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A replacement for a photodiode

yoel ben naim

Sep 28, 2014
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I am building a test circuit for the circuit shown bellow.

upload_2014-9-28_15-29-56.png
In my test circuit I need to replace the photodiode with a current source of a 10-40mA, 1-5uSEC pulse, and I need to attach it in a way that curret will flow through the Rf resistor.
Any idea?
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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Can you tell us more about what you want to do? "current source of 10~40 mA, 1~5 µs pulse" is not very specific. If you describe your whole project, we may be able to help.
 

yoel ben naim

Sep 28, 2014
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What I have is a logic circuit board with many features. One of those features is to detect a laser pulse of 60nSEC (I was mistaking earlier), this is the job of the photodiode.
The thing is I want to test this logic board without connecting it the laser system and this is why I need somthing that could simulate the operation of the photodiode.
I hope this explanation clearer.
 

Arouse1973

Adam
Dec 18, 2013
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Could you use a high power, narrow beam width LED driven from a pulse generator circuit? What photodiode are you using?
Adam
 

yoel ben naim

Sep 28, 2014
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This could be an option, but I was hoping that there will be a simpler solution to simulate the photodiode.
 

Arouse1973

Adam
Dec 18, 2013
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Ok look at using an adjustable regulator as a current sink, all you need is one resistor and the regulator.
Adam
 

yoel ben naim

Sep 28, 2014
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I thought of that but I have a problem with that.
The current direction on a photodiode is opposite to the diode direction (in my case left to R5). I am not sure that it will flow in that direction.
 

Arouse1973

Adam
Dec 18, 2013
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I thought of that but I have a problem with that.
The current direction on a photodiode is opposite to the diode direction (in my case left to R5). I am not sure that it will flow in that direction.

Yes current sink not source.
Adam
 

yoel ben naim

Sep 28, 2014
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Thank you Adam.
I looked for some examples but almost all of them have voltage on on side and gnd at the other.
I on the other hand have only gnd and I am affraid it won't work.
Could you show me how to connect a current sink instead of my photosiode?
 

Arouse1973

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Dec 18, 2013
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This is proving a little more difficult than I first thought. Leave it with me. I am trying to keep it simple.
Adam
 

Arouse1973

Adam
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Ok I have had a long think about this and have come up with two things that might work. The first one is to use an opto isolator. The second one is a bit off the wall and I haven't seen this done before but it seems to work.

It uses a 4 terminal FET as a diode. Current is controlled by a voltage supplied through the body terminal. It is very sensitive so will only take between 1-1.5 volts to get the current you require.

The only problem is the output is negative, but you can easily convert this with an opamp. I haven't looked into what FET to use, you will need to do this. Or the first option might be better.

Thanks
Adam

PDIODE.PNG
 
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