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THS4524 and THS4524EVM

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Replies: 7

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Is there any problem with an inductive circuit (magnetic-field coil) on the input to the THS4524?  

In the gain calculation, is the gain just Rf/Rg, where Rg is the magnitude of the impedance of the coil?  

7 Replies

  • Hello Ben,

    The Inductive circuit you mentioned in the EVM are actually transformers. On the input side of EVM, these transformers are not populated even if they are shown in the schematic. On the output side, these Transformers are used to convert the differential output of THS4524 to single ended outputs.

    The gain is calculated from Rf/Rg where Rf is the feedback resistor and Rg is the gain set resister. There is no Impedance of the coil coming in the gain equation since the input side of EVM is not using a transformer. Please refere to Sec 9.2.3 and 9.2.4 of THS452x Datasheet for various configurations of using THS4524.

    Thanks and regards,

    Eldho

  • In reply to Eldho George:

    Since we are operating in the kHz range, we are not using any transformers on this board.

    The input signal comes from a magnetic-field sensing coil, which has a substantial inductance.

    I understand that in the gain equation, Rg includes all resistance in the input circuit, including in this case, the resistance of our magnetic-field sensing coil.

    Should the inductive reactance of our input coil (2*pi*f*L) be included in a vector sum with the coil R in order to calculate the Rg that is used in the gain equation?

  • In reply to Ben Sternberg:

    Ben,
    If there is some inductance in series with the RG resistor then absolutely; that inductance will become part of the gain equation. You could try to use a standard single-ended opamp in non-inverting configuration to isolate that effect. Do you have a schematic so that we can better understand the circuit?
    -Samir
  • In reply to Samir Cherian:

    The input is very simple. We do not have any resistor on the EVM board in front of the amplifier, there is just a short across the pads where the resistor could be. The input is simply a coil of wire, with a DC resistance of 12 Ohms and an inductance of 7.4mH.

    Do we simply take the vector sum of the coil R and jwL, and use the magnitude of this value as the Rg resistor in the gain equation?

    The coil is set up for a fully differential input to the amplifier with a center tap from the coil to the amplifier ground. For testing, we are using a single input with the other side terminated in 50 Ohms. We are not getting the gain that we expect .
  • In reply to Ben Sternberg:

    Ben,

    Yes you would just use the vector sum. You can simulate this in TINA-TI a Spice tool. Its a lot easier. The link is below. You should have some resistance on that resistor pad to isolate the PCB and trace capacitance from the input of the amplifier.

    http://www.ti.com/product/THS4524/toolssoftware

  • In reply to Samir Cherian:

    Thank you for this information.

    How much resistance would you recommend on this input pad? We would like to keep the input resistance reasonably low to minimize noise.
  • In reply to Ben Sternberg:

    Ben, About 10 to 50 ohms should be enough for isolation.

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