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LPV802: Stability of LPV802 when it used as inverter amplifier

Ricardo Li
Intellectual 1045 points
Part Number: LPV802

Dears,

When I using LPV802 as inverter amplifier, I found it is unstable. The TINA simulation is as following:

The input capacitor is about 3pF. I calculated the zero of 1/β, is equal to 1/(2*π*3pF*5k)=10MHz. I think this circuit is stable. But the simulation shows that it is unstable. Why it is unstable? And how to make this circuit stable?

Here is my TINA file.

LPV802.TSC

Regards,

Ricardo

  • 0
    TI__Mastermind 23395 points
    Ricardo
    I will take a look at this tonight and get back to you.
    Chuck
  • 0 in reply to Chuck Sins
    TI__Mastermind 23395 points
    Ricardo
    I believe that I am following your analysis. Based on your calculation for the zero of 1/B, I completely agree with you that having a zero at 10MHz is not a problem since that is way beyond the 8kHz unity gain bandwidth of the LPV802. You are applying proper loop stability techniques.
    However, one thing you may not be modeling correctly is the capacitive load that is connected to the output of the amplifier when you built your inverting amplifier. If you notice in our datasheet, the LPV802 has a higher than normal open loop output impedance because it is a nanopower amplifier. It is actually around 90kohms. this means a much lower zero can be created when you combine this ro value with your capacitive load. My guess is that 1/2pi(ro)CL is lower than 8kHz. This means your rate of closure of your loop gain is 40dB/decade and you suffer instability.
    You can remedy this by putting a series resistor of 10k to 30k ohms between the output of the amplifier and your capacitive load. It is common to connect the node between the series resistor and the capacitive load to your feedback resistor as long as you don't need to drive high output current and are not driving close to the rail of the amplifier. this will give you the highest level of DC accuracy. Otherwise, you can connect the output of the amplifier directly to the feedback resistor. both of these techniques should solve your instability problem.
    Chuck
  • 0 in reply to Chuck Sins
    TI__Intellectual 1045 points

    Chuck,

    Thanks for your help.

    The Ro of LPV802 is 90k Ohms. The capacitive load in my design is 20pF. This two specifications will added an extra pole in Aol loop. The pole is 1/2pi(ro)CL is approximate 88kHz, so this pole will not affect the stability of LPV802.

    In my another simulation, I removed the capacitive load and increased the Rload to 100M Ohms. The result of AC analysis is as following. There have a huge overshoot at 900Hz.

    Maybe there have some specifications that I missed. 

    Regards,

    Ricardo

  • 0 in reply to Ricardo Li
    TI__Mastermind 23395 points
    Ricardo
    I will need to think about this a bit more in order to explain but try scaling resistors R1 and R2 to 1M and you should see significantly improved results. Being a nano-power amplifier, you don't want to use kohm resistor feedback. that is essentially the load on your output and is much too small for this device.
    Chuck
  • 0 in reply to Chuck Sins
    TI__Intellectual 1045 points

    Chuck,

    This is what I missed out. I didn't realize that a small feedback resistor will influence the LPV802.

    But I still want to know why the feedback resistor will influence the stability of LPV820. It seems that the stability analysis base on Aol and 1/β is incorrect. Could you help to explain this question? 

    Thank you very much for your help.

    Regards,

    Ricardo