• State TI Thinks Resolved
  • Locked Locked
  • Replies 7 replies
  • Answers 3 answers
  • Subscribers 33 subscribers
  • Views 2002 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.

[B17] PIN OPT8241 - VSUB_BIAS how to connect?

Russell Ak
Intellectual 625 points
Other Parts Discussed in Thread: LMR70503

Hi TI!
Please tell me how to connect [B17] PIN OPT8241

[B17] PIN OPT8241 - VSUB_BIAS how to connect?

from documents it is not clear what it is.
From the scheme SDK is not clear how the circuit with transistors and what it is.
Thank you.

  • 0
    TI__Expert 3800 points

    Hi Ruslan,

    Thanks for your interest in the product.

    VSUB_BIAS has to be connected to -5V (negative supply). The connections are given in the schematic files and they are as follows:

    As per the schematic, U1 and U2 are the transistors to be connected. Can you help me understand what connections you are asking about?

    Regards,
    Subhash Chandra

  • 0
    Intellectual 625 points

    VCC_5V0_MINUS_FSC -->VCC_5V0_MINUS-->VCC_OUT_5V0_NEG-->VOUT  ???

    If I use the power 3.3 volts for the circuit. What to apply for this port?
    Under the scheme, I do not understand how the work transistor unit.
    I passed on the circuit and came to Vot /
    To be honest, I do not know right now how to connect this pin.

  • 0 in reply to Russell Ak
    TI__Expert 3800 points
    Hi Ruslan,

    There are 2 portions to the whole circuit:
    LMR70503(U9) is a buck boost converter. All it does is to take 5V in and it gives out -5V. Once that is done, we connect the output of U9 to B1(ferrite bead) to remove any high frequency ripple in the power supply. "VCC_5V0_MINUS_FSC" net can be now used as a -5V voltage source.
    We need to connect -5V to B17 of OPT8241. It is ok to connect "VCC_5V0_MINUS_FSC" to B17 directly. However, U3 will draw current because of that all the time. The -5V supply is useful only during the time "ILLUM_EN" is high. In the CDK design we use a combination of a PMOS transistor(U1) and NMOS transistor (U2) to make sure the IC draws current only when "ILLUM_EN" is high.
    When "ILLUM_EN" is pulled high, "ILLUM_ENz" goes low. That turns on U1 and pulls the gate of U2 to 3.3V. This makes U2 to start conducting and "VSUB_BIAS" voltage goes to -5V. When "ILLUM_EN" is pulled low, U2 will stop conducting and the "VSUB_BIAS" connection will stop consuming power.
    As a summary, U9 generates -5V required for "VSUB_BIASU"; U1, U2 are only to optimise the current consumption. They connect "VSUB_BIAS" to "VCC_5V0_MINUS_FSC" only when required and reduce the amount of power consumed.
    Please let me know if you have any further questions. If you feel your question has been answered, please mark the thread as answered.

    Thanks and Regards,
    Subhash
  • 0 in reply to Subhash Chandra Sadhu
    Intellectual 625 points

    Hi Subhash
    I now understand how it works.
    I was left with two questions.
    1) receives 5 volts in the circuit are mandatory condition on the pin [B17] must be submitted at -5 V "ILLUM_EN" at the moment when there is no exposure to [B17] + 3.3V supplied. I understand correctly?
    2) I can directly [B17] to connect to -5, without the transistor unit. What will then be the consumption and the risk of a direct connection?

  • 0 in reply to Russell Ak
    TI__Intellectual 2030 points

    Ruslan,

    There is no functional risk in keeping the -5V on all the time. But depending on the ripple on the -5V, there may be a small noise degradation due to the fact that the negative supply is on even during the readout and ADC sampling of the sensor data (The effect negligible in most cases because of the differential nature of the whole signal pipeline). If you plan to connect directly, we suggest that you minimize the ripple to ~50mV level.

    Regards

  • 0 in reply to Bharath Patil
    Expert 1320 points
    Hi Bharath,

    I have question about SUB_BIAS Pin, In my design ,Can i connect this pin to ground,it means to '0'V,

    if i do this,how it is influence? or what is you advice?

    Regards

    None
  • 0 in reply to None_feiyu
    TI__Intellectual 2030 points
    Hi,

    As replied in another thread, there is no risk of connecting this pin to ground. It was connected to negative voltage in OPT8241-CDK to enhance the demodulation contrast.

    Regards