Transimpedance Amplifier vs Voltage follower/non-inverting amp combo

Discussion in 'Analog & Mixed-Signal Design' started by odm4286, Aug 9, 2016.

  1. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    I'm working with a photodiode in photovoltaic mode. In order to amplify the signal I've done the following

    [​IMG]
    Omitted power connections purposely. ​

    So this works great, it eliminated all the noise I'd get when I connected the photodiode directly to the non-inverting amplifier.
    My question is what advantage, if any, does a Transimpedance amplifier circuit have over this circuit? Thanks!
     
  2. OBW0549

    Well-Known Member

    Mar 2, 2015
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    One advantage a transimpedance amplifier has over your circuit is that it actually works: take a VERY close look at how you've connected U1 and the photodiode.
     
  3. odm4286

    Thread Starter Active Member

    Sep 20, 2009
    155
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    Hmm I'm not seeing it, as far as I could tell last night this worked fine. Roughly 300mV across my PD and about 3.3V on the output of U2. The output followed as I changed the light exposure to the diode as well. What am I'm doing wrong here?
     
    Last edited: Aug 9, 2016
  4. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    Ahh I think I've added an "extra" ground connection. Sorry, I drew this from memory. U1 inverting input should NOT be tied to ground, correct?

    [​IMG]
     
    Last edited: Aug 9, 2016
  5. OBW0549

    Well-Known Member

    Mar 2, 2015
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    Still not right.
     
  6. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    Could you tell me what is wrong?
     
  7. OBW0549

    Well-Known Member

    Mar 2, 2015
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    I could, but it's something glaringly obvious-- and fundamental-- that you really should be able to see for yourself (and it has nothing to do with power supply connections).
     
  8. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    ahhh my diode is backwards...I think. Still learning. Cathode to the non-inverting input?

    [​IMG]
     
    Last edited: Aug 9, 2016
  9. OBW0549

    Well-Known Member

    Mar 2, 2015
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    Nope. Which way your diode is pointing affects only the polarity of your output, and nothing else.

    Hint: what have you provided in your circuit to establish a reference potential for the inputs and output of U1? As you have it, they're undefined (i.e., floating).
     
  10. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    Ok, I've taken the diode out of the feedback loop. In hindsight that doesn't make much sense. How about this? Thank you for the help by the way!

    [​IMG]
     
  11. OBW0549

    Well-Known Member

    Mar 2, 2015
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    That's more like it!

    Now take a look at the circuit and ask yourself, why is the first opamp even necessary? Why not eliminate it, and connect the anode of the photodiode directly to the (+) input of the second opamp?
     
  12. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    I tried this and ended up with an extremely noisy signal. I'll give it another go tonight! So back to my original question, is there any reason NOT to do it this way? Also, I shouldn't have an issue with current flowing through my diode right? I'm wondering if I should add a series resistor to ground.
     
  13. OBW0549

    Well-Known Member

    Mar 2, 2015
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    Do you want an output that varies linearly with light level? Or do you want an output proportional to the log of the light level? As it is, your circuit will show a logarithmic response. To make it linear, you either have to shunt the photodiode with an appropriately-sized resistor, or switch to a transimpedance amplifier configuration.

    Also, a voltage-mode circuit such as this will have very poor response speed, especially at low light levels; a properly-designed transimpedance amplifier will be much, MUCH faster.
     
  14. crutschow

    Expert

    Mar 14, 2008
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    Adding an amplifier adds that amplifier's noise and offset to the signal.
    This will increase the amp noise by the √2 and can double the DC offset.
     
  15. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    odm4286 likes this.
  16. OBW0549

    Well-Known Member

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    What kind of noise? Broadband thermal (i.e., "white") noise? RF interference? Pickup from power mains? You've got some very high impedances in this circuit, and it will be quite sensitive to any kind of interference; good shielding is a must.
     
  17. odm4286

    Thread Starter Active Member

    Sep 20, 2009
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    Not sure of the type of noise, but instead of a clean DC signal i'd get something that looked sinusoidal. The output followed the input by introducing an offset, not by changing the amplitude of the "AC" output signal.
     
  18. OBW0549

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    Ummm... why? The second opamp is operating in an ordinary non-inverting configuration and has a high-impedance input. As in all opamp circuits, the inputs must have a DC path to ground (or to some other voltage within the common-mode range of the opamp); but the input to a non-inverting amplifier such as this does NOT need to be driven from a low-impedance source.
     
    Last edited: Aug 9, 2016
  19. OBW0549

    Well-Known Member

    Mar 2, 2015
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    Sinusoidal, but at what frequency? If it was 60 Hz, that's power-line pickup. Use more shielding.
     
  20. bertus

    Administrator

    Apr 5, 2008
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