transistor - 2

[logistic_guy]

For the transistor circuit in the figure, find \(\displaystyle I_B, V_{CE}, \text{and} \ v_o\). Take \(\displaystyle \beta = 200, V_{BE} = 0.7 \ \text{V}\).

 

[khansaheb]

For the transistor circuit in the figure, find \(\displaystyle I_B, V_{CE}, \text{and} \ v_o\). Take \(\displaystyle \beta = 200, V_{BE} = 0.7 \ \text{V}\).
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[logistic_guy]

Let us try KVL. It may work. It may not work.

First loop.

\(\displaystyle 3 - 6000I_1 - 2000(I_1 - I_2) = 0\)

\(\displaystyle 3 - 6000I_1 - 2000I_1 + 2000I_2 = 0\)

\(\displaystyle 3 - 8000I_1 + 2000I_2 = 0\)

 

[logistic_guy]

Second loop.

\(\displaystyle -2000(I_2 - I_1) - 0.7 - 400(I_2 - I_3) = 0\)

\(\displaystyle -2000I_2 + 2000I_1 - 0.7 - 400I_2 + 400I_3 = 0\)

\(\displaystyle -2400I_2 + 2000I_1 - 0.7 + 400I_3 = 0\)

 

[logistic_guy]

Loop Three.

\(\displaystyle -400(I_3 - I_2) + V_{CE} - 5000I_3 - 9 = 0\)


\(\displaystyle -400I_3 + 400I_2 + V_{CE} - 5000I_3 - 9 = 0\)


\(\displaystyle 400I_2 + V_{CE} - 5400I_3 - 9 = 0\)

 

[logistic_guy]

We have three equations but four unknowns! We have to figure out a way to solve this issue.

 

[logistic_guy]

Let us go back to the equation we got from the first loop.

\(\displaystyle 3 - 8000I_1 + 2000I_2 = 0\)

We know that \(\displaystyle I_2 = I_B\),

then,

\(\displaystyle 8000I_1 - 2000I_B = 3\)

 

[logistic_guy]

Second loop.

\(\displaystyle -2000(I_2 - I_1) - 0.7 - 400(I_2 - I_3) = 0\)

We analyze this again.

We know that \(\displaystyle I_2 - I_3\) is exactly the emitter current \(\displaystyle \longrightarrow I_E\) and \(\displaystyle I_2 = I_B\)

then,

\(\displaystyle -2000(I_B - I_1) - 0.7 - 400(I_E) = 0\)


\(\displaystyle 2000(I_B - I_1) + 0.7 + 400(I_E) = 0\)


\(\displaystyle 2000I_B - 2000I_1 + 0.7 + 400I_E = 0\)

 

[logistic_guy]

\(\displaystyle 2000I_B - 2000I_1 + 0.7 + 400I_E = 0\)

We know that:

\(\displaystyle I_E = (1 + \beta)I_B\)

Then,

\(\displaystyle 2000I_B - 2000I_1 + 0.7 + 400(1 + \beta)I_B = 0\)


\(\displaystyle 2000I_B - 2000I_1 + 0.7 + 400(1 + 200)I_B = 0\)


\(\displaystyle 2000I_B - 2000I_1 + 0.7 + 400(201)I_B = 0\)


\(\displaystyle 2000I_B - 2000I_1 + 0.7 + 80400I_B = 0\)


\(\displaystyle 82400I_B - 2000I_1 + 0.7 = 0\)

 

[logistic_guy]

\(\displaystyle 82400I_B - 2000I_1 + 0.7 = 0\)

\(\displaystyle I_1 = \frac{824I_B + 0.007}{20}\)

 

[logistic_guy]

\(\displaystyle 8000I_1 - 2000I_B = 3\)

\(\displaystyle 8000\left(\frac{824I_B + 0.007}{20}\right) - 2000I_B = 3\)


\(\displaystyle 400\left(\frac{824I_B + 0.007}{1}\right) - 2000I_B = 3\)


\(\displaystyle 329600I_B + 2.8 - 2000I_B = 3\)


\(\displaystyle 327600I_B + 2.8 = 3\)


\(\displaystyle 327600I_B = 0.2\)


\(\displaystyle I_B = \frac{0.2}{327600} = 0.0000006105 \ \text{A} =\textcolor{blue}{610.5 \ \text{nA}}\)

 

[logistic_guy]

\(\displaystyle I_E = (1 + \beta)I_B = (1 + 200)610.5 \times 10^{-9} = 0.0001227105 \ \text{A} = \textcolor{red}{122.7105 \ \mu \text{A}}\)

 

[logistic_guy]

\(\displaystyle v_o = 400I_E = 400\left(122.7105 \times 10^{-6}\right) = 0.0490842 \ \text{V} = \textcolor{blue}{49.0842 \ \text{mV}}\)

 

[logistic_guy]

\(\displaystyle I_C = \beta I_B = 200 \times 610.5 \times 10^{-9} = 0.0001221 \ \text{A} = \textcolor{indigo}{122.1 \ \mu\text{A}}\)