- #1

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## Homework Statement

I had a few questions about the npn BJT with regards to designing a few circuits. The questions and my attempts are below.

For the calculations, assume ##\beta = 140## and ##V_{BE} = 0.7V## for npn active mode.

1. Fixed bias circuit: Design the resistor values required to bias the circuit in the active region, such that ##I_C = 1 mA##, ##V_{CE} = 3V## (calculate ##R_C, I_B, R_B##):

I don't know why they are asking for ##R_C## in this circuit when they gave ##R_C = 12 k##. I think it may be a typo, but I'm not certain. I will solve for ##I_B## and ##R_B## assuming ##R_C## was a typo.

2. Self bias circuit: Design the self bias circuit for the BJT CE amplifier shown in question 3:

I assume they want me to find ##R_1## and ##R_2##.

3. BJT CE Amplifier: Design the CE amplifier (as per the specifications given). Derive the input impedance ##Z_i## as well as the open circuit voltage gain ##\frac{v_o}{v_b}##:

What will happen to the gain under loaded conditions?

What will happen to the gain if ##R_E## is bypassed?

What will happen to the input impedance if ##R_E## is bypassed?

## Homework Equations

## The Attempt at a Solution

1. Just to prove ##R_C## was a typo:

$$I_C = \frac{V_{CC} - V_{CE}}{R_C} \Rightarrow R_C = \frac{V_{CC} - V_{CE}}{I_C} = \frac{15V - 3V}{1 \times 10^{-3} A} = 12k \Omega$$

Now for active mode operation we know:

$$I_C = \beta I_B \Rightarrow I_B = \frac{I_C}{\beta} = \frac{1 \times 10^{-3} A}{140} = 7.143 \mu A$$

Now that we know the base current we can write:

$$I_B = \frac{V_{CC} - V_{BE}}{R_B} \Rightarrow R_B = \frac{V_{CC} - V_{BE}}{I_B} = \frac{15V - 0.7V}{7.143 \times 10^{-6} A} = 2.002 M \Omega$$

The circuit will be in active mode with the given specifications.

2. First, use Thevenin's theorem to simplify the base of the BJT. Short out ##V_{CC}## to find ##R_{th}##:

$$V_{th} = \frac{R_1}{R_1 + R_2} V_{CC}$$

$$R_{th} = R_1 || R_2 = \frac{R_1R_2}{R_1 + R_2}$$

So we will have ##V_{th}## and ##R_{th}## in the base of the BJT. Applying KVL to the BE loop now:

$$V_{th} = R_{th} I_B + V_{BE} + R_E I_E = \frac{R_{th}}{1 + \beta} I_E + V_{BE} + R_E I_E = \left(\frac{R_{th}}{1 + \beta} + R_E \right) I_E + V_{BE} $$

Now we must find ##R_{th}## and ##I_E##. We apply the stability criterion to select ##R_{th}##, namely we choose:

$$R_{th} = 0.1 \times (1 + \beta) R_E = 0.1 \times (1 + 140) (3.3 k \Omega) = 46.53 k \Omega$$

Now we must find the bias point so we can determine ##I_E##. Let the load line be given by:

$$V_{CE} = - (R_C + R_E)I_C + V_{CC}$$

The bias point should be in the middle of the load line, so:

$$V_{CE,Q} = \frac{V_{CC}}{2} = \frac{15V}{2} = 7.5V$$

$$I_{C,Q} = \frac{V_{CC}}{2(R_C + R_E)} = \frac{15V}{2(8.2k \Omega + 3.3k \Omega)} = 0.652 mA$$

Now going back to the previous KVL equation for ##V_{th}##, we find:

$$V_{th} = \left(\frac{R_{th}}{1 + \beta} + R_E \right) I_E + V_{BE} = \left(\frac{46.53 k \Omega}{141} + 3.3 k \Omega \right) (0.652 mA) + 0.7V = 3.067 V$$

This gives us two equations in the two unknowns ##R_1## and ##R_2##:

$$3.067 V = \frac{R_1}{R_1 + R_2} (15V)$$

$$46.53 k \Omega = \frac{R_1R_2}{R_1 + R_2}$$

The first equation gives:

$$0.204 = \frac{R_1}{R_1 + R_2}$$

$$R_2 = \frac{R_1}{0.204} - R_1 = 3.9R_1$$

Solving the second equation then yields:

$$46.53 k \Omega = \frac{R_1 (3.9R_1)}{R_1 + 3.9R_1} = \frac{3.9 R_1^2}{4.9R_1} = 1.256 R_1$$

$$R_1 = \frac{46.53 k \Omega}{1.256} = 37.05 k \Omega$$

Subbing back into the first equation then gives:

$$R_2 = 3.9R_1 = 3.9(37.05 k \Omega) = 144.5 k \Omega$$

Phew, that was a lot of work. I hope my answers to questions 1 and 2 look okay. If they do then I will try question 3.

If someone could let me know if my calculations look okay, then I will proceed to #3.

Thank you in advance.