OpAmp Integrator/LowPass Filter

D

drkidd22

Member

• Nov 20, 2014

• #1

Hello,

I'm working on a amplifier for a photo detector rated at 0.17A/W. I'm trying to figure out with a circuit that was used before on similar amplifier. I would like to understand what's the purpose of the Diode (D2). I think I got it clear that this is a LPF/Integrator but don't understand why the diode is there. Any references to app notes or ideas will be appreciated.

 

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ronsimpson

Well-Known Member

Most Helpful Member

• Nov 20, 2014

• #2

D2 will allow the output to go above "IS" voltage, but the output will not go below the IS voltage.

 

MikeMl

Well-Known Member

Most Helpful Member

• Nov 20, 2014

• #3

There is something wrong with using 200 Ohms for R1. If the thing is supposed to be a charge amplifier (integrator), R1 should be much, much higher, like 1 to 10meg?

 

alec_t

Well-Known Member

Most Helpful Member

• Nov 20, 2014

• #4

D2 will cause an asymmetric slope to the rising and falling edges of the opamp output in response to an input pulse at the inverting input of the opamp.

 

Dick Cappels

Active Member

• Nov 20, 2014

• #5

Besides acting like a clipper, it might be that the diode in the feedback is to make the response logarithmic. It all depends on the polarity of the photo detector and the expected output voltage.

 

audioguru

Well-Known Member

Most Helpful Member

• Nov 20, 2014

• #6

The opamp cannot produce enough output current to drive the very low value for R1 anyway.
The time constant for the 100 ohms for R2 and the 100 puffs capacitor is so small that maybe there is no such small amount of time (10 femto-seconds?).

 

D

drkidd22

Member

• Nov 20, 2014

• #7

Sorry guys, I should've mentioned that the value of R2 shown there is just me playing around with different resistances. And yes, the value of R2 will be much greater than 200ohm.