I have problem in op-amp can any help me please

Discussion in 'General Electronics Chat' started by ksalatawi, Apr 19, 2005.

  1. ksalatawi

    Thread Starter Member

    Nov 29, 2004
    21
    0
    I have question about op-amp can any one help me please.
    question in this attachment
     
  2. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    This looks like homework. If so, you should have been taught about op-amps before being set this excercise. So which part don't you understand?
     
  3. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    The two opamps on the left need not contribute any gain. They need to provide only an inversion of sign.

    The right hand one is the summer.

    Remember that the ratio of the feedbak R to the input R gives the voltage gain from an input to the output. This is true for eachof the three stages.
    You need to do the simple arithmetic for each of the three inputs.
     
  4. ksalatawi

    Thread Starter Member

    Nov 29, 2004
    21
    0
    thank you but
    how can i find the value of the resistors
    If i find that R2R8/R1R3=3 and
    R5R8/R4R6=2 and R8/R7=4
     
  5. dragan733

    Senior Member

    Dec 12, 2004
    152
    0
    Hi ksalatawi,
    1. The solution is that:
    R2R8/R1R3=3
    R5R8/R4R6=2
    R8/R7=4
    We have 8 unknown resistors and only three equations.

    2. Each op -amp. amplifies and inverts the input signal

    3.

    3V1+2V2-4V3=15
    2V1+2*3-4*1=15
    V1=4,33V
    3V1+2*3-4*1=-15
    V1=-5,67V
    Then the range of V1 for linear operation is: -5,67V until 4,33V
     
  6. ksalatawi

    Thread Starter Member

    Nov 29, 2004
    21
    0
    thaaaaaaaaank you very much dragan i missed you
     
  7. ksalatawi

    Thread Starter Member

    Nov 29, 2004
    21
    0
    can i find the value of each resistor
     
  8. dragan733

    Senior Member

    Dec 12, 2004
    152
    0
    Thank you ksalatawi. The answer I gave you also on your E-mail.
    1. The solution is that:
    R2R8/R1R3=3
    R5R8/R4R6=2
    R8/R7=4
    We have 8 unknown resistors and only three equations. That means we can select for example:
    R7=1K Ω , R1=1K Ω , R3=1K Ω , R4=1K Ω , R6=1K Ω then R8=4K Ω , R5=500 Ω, R2=750 Ω
    or for example:
    R8=4KΩ, R4=1K Ω , R6=2K Ω , R2=3K Ω ; R3=2K Ω then: R7=1K Ω , R5=1K Ω , R1=2K Ω
    Therefore there are infinity combinations
     
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