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Non-Inverting Operational Amplifier.

  1. Nov 27, 2007 #1
    Good evening.

    Right. My question is one of a simple nature, I think.

    Anyways, consider the following non-inverting Op-Amp design:


    Now, the basic formula for any non-inverting opamp is of course [itex]\frac{V_{O}}{V_{S}} = 1 + \frac{R_{2}}{R_{1}}[/itex]

    Meaning that [itex]V_{O} = V_{S} + \frac{V_{S}R_{2}}{R_{1}}[/itex]

    I'm trying to find [itex]V_{O}[/itex]

    My question, how do I take into account the[itex]R_{S}[/itex] In this situation? Does it have any bearing at all, if so, what?

  2. jcsd
  3. Nov 27, 2007 #2


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    Well, since the inputs are very high impedance, there's no current into them. Hence the R_s has no effect on the gain.
  4. Dec 2, 2007 #3

    Rs= 1.2k has nbo bearing at all. You cannot simply memorize the formulae for inverting and noninverting amps and solve every opamp question. You need to do it fron first principles. Heres how.

    1. V+= V- (thats a given)
    2. I= = I-= 0A (very little or no current in the order of pico amps enters intoi the terminals of the opamp ...it can be neglected)
    3. You have an expression for V- being V-= Vout(5k)/(15K) by voltage divider and you have V+= 2A (no V drop across Rs as I+= 0A)

    << final solution deleted by berkeman >>
    Last edited by a moderator: Dec 3, 2007
  5. Dec 12, 2007 #4


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    Points (1) and (2) that unplebeian mentions are often coined, the golden rules of op-amps and may be applied to these kinds of questions.
    Using, the rule: input to an op-amp draws (essentially) no current, you can easily deduce what happens to Rs.
    You can use the expression that unplebeian describes in point (3). However your own expression for Vo is equally valid and will agree in solution.
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