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TINA/Spice/OPA548: Problem with the OPA548 simulation

Part Number: OPA548
Other Parts Discussed in Thread: TINA-TI, ,

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  

  • Hi George,

    I have run the TNIA-TI simulation with your circuit and cannot see any amplitude error. The only I see is an offset voltage of about 5mV superimposed to the AC output signal.

    Kai
  • Hello Georgios,

    I have attached a TINATI schematic of the difference amplifier you in your post.  

    In transient analysis,  the output waveform is a 1:1 ratio with the input.  The offset referred to the output if roughly about 5.4mV.  This is because the offset if modeled as an internal dc source (Offset RTI for OPA548 model is approximately +/-2.7mV) connected to the input (+IN) of the op amp. 

    The Riso is used for stability compensation when the OPA548 is connected to a reactive load.

    Best regards,

    Errol Leon

    Texas Instruments Inc.

    Precision Amplifier Applications

     OPA548_Difference_Amp.TSC

  • Hello,


    thanks a lot for your helpful comments Errol and Kai. Another last question is, if the OPA 548 could be used as a current source solution and if so could the aforementioned solution work?If not what is your suggestion?

    Regards,
    George

  • What do you mean?

    Kai
  • Hi!!

    For example on page 18 of the datasheet there is a description of a Voltage source schematic. There are also solutions for current sources which means that i can set the current to a specific current value that i want to supply my circuit , for example to be able to provide current up to 1 A, ideally with a very good accuracy.

    Regards,
    George
  • Hello George,

    Okay, I took the voltage source concept shown on OPA548 datasheet Pg.18 and evolved it into a precision power current source. It uses the internal 4.75 V reference associated with the current limit circuit as the input voltage reference. You can see the results below.

    Since the OPA548 output stage cannot swing all the way to 0 V with a V- supply set to 0V, R2 is added to keep it within the linear range. The resistor is necessary but could dissipate a significant amount of power if made too high in value.

    Here's my TINA simulation file:

    OPA548_pwr_Isource_01.TSC

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • George

    We haven't heard back from you so we assume this answered your questions. If not, post another reply below, or create a new thread if this one has timed-out.

    Thanks
    Dennis
  • I wanted to create a current source of which the output current could be adjusted from the voltage of the generator. The signal is a sinus: 0,5sin(100πt) (V) and at its max it is supposed to be able to provide up to 1,2-1,5 Ampere. Unfortunately my circuit works as a voltage source.
  • Hello Thomas,

    unfotunately the circuit above cannot be used for my application because i need as i told you to adjust the output of the current adjusting the input signal which is a sinus signal up to 0,5V maximum

    Regards,
    George
  • Hello George,

    We have had a bit of a communications problem, but now I think I better understand what you need. Take a look at the Improved Howland Current Pump circuit. It provides the voltage-to-current (V-I) function that you need. An AC voltage can be applied to the OPA548 input, and a corresponding AC output current provided to a high current load. You just need to make sure that the power op amp has sufficient voltage compliance range for the expected current through the load.

    The OPA548 EVM document discusses the Improved Howland Current Pump in Section 7. The circuit can be designed for a specific input voltage range and output current. You can find that document here:

    www.ti.com/.../sbou132.pdf

    Additionally, there is quite a lot of information about the Improved Howland Current Pump on line. A search on the name should bring up a lot of interesting links.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • Hallo Thomas,

    i think that this is the answer i was looking for. I am going to try it and i will let you know. 

    Regards,
    George

  • Hi Georgios,

    what was wrong with the circuit with the voltage source? It was deadly simple...

    Kai

  • Hello Thomas,

    i have tried to make something which was of course not so accurate because i did not have resistors of high accuracy. However i tried to do the simulation with 1 Ohm (25W) (instead of the 150mOhm of my application) and the output should be up to 1A at 10V iput voltage. Unfortunately it could deliver up to 200 mA and the OPA 548 got warm, which was expected. I wonder if it had to do with the prototype that i made or with something that has to do with my design. Attached you will find my simulation. The resistor Rhcp is 10 Ohm and the Resistance Rf+ is 9990 Ohm. Could there be something wrong with that? In any case my idea is to test it with the Eval board from TI having the correct resistors, of high accuracy on it and the appropriate ground planes , believing that this was the only problem but i still have doubts because 200mA is far away from 1 A, even if it was only a prototype. What do you think?  I also do not know how can i improve the accuracy in y axis or how can i see some specific points and their values on the diagrams , when i do a simulation .CAL_OPA548_Howland_Current_Pump_Schematic_1_Ohm.TSC

  • Hi Georgios,

    the simulation works, I can see a current flow of 1A through R1.

    With the huge supply voltage of +/-20V your chip will heavily heat up, though. Up to 9W must be dissipated. Hopefully, you have a porper heatsink...

    Kai
  • Hello Kai ,

    thanks a lot for your help. I have prepared a "prototype board" using parts that i had already in my lab. As expected the IC was really hot but there was no heatsink at all . I am trying to find if there is something wrong or critical with my circuit hoping that the eval board OPA548EVM (and a very precise Rhcp resistor) will solve me the problems that i have. Do you see anything critical here?

    Regards George
  • Hi Geogios,

    in any case I would decrease the supply voltage to +/-15V. This will help a lot to decrease the heat dissipation. Nevertheless, you will still need an adequate heat sink!

    The Howland current source is well known for its stability issues. This comes from the fact that it contains a positive feedback loop. So, it's always wise to use some sort of phase lead compensation. This is especially true if your load contains some load capacitance. It's extremely important to know exactly what load capacitance is to be expected from the load. Because then you can properly compensate the introduced phase lag by a well chosen phase lead capacitance.

    What exactly is your load? Is it completely resistive?

    Kai
  • CAL_OPA548_Howland_Current_Pump_Schematic .TSCHello Kai,

    the output of my OPA548 "sees" ferrite of type WE 74279218 with DC resistance of 100mOhm  and a very presice resistor of 150 mOhm  and two filters.My IC measures the voltage of the 150 mOhm resistor, thus it is needed to be feasible to provide up to 1A and slightly more if needed. I realised too, that the circuit could be stable with 15V as well. See also my attached circuit

    George-

  • Hi Georgios,

    and is there even the least load capacitance? Can you estimate the maximum load capacitance? I give you an example: Sometimes chokes or solenoid valves are driven by a Howland current source and the designers are surprised of much interwinding capacitance can exist across the choke. Only by adding a suited phase lead capacitance the design can be made stable then.

    Kai
  • Hi Kai,

    there are 2 Capacitors (of course 2 resistors for the filters) in parallel with the rest circuit, which means : 2x 2,2nF so totally 4,4 nF.
    What do you mean with that :"Only by adding a suited phase lead capacitance the design can be made stable then." ???

    George
  • Hi Gergios,

    can you draw a schematic containing all the components of your circuit? Including all the filters, all the chokes and all the capacitances?

    Kai
  • The VM1 is supposed to be the voltage that my IC "reads".
  • Hi Georgios,

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

    Kai
  • Hi, yes it is WE74279218
  • 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

  • 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
  • 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

  • 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
  • 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 

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

  • Georgios

    We haven't heard back from you so we assume this resolved your issue. If not, post another reply below.

    Thanks
    Dennis
  • Hello guys,

    i have made the same configuration, using the eval board and to be honest everything works as expected, but there is something critical here. Using the Oscilloscope i realised that i have a very slight phase difference (9deg) between the input and the output signal when i use a 10 Ohm Resistor (Load Resistor) and at the moment i do not know how to eliminate this phase difference. Does anyone have any idea?

    Regards,
    George-   

  • Some more collateral you will find of interest for V to I circuits....

    V to I Circuits Link:

    https://e2e.ti.com/support/amplifiers/precision_amplifiers/w/design_notes/3600.high-current-v-i-circuits

  • Georgios

    Did this answer your questions? If not, post another reply below.

    Thanks
    Dennis