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### OP Amp as **Non-Inverting Amplifier**

Table of Contents

ToggleHere the output signal is fed back to the inverting input by a voltage divider. The output signal is divided by the voltage divider, which results in an output signal being amplified. The resistance ratio determines the gain factor.

Since the input resistance of an operational amplifier is very high, the input currents are negligible. Thus: I_{1} = I_{2} (1)

Since the differential voltage at the inputs of the operational amplifier is zero, the following applies: V_{R1} = V_{IN} (2)

Apply the mesh rule and you get for the output voltage V_{OUT}:

V_{OUT} = V_{R1} + V_{R2} or V_{OUT} = V_{IN} + V_{R2}

With I_{1} = I_{2} and Ohm's law you get:

#### Exercise - OP-Amp as **Non-Inverting Amplifier**

__Exercise 1:__

Calculate the output voltage with the following values: R_{1} = 2 kΩ, R_{2} = 22 kΩ, V_{IN} = - 500 mV

__Exercise 2:__

R_{1} = 2 kΩ, R_{2} = 22 kΩ, U_{e} = 1 V

- What voltage is measured between the inverting and the non-inverting input in the steady state ?
- Calculate the current flowing through R
_{2}? - What is the voltage drop across R
_{2}?

Solution exercise 1:

V_{out} = (1 + R_{1} / R_{2}) V_{IN} = (1 + 22 kΩ / 2 kΩ) (-) 0.5 V = - 6V

Solution exercise 2:

The difference between the two inputs of the Op-Amp is in the steady state: U_{diff} = 0V

Since the input current of the Op-Amp is negligibly small, the following applies: I_{1} = I_{2}

I_{2} = I_{1} = V_{R1} / R_{1} = V_{In} / R_{1} = 1 V / 2 kΩ = 0.5 mA

V_{R2} = I_{2} * R_{2} = 0.5 mA * 22 kΩ = 11 V