Op Amp ac circuit can also be analyzed with the same method with dc sources.

- Transform the circuit to the phasor or frequency domain.
- Solve the problem using circuit techniques
- Kirchhoff,
- Nodal,
- Mesh,
- Superposition,
- Source Transformation, and
- Thevenin and Norton

- Transform the resulting phasor to the time domain.

*Make sure to read what is ac circuit first.*

Make sure to read:

- What is phasor
- Impedance and admittance
- Kirchhoff’s laws for ac circuit
- Power calculation in ac circuit
- Three phase circuit

And its applications:

## Op Amp AC Circuits

Please remember, we will use ideal op amps here to ease the explanation. Take notice on the key of analyzing op-amp circuits those are :

- No current enters either of its input terminals
- The voltage across its input terminals is zero

Let us review the examples below :**1. Determine v_{o}(t) for the op amp in Figure.(1a) if v_{s} = 3 cos 1000t V.**

Figure 1. For example 1 : (a) the original circuit in the time domain, (b) its frequency domain |

*Solution :*

We first transform the circuit to the frequency domain, as drawn in Figure.(1b), where **V**_{s} = 3∠0^{o}, *ω* = 1000 rad/s. Applying KCL at node 1 we get

or

(1.1) |

At node 2, KCL gives

which leads to

(1.2) |

Substituting Equations.(1.2) to (1.1) gives

Thus

**2. Compute the closed-loop gain and phase shift for the circuit in Figure.(2). Assume that R _{1} = R_{2} = 10 kΩ, C_{1} = 2 uF, C_{2 }= 1 uF, and ω = 200 rad/s.**

Figure 2 |

*Solution :*

The feedback and input impedances are calculated as

Since the circuit in Figure.(2) is an inverting amplifier, the closed-loop gain is given by

Substituting the given values of R_{1}, R_{2}, C_{1}, C_{2}, and *ω*, we get

Hence, the closed-loop gain is 0.434 and the phase shift is 130.6^{o}.

Read also : voltage division rule