Power Dissipated Class B Amplifier

Discussion in 'Homework Help' started by AD633, Jul 12, 2013.

  1. AD633

    Thread Starter Member

    Jun 22, 2013
    96
    1
    Hi

    Considering the voltage VCEsat = 1.2V ,determine the maximum power that can be dissipated in a load RL = 8 Ohm

    Is it possible to answer this question without knowing the amplitude of Vi?

    Suposing that Vi=10 V sine wave ,the power dissipated could be calculated this way?

     PRL=(((10^2 V)/(sqrt(2))/(8 Ohm))=6,25 W

    Thanks
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    And what if Vi=11V? Would that dissipate even more power in the load? If so, then your answer isn't the maximum power than can be dissipated in the load.

    If the saturation voltage of the driving transistor is 1.2V, what is the largest voltage that it can place across the load? What is the power in the load at that point?
     
  3. AD633

    Thread Starter Member

    Jun 22, 2013
    96
    1
    That depends on the amplitude of Vi,but would be Vimax-VCEsat

    I forgot VCEsat it would be

    <br />
<br />
PRL=(((10^2 V -1.5 V)(sqrt(2))/(8 Ohm)))<br />
<br />
PRL= 6 W

    But i have always to know the value of the input signal Vi to calculate the dissipated maximum dissipated power right?
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    I have no idea how you are getting 6W from that mess.

    First, your units don't work out. Since you have 10^2, the units of that should be V^2. But then you are adding V^2 to V. As written, you have V/Ω, which is A and not W.

    CHECK THE UNITS!!!!!

    No matter what variation I use on the equation you gave, I can't get your answer.

    Forget the input voltage; what is the maximum voltage that you can get across the load?
     
  5. AD633

    Thread Starter Member

    Jun 22, 2013
    96
    1
    There was an operator clearly misising the above equation:

    <br />
PRL=(((10^2 V)/(sqrt(2))-1.5 V)))/(8 Ohm)<br />
<br />
PRL= 8,65 W

    But the maximum voltage across the load doesn't depends on the amplitude of Vi?If not i would say 15V-VCEsat,is the maximum value that we can get across the 8 Ohm resistor
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Why wouldn't the max voltage be 15V-Vcesat? Can you find an input voltage that will make that happen? Then use that as your input voltage. The problem places no constraints on Vi, so use whatever Vi will result in maximum power. You don't even have to know exactly what that Vi is, as long as you can say, "There exists a Vi that will result in Vmax appearing across the load where Vmax is equal to...."
     
    AD633 likes this.
Loading...