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INA149: Use INA149 to generate sinusoidal current.

Charlie Xiao
Expert 5840 points
Part Number: INA149
Other Parts Discussed in Thread: INA148

Hi Team,

Please take a look. Is it feasible to use INA149 to generate positive and negative sinusoidal currents with a maximum amplitude of 10mA? Is there any relevant information or practical application for our reference. Thanks.

Regards,

Charlie

  • +1
    Guru 140200 points

    Hi Charlie,

    according to figure 12 of datasheet the INA149 can deliver an output current of +/-10mA.

    But please explain why you want to choose the INA149 for this task. I ask because the INA149 is a special purpose difference amplifier being optimized for very high common mode input voltages.

    Please tell more about your application. Do you want to design a current pump, eventually the "improved Howland current pump"? What is your load? What is your signal bandwidth? What is your input signal? Please give many more details...

    Kai

  • 0 in reply to kai klaas69
    TI__Expert 5840 points

    Hi Kai,

    Thanks for your support.

    The customer is mainly looking for an AD chip to match their current MCU to output sine waves, the output amplitude can be adjusted, and the frequency can reach 100kHz. It should mainly be used in wearable medical products to generate positive and negative sinusoidal currents with a maximum amplitude of 10mA. Is this product suitable? Is there any more suitable product recommendation? Thanks.

    Regards,

    Charlie

  • +1 in reply to Charlie Xiao
    TI__Guru 57241 points

    Hi Charlie, 

    Charlie Xiao said:
    Is it feasible to use INA149 to generate positive and negative sinusoidal currents with a maximum amplitude of 10mA?

    Enclosed is an application note to generate positive and negative sinusoidal currents with a maximum amplitude of 10mA

    https://www.ti.com/lit/an/sboa437/sboa437.pdf?ts=1675719680369&ref_url=https%253A%252F%252Fwww.google.com%252F

    INA149 and INA148 are part of high common-mode voltage difference amplifiers. As Kai pointed out, we need the following information or design requirements in order to provide you with the product recommendation. 

    1. The application available power supply for the sinusoidal current source/sink application. 

    2. What is input common mode voltage? INA149 is able to handle Vcm range up to ±275V. With single supply rail, Vcm may be up to 550V.

    3. What types of load the current source is driving? Resistive, capacitive, inductive or complex load. 

    4. What is sinusoidal voltage driving amplitude at output?

    If you have additional questions, please let us know. 

    Best,

    Raymond

      

  • 0 in reply to Raymond Zhang1
    TI__Expert 5840 points

    Hi Raymoud,

    Thanks for your support.

    But what's the meaning of "The application available power supply for the sinusoidal current source/sink application. "? Thanks.

    Regards,

    Charlie

  • 0 in reply to Raymond Zhang1
    TI__Expert 5840 points

    Hi Raymoud,

    1. What power supply rails are available on the board?
    A: 5V

    2. What is the input common mode voltage?
    A: We hope below 5V. We hope that low voltage can be driven to generate -+10mA sine wave;

    3. What type of load does the current source drive? Resistive, capacitive, inductive or mixed loads.
    A: Current drive load up to 500k ohms

    4. What is the magnitude of the sinusoidal voltage drive at the output?
    A: The output terminal is a sinusoidal current, and the output voltage is the current multiplied by the load.

    In fact, the customer is looking for a solution to achieve: convert the sinusoidal voltage of 0 ~ 6.5V into a sinusoidal output current of 0 ~ -+5mA, and the output load is about 500k ohms. If the above devices and solutions are not suitable, please suggest best solution. Thanks.

    Regards,

    Charlie

  • +1 in reply to Charlie Xiao
    TI__Guru 57241 points

    Hi Charlie, 

    Charlie Xiao said:
    The output terminal is a sinusoidal current, and the output voltage is the current multiplied by the load.

    Per ±5mA constant sinusoidal current source and driving 500 kΩ load, we are talking up to ±2500 Vpk voltage swing across the 500 kΩ resistor. 

    We do not have any high voltage amplifier to meet the requirements. You may have to consider a discrete current source solution.

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    TI__Expert 5840 points

    Hi Raymoud,

    Yes, you're right.

    Do you have a discrete solution to achieve: convert the sinusoidal voltage of 0 ~ 6.5V into a sinusoidal output current of 0 ~ -+5mA, and the output load is about 500k ohms.? Thanks.

    Regards,

    Charlie

  • +1 in reply to Charlie Xiao
    TI__Guru 57241 points

    Hi Charlie, 

    To drive 500kΩ load with a constant 5mApk constant sinusoidal current source, the current source supply rail needs to be at least 2500V, more like ~3kV in order to drive the constant current generator. 

    I do not have a discrete solution for this. Please search google or internet for additional ideas.  We do not work with anything close to the voltage rail in PRAMPS team. 

    Best,

    Raymond