Measuring Current recieved from a Photodiode - Electronic Circuits Projects Diagrams Free

crush · 9th November 2006, 08:50 AM

Hi all,
I cant figure out how to calculate the current from a photodiode!
I am using a photodiode in reverse bias to detect and infrared signal incident on it from a transmitter.
Have built the recieving circuit by incorporating a gyrator circuit- as a constant current source to the LED.
when a signal of the frequency of interest in incident on the photodiode, the time constant of the photodiode is such that it cannot react fast enough, hence producing a low signal at the end of the photodiode. This low signal is then rectified and amplified by a transimpedance amplifier.

(For circuit diagra, - refer: http://generous.boy.googlepages.com/...cievingcircuit)
Have the circuit going. I was wanting to test the accuracy of my circuit, hence:

How do i calculate the amount of current that is top be produced by the photodiode? (for experimental pursposes keeping the height of the transmitter constant at 5 cm and moving the transmitter away from the reciever to 1 cm, 2 cm. 3 cm and so on)

the photodiode i am using is a BP104 (http://www.jaycar.co.nz/products_uploaded/ZD-1947.pdf)
the transmitter generic 5mm (http://www.jaycar.co.nz/productView....Max=&SUBCATID=)

I am driving the LED for short bursts at 1A.

any ideas?

Optikon · 9th November 2006, 04:12 PM

Look at the spectral sensitivity specification for your part. This spec says that given an amount of incident optical power falling on the surface, a certain amount of current will be generated. The tough part is figuring out how much optical power is falling on surface. It is a geometry & unit conversion nightmare. But you can probably find some help on the web. Have you googled lately?

justDIY · 9th November 2006, 05:47 PM

place a resistor of known value in parallel with the diode, and meausre the Vdrop across the resistor ... it'll need to be a LARGE value resistor otherwise it'll just kill the photovoltaic effect - there is very very little current

audioguru · 9th November 2006, 06:39 PM

The photodiode is used in the reverse-biased mode, not in the photo-voltaic mode. So it "leaks" current when light shines on it.
To measure its current just measure the voltage drop across the resistor that is feeding it the reverse bias.

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9th November 2006, 08:50 AM	(permalink)
crush	Measuring Current recieved from a Photodiode Hi all, I cant figure out how to calculate the current from a photodiode! I am using a photodiode in reverse bias to detect and infrared signal incident on it from a transmitter. Have built the recieving circuit by incorporating a gyrator circuit- as a constant current source to the LED. when a signal of the frequency of interest in incident on the photodiode, the time constant of the photodiode is such that it cannot react fast enough, hence producing a low signal at the end of the photodiode. This low signal is then rectified and amplified by a transimpedance amplifier. (For circuit diagra, - refer: http://generous.boy.googlepages.com/...cievingcircuit) Have the circuit going. I was wanting to test the accuracy of my circuit, hence: How do i calculate the amount of current that is top be produced by the photodiode? (for experimental pursposes keeping the height of the transmitter constant at 5 cm and moving the transmitter away from the reciever to 1 cm, 2 cm. 3 cm and so on) the photodiode i am using is a BP104 (http://www.jaycar.co.nz/products_uploaded/ZD-1947.pdf) the transmitter generic 5mm (http://www.jaycar.co.nz/productView....Max=&SUBCATID=) I am driving the LED for short bursts at 1A. any ideas?

9th November 2006, 04:12 PM	(permalink)
Optikon	Look at the spectral sensitivity specification for your part. This spec says that given an amount of incident optical power falling on the surface, a certain amount of current will be generated. The tough part is figuring out how much optical power is falling on surface. It is a geometry & unit conversion nightmare. But you can probably find some help on the web. Have you googled lately?

9th November 2006, 05:47 PM	(permalink)
justDIY	place a resistor of known value in parallel with the diode, and meausre the Vdrop across the resistor ... it'll need to be a LARGE value resistor otherwise it'll just kill the photovoltaic effect - there is very very little current __________________ If you don't have a planet, what good are gold bars? want to contact me directly? gmail gordonthree check out my project website: http://projects.dimension-x.net Favorite numbers: 09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0

9th November 2006, 06:39 PM	(permalink)
audioguru	The photodiode is used in the reverse-biased mode, not in the photo-voltaic mode. So it "leaks" current when light shines on it. To measure its current just measure the voltage drop across the resistor that is feeding it the reverse bias. __________________ Uncle $crooge