amplifying a photodiode signal

The data sheet (from Digikey):

Says that with a 5 volt reverse bias, the capacitance is 3.5 pf.

At 500 MHz that has an impedance of about 90 ohms. That is in parallel with your 100k pull up resistor.

The RC 3db down corner for 100k and 3.5 pf is 450 kHz. Then you have to parallel all the other stray capacitance of what else is connected to that node. Probably a few more pf, at least.

--
John Popelish

Hi,

I have a photodiode with the anode connected to a 100K pullup to 12V, cathode to ground, and an RC highpass filter (47pF, 2.2Mohm) driven from the photodiodes anode, what is the best way to amplify the signal to 1V peak to peak from the highpass filter assuming a frequency of up to

500MHz at 0.1mV amplitude? I am thinking of an MMIC RF amplifier, or a high speed comparator but I am not sure what would be best. I would like to feed this signal into a high speed serial to parallel converter to interface it to 5Volt logic.

cheers, Jamie

I am not sure about the capacitance, it isn't in the 2page datasheet I have.

It is used in the Ronja optical datalink project:

400MHz 820nm photodiode digikey part#: 425-1029-5-ND manufacturer part#: PD101SC0SS

cheers, Jamie

I read in sci.electronics.design that Jamie Morken wrote (in ) about 'amplifying a photodiode signal', on Thu, 6 Jan 2005:

500MHz and 2.2 Mohm are incompatible. You will not get the filtering you expect. The stray capacitance across your 100 kohm also attenuates such fast signals severely.

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Regards, John Woodgate, OOO - Own Opinions Only. 
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Hi, you won't get anything like 500MHz from such a simple circuit. There is good site somewhere that tells you how to do it, I can't remember where but it didn't take me long to find using the search engines. Your other option is to read the diodes application notes, usually the best place to start anyway.

Jamie, you need to make the diode fast by a moderate reverse voltage of say 12V or so. What photodiode do you have and what is its rise time ? Then how much light do you have ? And how much jitter can you tolerate ?

Rene

Jamie Morken wrote:

Go get yourself a copy of Jerald Graeme's book "Photodiode Amplifiers: Op Amp Solutions" It's got lots of pretty equations that you can use for the first stage transimpedance amp. All of the posts regarding the fact that you won't be able to get that kind of signal swing from a passive I/V converter are dead on the money.

That said, how much current are you expect> Hi,

Hi,

I found a good site that goes through this topic very well:

cheers, Jamie

Rob Gaddi wrote:

I am thinking of using a circuit like this:

I am looking for a photodiode with a high A/watt light sensitivity rating and also a low capacitance - and a max price of $10! This is one possible photodiode: digikey part#: 425-1029-5-ND manufacturer part#: PD101SC0SS

This is the current to voltage circuit I am looking at:

I don't fully think this is the best circuit to use as I would like to get rid of any DC current coming out of the photodiode from ambient light. I would like a circuit that can detect a very small, high frequency, lightsource from the photodiode. Would it be possible to AC couple the photodiode right to the input of an RF low noise amplifier and get something useful out of it? :)

cheers, Jamie

I tend to doubt you're going to reach 500MHz with a 10$ diode. After having had a look at the datasheet, it is specified for 400MHz.

Sort of, provided you have sufficient light. The source impedance for an RF amp is not to be much more than say 100 Ohms.

Rene

Yes, you can AC couple the photodiode and use an RF amplifier. Typically, you'd use a 50Ohm load. However, photodiode will have to be quite small to have sufficiently small capacitance for 500MHz operation, AND, you'll have to begin by understanding that the CATHODE is positive for reverse bias!

Paul Mathews