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OPA657 as Transimpedance Amplifier

Other Parts Discussed in Thread: OPA657, LMH6629, TINA-TI

Hi,

I am having a few problems designing the OPA657 as a transimpedance amplifier.

Using the 3dB frequency equations outlined in the datasheet for a gain of 1G (Cf = 8.92fF (not shown) with photodiode shunt capacitance 400pF) we should achieve 25.231kHz bandwidth?

However, running a simulation in LT Spice the bandwidth achieved is much lesser than expected. This variation from the relationship shown in the datasheet seems to worsen with increasing gain.

The bandwidth required from the system is 20kHz.

 

The feedback loop is in a T-circuit to minimise resistor values for the very high gain and minimise johnson noise. Gain is also switchable between 10M and 1G in this configuration. At 10M the bandwidth is also lesser than expected although i have achieved the specification of 20kHz and so this is less of a problem.

Is there any reason why the achievable bandwidth does not match the applications example in the datasheet?

I hope this is clear.

Thanks in advance,

Michael

 

 

  • Hi Michael,

    The T-circuit arrangement that you have shown (to reduce the feedback resistor value), has the side-effect that you have noted (less bandwidth that you could otherwise achieve with a simple feedback resistor).

    I  think the simulated bandwidth correctly estimates the speed / bandwidth.

    Regards,

    Hooman

  • Hi Hooman,

    Thanks for the reply.

    I also carried out the analysis using a simple feedback resistor and view the same results. Attached is the AC response of transimpedance gain when a simple feedback resistor is used. There must be something else causing the poor bandwidth but i am unsure what.

    The Green plot shows the TI Gain.

    I should say that almost exactly the same response is observed when using the T-Circuit.

    Thanks again,

    Michael

     

  • Hi Michael,

    The OPA657 datasheet expression for bandwidth is an estimate with typical feedback resistors assuming that the Noise Gain intercept with the amplifier open loop gain occurs at the -20 dB/decade region (see LMH6629, Figure 64). This is not the case with your values.

    With the super large RF value of 1Gohm, the intercept occurs at 2.2kHz (which is below the 1st dominant pole of of the OPA657 of around 300kHz!). Therefore, the -3dB Bandwidth estimate does not apply in your case. The reason for the low frequency intercept is the Noise Gain "zero" occurring at a low frequency due to large RF and large diode capacitance.

    Here is the TINA-TI simulation file I have used for above:

    2072.OPA657 TIA Open Loop Analysis E2E large RF Hooman 9_2_14.TSC

    To estimate the bandwidth in your case, you could use this expression instead:

    f_3dB= ~ (f_zero) x A_OL

    where:

    A_OL= 75dB = 5.62 kV/V

    f_zero = 1/ (2x pi() x RF x C_diode)= 0.4Hz (with C_diode= 400pF, RF= 1Gohm)

    or:
    f_3dB (in your case) = 0.4Hz x 5.62 kV/V = 2.24 kHz

    Regards,

    Hooman

  • Hi Hooman,

    Thank you very much. It is all starting to make more sense.

    I have just one more query on the back of the analysis you describe. I looked into the minimum frequency at which the noise gain "zero" can be present such that we will not intercept the flat frequency area of the open loop gain at 75dB.

    Approximating this to 100Hz then using a gain of 10Megs i should be able to optimise the feedback capacitance value to provide a pole where the noise gain would intercept the open loop gain with the diode capacitance of 400pF.

    I have two questions based on this:

    1) The optimisation formulas for feedback capacitance do not provide a pole at greater than 300kHz which would be expected as there derivation is based on the intercept between the frequency where this pole sits and the open loop gain slope? Abusing this derivation and implementing a pole greater than 300kHz reduces the frequency where the noise gain pole is dictated by GBW to below 300kHz. Why do these values not seem to make sense?

    2) Secondly, again based on the locations of the two noise gain poles, the noise gain would never intercept the open loop gain in many situations (including the situation described above) due to the high gain and feedback capacitance reducing the frequency of these poles. In this case, what dictates the bandwidth of the system?

    I suppose the only two options here are to reduce the gain/photodiode capacitance or identify a new high speed precision op amp with an open loop pole of less than 300kHz.

     

    Thanks again for your help,

    Michael

  • Hi Michael,

    Sounds like you like to optimize TIA bandwidth by varying RF and the diode capacitance when you use the OPA657. The TIA bandwidth will be the lower of the following:

    1. The intercept of the noise gain with OPA657 open loop gain

    2. The frequency 1/ (2 x pi() x RF x CF)

    With OPA657's 350kHz dominant pole, when the noise gain zero (f_zero) is 62Hz or less, you will intercept the Aol curve flat region. In this region, you won't need a capacitor across feedback (if you have one, it would lower your bandwidth) and your bandwidth is limited to what I pointed out yesterday:

    f_3dB= ~ (f_zero) x A_OL

    Keep in mind that board and component parasitic capacitance comes to play when you are dealing with discrete values close to 1pF or less, such that simulated results will be affected by these.

    Here is a tabulation of some simulation results for various Rf for both 400pF and 200pF diode capacitance:

    2543.TIA OPA657 Simulation Bandwidth E2E various C_diode Hooman 9_3_14.xlsx

    Here is the image for one of the conditions (RF= 500k, CF= 0.2pF, for C_diode= 400pF):

    Here is the TINA-TI simulation file I have used for the same condition as above (you can vary the circuit to see other scenarios):

    7024.OPA657 TIA Open Loop Analysis E2E large RF Hooman 9_3_14.TSC

     

    Regards,

    Hooman

  • Hi Hooman,

    I see what the limiting factors are for bandwidth now.


    Thanks for all your help.

    Regards,

    Michael