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:
We are glad that we were able to resolve this issue, and will now proceed to close this thread.
If you have further questions related to this thread, you may click "Ask a related question" below. The newly created question will be automatically linked to this question.
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.
Precision Amplifiers Applications Engineering
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