• State TI Thinks Resolved
  • Locked Locked
  • Replies 5 replies
  • Answers 3 answers
  • Subscribers 30 subscribers
  • Views 735 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

LDC1614: Measuring Inductance

Simon Gast
Prodigy 80 points
Part Number: LDC1614

Hello Team,

My first question is due to the measuring the resonanz frequenzy of the sensor LDC1614.
I want to use the IC to measure the change in the magnetic permeability near a coil with the changing Inductance.

1.)
Im not sure how exactly the resonant frequenzy is measured (or even if the measured frequenzy is the eigenfrequenzy of the LC-tank).
Do the "resonant circuit driver" in the block diagram vary/tune the frequenzy and what exactly is measured to obtain the resonanz freq. with the closed loop of the amplitude?
I think it is the similar question: but what does the frequenzy selective circuit (resonator) exactly do?

2.)
Is it possible to use the IC in combination with a much slower generated AC (about 10 Hz, higher amplitude) driving the same coil? This coil shall be part of a magnetic circuit (some kind of electromagnet) that is generating higher flux densities (not a planar coil but a coil with some more wire turns).
Alternativly i will try to seperatly measure the magnetic permeabilty with another coil and not the driving one.

Thanks for your help in advance.
with kind regards from Germany

Simon

  • 0
    TI__Mastermind 23810 points
    Simon,

    The LDC1614 drives the circuit at its resonant frequency by using a current source and then switching the drive direction based on the zero crossing of the INx signal.

    I think that driving the coil with a 10Hz signal as well will mess up the resonant driver and as a result mess up the frequency measurement.

    Thanks,

    -Clancy
  • 0 in reply to Clancy Soehren
    Prodigy 80 points

    Hello Clancy,

    Im not sure if i understood you right since my english is not the best and im note sure what is meant by "switching the drive direction".
    I interpret your answer as following: the driver or core can alternate the frequenzy of the signal that is driving the L-C-Tank. At the point where the impedanz of L and C of the L-C-Tank  cancel each other is where the phase angle is 0. This is also the resonanz-frequenzy to drive the coil.

    Please correct me if i'm wrong.

    best regards
    Simon

  • 0 in reply to Simon Gast
    TI__Mastermind 23810 points
    Simon,

    I am saying that a coupled 10 Hz signal will impact the LDC device driver.

    Note that the LC tank is a bandpass filter, so if the 10Hz signal is not actually observed at the INx inputs, then you might be OK.
  • 0 in reply to Clancy Soehren
    Prodigy 80 points

    Hello Clancy,

    Thanks for your answer. I'm sorry but i did not understand it in depth yet.

    Actually i just try to understand how the sensor finds the resonant frequenzy (if it does) disregarding the second coil i mentioned at the beginning (the one with 10Hz for example).

    In the "ldc1614.pdf" datasheet under 8.2.4 point 1.) a) the divider is set to "1". I guess this is done so the resonant driver finds the resonant frequenzy faster!? So the resonant freq. itself will be not be calculated and entered into any register by the user? (please correct me here if i'm wrong)
    In your last post i did not understand "then switching the drive direction based on the zero crossing of the INx signal".

    1.) Does the IC generate a jump/step function and then measure the oscilation period with the 0-crossing of the signal?

    2.) Does the IC alternate the frequenzy and measure something else till the resonant freq. is obtained?

    Sorry if these are silly questions. I try to understand in depth how the LC-Tank is driven by its resonant frequency, and if the IC can do this all by itself.

    Best regards


  • 0 in reply to Simon Gast
    TI__Mastermind 23810 points
    Simon,

    The IC can do it all by itself.

    1.) You can think of the driver in this way. It only needs to wait for a single oscillation before it is able to keep driving the circuit.