1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

Photodiode current measurements

Discussion in 'Homework Help' started by DX400, Oct 26, 2017.

  1. DX400

    DX400 New Member

    Joined:
    Oct 26, 2017
    Messages:
    3
    Likes:
    0
    I am trying to build transconductance amplifier circuit to obtain values of the reverse biased photodiode (PD) current in order to estimate some of the LED characteristics.


    At first, I have used the scheme named “Recommended Zero Biased Circuit” from page http://home.sandiego.edu/~ekim/photodiode/pdtech.htm (placed at the bottom of the page, the fourth one figure from the bottom ). The photodiode is Vishay BPW21R, OpAmp is LM358n, the value of the Rf control resistor of the negative feedback is 2 kOhm. The output voltage is measured by the multimeter. To obtain some signal, I am using flash light with 3W cool white LED like this https://www.sparkfun.com/products/13105. The flashlight is placed near in front of the photodiode to obtain the same luminous flux at each try.

    When flashlight is off (no light) the output voltage (Uo) is about 0.

    I have found that, the variation of the Rf (the gain of the amplifier) from 300 Ohm to 500 kOhm leads to not quite proportional change in output voltage (Uo). So, the values of the PD current are slightly different for the different Rf. What is the cause of it?

    For example for the Rf of 2 kOhm I have obtain Vout of about 3.6 V and the PD current of 1.8 mA.


    Then, I have used the scheme named “Positive bias circuit” from page http://home.sandiego.edu/~ekim/photodiode/pdtech.htm (placed at the bottom of the page, the second one figure from the bottom ). I have used 10 V which supply the OpAmp and to make photodiode reverse biased.

    The value of Vout for the Rf of 2 kOhm decreases to 0.7 V for the same illumination.


    So the problems:

    1. What is the reason of output voltage drop.

    2. I have something like overflow. When flashlight is off (no light) the Uo is about 0. Under flash The variation of the Rf, which defines the amplification gain of the circuit, from 1 to 100 kOhm does not change the value of the output voltage Uout of about 0.7.

    3. The increase of the Rf up to 0.5-1 MOhm leads to much decrease of the Uo.

    4. The variation of the Rf also leads to not proportional change of the output voltage. So the Rf does’t work as gain.

    5. I don’t understand how to be certain about obtained results? How to calibrate the results of the PD current? How to make a calibrated light source for such measurements to be sure of correct PD current?


    I should be very obliged for help to solve this problem.
     
  2. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,099
    Likes:
    562
    Location:
    AZ 86334
    Post your circuits. Your links dont work.

    What is the supply voltage for the opamp?
     
  3. DX400

    DX400 New Member

    Joined:
    Oct 26, 2017
    Messages:
    3
    Likes:
    0
  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. DX400

    DX400 New Member

    Joined:
    Oct 26, 2017
    Messages:
    3
    Likes:
    0

    The supply voltage of the OppAmp is 10 volts unipolar in all cases.
     
  6. audioguru

    audioguru Well-Known Member Most Helpful Member

    Joined:
    Mar 16, 2004
    Messages:
    32,518
    Likes:
    942
    Location:
    Canada, of course!
    ONLINE
    You used a single positive supply then how in earth can the output go negative as shown on the diagrams?? Their opamps probably have a dual polarity supply that is stupidly not shown.
     
  7. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,099
    Likes:
    562
    Location:
    AZ 86334
    V(out) for Rf= 3KΩ, 6KΩ, 12KΩ,..., 96KΩ

    D1 is needed only for simulation. I1 represents the photodiode current:

    pd3.png

    pd4.png

    When you understand the plot, you might have some insight into what your circuit is doing...

    Questions you should be asking?

    What is the input bias current of the opamp?
    What is the output swing of the opamp?
    What is the input common-mode range of the opamp?
    What is the noise floor of the opamp?
     
    Last edited: Oct 26, 2017
  8. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

    Joined:
    Oct 30, 2010
    Messages:
    9,938
    Likes:
    1,099
    RC needs to be the same for your tests, otherwise your changing the response of the amplifier.

    Be careful, if the OP amp cannot drive a capacitive load, sometimes the OP amp will oscillate. One range of an I-V converter I designed was very unstable at say 20 mA from a solar cell. That result scared me to death.

    The company I worked for designed an I-V converter for in house use and the IC selection is problematic. Their initial design used the CA3140. I changed it to an LF351 which had better performance.

    My "Holy grail" design did have some bugs, but it outperformed anything else we ever had for low level AC measurements as the front-end to a lock-in amplifier.
    I had intended to be able to zero the amplifier automagically and I incorporated no mechanical zero adjustments. When I set a D/A to zero, it;s output was non-zero. In order for the algorithm to work, I would need a "hard zero", but 40 pA of offset current was no bg deal for us in the DC mode. i was told the project was "done".

    The "Holy grail" was also a true 4-terminal device with 4 ranges, biasing and suppression, For giggles it had an open circuit voltage mode and a 2T/4T mode.
    Zero Check and Zero Correct didn't work.

    Ib, Vos and the temperature dependence of Vos. With high values of the FB resistor, Vos is is very important without offset-nulling,

    0.7 V is a magic number.
     

Share This Page