Op-amp circuit

Discussion in 'Homework Help' started by boks, Nov 2, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
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  2. thingmaker3

    Retired Moderator

    May 16, 2005
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    No current goes into the inputs of an ideal op-amp. The output will source or sink as much current as needed for the ideal circuit.
     
  3. boks

    Thread Starter Active Member

    Oct 10, 2008
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    OK, but how should I start here?

    0.8 mA divides between the 10 and 6.4 ohms resistances. Some goes back to the current source through the 12 ohms resistance, while the rest goes back through the op-amp?
     
    Last edited: Nov 2, 2008
  4. thingmaker3

    Retired Moderator

    May 16, 2005
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    Since the op-amp i/p are high impedance, all of the 0.8 mA is through the 12K resistor. This tells the E drop across same resistor.
     
  5. boks

    Thread Starter Active Member

    Oct 10, 2008
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    The output voltage is then 0.8 mA * 12 kOhms = 9.6 V.

    The current through the 10 kOhm resistor is 9.6 V/10 kOhms = 0.96 mA, and the current through the 6.4 kOhms resistor is 9.6 V/6.4 kOhms = 1.5 mA.

    The currents go from the 9.6 voltage to ground, so that I 0 is the sum of 0.8, 0.96 and 1.5 mA (3.26 mA), but in the opposite direction from that shown in the figure.
     
    Last edited: Nov 2, 2008
  6. thingmaker3

    Retired Moderator

    May 16, 2005
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    There are 9.6V across the 12K resistor, yes. But is the - i/p of the op-amp at ground?
     
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