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OPA404: OPA404_circuit understanding

NAFISA A
Prodigy 60 points
Part Number: OPA404
Other Parts Discussed in Thread: OPA637, TINA-TI

Hi,

Need the below circuit understanding with input 5V fed to the opamp OPA404.

 

  • 0
    Guru 259460 points

    Are you sure that the inputs have different voltages, i.e., that pin 2 is at 5 V without an intermediate resistor?

  • 0
    TI__Guru 61900 points

    NAFISA,

    You cannot apply a voltage source directly to an inverting input (pin 2) of an op amp unless you want to use it as a comparator.  Under such condition, the output will be pushed close to its negative (V-).

  • 0 in reply to Clemens Ladisch
    Prodigy 60 points

    Attached the actual circuit image. The input to the SUM circuit is from the trans impedance amplifier(OPA637) output SCAT_DC of voltage 12.2V as per the simulation result with current input of -10nA at IS1.

  • 0 in reply to NAFISA A
    TI__Guru 61900 points

    Nafisa,

    Let me first say, your schematic is very difficult to read so there may be some errors in the component values I used below.

    From the DC perspective, lower circuit (OPA637) plays no role since it is isolated by C14 cap - thus, the upper op amp is simply in an inverting gain of G=-1/425.5 resulting in the VF1 = V1*Gain= -15V/425.5 = -35.25mV - see below.

    You say that with current input, IG1, of 10nA, you read SCAT_DC voltage of 12.2V BUT you cannot have DC current that is AC coupled (C10) like you show below because this would cause the input of IOP1 (OPA637) to saturate at the positive rail (a non-linear operation). 

    Thus, the circuit could only work with no C10 cap as shown below.

    To say anything about the circuit response beyond its DC operating point shown above, one must provide information regarding IG1 AC input signal.

    For example, using IG1 of 1uA at 1kHz, results in the output shown below.

    Below I have attached Tina-TI schematic for your own simulation.

    NAFISA OPA404.TSC

  • 0 in reply to Marek Lis
    Prodigy 60 points

    Attached the actual SUM circuit and its response. Please tell how this opamp reponds with Schottky diode and Capacitor at the feedback. Is this an integrator circuit?

    Also, regarding the below circuit input, Its both DC and pulsed.

    Attached the application circuit for reference.

    The output of the photodiode will be proportional to the input light, so constant light would produce a DC signal, and oscillating light would produce an oscillating signal. So this will depend on laser diode.

     

  • 0 in reply to NAFISA A
    TI__Mastermind 18037 points

    Nafisa,

    1. The circuit you show at the top of the page has a gain of 47V/V.  See the slope of curve 2.
    2. The 15V and 20Meg input shift the gain curve by -35mV.  (15V/20M)(47k) = 35mV.  See offset of curve 2.
    3. The Schottky diodes limit the output range to +/- one diode drop (see curve 1). 
    4. Thus, the only linear response of the amplifier is for inputs of about -3.5mV to +3mV.  This corresponds to an output swing of 127mV to -166mV.
    5. The SUM aspect of the circuit is to add the offset from the 15V x 20M to the input signal x 1k.
    6. The capacitor is used to filter the output fc = 1/(2pi(47kohm)(100pF)) = 33kHz.

    I 

    sum witth diode.TSC

    I hope this was helpful to you.

    Best regards,

    Art