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A common-emitter amplifier is built with a supply of 12 V and a collector resistor of 3.0 kΩ…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A common-emitter amplifier is built with a supply of $12$ V and a collector resistor of $3.0\ \text{k}\Omega$ . The transistor has $\beta = 100$ and an input resistance $r_i = 1.5\ \text{k}\Omega$ . The designer wants the resting collector voltage to be $6.0$ V, so that the output can swing about the middle. Take $V_{BE} = 0.7$ V and $V_{CE(sat)} = 0.2$ V. Find the resting collector current, the resting base current and the base resistor that gives it, when the base is fed from the $12$ V supply. Find the voltage gain. Then find the largest amplitude of the input signal that the amplifier can handle with no clipping. A tester applies a signal of amplitude $40$ mV. Find the highest and the lowest output voltage that he sees.

Need a reminder? The key ideaShow

The voltage gain of a common-emitter amplifier is $A_V = -\dfrac{\beta R_C}{r_i}$ , the current gain is $\beta$ and the power gain is $\beta |A_V|$ . The output $V_{CE}$ stays between $0.2$ V (saturation) and $V_{CC}$ (cut-off). A transistor switch saturates when $I_B \ge \dfrac{V_{CC} - 0.2\ \text{V}}{\beta R_C}$ .

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. $V_{CC} = 12$ V, $V_{CEQ} = 6.0$ V, $R_C = 3.0\ \text{k}\Omega$ , $\beta = 100$ , $r_i = 1.5\ \text{k}\Omega$ .

Step 1. The resting collector current. $I_{CQ} = \dfrac{V_{CC} - V_{CEQ}}{R_C} = \dfrac{12\ \text{V} - 6.0\ \text{V}}{3.0\ \text{k}\Omega} = 2.00$ mA.

Step 2. The resting base current. $I_{BQ} = \dfrac{I_{CQ}}{\beta} = \dfrac{2.00\ \text{mA}}{100} = 20.0\ \mu\text{A}$ .

Step 3. The base resistor. $R_B = \dfrac{12\ \text{V} - 0.7\ \text{V}}{20.0\ \mu\text{A}} = 565\ \text{k}\Omega$ .

Step 4. The gain. $|A_V| = \dfrac{\beta R_C}{r_i} = \dfrac{100 \times 3.0\ \text{k}\Omega}{1.5\ \text{k}\Omega} = 200$ .

Step 5. The room for the signal. The output can fall by $6.0\ \text{V} - 0.2\ \text{V} = 5.8$ V and rise by $12\ \text{V} - 6.0\ \text{V} = 6.0$ V. The smaller of the two, $5.8$ V, limits the swing. So $v_{i,max} = \dfrac{5.8\ \text{V}}{200} = 0.029\ \text{V} = 29.0$ mV.

Step 6. A signal of $40$ mV. The ideal output swings $200 \times 40\ \text{mV} = 8.0$ V about $6.0$ V, which is from $-2.0$ V to $14$ V. The real output is cut at $0.2$ V and at $12$ V.

Answer: $2.00$ mA, $20.0\ \mu$ A, $565\ \text{k}\Omega$ and $29.0$ mV. The output of the $40$ mV signal goes from $0.2$ V to $12$ V, and both peaks are clipped.

Check. At $29.0$ mV the output swings exactly between $0.2$ V and $11.8$ V, just inside the limits.

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