Solved transistor exercises: biasing, small-signal amplifiers and logic gates

Twelve exercises with solutions: divider design, pnp, sensitivity to beta, JFET and MOSFET, common emitter with degeneration, follower, common source, Miller effect and cascode, bypass capacitor, TTL and RTL fan-out.

ielettronica 26 October 2026 9 min read 1 comment

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Transistor exercises of the kind found in electronics exams, at three levels: basic, with the calculation of the operating point (npn and pnp BJT, JFET, MOSFET); intermediate, with small-signal amplifiers; exam, with frequency response and logic gates. Unless stated otherwise $V_{BE} = 0.7$ V and $V_T = 25$ mV are used. The result is given straight away; the complete step-by-step solution opens with a click. The proofs of the formulas are in the lessons on biasing, on small-signal amplifiers and on BJT logic gates.

Basic level

Exercise 1. Designing a divider bias

An npn BJT with $\beta$ between 80 and 300 (mean value 150) must work with $I_C = 1$ mA, a 15 V supply, 1.5 V on the emitter and 9 V on the collector. Size $R_E$ , $R_C$ , $R_1$ and $R_2$ with the rule $R_{TH} = 0.1\,\beta_{min} R_E$ , choose standard values and check the current with $\beta$ = 80, 150 and 300.

Result: $R_E = 1.5$ kΩ, $R_C = 6$ kΩ, $R_1 = 82$ kΩ, $R_2 = 15$ kΩ; $I_C$ = 0.966, 1.016 and 1.047 mA.

Solution

1. From the desired voltages: $R_E = 1.5\text{ V}/1\text{ mA} = 1.5$ kΩ and $R_C = (15 - 9)\text{ V}/1\text{ mA} = 6$ kΩ (standard values: 5.6 or 6.2 kΩ; 6 kΩ is kept here to follow the calculations).

2. Stability rule: $R_{TH} = 0.1 \cdot 80 \cdot 1.5\text{ k}\Omega = 12$ kΩ. The Thévenin voltage must provide the drop across $R_E$ , $V_{BE}$ and the drop of the base current across $R_{TH}$ (computed with the mean $\beta$ ): $V_{TH} = 1.5 + 0.7 + (1\text{ mA}/150) \cdot 12\text{ k}\Omega = 2.28$ V.

3. From the divider: $R_1 = R_{TH} V_{CC}/V_{TH} = 78.9$ kΩ and $R_2 = R_{TH} V_{CC}/(V_{CC} - V_{TH}) = 14.2$ kΩ. Choosing 82 kΩ and 15 kΩ gives $V_{TH} = 2.32$ V and $R_{TH} = 12.7$ kΩ.

4. Check with $I_E = (V_{TH} - V_{BE})/(R_E + R_{TH}/(\beta + 1))$ : with $\beta = 80$ , 0.966 mA ( $V_{CE} = 7.74$ V); with 150, 1.016 mA (7.37 V); with 300, 1.047 mA (7.14 V). A change of almost four times in $\beta$ moves the current by 8 %.

Exercise 2. A pnp

A pnp with $\beta = 150$ is powered from 9 V. The emitter goes to the supply through $R_E = 2.2$ kΩ, the collector to ground through $R_C = 1.8$ kΩ; the base is on a divider with 10 kΩ towards the supply and 33 kΩ towards ground. Find $I_C$ , the emitter and collector voltages and $V_{EC}$ .

Result: $I_C = 0.615$ mA, $V_E = 7.64$ V, $V_C = 1.11$ V, $V_{EC} = 6.53$ V.

Solution

1. With a pnp it is convenient to measure voltages downwards from the supply. The divider brings the base $V_{TH} = 9 \cdot 10/43 = 2.09$ V below the supply, with $R_{TH} = 10 \parallel 33 = 7.67$ kΩ.

2. The loop supply, $R_E$ , emitter-base junction, $R_{TH}$ gives the same formula as the npn: $I_E = (2.09 - 0.7)/(2.2\text{ k}\Omega + 7.67\text{ k}\Omega/151) = 0.619$ mA, and $I_C = 150/151 \cdot I_E = 0.615$ mA.

3. The emitter sits $I_E R_E = 1.36$ V below the supply: $V_E = 7.64$ V. The collector sits $I_C R_C = 1.11$ V above ground. $V_{EC} = 7.64 - 1.11 = 6.53$ V: the transistor is in the active region, with a wide margin.

Exercise 3. Sensitivity to beta

In the divider circuit of the lesson ( $R_E = 1.2$ kΩ, $R_{TH} = 13.0$ kΩ, $\beta = 200$ ) compute the relative sensitivity of the current to $\beta$ and use it to estimate the change in current when $\beta$ goes from 200 to 220. Compare with the exact calculation.

Result: $S = 0.051$ ; estimate +0.51 %, exact value +0.47 %.

Solution

1. $R_{TH}/(\beta + 1) = 13\,015/201 = 64.8$ Ω, and $S = 64.8/(1200 + 64.8) = 0.0512$ .

2. A 10 % increase in $\beta$ (about 10 % in $\beta + 1$ too) produces a relative change in the current of $S \cdot 10\% = 0.51$ %.

3. The exact calculation, with $I_E = (V_{TH} - V_{BE})/(R_E + R_{TH}/(\beta + 1))$ for 200 and 220, gives +0.47 %. The first-order estimate is slightly pessimistic because the sensitivity decreases as $\beta$ grows. With fixed bias the current would rise by 10 %.

Exercise 4. Self-biased JFET

An n-channel JFET with $I_{DSS} = 8$ mA and $V_P = -3$ V must work at $I_D = 2$ mA with self-bias. Find $V_{GS}$ , $R_S$ and the transconductance. What current is obtained with the same $R_S$ by fitting a unit with $I_{DSS} = 12$ mA and $V_P = -4$ V?

Result: $V_{GS} = -1.5$ V, $R_S = 750$ Ω, $g_m = 2.67$ mS; with the other unit 2.77 mA.

Solution

1. From Shockley’s law: $1 - V_{GS}/V_P = \sqrt{I_D/I_{DSS}} = 0.5$ , hence $V_{GS} = -3 \cdot 0.5 = -1.5$ V.

2. With self-bias $V_{GS} = -I_D R_S$ : $R_S = 1.5/2\text{ mA} = 750$ Ω.

3. Differentiating Shockley’s law: $g_m = \frac{2I_{DSS}}{|V_P|}\left(1 - \frac{V_{GS}}{V_P}\right) = \frac{16\text{ mA}}{3\text{ V}} \cdot 0.5 = 2.67$ mS (which is also $2I_D/(V_{GS} - V_P) = 4\text{ mA}/1.5\text{ V}$ ).

4. With the other unit one solves $I_D = 12\text{ mA}\,(1 - I_D \cdot 750/4)^2$ , looking for the root with $V_{GS}$ between −4 V and 0: $I_D = 2.77$ mA. The current changes by 39 % for a 50 % change in $I_{DSS}$ : self-bias helps little.

Exercise 5. MOSFET with drain feedback

An n-channel MOSFET with $K = 1$ mA/V² and $V_t = 1.5$ V has a 1 MΩ resistor between drain and gate and $R_D = 3.3$ kΩ to the 10 V supply; the source is grounded. Find $V_{GS}$ , $I_D$ and $g_m$ , and check that the transistor is in saturation.

Result: $V_{GS} = V_{DS} = 2.96$ V, $I_D = 2.13$ mA, $g_m = 2.92$ mS.

Solution

1. No current flows in the gate: $V_{GS} = V_{DS} = 10 - 3.3\text{ k}\Omega \cdot I_D$ .

2. Letting $u = V_{GS} - V_t$ : $u = 10 - 1.5 - 3.3 \cdot u^2$ (with $I_D$ in mA and $K = 1$ ), that is, $3.3u^2 + u - 8.5 = 0$ , whose positive root is $u = 1.461$ V. Hence $V_{GS} = 2.96$ V and $I_D = u^2 = 2.13$ mA.

3. $g_m = 2Ku = 2.92$ mS.

4. Saturation: $V_{DS} = 2.96$ V > $V_{GS} - V_t = 1.46$ V. As proved in the lesson, with this connection it is always true.

Intermediate level

Exercise 6. Common emitter with and without degeneration

A common-emitter stage has $I_C = 1$ mA, $\beta = 150$ , $R_C = 4.7$ kΩ and load $R_L = 10$ kΩ. Compute the voltage gain and the input resistance seen from the base with the emitter fully bypassed, and then with 220 Ω unbypassed.

Result: bypassed: $A_v = -128$ , $R_{in} = 3.75$ kΩ; with 220 Ω: $A_v = -13.0$ , $R_{in} = 37.0$ kΩ.

Solution

1. $g_m = 1\text{ mA}/25\text{ mV} = 40$ mS, $r_\pi = 150/40\text{ mS} = 3.75$ kΩ, $R_L' = 4.7 \parallel 10 = 3.20$ kΩ.

2. Bypassed emitter: $A_v = -g_m R_L' = -128$ , $R_{in} = r_\pi = 3.75$ kΩ.

3. With $R_E = 220$ Ω: $A_v = -\beta R_L'/(r_\pi + (\beta + 1)R_E) = -150 \cdot 3197/(3750 + 151 \cdot 220) = -13.0$ ; the approximate formula $-R_L'/(r_e + R_E) = -3197/245 = -13.05$ is almost identical. $R_{in} = 3750 + 151 \cdot 220 = 37.0$ kΩ.

4. Degeneration divides the gain by about 10, multiplies the input resistance by 10 and makes the gain independent of $g_m$ (and therefore of current and temperature): $-R_L'/R_E \approx -14.5$ if $r_e$ is neglected too.

Exercise 7. The follower as an impedance converter

A source with an internal resistance of 10 kΩ must drive a 100 Ω load. Compute the signal reaching the load when connected directly, and through an emitter follower with $\beta = 100$ , $I_C = 5$ mA and $R_E = 1$ kΩ (ignore the bias divider). Also find the output resistance of the follower.

Result: directly 0.99 %; with the follower 46.7 %; $R_{out} = 94$ Ω.

Solution

1. Direct connection: $100/(10\,000 + 100) = 0.0099$ . The signal arrives reduced a hundredfold.

2. Follower: $r_\pi = 100/(5\text{ mA}/25\text{ mV}) = 500$ Ω; on the emitter $R_L' = 1\text{ k}\Omega \parallel 100\ \Omega = 90.9$ Ω. $R_{in} = 500 + 101 \cdot 90.9 = 9.68$ kΩ and $A_v = 101 \cdot 90.9/9682 = 0.948$ .

3. With the source: $v_{out}/v_s = 0.948 \cdot 9682/(9682 + 10\,000) = 0.467$ . The signal on the load is 47 times larger than with the direct connection.

4. $R_{out} = R_E \parallel (r_\pi + R_s)/(\beta + 1) = 1\text{ k}\Omega \parallel 104\ \Omega = 94$ Ω: the source resistance, seen from the load, is divided by $\beta + 1$ .

Exercise 8. Common source

A MOSFET with $K = 2$ mA/V² works at $I_D = 1$ mA, with $R_D = 5$ kΩ and a 20 kΩ load. Compute $g_m$ , the overdrive voltage and the gain with the source bypassed and with 500 Ω unbypassed. Compare with a BJT at the same current.

Result: $g_m = 2.83$ mS, $V_{ov} = 0.71$ V; $A_v = -11.3$ and $-4.7$ ; the BJT would have $g_m = 40$ mS and $A_v = -160$ .

Solution

1. $g_m = 2\sqrt{K I_D} = 2\sqrt{2 \cdot 1}$ mS $= 2.83$ mS; $V_{ov} = V_{GS} - V_t = 2I_D/g_m = 0.71$ V.

2. $R_L' = 5 \parallel 20 = 4$ kΩ. Bypassed: $A_v = -g_m R_L' = -11.3$ . With $R_S = 500$ Ω: $-g_m R_L'/(1 + g_m R_S) = -11.3/2.41 = -4.7$ .

3. A BJT at 1 mA has $g_m = I_C/V_T = 40$ mS, fourteen times more, and $A_v = -160$ . The ratio between the transconductances is $V_{ov}/(2V_T) = 0.71/0.05 = 14$ : at the same current the BJT always wins, the MOSFET wins on input resistance and linearity.

Exam level

Exercise 9. The Miller effect

A common-emitter stage has $C_\pi = 20$ pF, $C_\mu = 3$ pF, a base-to-collector gain of $-100$ and an equivalent source resistance seen from the base of 1 kΩ. Estimate the upper cut-off frequency with the Miller approximation, and compare it with that of a cascode stage, in which the gain at the collector of the input transistor is about −1.

Result: $C_{in} = 323$ pF, $f_H \approx 0.49$ MHz; with the cascode $C_{in} = 26$ pF and about 6.1 MHz.

Solution

1. Miller’s theorem brings $C_\mu$ to the input multiplied by $1 - A = 101$ : 303 pF. With $C_\pi$ , $C_{in} = 323$ pF.

2. The input pole: $f_H = 1/(2\pi \cdot 1\text{ k}\Omega \cdot 323\text{ pF}) = 0.49$ MHz.

3. In the cascode the collector of the input transistor is connected to the emitter of a common-base transistor, which has resistance $r_e$ : the base-to-collector gain is about −1, and $C_\mu$ is multiplied only by 2. The input capacitance becomes $C_\pi + 2C_\mu = 26$ pF and the bandwidth $1/(2\pi \cdot 1\text{ k}\Omega \cdot 26\text{ pF}) = 6.1$ MHz: twelve times more, with the same overall gain. A 3 pF capacitance, the smallest in the transistor, alone decides the bandwidth of the common-emitter stage.

Exercise 10. The bypass capacitor

In a common-emitter stage with $R_E = 1$ kΩ fully bypassed, $r_\pi = 2.5$ kΩ, $\beta = 100$ , and a source resistance in parallel with the divider of 900 Ω, what must $C_E$ be for its pole to be at 20 Hz?

Result: resistance seen 32.6 Ω; $C_E \ge 244$ µF (270 µF is chosen).

Solution

1. The capacitor sees $R_E$ in parallel with the resistance looking into the transistor’s emitter: $(r_\pi + R_s')/(\beta + 1) = 3400/101 = 33.7$ Ω. In parallel with 1 kΩ: 32.6 Ω.

2. $C_E = 1/(2\pi \cdot 20 \cdot 32.6) = 244$ µF; the next standard value is 270 µF.

3. Coupling capacitors see resistances of kΩ and a few µF are enough; the bypass capacitor sees a few tens of ohms and must be a hundred times larger. This is why part of $R_E$ is often left unbypassed: it raises the resistance seen and reduces $C_E$ .

Exercise 11. Fan-out between TTL families

A standard 74 TTL output ( $I_{OL} = 16$ mA, $I_{OH} = 400$ µA) drives 74LS inputs ( $I_{IL} = 0.4$ mA, $I_{IH} = 20$ µA). How many can it drive? And how many standard 74 inputs ( $I_{IL} = 1.6$ mA, $I_{IH} = 40$ µA) can a 74LS output ( $I_{OL} = 8$ mA, $I_{OH} = 400$ µA) drive?

Result: 20 in the first case; 5 in the second.

Solution

1. The fan-out is the minimum of the low-level and the high-level fan-outs. 74 → 74LS: $16/0.4 = 40$ at the low level and $400/20 = 20$ at the high level: 20.

2. 74LS → 74: $8/1.6 = 5$ at the low level, $400/40 = 10$ at the high level: 5. The limit is the current flowing out of the low standard inputs, which the LS output must sink.

3. Mixing families is possible because the voltage levels are compatible, but the fan-out must be recomputed with the currents of each.

Exercise 12. RTL fan-out

An RTL gate has $V_{CC} = 3.6$ V, $R_C = 640$ Ω, $R_B = 450$ Ω and $V_{BE} = 0.75$ V. How many gates can it drive if the high level must not fall below 1 V?

Result: $N \le 7.3$ : at most 7 gates.

Solution

1. From the formula of the lesson, $V_{OH} = V_{BE} + (V_{CC} - V_{BE})\frac{R_B/N}{R_B/N + R_C} \ge 1$ V.

2. Isolating $N$ : $\frac{R_B/N}{R_B/N + R_C} \ge \frac{0.25}{2.85}$ , that is, $1 + \frac{N R_C}{R_B} \le \frac{2.85}{0.25} = 11.4$ , and $N \le \frac{10.4 \cdot 450}{640} = 7.31$ .

3. With 7 loads $V_{OH} = 1.01$ V, with 8 it drops to 0.98 V. Even with 7 loads the margin above the conduction threshold (0.75 V) is only 0.25 V: this is why RTL was replaced by DTL and TTL.

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Basic level

Exercise 1. Designing a divider bias

Exercise 2. A pnp

Exercise 3. Sensitivity to beta

Exercise 4. Self-biased JFET

Exercise 5. MOSFET with drain feedback

Intermediate level

Exercise 6. Common emitter with and without degeneration

Exercise 7. The follower as an impedance converter

Exercise 8. Common source

Exam level

Exercise 9. The Miller effect

Exercise 10. The bypass capacitor

Exercise 11. Fan-out between TTL families

Exercise 12. RTL fan-out

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