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OPA388: Regarding stability of differential attenuator of OPA388

JK
Prodigy 120 points
Part Number: OPA388
Other Parts Discussed in Thread: TINA-TI

Tool/software:

Hello TI support,

There are couple of questions on operating OPA388 as attenuator in differential mode. 

1. Can we operate OPA388 as attenuator (Gain=0.545) in differential mode? We are using attached circuit & quick simulation shows it to be stable. Though simulation are positive, I just wanted to get TI's review so we don't miss anything before going forward. We are looking for bandwidth of near 5KHz. 

2. Are there any limitations on lower gain (G<1)? Like can we go for G=0.05, 0.1 or so?

Regards,

JK

  

  • 0
    TI__Guru 68561 points

    JK,

    You may operate the circuit in any attenuation you wish BUT its maximum bandwidth is controlled by the non-inverting gain: BW = GBW/(1+24k/44k) = 10MHz/1.545 =~6.5MHz.

    However, the circuit as shown has the -3dB bandwidth of around 31kHz set by 22kohm input resistors and 220pF differential input capacitance: fc = 1/(6.28*22k*220pF) 

    As far as stability goes, the circuit is only marginally stable with 37 degrees phase margin (minimum recommended is 45 degrees) -  see below.

    Running transient simulation shows the small-signal overshoot of 26% - see below.

    Adding 12pF across the feedback resistors improves the phase margin to 97 degrees - see below.

    The said addition of 12pF caps also improves stability by decreasing the small-signal overshoot down to 4% - see below.

    The overall bandwidth of the configuration remains constant at around 31kHz - see below.

    For your convenience below I have attached Tina-TI simulation files.  You mau download a free copy of Tina-TI simulator by clicking on following link: 

    https://www.ti.com/tool/TINA-TI

    OPA388 diif amp bandwidth.TSC

    6558.OPA388 AC Stability.TSC

    6558.OPA388 Transient Stability.TSC

  • 0
    Prodigy 120 points

    Hello Marek,

    Thanks for your detailed answer & explanations. I have taken your feedback regarding 12pF. However, I would appreciate clarification of below queries,

    1. .AC response doesn't shows the peaking in LTSpice but even though circuit is marginal stable. Any idea about the reasons for this?

    2. This point is similar to point-1 where I tried pulse inputs with 100ns rise/ fall times (differentially) at 5KHz BW & output doesn't shows any overshoots, so am I doing something different in accessing stability w.r.t one you did?

    3. I see you have used 100KHz current source as load while pulse circuit in LTSpice is having 10K load, any advantage with 100KHz current load?

    Regards,

    JK

          

  • 0 in reply to JK
    TI__Guru 68561 points

    JK,

    1.  You do not see gain peaking ONLY because of the low-pass filters you placed in front of the difference amplifier, which roll-off the gain at much lower frequency.  However, if you remove your LP filters at the input, you will see gain peaking - see below.

    However, adding CF= (Rin/Rf)*(Cin_cm+Cin_diff) = (44k/24k)*(4.5pF+2pF) = ~12pF feedback cap cancels the zero in the close-loop transfer function created by the interaction of Rin and Cin, which eliminates the gain peaking - see below.

    2.  In order to see output overshoot, you must stimulate the op amp with a SQUARE input signal. Even though you apply the square input signal at VF1 and VF2, it is being rounded off by your LP input filter before it is seen by op amp - see Vin+ and Vin-.  Thus, the reason you do not see the  output overshoot is that you apply the square input waveform at the input, which gets filtered out before reaching op amp, whereas I applied it directly at the output.

    All in all, for the same reason you did not see the gain peaking, if you remove your input low-pass filter, you will see the output overshoot - see below.

    3. The reason for pulsing the output instead of input with a small-signal current source was to stimulate the op amp with square waveform without the need to remove the input low-pass filter as discussed above in 1) and 2).

  • 0 in reply to Marek Lis
    Prodigy 120 points

    Hello Marek,

    I got the point, you are trying to check stability without LPF & trying to cancel out the pole made by feedback resistor & input capacitance. This is interesting & I shall further appreciate few more questions,

    1. Is it like LPF not being part of pole cancellation can't be considered in stability? 

    2. Could you explain the need of output stimulus in accessing stability criteria. Do you recommend this in general for any opamp circuit?

    3. General literatures talk of stability by giving pulse/ square waveforms at input & seeing output AC response or seeing the overshoots in time domain. I would like to understand how do I check stability of any opamp circuit  having combination of RC filters at input, do we remove those before applying stimulus? Or provide output stimulus?

    Yes, could see overshoot in output with 100KHz current load & reduce those as well with 12pF. 

    Regards,

    JK

            

  • +1 in reply to JK
    TI__Guru 68561 points

    Please see my answers below:

    1. Correct - LPF in front of the op amp does not cancel a zero formed by interaction of input capacitance with input resistors.

    2. In order to see the percent overshoot, the output must be stimulated with a square waveform.  Since applying the square signal at the input of circuit with LPF would not result in step stimulus at the output, in such case the signal mudy be applied directly at the output to bypass LPF rounding off effects.

    3. In order to see output overshoot with RC input you do NOT remove LPF but instead must apply a square signal directly at the output as discussed in 2).