Non-Inverting Operational Amplifier.

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The discussion focuses on understanding the effect of source resistance (R_s) in a non-inverting operational amplifier (Op-Amp) configuration. It is clarified that due to the high input impedance of the Op-Amp, R_s does not influence the gain or output voltage (V_O). Participants emphasize the importance of applying fundamental principles rather than memorizing formulas, highlighting the "golden rules" of Op-Amps, which state that input terminals draw negligible current. The voltage divider principle is used to derive expressions for the output voltage. Overall, the consensus is that R_s can be neglected in this context, allowing for straightforward calculations of V_O.
MushManG
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Good evening.

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

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

http://www.freewebs.com/mushg/FWThumbnails/OpAmp.bmp

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

Meaning that V_{O} = V_{S} + \frac{V_{S}R_{2}}{R_{1}}

I'm trying to find V_{O}

My question, how do I take into account theR_{S} In this situation? Does it have any bearing at all, if so, what?

Thanks
 
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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.
 
Hi,

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 >>
 
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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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