A student measures a silicon transistor in the common-emitter circuit…

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A student measures a silicon transistor in the common-emitter circuit. At $V_{CE} = 6.0$ V, she finds the values in the table. Then, with the base current held at $30\ \mu$ A, she raises $V_{CE}$ from $6.0$ V to $10$ V, and the collector current rises from $2.88$ mA to $2.95$ mA. Use the table to find the input resistance, the dc current gain at $I_B = 60\ \mu$ A, the ac current gain between the two rows and the output resistance. Say why the output resistance is so much larger than the input resistance.

$V_{BE}$ (V)	$I_B$ ( $\mu$ A)	$I_C$ (mA)
0.70	30	2.88
0.72	60	5.82

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The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. The table, and $I_C$ at $I_B = 30\ \mu$ A: $2.88$ mA at $6.0$ V and $2.95$ mA at $10$ V.

Plan. Use the changes for the resistances and the ac gain, and one row for the dc gain.

Step 1. The input resistance. $r_i = \dfrac{\Delta V_{BE}}{\Delta I_B} = \dfrac{0.72\ \text{V} - 0.70\ \text{V}}{60\ \mu\text{A} - 30\ \mu\text{A}} = \dfrac{0.020\ \text{V}}{30 \times 10^{-6}\ \text{A}} = 667\ \Omega$ .

Step 2. The dc gain. $\beta_{dc} = \dfrac{I_C}{I_B} = \dfrac{5.82\ \text{mA}}{0.060\ \text{mA}} = 97.0$ .

Step 3. The ac gain. $\beta_{ac} = \dfrac{\Delta I_C}{\Delta I_B} = \dfrac{5.82\ \text{mA} - 2.88\ \text{mA}}{0.060\ \text{mA} - 0.030\ \text{mA}} = \dfrac{2.94\ \text{mA}}{0.030\ \text{mA}} = 98.0$ .

Step 4. The output resistance. $r_o = \dfrac{\Delta V_{CE}}{\Delta I_C} = \dfrac{10\ \text{V} - 6.0\ \text{V}}{2.95\ \text{mA} - 2.88\ \text{mA}} = \dfrac{4.0\ \text{V}}{0.070 \times 10^{-3}\ \text{A}} = 57.1\ \text{k}\Omega$ .

Answer: $667\ \Omega$ , $97.0$ , $98.0$ and $57.1\ \text{k}\Omega$ .

Why $r_o$ is large. In the active region, the collector current hardly depends on $V_{CE}$ , so the curves are nearly flat. A large change of voltage gives a small change of current. The input junction is forward biased, so a small change of voltage changes the current a lot, and $r_i$ is small.

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