Agreed, it looks like the OPA354 is part of the problem. Maybe if it starts working well the LM2904 might start looking better.

Is Cd are real capacitor or just representative of the capacitance of the photodiode?

A guess is that the OPA354 is operating too close to ground. Please try this:

1. Remove the LM2904 from the circuit.

2. Disconnect the photodiode and if Cd is a discreet capacitor, remove it too.

3. Make sure there is a 47 pf capacitor across RF

4. Bias the non-inverting input to the OPA354 to half of Vs as shown in the circuit below, but also put a 10 uf capacitor from the noninverting input to ground.

What do you see? The answer should be +2.5 volts and not much else.

 

งานรับสมัครนักศึกษาใหม่ ระดับบัณฑิตศึกษา

Agreed, it looks like the OPA354 is part of the problem. Maybe if it starts working well the LM2904 might start looking better.

Is Cd are real capacitor or just representative of the capacitance of the photodiode?

A guess is that the OPA354 is operating too close to ground. Please try this:

1. Remove the LM2904 from the circuit.

2. Disconnect the photodiode and if Cd is a discreet capacitor, remove it too.

3. Make sure there is a 47 pf capacitor across RF

4. Bias the non-inverting input to the OPA354 to half of Vs as shown in the circuit below, but also put a 10 uf capacitor from the noninverting input to ground.

What do you see? The answer should be +2.5 volts and not much else.

Dear,

Before trying with your idea, I have tried to improve more energy for LED, a little bit. and now the problem seems to be fixed @@
I will check more to ensure the problem is power of LED
Thanks a lot.

 

Agreed, it looks like the OPA354 is part of the problem. Maybe if it starts working well the LM2904 might start looking better.

Is Cd are real capacitor or just representative of the capacitance of the photodiode?

A guess is that the OPA354 is operating too close to ground. Please try this:

1. Remove the LM2904 from the circuit.

2. Disconnect the photodiode and if Cd is a discreet capacitor, remove it too.

3. Make sure there is a 47 pf capacitor across RF

4. Bias the non-inverting input to the OPA354 to half of Vs as shown in the circuit below, but also put a 10 uf capacitor from the noninverting input to ground.

What do you see? The answer should be +2.5 volts and not much else.

Dear,

I've already tested with new level of LED power. The problem is exactly LED power. But It is not my final goal.
My final goal is:
1. Measure received voltage to know how far between led and receiver.
2. Measure period and dutycycle of received signal to know frequency of LED.

So, the problem I met now is how to measure period and dutycycle when the receiver move far away from LED.
I intend to use comparator Opam (LM393/311) to force the received signal to square shape. Is this solution ok or not?
If you have better solution, pls help me.
Thanks a lot!

 

The circuit below should work well for clipping this kind of signal.

The signal swings below and above ground so you will have to either add a negative power supply or bias the non-inverting input up from ground. You will also do well to AC couple the signal so that the clipping is always symmetric.


The circuit above and a much more complex one are discussed at the link below.
https://electronics.stackexchange.com/questions/45439/clip-signal-with-very-low-threshold

If your duty cycle is going to be shifting around, that could affect the apparent duty cycle, particularly if the signal has long rise and fall times compared to the signal's period. To get around that limitation, A alternative to a clipping circuit to the above is to let the opamp's rail-to-rail characteristics do the clipping if that doesn't cause problems with rise and fall times. Just adding one or two AC coupled high gain stages and take the output of the last one.