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power dissipated for op amp

Mister
Prodigy 40 points

Hi,

With my understood, the PD( Total Power Dissipated ) has below formula: 

TJ= PD*( θJA) + TA, PD = PIQ + POUT=Vs*IQ + Vout*Iout

But I also saw below formula from some documents,

PD= Vs*IQ + (Vs-Vout)*Iout

 

Could you please explain more details about the different?

 

Thank you.

 

 

  • 0 in reply to Mister
    TI__Mastermind 22825 points

    Hello,

    The load related power dissipated by an op amp is the current through its output stage (mostly dominated by the load current) multiplied by the voltage across the output device (which is different for when the device is sourcing current vs. when it is sinking current) delivering the current:

    Sourcing: (V+ - Vout)

    Sinking: (Vout - V-)

    So, to properly calculate power dissipation when the output stage is sourcing and sinking current, different expressions need to be used. Of course the average power is the integration of the instantaneous power over one full cycle (assuming a repetitive waveform).

    Hope this was helpful.

    Regards,

    Hooman

  • 0 in reply to Hooman Hashemi
    Prodigy 40 points

    Hi Hooman,

       Thanks for your reply.

       So if it is PWM input to a follower as below. 

       Assuming |V+| ≠ |V-| 

       The PD= (|V+| + |V-|)*IQ + [(V+) - A]*(500/A)  + [-A - (V-)]*(500/A)   , right?

      

     

  • 0 in reply to Mister
    TI__Mastermind 22825 points

    Hi Mister,

    The instantaneous power will be:

    A = Vout amplitude


    P_inst_pos = (V+ - V-) * IQ + (V+ - A) * (A / 500ohm)               For the positive cycle

    P_inst_neg = (V+ - V-) * IQ + (-A - V-) * (A / 500ohm)               For the negative cycle

    To get the average power dissipation, integrate the expressions above over the PWM active time for each cycle and divide by the period.

    For example, assume:

    V+= 5V

    V-= -4V

    A= 2V

    IQ= 2mA

    PWM duty cycle = 30%

    Then:

    P_pos_cycle = (5 - (-4)) * 2mA +(5 - 2) * (2/ 500ohm) * 0.3 = 18mW + 12mW * 0.3 = 21.6mW

    P_neg_cycle = (5 - (-4)) * 2mA + (-2 - (-4)) * (2 / 500ohm) * 0.7= 18mW + 8mW * 0.7= 23.6mW

    P_average = (21.6mW + 23.6mW) /2 = 22.6mW

    Regards,

    Hooman

  • 0 in reply to Hooman Hashemi
    Prodigy 40 points

    Very Thanks Hooman,

    Sorry for late reply.

    But I have a question about the 'P_average', why it need divided by 2? The period is 2 sec?

    Because I think which P_pos_cycle with '*0.3' and P_neg_cycle with '*0.7' should mean T=1 sec.

    So P_average = (21.6mW + 23.6mW) /1 = 45.2mW

     

    Thank you.

    Mister

  • 0 in reply to Mister
    TI__Mastermind 22825 points

    Hello Mister,

    1. 0.3 and 0.7 justification: Those are percentages (30% and 70%). The period / frequency is irrelevant. Only duty cycle matters.

    2. Averaging (divide by 2): So, I'm just taking the average of two numbers (which involves adding them together and then dividing the result by 2). It is not the period of the waveform.

    Regards,

    Hooman