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Given the following circuit:

enter image description here

I don't really understand it. The potential at the (+) input of op-amp is 0 thus no current is going through the \$R_s\$. So it's basically just \$v_1\$ existing there, and I don't know why "some portion or some wire" going from the (-) of the op-amp back to the \$v_O\$ and to the \$R_L\$ resistor. I think no current is going through that resistor; that's the only reason to make \$v_O = v_1\$, is this valid logic?

JYelton
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HellBoy
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2 Answers2

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For an ideal op-amp, no current flows in (or out) of the inputs. So no current flows through Rs and therefore the voltage at the non-inverting input = vi (whatever that may be, and generally it will be non-zero and a function of time).

The way an ideal op-amp works, with extremely high gain, it will drive the two inputs to be very close to being equal. The rest follows.

Spehro Pefhany
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The assumption is that no matter what the voltage Vi is, there will be no current throug resistor Rs.

The Vi and Rs may resemble the voltage and source resistance of some sensor, so it does not relate to the op-amp in any way.

Justme
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  • so are my assumptions valid too that there is no current going through the $ R_L $ resistor? – HellBoy Oct 22 '23 at 19:56
  • Yes, that part was correct but you also assumed voltage being 0 so that's not correct. – Justme Oct 22 '23 at 19:59
  • isnt the voltage at the (+) gate of an ideal-op-amp 0? – HellBoy Oct 22 '23 at 20:00
  • No, except when it is 0. In some circuits it could be 0 but not in this circuit. Unless, of course it is 0 when Vi is 0. – Justme Oct 22 '23 at 20:03
  • ah so in this circuit $ V^+ = V_i = V^- = V_o $? – HellBoy Oct 22 '23 at 20:05
  • Your opamp has no part number then we cannot see on its datasheet how close to the positive or negative supply its inputs still work or can its output become. Your opamp has no power supply so we cannot see if it is positive and negative and if its output can go negative or not. Many opamps have inputs and outputs that need a positive and negative supply. – Audioguru Oct 23 '23 at 23:30
  • "isnt the voltage at the (+) gate of an ideal-op-amp 0?" - No. You've said that twice, and have been corrected. Why do you think that is the case? The voltage at the + input pin is equal to Vi. If Vi is a 10 Vrms sinewave, then the voltage at the + pin oscillates between +14.14 Vp and -14.14 Vp, – AnalogKid Nov 23 '23 at 01:38