simple op amp

Discussion in 'Homework Help' started by jstrike21, Feb 3, 2010.

  1. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    The assignment is attached.
    Problem 6 is the one i am struggling with. Im just not sure how to solve this problem. Any suggestions?

    I set up an equation for the current at the node going into the negative terminal.
    i1=(Vi-V1)/(1/sc)+(Vi-V1)/(10k)
    I then set V1 to zero because V2(voltage at positive terminal) - V1=0 and V2 has to equal 0 so V1 is also 0.
    THen i got the equation:
    i1= (sc+1/10k)Vi

    Is this right so far?



    Also I need help on number 5.... It has an input that goes into the + and an input the goes into the - im not sure what to do.
     
    Last edited: Feb 3, 2010
  2. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    Can't help with problem 6 I'm afraid bit complicated for me, but 5 is just a non-inverting amplifier.

    The inverting input will have a voltage which is a voltage divider function of the two resistors. Set that equal to Vi, and you will have Vo/Vi = 1 + R1/R2, i'll leave you to figure out which resistor value goes where.
     
  3. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    thats what i did at first
    Vo/Vi=1+500k/10k
    Vo=51Vi

    but this seems wrong because i didnt use V2
     
  4. thyristor

    Active Member

    Dec 27, 2009
    94
    0
    This is just a simple inverting amplifier with a capacitor across the input resistor. You're on the right lines but you have made an algebraic slip up.

    If X is the parallel combination of the 10K resistor and the capacitor, then:

    i = Vin/X and also i = -Vout/500K since the input terminal is at ground potential for the reasons you outlined.

    so Vout/Vin = - 500K/X

    I'll leave you to calculate X
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Yep. You need to reexamine your solution for number 5. Your answer should contain the term V2 in it.

    If you can post your work in solving number 5 we could be of grater assistance.

    hgmjr
     
  6. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    I dont see whats wrong with the algebra....
    Vo/Vi= -(sc+(1/10k))*(500k)
    right?
     
  7. thyristor

    Active Member

    Dec 27, 2009
    94
    0
    Actually, that's not correct because the resistor chain doesn't go to 0v but, instead, to V2 as shown, so the calculation is a little more complex.

    Calculate the input voltage at the -ve terminal by simple voltage divider maths (call this V3) and then Vout must be V1 - V3
     
  8. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    My book has no examples with two inputs like this problem. I dont know how I am supposed to start this problem.
     
  9. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    I calculated the current at the node and that was i1=(sc+(1/10k))V1) so wouldn't I just multiply this by 500k to get Vout?
     
  10. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Like thyristor has indicated, you can start by calculating the voltage at the negative input to the opamp. This equation will contain Vout and V2. Can you at least give that a try and then post you result.

    hgmjr
     
  11. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    The current is not what you need to calculate. You need to calculate the voltage at the negative input.

    hgmjr
     
  12. thyristor

    Active Member

    Dec 27, 2009
    94
    0
    Apologies, yes that is correct. I was flicking between qs. 5 & 6 and confused myself.
     
  13. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    If i make V3 voltage at -ve input then say:
    V3=V2(500k/510k)
    but this doesnt use Vout
     
  14. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    Ok what really confused me about this problem is the input of 2u(t) i'm not quite sure how to use this for the input.
     
  15. thyristor

    Active Member

    Dec 27, 2009
    94
    0
    That's incorrect.

    What is the voltage across the 10K and 500K resistor in total. That should start you on the right lines. Come back and tell us.
     
  16. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    That is because your expression for V3 (which you have equated to V-) is in error.

    hgmjr
     
  17. thyristor

    Active Member

    Dec 27, 2009
    94
    0
    Just substitute 2u(t) for Vin.

    2u(t) merely means that Vin is a time dependent function, ie: its frequency may change and we know that the amplifier in question has a frequency dependent transfer function.
     
  18. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    how about Vout=((1/10k + 1/500k)^(-1))*500k*V2
     
  19. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    2u(t) is the expression for a "step" function. It says that for t < 0, the signal amplitude is 0. For t >= 0, the function has an amplitude of 2.

    hgmjr
     
  20. jstrike21

    Thread Starter Member

    Sep 24, 2009
    104
    0
    so is it proper use to put my answer as Vo=-100.1u(t) or do i just leave the u(t) off?
     
Loading...