analysis circuit photo diode preamp

[asadi.siyavash]

Hi, I found Circuit below in Linear Technology solutions, but I can't understood its analysis. certainly I can't calculate JFET's Id. I simulate it in TINA, its work good. one other problem that I have is that it used positive feedback and it works linear.
please help me for understanding this circuit.

 

[Audioguru]

The Jfet is an inverting transistor and the opamp is non-inverting causing the entire amplifier to have negative feedback.
The (-) input of the opamp is at +2V so when there is no light and the photo-diode is doing nothing then the opamp output will go high enough to cause the (+) input of the opamp to also be at +2V.
If the gate of the Jfet is at +1.5V then it is turned on very hard forcing the (+) input of the opamp to be at +1.5V which will cause its output voltage to drop low enough (to about +0.5V) that the Jfet is turned off a little until the (+) input of the opamp is at +2V.

When there is light then the photo-diode is a photo cell and generates a negative current that turns the Jfet off a little which causes the (+) input voltage of the opamp to increase which causes its output voltage to increase a little.

 

[asadi.siyavash]

Hi, Really thanks I understood. the key was negative feedback.
I know J-FET has very low current noise but my problem is what is the benefit of this circuit vs common circuits with negative feedback op-amp. like use LMP7721 in below picture.

 

[Audioguru]

what is the benefit of this circuit vs common circuits with negative feedback op-amp, like use LMP7721 in below picture?

Simply look at a description of a photo-diode's extremely low output current in this circuit that gives it no reverse bias voltage and read the datasheet for the modern LMP7721 opamp that has Mosfet inputs for the industry's guaranteed lowest input current.
I think your "common circuits" used very old opamps so they needed an external Fet to get a low input current.

 

[D.A.(Tony)Stewart]

Hi, I found Circuit below in Linear Technology solutions, but I can't understood its analysis. certainly I can't calculate JFET's Id. I simulate it in TINA, its work good. one other problem that I have is that it used positive feedback and it works linear.
please help me for understanding this circuit.
View attachment 112242

Photodiodes have a capacitance that reduces frequency response and reduces with reverse bias away from 0V.
SFH215 Capacitance Co 11 pF (VR = 0 V, f = 1 MHz, E = 0)

Op Amps also have input current and noise which limits level useful range and frequency response.

J-Fet Id is determined by V(-) bias of 2V which determines V(+) level of 2V and thus current thru Drain R2. of 1V/1k = 1mA.
The OA negative DC feedback guarantees V(+)=V(-) in the linear range .

The capacitive feedback is selected to reduce the diode RC time for compensation like your simple example but lacks the low input capacitance and low noise advantages.

Another option is an integrated optical sensor with FET buffer inside.