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THS4032: Slew rate calculation

Shibin Varghese
Prodigy 185 points
Part Number: THS4032

Hi Team, 
I am using THS4032 as a trans-impedance amplifier in one of my applications.
My input signal is a 250kHz, +/-10V signal.
How to calculate the slew rate and gain-bandwidth product required for the op-amp for such a signal.
Actually, TI's experts in E2E suggested this op-amp. But  I am curious to know how to calculate slew rate and gain-bandwidth product.
Hope you can help me?.


Also, What does VOC referring to?.

  • 0
    TI__Expert 3620 points

    Hi Shibin,

    This is not a simple topic, so lets walk through a design approach for this device's behavior specifically using the datasheet.

    VOC here is the peak amplitude (above the common mode) so 10V for your case. You'll notice that the slew rate is dependent on step size:

    You might also notice this if you check the math for the full power bandwidth using the equation in your post:

    Using FPBW = slew/(sqrt(2)*pi*Vpeak) you should get 2.25MHz using 100V/us for 10Vpeak, and 7.20MHz using 80V/us for 2.5Vpeak.

    This behavior is also evident in the charts:

    If you are designing around a pulse or square wave input and are concerned about rise-time, my advice would be to simulate the response using TINA or PSpice for TI. I can help you set that up but I will need more information about the system, ideally a schematic and the input waveform characteristics and output loading.

    Best,
    Sam

  • 0
    Guru 140200 points

    Hi Shibin,

    as Sam already mentioned, there's no simple answer. The detector capacitance, the feedback capacitance, the feedback resistance, the transimpedance, the input signal shape and amplitude would be needed to find a proper OPAmp.

    The minimum slew rate for a sine of 250kHz and 10V amplitude at the output of TIA would be

    2 x pi x 250kHz x 10V = 15.7V/µs

    You get this formula when taking the time derivative of

    10V x sin (w x t)

    assuming

    w = 2 x pi x 250kHz

    Kai