Photodiode to amplifier connection?

I have a large area silicon photodiode connected to pre-amp. I want to replace it with a new one. Original one is discontiued and I have found a new one with almost all electrical parameters the same as the original one but it is smaller and active area is half. Since active area is half phtodiode will produce half the current when light shine on it but since pre-amp OP amp usually is a very high imput impedance one I though it may work but I am not sure. Do I need to find a higher input impedance OPAmp chip or just connect the new photodiode as it is?

thx

Hi,

You need to know the amplifier circuit.

And in electronics we need specifications = values with units. We can't calculate "text".

Klaus

steve_rb said:

...but it is smaller and active area is half. Since active area is half phtodiode will produce half the current when light shine on it but since pre-amp OP amp usually is a very high imput impedance one I though it may work but I am not sure. ...

Click to expand...

Photodiode is measuring some light and, in all likelihood, the light is focused on the photodiode by some lens. Usually the focus is not "pin-point", it is supposed to distribute the total available light over the active surface area of the photodiode. If the active area is reduced, the amount of light reaching the photodiode will be correspondingly reduced BUT you can perhaps adjust the optics to refocus on the current active area of the photodiode.

I wonder why the original designers opted for a large area silicon photodiode because large area low noise photodiodes tend to be expensive. Perhaps the beam is not amenable to focusing. Perhaps there may be other reasons I cannot guess.

Photodiode is used for precise alignment similar to guiding systems. A very narrow laser beam drops on the photodiode. Photodiode active area is circular. When beam drops on the center signal has highest amplitude but when beam drops close to periphery signal amplitude goes towards zero (as shown in the attached picture). Comparing signal amplitude at the center and offcenter it is possible to find error signal and how much system is misaligned. I have attached pre-amp stage here. Since slop of the red line changed to the slop of the green line now in order for the system to do the same precise alignment with the new smaller phtodiode It seems some modifications should be done to the elements of the pre-amp. Any idea which elements and how?

As you mention that the new photodiode has about 1/2 active area, the radius will be about 70% of the original. Therefore the green line corresponds to the photodiode with 50% of the surface area of the red line.

Typical laser beam has an approximately gaussian profile of intensity distribution. However, in the graph it has been implicitly assumed that the laser beam is having a rectangular cross section. (you can slide a narrow slit over a circle and you can get the response profile).

If you just increase the gain of the photodiode by a factor proportional to R2/R1, the system will act as if a photodiode with the larger area has been used. I cannot see any further effect.

R2/R1 ? Which resistors do you mean exactly referring to the FET circuit?

steve_rb said:

R2/R1 ? Which resistors do you mean exactly referring to the FET circuit?

Click to expand...

R2/R1 refers to the second diagram in post #4; large photodiode naturally gives higher current.

If you decrease the RL (first diagram in #4), the photodiode current will increase and the curve in second diagram (same post) will shift towards right.

There is another problem worries me. I have attached picture showing two possible case. Case 1: green line and red line meet each other at one maximum signal amplitude point and case 2: green line and red line are parralel. Case 2 is more prabable because when phtodiode is smaller and laser beam drops at the center it will give smaller maximum amplutude because more carriers will be lost through photodiode periphery because photodiode is smaller but this is also depend of the mean free path of the carriers which is not simple to figure out.
Cosidering case 2 is what is hapening changing photodiode gaine simply by a proportion of R2/R2 can fix the problem as c_mitra mentioned but even with fixing the gain still one other problem will remain which is system higher stifness with regard to the large photodiode case and still it will remain how to over come higher stifness problems otherwise system may go ito oscillations?

If case 1 is the true case then we can't just simply fix the problem by adjusting the gain through R2/R1 because close to the center we need zero gain change but at far from the center we need more change in the gain. So more complicated change in the gain is neccessary than simple proportion of R2/R1. Even if we find a proper nonlinear fix for the gain somehow still problem of higher stifness of the system remains to solve.

You need to make a model of what you are trying to say. Let us consider the ideal case first:

1. If the source (light beam) is a point source, the response will be constant as long as it falls on the photodiode; when it moves out of the active area, the response will fall to zero. The graph will be a step function in that case.

2. If the light source is in the form of a beam (very narrow rectangle: usual outputs from a monochromator), and the beam height is smaller than the photodiode diameter, the response will fall off as the beam falls off centre.

3. If the light source is gaussian (2-D), then the response will be max at the centre but will fall off as the beam goes near the edge. This is the normal mode used in position sensors.

It is a recommended practice to see that the photodiode is more or less uniformly illuminated. If a point source of light falls in the centre there will be local saturation and the relationship will be hard to predict. The photodiode may be damaged also.

Photodiode is used for precise alignment. Beam is Gaussian. Response is max at the center and falls off as beam goes near the center. This is shown on the picture on post #8 by both green and red lines. Comparison between the signal amplitude at any off center point with signal amplitude at max gives the error signal. System corrects itself according to this error signal in order for the beam gets back to the center which means system is precisely aligned.

For alignment purposes, a four quadrant photodetector with 4 outputs are common. The four outputs are commonly processed as X,Y and SUM. X signal suggests movement on the X-axis and so on. In the present case, the servo will not know which direction to turn because that information is insufficient.

System is very old. AT that time there were no quadrant detector. That is why they have used single element detector. There is another device sends direction signal to the servo. It is attached to the system and moves with the system movements.

Four quadrant photodiodes are not new, but that is a different story. You need three signals: horizontal shift (with direction sense), vertical shift (with direction sense) and total intensity (to scale the shift values).

Please tell us what is the signal that tells us about the horizontal shift? And also how to you sense the direction of the shift?

System is a rotating one. Then x and y shift and also direction is not our concern. Shift in the r is the one should be considered instead. Another thing is photodiode current will not change by changing RL since photodiode is at the reverse bias and current only will change if light droping on it increase. We have to increase RL instead of decreasing. This way for the same current signal to the FET will increase which means green line is shifted towards the red line. right?

steve_rb said:

since photodiode is at the reverse bias and current only will change if light droping on it increase...

Click to expand...

This is the most common mode; within limits the photodiode acts like a current source and the magnitude of this current is directly proportional to the light intensity.

RL (in your circuit diagram) basically acts like a current to voltage converter. The higher the value, the greater is the voltage produced. The same current also charges the capacitor. You may also need to see the FET stays within a linear region.

Conclusion:

what finally I should do? I know a little bit of physics but not much electronics. I need expert opinion to get over this problem and replace this smaller photodiode and make necessary modifications in order to make the system to work successfully. More detail explanation will be appreciated since as I said I don't know much of electronics.

I guess that the photodiode can be replaced as such; because the beam is focused, the light (almost 100%) will fall on the photodiode whether it is small or large (size does not matter). The limit will appear as the beam shifts off-centre; if the point source assumption is valid, we will not see much change till the beam moves out of the active area.

I do not believe the original designer made the detector output (the output will be a voltage in this case) critically sensitive to the photodiode characteristics. If the system shows instability you will need to increase the RL (the by about 20-30% but surely not 40%) because that may change the response time constant (smaller photodiode may have smaller capacitance) but still may stay within limit.