# op-amp circuit explanation [closed]

Hi everyone I'm studying Op-Amp but I can't understand this point in the AC circuit :

In the first picture it says Vo = Ic * Rc, and in the second picture I have the value of Ic, when I try to apply this in the given formula before the answer is wrong. Why?

This second formula is used instead of the first one?

Is there is a difference ?  ## closed as unclear what you're asking by Andy aka, Voltage Spike, Dave Tweed♦Sep 30 '17 at 14:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

## 1 Answer

You're first picture has no component called $R_c$, but we can clearly see that $V_o = I_c R_c$ does not apply for the second picture, but instead it is $$V_{R_c} = I_c R_c \qquad V_o = V_{cc} - I_c R_c$$

But when doing a AC-Analysis $V_{cc}$ is removed, as it is a constant voltage source, and therefor $R_c$ is short-circuited to ground. Then you actually have $V_o = I_c R_c$.

Attention: AC-Analysis vs. DC-Analysis

We use DC-Analysis to find the bias-/working-point. With the small-signal-model we can use, altough being restricted to a bounded input voltage, a linear model for easier analysis of the circuit, despite the nonlinear behavior of a transistor.

Microelectronic circuits from Sedra&Smith is a very good source of information if you're new to electronics.