Op-Amp average power

Discussion in 'Homework Help' started by ihaveaquestion, Oct 4, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    http://img24.imageshack.us/img24/6531/imgot.jpg

    I'm supposed to find the average power through the 20k resistor.

    The plan is to use nodal analysis to find V1 and then find the current going through the 20k resistor.

    My equation at the bottom (the way it looks) will not give me a value of V1 though, what's up?

    Thanks in advance.
     
  2. t_n_k

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    Mar 6, 2009
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    If the opamp is ideal, the voltage at the -ve input is Vs. As you have correctly shown on your diagram.

    From this you may find the current in Z1, where I_Z1=Vs/Z1.

    The current in Z1 and Z2 are equal for the ideal amplifier case.

    You can then find the voltage drop in Z2.

    You add this voltage drop in Z2 to the value Vs (using complex addition) to give you V1.

    If you know V1 you can then find the current in Z3 and finally the power in the 20k.
     
  3. ihaveaquestion

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    Thanks tnk... I see what you did... I have yet to see a scenario where the node method simply 'can't' be applied though... after double checking my work with it, it looks right to me... anyone see what is wrong with what I did etc?
     
  4. t_n_k

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    I realy don't understand what you've done with your initial equations.

    It might also be helpful to realise that the gain of a non-inverting amplifier of the type shown is simply

    Av=(Z1+Z2)/Z1 where Z1 & Z2 are your notations

    Hence V1 is given by

    V1=Vs*(Z1+Z2)/Z1

    That's all you need to find V1 - after that you can find the current and power in the 20k. Z3 plays no part in the circuit gain.
     
  5. ihaveaquestion

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    You don't understand HOW i got my initial equations or what I did with them?
     
  6. t_n_k

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    I think I can see what you are trying to do - write a set of nodal equations to solve for V1. I'd refer you the discussion on your other post which I think deals with handling an ideal operational amplifer in the circuit.

    If I1 is the current flowing in Z1, I2 is the current in Z2 and I3 is the current in Z3 I would write the equations as follows

    I1=I2 (provided the opamp is ideal)

    I1=Vs/Z1
    I2=(V1-Vs)/Z2
    I3=V1/Z3

    Since I1=I2
    Vs/Z1=(V1-Vs)/Z2

    Hence Vs/Z1+Vs/Z2=V1/Z2

    V1=Z2*(1/Z1+1/Z2)*Vs

    or

    V1=(Z1+Z2)/Z1*Vs

    I3=V1/Z3=(Z1+Z2)/(Z1*Z3)*Vs

    If the opamp were not ideal the problem becomes rather more difficult but I guess that's not an issue here. Judging by the discussion on the other post I believe you are having difficulty constructing the nodal equation at the amplifier output node - which doesn't follow what one might call normal convention, in this case. An ideal amplifier has infinite gain-bandwidth and input resistance.
     
  7. ihaveaquestion

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    May 1, 2009
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    I agree with your reasoning, but I'm confused as to why you ADD a voltage 'drop' to Vs to find V1...

    Also, this might even be related to the question above, but does it matter how I specificy my equation... apparently it does because I put (Vs-V1)/Z2 = i which gives me a different value for v1 after solving then what I get with your method
     
  8. t_n_k

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    The same current flows in Z1 and Z2 so their individual voltage drops are additive. The 'natural' assumption would be that the conventional current flows from the node at potential V1 to ground via Z2, node Vs and Z1.

    Hence V1 = VZ1 + VZ2
    and since
    VZ1 = Vs
    one gets
    V1=Vs+VZ2

    Yes it does relate to the previous question because the implication in your rendering

    (Vs-V1)/Z2=i

    is that the current polarity in Z2 is opposite to that of the current in Z1. You would have to write Vs=-i*Z1 for your approach to be consistent.
     
  9. ihaveaquestion

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    http://img694.imageshack.us/img694/6043/12443122.jpg

    A different approach as posted earlier:

    I can simply use voltage division from the negative terminal of the op amp with the output voltage node:

    vs=v-=1.73+i

    vo=(v-)(20000-12000i)/(30000-6000i)

    then find the current I=vo/(20000-12000i)

    And the average power across the resistor that it's asking as(|I|^2*20000)/2

    Don't you agree?

    I get 4.27*10^-5 W or 427 mW
     
    Last edited: Dec 12, 2009
  10. t_n_k

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    I don't get your answer.

    My value was 1.794mW.

    Also 4.27*10^-5 isn't 427 mW - it's 42.7uW.

    My approach was work to out the gain, then the output voltage and then the 20kΩ current and hence the power.

    As a start the gain I determined was ....

    Av=(12k + j10k)/(2k + j4k)=3.2-j1.4=3.493 @ -23.63°
     
  11. ihaveaquestion

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    Voltage division as I have it shown should come out correct though right?
     
  12. ihaveaquestion

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    Anyone sure?
     
  13. t_n_k

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    Sure - you can use the voltage divider to relate vs,v- and vo.

    But your math is wrong.

    v-=vo*(2k+j4k)/(12k+j10k)=(0.2623-0.1148j)*vo=vs

    vo=vs/(0.2623-0.1148j)=(3.2-1.4j)*vs
     
  14. ihaveaquestion

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    I don't quite see how my math is wrong. I see what you're doing, but I'm simply doing the reverse.... i.e. you're relating the input voltage vs to the output.... I'm relating the output to the input...
     
  15. t_n_k

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    Well then your understanding of the circuit requires some extra thought.

    Consider your relationship .....

    vo=(v-)(20000-12000i)/(30000-6000i)

    You appear to have values including the 20k-j12k load determining the output voltage, when in fact the output voltage is completely independent of the load.

    This circuit is essentially a non-inverting amplifier with the overall gain determined by the relationship I gave in the earlier post.

    However you approach the solution, it's essential to understand which parts of the circuit contribute to the input signal amplification and which don't.
     
  16. t_n_k

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    It may also be worth placing a reference or earth node on the circuit diagram to aid in understanding and derivation of the correct solution - the most obvious point being the bottom connection rail on the circuit.
     
  17. ihaveaquestion

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  18. t_n_k

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    No.

    Vo=V1+(V2-V1)*R1/(R1+R2)

    Hence after simplification

    Vo=[V1*R2+V2*R1]/(R1+R2)
     
  19. ihaveaquestion

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  20. kdillinger

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    Jul 26, 2009
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    Yes.

    What your missing is one term will be 0 because R2 is ground. When writing equations always think that ground is a potential that nodes are referenced to. If you actually write out the full equation TNK's equation is correct, but since V2 is ground then that term will be 0 and drops out leaving the classic Vx*Rb/(Rb+Ra).
    What TNK is using is superposition. You short out (deactivate) V2 and determine the voltage at the Vout node: V1*R2/(R2+R1). Then you do the same with V1; deactivate V1 and calculate the voltage at Vout. This time it will be V2*R1/(R1+R2). Then you add these two equations together:

    V1*R2/(R2+R1) + V2*R1/(R1+R2)

    or simplifying like TNK has done:

    (V1*R2 + V2*R1)/(R1+R2)
     
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