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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 VIN to the input, a current I1 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 I1 must flow via R2.
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, VIN = 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, UIN = -1 V
- What is the output voltage VOUT ?
- Which statement is true?
IR1 < IR2
IR1 = IR2
IR1 = 10-IR2 - What current flows through R1 ?
Exercise 1:
- The difference between the two inputs of the Op-Amp is in the steady state: Udiff = 0V
- In the stationary state, the inverting input of the OPV is virtually at 0 V.
- I1 = V1 / R1 = 1V / 2 kΩ = 0,5 mA
Exercise 2:
- Vout = - R2 / R1 * Vin = - 200 kΩ / 20 kΩ (-1V) = 10 V
- Since the input current of the Op-Amp is negligibly small, the following applies: I1 = I2
- I1 = V1 / R1 = - 1V / 20 kΩ = - 0,05 mA