notes to unterstand this.

If you follow my suggestions you surely get a working system.

BPW104 is a standard device used by the big TV manufacturers - so it must be cheap and reasonable good.

I can make a development for you if you like. Just contact me snipped-for-privacy@gmx.net

cheers - Henry

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I don't recall saying the short circuit current and A/W were the same.

I've posted links to references that don't agree with your assesment of A/W, but would be glad to read any references you have. There are A/W plots relative to wavelength, but this is not the same as the datasheet parameter of A/W done at a specific wavelength.

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A/W

input

I'm not an expert here. I'm sure if you browse the various app notes from the manufacturers you will get all info you need.

As I recall of my memory: Absolute spectral senitivity is normalized to one area unit of surface of the photo diode. Maybe the definitions depend on the specific manufacturer. This is the definition used by Vishay/Temic and Infineon/Siemens. They are surely the greatest quantity manufacturers in the world for those devices.

BTW: You must understand that at least three different angle-less measurement systems exist: physical watts, lumen and candela Additional there are all a second time in depency of angle. So you have at least six systems!

For your design use simply the physical radiation power without angle. This works good with ir diodes.

A good ir transmitter diode have 25mW optical power output at 100mA. Think of the area the light will be distributed, then think of the reflection coefficient of the walls. The walls are primary mirrors. Accumulate all the light coming to the receiver diode. The photo diode have a efficiency of about the A/W value e.g. 0.62 if transmitter and receiver diode is of equal wavelength. You now the incoming optical power and the area of the photo diode. The three values give you the photo current!!! Reasonable wavelengths are 870nm and 950nm. 870nm systems have more speed.

950nm have more power. You cannot efficency mix them! For optimum pre-amp amplification the Zout of photodiode and Zin of amp must be mostly equal. That is not exactly true but reasonable (You must it noise adapt in theory). So calculate your Zout and make a pre-amp. LF357 or NE5534 is a good start.

Don't hack on the theory. Just use reasonable devices and build/test it.

regards - Henry

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snipped-for-privacy@sushi.com wrote:

A larger area means more Watts of IR power get absorbed by the photodiode. More Watts of IR mean more Amps out of the photodiode. Amps-per-Watt does not change.

(Assuming the IR is not being focused to an area less than the photodiode area.)

Mark