Transimpedance amp

Discussion in 'The Projects Forum' started by gvi70000, Nov 15, 2009.

  1. gvi70000

    Thread Starter Active Member

    Oct 4, 2009
    34
    0
    Hello,
    I'm trying to make a high sensitivity light/dark sensor using a bpw34 photodiode. So far i came up with this circuit...do you think will work?

    The output will go into a delay circuit showed in 556.png

    Thanks for helping me out
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You DO know that the BPW34 is most sensitive to infrared in the 850nm-990nm wavelength range, right? Below around 430nm, you won't get any response at all; between around 440nm to 850nm the response increases rapidly.
    Datasheet: http://www.vishay.com/docs/81521/bpw34.pdf
    See figure 7, on the 3rd page.

    [eta]
    Seems that you haven't provided a means to set a threshold level for the noninverting input of U2 (pin 3).

    You might use a 120k resistor from your 4.5v Vcc in series with a 10k pot to ground. Connect the pots' wiper to pin 3, with a small cap (say, 220pF to 10nF) from the wiper to ground to keep things quiet.
     
    Last edited: Nov 15, 2009
  3. gvi70000

    Thread Starter Active Member

    Oct 4, 2009
    34
    0
    Hello, thanks for replying
    After i've look at some examples for this kind of amp's i saw that the non inverting input is connected directly to ground.
    I'll make a simulation with yours suggestions and see what's happening
    This circuit should work in day light to detect lightnings and also at the light room.
    Thanks again
     
  4. gvi70000

    Thread Starter Active Member

    Oct 4, 2009
    34
    0
    hi,
    I simulate the circuit with changes made according with your specifications
    The only difference is that the output of the first stage will oscillate with an offset V made by the 120K and 10k. If the second resistor(10k )value will be smaller then the minimum peak will be near to ground, so i don't see any reason to put these 2 resistor in the circuit...but as I'm a begginer i got 99% chances to be wrong. Can you please explain that to me if you have the time?
    I've attached the simulation result Sim2 is the result with the 2 resistors you recommended and Sim the circuit in the post

    Here http://www.euronet.nl/~mgw/background/opamps/uk_opampgain_1.html they say "This type of amp is used for amplification of devices that behave more like a current source than a voltage source. The opamp is used to transform the current input in a voltage output. Since the non-inverting input of the opamp is grounded, at the inverting input the will be no current. The current coming from the input device will therefore go through the feedback circuit connected between the opamp output and the inverting input. The op-amp provides the output voltage necessary to equalize these currents with the inverting input voltage held equal to the non-inverting input voltage (which is zero since it's connected to ground).
    V_out = V_fb = - I_source * R2.
    In case of a MC cartridge in the inverting input of an opamp, the cartridge is effectively shorted (zero impedance). As a result, the current is much higher than with the traditional voltage amp where the cartridge is loaded."
     
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    • Sim2.PNG
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  5. Darren Holdstock

    Active Member

    Feb 10, 2009
    262
    11
    Your transimpedance amp is oscillating because of the parasitic capacitance of the PD forming a pole with the feedback loop. Get around this by lowering R1 (at the expense of gain), raising C2 (at the expense of bandwidth), or using a PD with a lower junction capacitance. A slower op-amp won't be quite so frisky too - you don't need all that speed, and a more leisurely device will be more forgiving.
     
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