**Home ⇒ Overview Courses ⇒ Electronics ⇒ Operational amplifier ⇒ Inverting OP-Amp**

### Inverting Operational Amplifier - Inverting Op Amp

This type of op-amp circuit produces an output which is 180° out of phase with its input.To explain this, let us again think of the virtual zero point at the inputs of the operational amplifier. If we now apply a voltage V_{IN} to the input, a current I_{1} must inevitably flow:

Since the input resistance of the operational amplifier is very high, the input current of the OPV must also be negligibly low. This means that the current I_{1} must flow via R_{2}.

Combining both equations, taking into account the direction of the current and you get:

#### Exercises - Inverting Op Amp

**Exercise 1:**

R1 = 2 kΩ, R2 = 22 kΩ, V_{IN} = 1 V

- What voltage is measured between the inverting and the non-inverting input in the steady state ?
- At what potential is the inverting input in the steady state ?
- Calculate the current flowing through R2 ?

**Exercise 2:**

R1 = 20 kΩ, R2 = 200 kΩ, U_{IN} = -1 V

- What is the output voltage V
_{OUT}? - Which statement is true?

IR1 < IR2

IR1 = IR2

IR1 = 10-I_{R2} - What current flows through R1 ?

**Exercise 1:**

- The difference between the two inputs of the Op-Amp is in the steady state: U
_{diff}= 0V - In the stationary state, the inverting input of the OPV is virtually at 0 V.
- I
_{1}= V_{1}/ R1 = 1V / 2 kΩ = 0,5 mA

**Exercise 2:**

- V
_{out}= - R_{2}/ R_{1}* V_{in}= - 200 kΩ / 20 kΩ (-1V) = 10 V - Since the input current of the Op-Amp is negligibly small, the following applies: I
_{1}= I_{2} - I
_{1}= V_{1}/ R_{1}= - 1V / 20 kΩ = - 0,05 mA