Hi everyone

I have been googling this for several days and it is driving me crazy...

I want to recover 1kHz-10kHz signal from a photo-diode. The signal is very small comparing with the DC part (180nA Ip-p/ 5.4uA Ip-p).

The current plan is to amplify the current into voltage first(0.16V/5V) and use a capacitor to block the DC part. The voltage(0.16ac/5DC) will turn into 0.16ac.

Then use another opamp to further amplify the signal.


However, I do not really like to use two-stage amplifiers, since the other one may bring noise.

I need the signal to have high precision and very low noise.
Any suggestion is appreciated.


Best regards
Richie

 

Check out this clever circuit that gets rid of DC
http://forum.allaboutcircuits.com/showthread.php?t=62281

 

Changes you make to the circuit to remove the DC are likely to increase noise and/or affect the stability of the circuit. It actually makes more sense to use a 2nd amplifier. Use a low-noise opamp and low-noise design/layout. You can even add a filter between the stages to forther suppress noise if your signal has a well defined spectrum.

 

With sufficient gain in a low nose first stage, and a reasonably low noise second stage, the noise of the second stage will cause a negligible addtion to the total noise level.

 

Check out this clever circuit that gets rid of DC
http://forum.allaboutcircuits.com/showthread.php?t=62281

Thanks Jimkeith and everyone!

I checked the circuit you provided above but have something not clear with.

The information part of the current generated by photodiode is very small actually(5.4uA). Will it pass through the Capacitor C2 which is connected with the ground? And if the signal frequency is about 1kHz, it may also pass through the inductor I guess...

What do you think about the ideas given by other two friends?


Richie

 

Thanks Jimkeith and everyone!

I checked the circuit you provided above but have something not clear with.

The information part of the current generated by photodiode is very small actually(5.4uA). Will it pass through the Capacitor C2 which is connected with the ground? And if the signal frequency is about 1kHz, it may also pass through the inductor I guess...

What do you think about the ideas given by other two friends?

Richie

Yes, you are correct--you may wish to avoid the shunt capacitor altogether and yes, the inductive reactance will tend to shunt some of the lower frequency components of the signal. You may have to tailor the inductor for your application. The beauty of this technique is that the amplifier is AC coupled, therefore has extreme DC stability even with high gain.

Using other techniques--If the DC component is not nulled automatically, the output will have problems drifting into saturation.

Regarding Brownout and crutshow's contribution, I agree--a 2nd stage should not introduce additional noise. However, I think you may be able to get by with only one op amp.

 

What if adding an inductor and a capacitor at the input of the amplifier?


Richie

 

My apology--I think that I've led you down the wrong path--just going through the numbers, it looks like you need an inductive reactance of about 100K to be able to extract your signal efficiently. This works out to a ridiculously high inductance in the order of 10 to 20H. It seems that the shunt inductor is only good for extracting the edges of a rectangular waveform containing high harmonic frequencies.

I believe that the OPA129 is overkill--your impedances are not nearly that high

Check out the circuit--there are many acceptable op amps with FET inputs--I indicated the LM444 only because I have used it before.

The constant current through R2 nulls the DC component coming out of the pin diode so no coupling capacitor is required. U1C is an integrator that reacts slowly so as to not interfere with the desired signal. There are two paths for the integrator source current--the first through R6 provides a low rate when everything is in equilibrium-- the 2nd has series diodes that conduct when the error is high to get it into equilibrium quickly.

I have used this technique with photo-transistors successfully--it should basically work, but will require tweaking...

 

I quite not understand actually...

Why not just put a blocking capacitor between the photodiode and the input of the opamp?

The capacitor will block the DC off and let the DC go...

 

Lets start over--
The object is to operate the photodiode at zero bias (0V). At zero bias, diode capacitance issues may be neglected as there can be no charge in the junction. All the op amp has to do is to counter the diode current through its feedback resistor.
This is how the original circuit you posted works, using the OPA129 op amp.
However, the current you indicated (5.4ua) is quite high--well within the range of more conventional FET or even some bipolar op amps.

Input bias current for a few op amps:
OPA129..............0.1pA
TL082.................30pA
LM444.................50pA
LM324...........20,000pA

To be able to truly neglect errors caused by input bias current it must be below the node current by a factor of 100:1 to 1000:1.
Bias current is the input current for bipolar op amps or leakage current for FET input op amps.

Dividing your 5.4ua by 1000 equals 5.4nA or 5,400pA--so the op amp bias current should be equal or lower than this figure--all pass this OK except the LM324 bipolar op amp--however, the LM324 would easily pass if we made the criteria 100:1

So we can see that there is no compelling need to use the relatively expensive OPA129 if bias current is the only issue.

Calculating the feedback resistor for 5V output: R = E/I = 5/5.4uA ≈ 1M
Increasing this resistance beyond 1M risks putting the op amp into saturation.
Note how much lower this is compared to 10 ^10Ω in the original sketch.

The DC blocking capacitor cannot be placed at the photodiode because this would violate the zero bias requirement, so the capacitor must be placed at the output of the op amp. A 2nd op amp may then be used to amplify the AC component to the desired level. I do not think that noise will be an issue here, but try it and see how well it works. Published papers on this photodiode place great emphasis on the sources and types of noise encountered.

 

Thanks Jimkeith

Your explanation about the choice of OpAmp part is quite clear.
But what do you mean by "this would violate the zero bias requirement"?

Do you mean the current is not sufficient to pass the node current by 1000 times?

 

Oh!!
You mean if the input bias current is AC only, then at certain point the bias current will be zero and the requirement will not violated?

 

The normal intention is to operate the photodiode with zero volts across it--placing it at the inverting input of a summing amplifier does this because this node operates at a virtual ground potential. Adding a coupling capacitor here would allow the diode to generate voltage rather than simply current.

This is where it is confusing to me too--the photodiode has two modes of operation--photovoltaic (solar cell) and photodiode. The solar cell actually generates current based upon light intensity, while the photodiode is reverse biased and has a (much lower) leakage current that is a function of light intensity. The photodiode speed performance is orders of magnitude higher than photovoltaic mode performance. However, operating the photovoltaic cell at zero volts, the speed is also increased.

I believe that the device you are using is applied in the photovoltaic mode because there in no external reverse bias voltage source.

Does this agree with anything you know? I am not expert here.

 

If it is applied in the photovoltaic mode generating voltage, it is problematic because how much voltage will it generate? Certainly it is possible to have it generate a low voltage (e.g. 100mV) and have a capacitively coupled amplifier amplify the AC component.

Before this can be attempted, you must measure the open circuit voltage of the photodiode--this will help us to select a suitable load resistance--then you can simply amplify your AC signal with a single high input impedance (non-inverting) amplifier.

The frequency range you are working with does not even come close the speed limits of either mode of operation.

 

What you said matches my limited knowledge.
The photodiode I am using is in photovoltaic mode actually.

Quite interesting! What will happen if a capacitor is added.... How large will be the voltage accumulated on the capacitor?

I will have a try now.

 

When a capacitor is placed between the photodiode and the inverse port of OpAmp, the voltage of P1 is measured. It is -400mV.

When the capacitor is removed the voltage P2 is measured and it is -350mV.

It seems that a voltage of -50mV is generated by the capacitor charged by photodiode.
And -350mV is actually generated by the photodiod itself....

The OpAmp is not grounded? Or maybe the oscilliscope I am using to detect voltage has small impedance and drew all the current.

 

I do not think your voltage measurement with the op amp connected is valid. The difference in voltage between the two may be caused by the op amp bias current interacting somehow.

However, 400mV is a useful number.

To calculate a useful load at 200mV:
R = 0.2V / 5.4ua = 37K (use standard 5% value 33K)

Now the AC component
E = 0.18ua * 33K ≈ 6mV

An AC gain of 1000 will bring this level up to 6V.

Wait for circuit suggestion

 

OK finally found where the link was hidden

 

Good Morning Jimkeith

Are you trying to transfer the current into voltage firstly and then amplify the voltage?
Why in this way? Instead of just put a blocking capacitor between photodiode and inverse port of OpAmp.

 

Good Morning there--good evening here 10.27PM

In my circuit, the photodiode has a load resistor, is AC coupled, and the amplifier has a gain of 1000--to get this kind of gain without loading the photodiode excessively via an inverting amplifier, a non-inverting amplifier is used.

Your circuit has problems--needs a current path for the diode current