Can anyone explain to me how this current-voltage converter work?

Discussion in 'Analog & Mixed-Signal Design' started by thanhvu94, Feb 3, 2017.

  1. thanhvu94

    Thread Starter New Member

    Oct 1, 2016
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    Hello everyone,

    I'm working on a project with a sensor which has an analog circuit to convert output current signal of photodiode to voltage like attached in the file. I'm not so specialized in analog circuit but I really want to understand how this circuit can convert from current to voltage. Can you explain in detail how does it work (the sub-component linear amplifier and non-inverter amplifier)?

    Here is the only information written in the report about this circuit but I can't understand well:
    - The linear amplifier part is an analog low-pass filter with a cutoff frequency equal to 106kHz. Considering the high gain given by the resistance of 150kOhm, the linear amplifier aims at preventing signal from oscillations.

    Thank you for your explanation.

    P/S I also attach the paper of the sensor.
     
  2. ericgibbs

    Moderator

    Jan 29, 2010
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  3. thanhvu94

    Thread Starter New Member

    Oct 1, 2016
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  4. ericgibbs

    Moderator

    Jan 29, 2010
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    hi,
    It amplifies the low level signal output from the Trans/amp, by a factor of 1+ (100k/4k7)= ~ 22 times.
    The non inverting mode means that the output signal from the trans/amp is not inverted

    E
     
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  5. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
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    Simply DC gain.

    Both amplifiers are linear. The first is a "current-to-voltage" converter, where 1 uA of photodiode current produces an output of 0.15 V. The second is a DC coupled gain stage with a gain of 22.3. These values are based on my guess of the resistor values in the schematic. If there are decimal points in the resistor values, I can't see them.

    ak
     
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