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## Homework Statement

## Homework Equations

## The Attempt at a Solution

I was not too sure about the analysis and thought I would ask some questions.

Do I short out the AC source ##v_i## when doing the DC analysis?

The input capacitor with ##C = \infty## is designed to couple the AC signal and behaves as a short circuit.

Then I thought to simplify the base of the transistor. The Thevenin voltage and resistance are given by:

$$V_{th} = \left[\frac{100k}{200k} \right] (5 V) = 2.5V$$

$$R_{th} = 100k || 100k = 50k$$

Now there would be a DC voltage of ##2.5V## and a ##50k## resistor in the base (assuming ##v_i## is shorted).

Does this sound reasonable so far?

Now If the transistor is active, does that imply ##V_{BE} = 0.7V##?

If so I'm guessing the next step would be:

$$I_B = \frac{2.5V - 0.7V}{50k} = 0.036 mA$$

I am not entirely sure how to proceed from here and I'm wondering if this is even correct.