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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 Georgios Kefis:

    Hi Georgios,

    when you decrease R5, the output voltage of OPA548 goes into saturation. This makes that the heat dissipation of OPA548 goes down a bit, because less voltage is dropping across the OPA548. So, the chip is heating up slowlier and the thermal shutdown occurs a bit later. This makes your circuit appear to work, at least for a brief period. But of course, the OPA548 no longer works properly because the topology of Howland current source is destroyed by changing R5.

    Please provide an adequate heat sink! All the measurements do not make much sense, if the chip is running into thermal shutdown over and over again.

    Mount a thick piece of metal to the cooling fin of OPA548. This could also work as a heat sink. At least, you can profit from the heat capacity of this piece of metal.

    Kai
  • In reply to Georgios Kefis:

    Hello Georgios,

    Keep in mind that the Improved Howland Current Pump employs both negative and positive feedback. That is apparent by the circuit schematic for the pump. The design constraints for the pump should result in a predictable V-to-I behavior and be stable with a resistive load. However, if one deviates from circuit constraints then its behavior and stability can be compromised.

    Below I provide some details about a basic Improved Howland Current Pump. If you substitute the values you are using in your circuit into the IL equation, it should yield the current you can expect from the circuit. That said changing the input resistor R1 from 10 k, to 5 k, is a major departure from the standard current pump configuration where R1 is equal to RZ. 

    At some point we are going to need to complete this E2E thread. If you have all the information you need to move forward with your project, it would be a good time to close this thread.

    Regards, Thomas

    Precision Amplifiers Applications Engineering