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  • TI Thinks Resolved

TINA/Spice/OPA548: Problem with the OPA548 simulation

Part Number: OPA548

Tool/software: TINA-TI or Spice Models

Hello,

i am trying to do a calibration and for this reason i have to use the sinus signal of a function generator. My generator can provide a signal up to 200 mA @10V p, but of course i will try to use less current if possible.  My "electrical load: 0,35 Ohm" needs current up to  +-1,2 A which corresponds to +-0,5V (that's the voltage limit) at 50 Hz. So i the first idea was to use the LM 1875 which is connected with an electrical transformer 1/20 and then using the generator i would be able to set the voltage where i want to . This idea is ok and it works  fine, at least at a simulation level. The next idea is to use the OPA548 having a gain 1.  For this reason i have done a simulation using TINA which  does not give me the results i would like to see and in order to be more exact the output seems to be a bit higher than the input. In this case the IC that gets this signal would be destroyed(only voltages up to +-0,5V are allowed. Could someone tell me if there is something wrong regarding my simulation. 
Here is my circuit and the inputs/outputs as well.

 

As you see the VM2 output is slightly bigger, than the VG1 which is not supposed to happen.
And something more-->In the Datasheet ,Figure 26 can be found the application example i simulated and it is practically what i need. There is also a Resistor (Riso). Why is this resistor used for?

Regards,
George  

  • In reply to kai klaas69:

    CAL_OPA548_Howland_Current_Pump_Schematic_Out to ADE7880.TSCHi Kai,

    attached you will find the circuit.

    George-

  • In reply to Georgios Kefis:

    The VM1 is supposed to be the voltage that my IC "reads".
  • In reply to Georgios Kefis:

    Hi Georgios,

    ah, ok. R1 is the ferrite bead, right?

    Kai
  • In reply to kai klaas69:

    Hi, yes it is WE74279218
  • In reply to Georgios Kefis:

    Hi Gergios,

    I have made some simulations. It turns out that the ferrite bead destabilizes the circuit:

    The first simulation shows the step response of the circuit. The excitation signal is a 10kHz square wave with 10V amplitude.

    The second simulation shows a simplified phase stability analysis. As you can see, the phase margin is heavily eroded.

    I would recommend to omit the ferrite bead:

    Then I would add a phase lead capacitance of 47pF to furtherly restore the phase margin:

    There are also other ways to increase the stability (changing the feedback resistor ratio, increasing the load resistance, etc.). But this isn't simulated here.

    Kai

  • In reply to kai klaas69:

    Hello Kai,

    at first i have to thank you and secondly i do not know if it is important to mention that the signal is up to 50Hz.

    Regards,
    George
  • In reply to Georgios Kefis:

    Hi Georgios,

    if the circuit shows any instability you go and tell the circuit: "Hey, circuit, you are only a 50Hz application. You must not oscillate at 500kHz!" and the chip immediately stops oscillating. Hahaha...

    Even if it was a plain DC application the circuit can start to oscillate if the phase margin is eroded at a frequency where the OPAmp has a gain > 1. The instability can be excitated by any signal which has energy at the frequency where the circuit is unstable. This can be the power-up of supply voltage or even noise. Oscillation can start from even the smallest excitation.

    The step response is a common way to visualize such an instability.

    Kai

  • In reply to kai klaas69:

    Hi Georgios,

    Kai is providing very good guidance on assuring the Improved Howland Current Pump circuit remains stable in your circuit. And he is absolutely spot on regarding the need for proper heat sinking of the OPA548. If the OPA548 gets too hot it will go into thermal shutdown.

    If it looks like I can contribute anything to the discussion I will jump back in!

    Regards, Thomas
    Precision Amplifiers Applications Engineering
  • In reply to Thomas Kuehl:

    Hi guys,

    thanks a lot for your help. Today i tried something different and at the end i took the current i wanted. I replaced the resistor R5  between the signal generator and  OPA548 Pin (1) with 0Ohm at the beginning and with 0 Ohm afterwards and it worked as expected. Of course the device was after some moments not stable because i cannot install any heat sink here. In the figure 10, page 16 of the datasheet of the evalboard EVM OPA548 is a circuit. Playing with different values of the resistor R5 results in : functionality of my prototype , exactly as expected but i have to use cold air in order to keep its temperature low, otherwise it is not stable and secondly when i simulate the circuit i get the proper outputs only when the value of the resistor R5 is 10k. Why is this happening? I have attached the circuit with 5k.

    Regards,
    George 

  • In reply to Georgios Kefis:

    I mean that it should not work at all but it works...

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