• State
  • Locked Locked
  • Replies 1 reply
  • Subscribers 47 subscribers
  • Views 963 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.

ISO 121 BG Iso Amp

Kevin Whitton
Prodigy 10 points
Other Parts Discussed in Thread: ISO121

Hi,

Im working with the ISO 121BG isolation amplifier and I was wondering if someone could tell me the power required from a voltage supply necessary to power this isolation amplifer? I couldn’t find any attributes in the datasheet along the lines of max internal power dissipation or max current load on the power supply (15 V supply being used.).

Thanks for the help

Kevin Whitton 

  • 0
    TI__Guru* 93105 points

    Hi,

    The ISO121 doesn’t consume much power during normal operation and the power dissipation usually isn’t a concern. Using some of the datasheet electrical characteristics, the power dissipation can be determined.

    • Supply range, Vs is ±4.5 to ±18 V, with ±15 V typical (30 V total)

     

    • Input stage operating current for Vs1, ±4 mA typical, ±5.5 mA max.

     

    • Output stage operating current for Vs2, ±5 mA typical, ±6.4 mA max.

     

    • Using this information the typical and quiescent power dissipation can be determined

     

    • Pd(QUIESCENT TYP) = (Vs1) ( Is1) + (Vs2) ( Is2) = (30 V) ( 4 mA) + (30 V) (5 mA) = 0.27 W

     

    • Pd(QUIESCENT MAX) = (36 V) ( 5.5 mA) + (36 V) (6.4 mA) = 0.43 W

    If the output is sourcing, or sinking, current to a load then the internal power dissipation increases beyond the quiescent condition. How it increases depends on how much current the output must deliver, the output voltage swing and characteristics of the output waveform. The worst-case dissipation occurs when the output is required to deliver a dc level, halfway between 0 V and Vs. The power generated in the output will then be in addition to the internal operating power. For most cases, the internal power will be about the same as when it is driving load current.

    • Io = ±5 mA min, ±15 mA typical

     

    • P(MAX-TYP) = Pd(QUIESCENT TYP) + Po(TYP) = Pd(QUIESCENT TYP) + ((Vs2 – Vo)/2) ( Io)

     

    • P(MAX-TYP) = 0.27 W + ((15 V – 0 V)/ 2) (5 mA) = 0.31 W

     

    • P(MAX) = Pd(QUIESCENT MAX) + Po(MAX)            (Use Vs2 max of 18 V)

     

    • P(MAX) = 0.43 W + ((18 V – 0 V)/ 2) (15 mA) = 0.57 W

    The calculations become more complex when the output is delivering a sine wave, or other ac waveform, but the dc 1/2 Vs case is the worst-case.

    Regards, Thomas

    PA – Linear Applications Engineering