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Why is -Vout equal to the voltage going through Rf?

here is a link to a site which explains the inverting amplifier circuit but I couldn't understand why this is.

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  • \$\begingroup\$ Current going through the resistor produces voltage across resistor. Voltage doesn't go through resistor. It's unclear what you're asking. \$\endgroup\$ – Nick Alexeev Sep 30 '13 at 1:44
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Why is -Vout equal to the voltage going through Rf?

(1) The voltage at the inverting terminal is 0V* so, by KVL,

$$v_{OUT} = 0V - i_{R_F}R_F $$

where the reference direction for \$i_{R_F}\$ is from left to right through \$R_F\$

(2) By KCL,

$$i_{R_i} = i_{R_F} $$

since the inverting input is an open circuit.

(3) By Ohm's Law,

$$i_{R_i} = \dfrac{v_{IN}}{R_i}$$

(remember, the voltage at the inverting terminal is 0V).

Thus:

$$v_{OUT} = 0V - i_{R_F}R_F = -i_{R_i}R_F = -\dfrac{v_{IN}}{R_i}R_F$$

or

$$\dfrac{v_{OUT}}{v_{IN}} = -\dfrac{R_F}{R_i}$$

*For an ideal op-amp with negative feedback, the inverting and non-inverting input voltages are equal.

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The schematic you show is of a classic inverting gain stage.

Recognize that, because the infinite impedance of the op amp is infinite, and the gain is infinite, ANY CURRENT AT ALL between the op amp inputs induces a HUGE voltage swing at the op amp output. This means, for the stage to be "quiescent", no current can enter the inverting input, and the inverting input must be at ground potential. This is called a "virtual ground".

Because there is a voltage difference across the input resistor, there must be current flowing through the input resistor. Because that current cannot enter the op amp, it must flow through the feedback resistor. Ohm's Law then gives you the voltage that must appear at the other end of the feedback resistor, which must be the output voltage of the op amp.

This is one of those times where you really do want to set up the equations and solve them, so you understand IN DETAIL what is going on.

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  • \$\begingroup\$ Where you wrote "ANY CURRENT AT ALL", did you mean "any potential difference at all"? \$\endgroup\$ – The Photon Sep 30 '13 at 2:19
  • \$\begingroup\$ @ThePhoton: Same same. No practical difference. If there is potential difference between the inverting and non-inverting inputs, there will be current entering the op amp. If there is current entering the op amp, it will develop a voltage drop across the input resistance, which develops a potential difference between the two inputs. \$\endgroup\$ – John R. Strohm Sep 30 '13 at 5:49

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