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RC4558 maximum voltage swing

Expert 1855 points

Replies: 4

Views: 583

Hi, 


May I know what is the maximum voltage swing for RC4558 with below condition? I can't find it from the datasheet. Need advise. Thank you very much!

Vin= +/- 5V

Load= 10kohms

Thanks and best regards,

KH

  • Look in the datasheet on page 5 in the table 6.5 the line VOM. Then you can calculate it with your condition.
    Gerald
  • Kum Hoo

    Output will be typically +/-4V. It can be a little less at cold temperature, a little more at high temperature.
    I expect worst case would be > +/-3.5V

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Hi Ron,

    Thanks for the reply, may I know is there anyway to do the calculation to get the output you have mentioned?

    Best regards,
    KH
  • In reply to Kum Hoo Lee:

    KH,

    I looked at table 6.5 typical 10k VOM, schematic on page 9, and typical chart, figure 5 (2k chart).
    At +/-5V there will be less load current, so VOM will be closer to rail than supply at +/-15V.
    Temperature effect is mostly base emitter voltage which is -2mV/C. At -40C, output will be +/-130mV less. (than 25C)
    +/-3.5V worst case estimate is conservative value based on the known factors and margin for variance.

    Now there is a decision point. Is this output range acceptable?
    If not, consider higher supply voltage or another device with rail to rail output.

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

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