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## Homework Statement

I want to plot ##I_C## versus ##V_{CE}## curves for the following:

To do so, I need to find appropriate values for ##R_C, R_B## and ##R_1## in the active region.

Note that ##R_2## is a potentiometer.

## Homework Equations

##I_C = 1 mA##

##V_{BE} = 0.7 V##

##\beta = 150##

##R_x = R_y = 500 \Omega##

##I_{R1} = 10I_B##

##V_{CEsat} = 0.5 V##.

## The Attempt at a Solution

First I need to find appropriate values for the resistors, but I've hit a snag. The analysis proceeds like so:

1) ##v_{Cmax} = V_{CC} = 15V## and ##v_{Cmin} = V_{B} = V_{E} + V_{BE} = 0.7V##.

Therefore ##V_{C} = \frac{1}{2} (15V + 0.7V) = 7.85 V##.

2) ##R_C = \frac{V_{CC} - V_C}{I_C} = \frac{15V - 7.85V}{1 mA} = 7.15 k \Omega##.

This ##R_C## produces a ##V_{CE} > 0.5 V##, so we are in active mode.

3) Now that we have ##R_C##, we need to find ##R_B## and ##R_1##. Let ##I_B = \frac{I_C}{\beta} = \frac{1 mA}{150} = 6.68 \mu A##.

Then ##I_{R1} = 10I_B = 66.8 \mu A##. This will be the current flowing across the series combination of ##R_1## and ##R_x##.

Now ##I_{Ry} = I_{R1} - I_{B} = 60.12 \mu A##. This will be the current flowing across ##R_y##.

4) Here is where I start encountering results that don't add up. I want to find the voltage at the node between ##R_1 + R_x##, ##R_B## and ##R_y## (call it ##V_{BB}##), so I used:

$$\frac{V_{BB} - 0}{500 \Omega} = 60.12 \mu A \Rightarrow V_{BB} = 0.03 V$$

This result makes no sense because the voltage at the base is ##0.7V##.

What went wrong here?