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Sci.Electronics.Basics -> balanced photodiode
There are 2 messages in this thread.
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Author: sperelatDate: 03:33 02-10-07
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My professor gave me a circuit schematics of Balanced photodiode.
This is my first time to study Balanced photodiode.
It is composed of Photodiode pair( picture)
---------------|>|------------------
| | op
amp pin 2( - )
g-------- ---------------------
n | |
d ---------------|<|------------------
PD
1Mohm resisitor and 33 pf capacitor are connected between pin 2 and
output of op amp.
(LF 356)
I understand how PDs in above picture measure light intensity
difference.
What I don' know is the function of OP-amp after PDs.
(I don't know how to draw OP amp here)
It seems to be filter or integrator or differentiator.
The problem is that there is no resistor or capacitor before pin 2.
So I don't know the exact function of OP-amp in this circuit.
I just guess that capacity of Photodiode works for this circuit.
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Author: John PopelishDate: 08:33 02-10-07
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sperelat wrote:
> My professor gave me a circuit schematics of Balanced photodiode.
>
> This is my first time to study Balanced photodiode.
>
> It is composed of Photodiode pair( picture)
>
>
> ---------------|>|------------------
> | | op
> amp pin 2( - )
> g-------- ---------------------
> n | |
> d ---------------|<|------------------
> PD
>
> 1Mohm resisitor and 33 pf capacitor are connected between pin 2 and
> output of op amp.
> (LF 356)
>
> I understand how PDs in above picture measure light intensity
> difference.
>
> What I don' know is the function of OP-amp after PDs.
>
> (I don't know how to draw OP amp here)
>
> It seems to be filter or integrator or differentiator.
>
> The problem is that there is no resistor or capacitor before pin 2.
>
> So I don't know the exact function of OP-amp in this circuit.
>
> I just guess that capacity of Photodiode works for this circuit.
Consider each of the photo diodes to be current sources.
Two current sources in parallel just add their outputs
together. Turn one around and they subtract from each other.
This difference current (proportional to the difference of
the two illuminations) is a current input to the inverting
input of the opamp. the feedback path from opamp output to
that input must carry away all the resultant current these
two sources deliver (since the input can pass no current).
The opamp finds what output voltage it takes to force the
feedback elements to pass that current by keeping the
voltage across the photo diodes (and its two inputs)
balanced at zero volts at all times.
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