operational amplifier and measuring voltage drop

Discussion in 'Homework Help' started by logearav, Aug 21, 2011.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
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    Dear Members,
    I have attached the image of inverting operational amplifier. I have the following doubts
    For measuring current input current Iin we take voltage drop across Ri as
    Vin - Va
    Similarly voltage drop across Rf as Va - Vout for the current If
    My question is why not we take as Va - Vin, for the first case and
    Vout - Va for the second case
    The current flows from left to right in this diagram. Please help revered members
     
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  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    As you said the current flow form left to right (from input source to the output of the opamp) . If so, the voltage higher potential at Vin then Va and Va has a higher potential then Vout.
    See the example
    [​IMG]
     
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  3. steveb

    Senior Member

    Jul 3, 2008
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    You can do that if you like. You are free to define the voltage drops and associated current directions arbitrarily. However, once you define them, you have to correctly apply Kirchoff's Voltage Law and Kirchoff's Current Law using those definitions.

    I agree with Jony130. Once you make this statement, you have effectively defined the direction and the polarity, and your formulas are a consequence of this decision.
     
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  4. logearav

    Thread Starter Member

    Aug 19, 2011
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    Thanks for the reply Mr. Jony130 and Mr. Steveb.
    Mr. Jony, you said Va is at higher potential than Vin and similarly Va is at higher potential than Vout. But in your diagram a) Vin is -1 volt and Va is 0 Volt. Does this mean -1 is greater than O volt. Also Vout is 2 volt. Vout is higher than Va but how
    0 ( Va) > 2 ( Vout)?
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Diagram a) show the situation when the current flow from "right to left".
    So in this case we have Vout>Va and Va>Vin

    And diagram b) show the current flows from "left to right".

    Simply, the opamp output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
    So if you have 0V at the "+" input and the the opamps wants to have the same voltage at it's "-" input (0V) the only way to do this is by increasing it's output voltage to 2V if input voltage is -1V.
    And for case b) to make the voltage difference between the inputs equal to zero. The opamp will decrease it's output voltage to -4V.
    Because only when the output voltage reach -4V the difference between the "+" input and "-" input is equal to 0V for +2V input voltage.
     
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  6. logearav

    Thread Starter Member

    Aug 19, 2011
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    Thanks John130. " the op amp attempts to do whatever is necessary to make the voltage difference between the inputs zero"
    1)Does this apply only to inverting amplifier alone or all types of amplifier, say non inverting amp, summing amp, difference amp etc?
    2) How the voltage becomes 0 at the junction between the two resistors R1 and R2?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    This is general case rule apply for all op amp circuit.
    And we call it "virtual short" ( linear region)

    Sometimes when op amp was unable to "bring" inputs to zero. The op amp will entering the saturation mode. The output voltage is equal the power supply voltage.
    [​IMG]

    In real life the op amp alone (without any feedback loop) is nothing more then a differential amplifier.
    The output voltage is the difference between the + and - inputs multiplied by the open-loop gain:
    Vout = (V"+" − V"−")*G_open_loop.

    [​IMG]

    So when the voltage at the input "+" (non-inverting) rise,output voltage is also increases.
    Increase voltage at "-" input (inverting input) causes, decrease the output voltage.
    Decrease voltage at input "-" increases the output voltage.

    [​IMG]

    But in circuit theory we assume that the open loop gain is infinitely large.
    And we add the feedback network.
    So for example in case a) we apply 1V into the input terminal of a Non-inverting amplifier. The op amp output voltage will start rising because we have 1V at V"+" input and 0V at V"-".
    So the output voltage rise until non-inverting input voltages (+) will be equal to inverting input voltages (-)
    [​IMG]

    We have similar situation in Inverting amplifier.

    [​IMG]


    For example in case b)
    We apply 2V at inverting input so the op amp see the potential difference between his inputs. So the output voltage start to starts to decrease. And it will be decrease until V"+" = V"-". As you can see nothing magical is happening here.
    When we apply positive voltage at inverting input, the op amp output voltage start to decrease.
    But thanks to R2 some part of the output voltage is feedback to inverting input.
    And when V"+" = V"-" the op amp see 0V difference between his inputs terminals.
    And output voltage stop changing.
    From the viewer perspective op amp "magically" brought his input terminal differential voltages to 0.

    The more detailed explanation
    Can you solve for Va in this simply circuit ??

    [​IMG]
     
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    Last edited: Aug 23, 2011
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  8. logearav

    Thread Starter Member

    Aug 19, 2011
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    Since Va is grounded, it is zero. Am i correct?
    Also, you said the output is open loop gain multiplied by voltage difference between the inputs. You have specified the open loop gain as 100 000. In the a) diagram the difference between 10,001 and 10,000 is 1 volt and if this multiplied by 100 000 we ll get 100 000 as output. But the output is specified as 10 volt. I am confused.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Va has no physical connection to the ground.
    But yes, Va voltage for this case is equal 0V.

    Va = 2V - I*R1 or Va = -4V + I*R2


    The total current is equal to

    Itot = ( 2V + 4V) / 30KΩ = 200μA

    then

    Va = 2V - 200μA*10KΩ = 2V - 2V = 0V

    Or

    Va = -4V + 200μA*20K = -4V + 4V = 0V

    So, only when
    V2/V1 = - R2/R1 then Vo will be equal to 0V

    Sorry for confusion but in my country we use often comma as a decimal separator.
    So for example
    1,1 = 1 point 1 = 1.1 = 1+1/10 = 1 + 0.1
    And if the input voltage is equal to
    1.00000V = 1V
    1.00005V = 1V + 5/(100 000) = 1V + 0.00005V = 1V + 50μV


    And I change the imagine, and now everything should be oki.
     
    Last edited: Aug 23, 2011
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  10. logearav

    Thread Starter Member

    Aug 19, 2011
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    Thanks for a very detailed explanation. I got it now. In deriving the equation for Va, i think this is the step.
    I = (2V - Va) / R1 for the first case
    I = Va - (-4V)/ R2 for the second case
    Am i correct sir?
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, that is also the correct approaches.
    You simply wrote the Nodal equation
    (2V - Va) / R1 = (Va - (-4V))/ R2
     
  12. logearav

    Thread Starter Member

    Aug 19, 2011
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    Thanks sir. Any other way to arrive at the equation other than the way i mentioned?
     
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