I am working on a project using IR LED (1200nm peak response) and a photodiode (G10899-01K) 500-1700nm spectral response. I used photovoltaic mode for Transimpeandance amplifier. The output ranges from 100mV to 400mV and varies with application of sample in between IR LED and photodiode. When I use a filter 1mega ohm and 10nF at the feedback path, the signal is attenuated, I get 1.8mV which and not varying as it is should even after amplifying to 3V. What could be the problem? Kindly assist on the design of how to eliminate noises. I am attaching the data sheet for photodiode, I may be wrong in interpreting. Any help is highly appreciated.

 

Can you provide a schematic of your circuit? I am havin trouble picturing it from the description.y

 

Hi k854,
As requested, please post your circuit diagram.
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Thanks all.
This is the circuit diagram I am working on it. It is using LM358 op amp with 5V dc supplyIR LED is biased by 1.5V.

 

I need a circuit that filter noise, amplify and faithfully transmits the signal such that any variation with the input signal, there is a corresponding variation in the output signal.

 

hi k,
Have you considered that using only a single power supply will limit the Vout of the OPA's to above zero volt output.?
I will post a LTSpice circuit of your circuit.
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hi k,
This is what your circuit shows in simulation, using a sine wave source for the photo detector.

Is the IR beam connected continuously to the detector Photo diode, or is being interrupted to produce a pulse.?
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Hi k,
This sim is using a 10uSec pulse , also the 10uF has been added.
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Added: exactly what is the signal at the photo diode .?

 

Hi ericgibbs,
Thank you very much for your response.
The IR beam (at 1200nm wavelength) is interrupted (i.e allowed to pass through the sample) to produce a pulse to photodiode
When the sample is place in between IR LED and photodiode, absorbance of IR light takes place.
This means that the sample has taken in some IR light.
The IR light that through to the photodiode give us information of the sample inserted (glucose in this case) using the relationship: Absorbance is proportional to concentration.
Where have u added the 10uF capacitor?

 

hi k,
Your original post said, 10nF at the feedback, as shown in your circuit.
I have added a 10nF across the 1meg, post #8., not 10uF
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The IR LED uses max 1.5V for forward biasing. photodiode uses 5V. Therefore single power supply will not apply at the moment since I did not want to change that IR LED. Any way out is highly appreciated.

 

Hi k,
This sim is using a 10uSec pulse , also the 10uF has been added.
E


Added: exactly what is the signal at the photo diode .?

View attachment 292618
The signal at the photodiode is current(photocurrent). This is converted by the transimpendance amplfier to voltage

 

hi k,
In an attempt to show what happens if you insert a substance between the IR source and photodetector
I have swept the photodiode signal down from +10mV to 0mV.

You can see your amplifier is saturated for most of the time except for a small region at approx 5mV signal

Basically, your amplifier has far too much gain.

E
.

 

hi k,
In an attempt to show what happens if you insert a substance between the IR source and photodetector
I have swept the photodiode signal down from +10mV to 0mV.

You can see your amplifier is saturated for most of the time except for a small region at approx 5mV signal

Basically, your amplifier has far too much gain.

E
View attachment 292619.

Thanks for the response Ericgibbs, and pointing out the problem. Can we get a suitable amplifier by manipulating the values in LTSpice? I will appreciate if you set and give reasonable range

 

hi k,
A rough approximation would be this circuit.
The IR emitter is set to give only approx 3mV at the photodiode when nothing is inserted.
So your 1.5V bias on the IR emitter may have to be reduced.
Also I have reduced the overall gain by a large factor.

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hi k,
A rough approximation would be this circuit.
The IR emitter is set to give only approx 3mV at the photodiode when nothing is inserted.
So your 1.5V bias on the IR emitter may have to be reduced.
Also I have reduced the overall gain by a large factor.

E
View attachment 292620

Thank you Ericgibbs for your response. This is very helpful.
Will the capacitor remain to be 10nF parallel with R3 at U2.
What is the output of the photodiode with this circuit.
Let me do it practically and see what happens.

 

Attached is the datasheet and screenshot of the data sheet for the photodiode. This might help in value determination

 

hi K,
I would build the test circuit as I posted, then design the voltage bias for the IR emitter so that you can vary the drive voltage to the IR.
Insert a test sample and then adjust the IR drive voltage so that you get a voltage change on the final OPA output Vout
Make a record of the IR drive and the Vout for a range of inserted sample types.

This should give you an idea of a rough calibration of the sample types versus the IR drive and Vout.

If you can do this, post a short list of your results, so that we can compare it with the simulation.

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Thank Ericgibbs,
Let me do it. You have given me a way out.
Sure I will share the outcome