BJT Biasing Circuits and Stability

About this problem set

Q-point calculations for the four standard BJT bias topologies — fixed, emitter, voltage-divider and collector-feedback — together with the stability factor S(ICO) that ranks them, and a full four-resistor bias design worked to standard resistor values.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

PROBLEM 01

Fixed-Bias Q-Point and Load-Line Limits

Problem Statement

An npn silicon transistor is fixed-biased from a \(V_{CC} = 12~\text{V}\) rail: a 470 k\(\Omega\) resistor runs from the +12 V rail to the base, and a 2.2 k\(\Omega\) collector resistor runs from the same rail to the collector. The emitter is grounded. Take \(\beta = 100\) and \(V_{BE} = 0.7~\text{V}\).

Find the Q-point \( (I_{BQ},\, I_{CQ},\, V_{CEQ}) \).
Locate the two ends of the DC load line and state how far the Q-point sits from saturation.
The transistor is replaced by one with \(\beta = 200\). Recompute \(I_C\) and \(V_{CE}\) and comment.

Solution

\(V_{CC} = 12~\text{V}\), \(R_B = 470~\text{k}\Omega\), \(R_C = 2.2~\text{k}\Omega\)
\(\beta = 100\), \(V_{BE} = 0.7~\text{V}\) (silicon, active region assumed)

Part (a) — base loop, then collector loop

The base loop contains only \(V_{CC}\), \(R_B\) and the base–emitter junction, so \(I_B\) is fixed by the supply and is independent of \(\beta\):

\[\begin{aligned} I_B &= \frac{V_{CC} - V_{BE}}{R_B} = \frac{12 - 0.7}{470~\text{k}\Omega} = 24.04~\mu\text{A} \\ I_C &= \beta I_B = 100 \times 24.04~\mu\text{A} = 2.404~\text{mA} \\ V_{CE} &= V_{CC} - I_C R_C = 12 - (2.404~\text{mA})(2.2~\text{k}\Omega) = 6.711~\text{V} \end{aligned}\]

\(I_{BQ} = 24.0~\mu\text{A}\), \(I_{CQ} = 2.40~\text{mA}\), \(V_{CEQ} = 6.71~\text{V}\)

Part (b) — load-line end points

Saturation is \(V_{CE} = 0\); cutoff is \(I_C = 0\):

\[\begin{aligned} I_{C(\text{sat})} &= \frac{V_{CC}}{R_C} = \frac{12~\text{V}}{2.2~\text{k}\Omega} = 5.455~\text{mA} \\ V_{CE(\text{cutoff})} &= V_{CC} = 12~\text{V} \end{aligned}\]

The Q-point sits at \(I_{CQ}/I_{C(\text{sat})} = 44.1\%\) of the saturation current — close to mid-line. The undistorted output swing is limited by the smaller of \(V_{CEQ}\) and \(V_{CC} - V_{CEQ}\):

Load line runs from \((12~\text{V},\,0)\) to \((0,\,5.45~\text{mA})\); maximum symmetrical swing \(\approx 5.29~\text{V}\) peak.

Part (c) — the same circuit with \(\beta = 200\)

\(I_B\) does not change, so \(I_C\) scales directly with \(\beta\):

\[\begin{aligned} I_C' &= 200 \times 24.04~\mu\text{A} = 4.809~\text{mA} \\ V_{CE}' &= 12 - (4.809~\text{mA})(2.2~\text{k}\Omega) = 1.42~\text{V} \end{aligned}\]

\(I_C\) changes by 100\% and \(V_{CE}\) falls to \(1.42~\text{V}\) — the stage is driven to the edge of saturation.

Doubling \(\beta\) doubles \(I_C\) exactly, because nothing in the base loop feeds back from the collector. That 1:1 sensitivity is why fixed bias is never used where \(\beta\) spread or temperature matters.

PROBLEM 02

Emitter-Bias Q-Point and Beta Sensitivity

Problem Statement

An emitter-biased stage runs from \(V_{CC} = 20~\text{V}\). A 430 k\(\Omega\) resistor connects the +20 V rail to the base, a 2 k\(\Omega\) resistor connects the rail to the collector, and a 1 k\(\Omega\) resistor connects the emitter to ground. \(\beta = 50\), \(V_{BE} = 0.7~\text{V}\).

Find \(I_B\), \(I_C\), \(I_E\) and \(V_{CE}\).
Find the node voltages \(V_B\), \(V_E\) and \(V_C\).
Repeat (a) for \(\beta = 100\) and compare the \(I_C\) sensitivity with fixed bias.

Solution

\(V_{CC} = 20~\text{V}\), \(R_B = 430~\text{k}\Omega\), \(R_C = 2~\text{k}\Omega\), \(R_E = 1~\text{k}\Omega\), \(\beta = 50\)
Active region, constant \(\beta\), \(I_E = (\beta+1)I_B\).

Part (a) — base loop with the emitter resistor reflected

Walking the base loop, \(R_E\) is crossed by \(I_E = (\beta+1)I_B\), so it appears in the base circuit as \((\beta+1)R_E\):

\[\begin{aligned} I_B &= \frac{V_{CC} - V_{BE}}{R_B + (\beta+1)R_E} = \frac{19.3~\text{V}}{430~\text{k}\Omega + 51(1~\text{k}\Omega)} = \frac{19.3}{481.00~\text{k}\Omega} = 40.12~\mu\text{A} \\ I_C &= \beta I_B = 2.006~\text{mA}, \qquad I_E = (\beta+1)I_B = 2.046~\text{mA} \\ V_{CE} &= V_{CC} - I_C R_C - I_E R_E = 20 - 4.012 - 2.046 = 13.94~\text{V} \end{aligned}\]

\(I_B = 40.1~\mu\text{A}\), \(I_C = 2.01~\text{mA}\), \(I_E = 2.05~\text{mA}\), \(V_{CE} = 13.9~\text{V}\)

Part (b) — node voltages

\[\begin{aligned} V_E &= I_E R_E = (2.046~\text{mA})(1~\text{k}\Omega) = 2.05~\text{V} \\ V_B &= V_E + V_{BE} = 2.05 + 0.7 = 2.75~\text{V} \\ V_C &= V_{CC} - I_C R_C = 20 - 4.012 = 15.99~\text{V} \end{aligned}\]

\(V_B = 2.75~\text{V}\), \(V_E = 2.05~\text{V}\), \(V_C = 16.0~\text{V}\) (check: \(V_C - V_E = 13.9~\text{V} = V_{CE}\)).

Part (c) — \(\beta = 100\)

\[\begin{aligned} I_B &= \frac{19.3}{430~\text{k} + 101(1~\text{k})} = 36.35~\mu\text{A} \\ I_C &= 3.635~\text{mA}, \qquad V_{CE} = 9.060~\text{V} \end{aligned}\]

Quantity	\(\beta = 50\)	\(\beta = 100\)	Change
\(I_C\) (mA)	2.01	3.63	\(+81.2\%\)
\(V_{CE}\) (V)	13.9	9.06	\(-4.88~\text{V}\)

Doubling \(\beta\) raises \(I_C\) by only 81.2\%, against 100\% for fixed bias.

The emitter resistor provides series current feedback: a rise in \(I_C\) raises \(V_E\), which cuts \(V_{BE}\) and pushes \(I_B\) back down. The larger \((\beta+1)R_E\) is compared with \(R_B\), the stiffer that feedback.

PROBLEM 03

Voltage-Divider Bias: Exact versus Approximate

Problem Statement

A universal-bias stage has \(V_{CC} = 22~\text{V}\), divider resistors \(R_1 = 39~\text{k}\Omega\) (rail to base) and \(R_2 = 3.9~\text{k}\Omega\) (base to ground), \(R_C = 10~\text{k}\Omega\) and \(R_E = 1.5~\text{k}\Omega\). The transistor is silicon with \(\beta = 140\).

Solve the Q-point exactly using a Thevenin equivalent of the divider.
Solve it again by the approximate (stiff-divider) method, after checking that the method is legal.
Tabulate the two results and quote the percentage error.

Solution

\(V_{CC} = 22~\text{V}\), \(R_1 = 39~\text{k}\Omega\), \(R_2 = 3.9~\text{k}\Omega\)
\(R_C = 10~\text{k}\Omega\), \(R_E = 1.5~\text{k}\Omega\), \(\beta = 140\), \(V_{BE} = 0.7~\text{V}\)

Part (a) — exact solution via Thevenin

Look back from the base with \(V_{CC}\) shorted to ground for the resistance, and open-circuit for the voltage:

\[\begin{aligned} R_{Th} &= R_1 \parallel R_2 = \frac{(39)(3.9)}{39 + 3.9}~\text{k}\Omega = 3.5455~\text{k}\Omega \\ E_{Th} &= \frac{R_2}{R_1 + R_2} V_{CC} = \frac{3.9}{42.9}(22~\text{V}) = 2.000~\text{V} \end{aligned}\]

The base loop is now identical in form to emitter bias:

\[\begin{aligned} I_B &= \frac{E_{Th} - V_{BE}}{R_{Th} + (\beta+1)R_E} = \frac{1.300~\text{V}}{3.545~\text{k} + 141(1.5~\text{k})} = 6.045~\mu\text{A} \\ I_C &= \beta I_B = 0.8463~\text{mA}, \qquad I_E = 0.8524~\text{mA} \\ V_{CE} &= V_{CC} - I_C R_C - I_E R_E = 22 - 8.463 - 1.279 = 12.26~\text{V} \end{aligned}\]

Exact: \(I_{CQ} = 0.846~\text{mA}\), \(V_{CEQ} = 12.26~\text{V}\)

Part (b) — approximate method

The approximation treats the divider as unloaded, which is safe when the reflected input resistance swamps \(R_2\), i.e. \(\beta R_E \ge 10R_2\):

\[\beta R_E = 140 \times 1.5~\text{k}\Omega = 210.0~\text{k}\Omega \;\ge\; 10R_2 = 39.0~\text{k}\Omega \quad \checkmark\]

\[\begin{aligned} V_B &\approx \frac{R_2}{R_1+R_2}V_{CC} = 2.000~\text{V} \\ V_E &= V_B - V_{BE} = 2.000 - 0.7 = 1.300~\text{V} \\ I_E &= \frac{V_E}{R_E} = \frac{1.300~\text{V}}{1.5~\text{k}\Omega} = 0.8667~\text{mA} \approx I_C \\ V_{CE} &= V_{CC} - I_E(R_C + R_E) = 22 - (0.8667~\text{mA})(11.5~\text{k}\Omega) = 12.03~\text{V} \end{aligned}\]

Approximate: \(I_{CQ} \approx 0.867~\text{mA}\), \(V_{CEQ} \approx 12.03~\text{V}\)

Part (c) — comparison

Quantity	Exact (Thevenin)	Approximate	Error
\(I_{CQ}\) (mA)	0.8463	0.8667	\(+2.40\%\)
\(V_{CEQ}\) (V)	12.26	12.03	\(-1.83\%\)

The approximation is within 2.40\% on \(I_C\) and 1.83\% on \(V_{CE}\) — entirely adequate for design.

Note that the exact \(I_C\) is the smaller one: real base current loads the divider and pulls \(V_B\) down. The error grows quickly once \(\beta R_E\) drops towards \(R_2\).

PROBLEM 04

Collector-Feedback Bias and Its Self-Correction

Problem Statement

In a collector-feedback stage the base resistor returns to the collector rather than to the supply. Here \(V_{CC} = 10~\text{V}\), \(R_F = 250~\text{k}\Omega\) from collector to base, \(R_C = 4.7~\text{k}\Omega\) from the rail to the collector and \(R_E = 1.2~\text{k}\Omega\) from emitter to ground. \(\beta = 90\), \(V_{BE} = 0.7~\text{V}\).

Derive the base-loop equation and find \(I_B\), \(I_C\) and \(V_{CE}\).
Recompute for \(\beta = 180\) and quote the percentage change in \(I_C\).
Explain in one sentence the physical mechanism that limits that change.

Solution

\(V_{CC} = 10~\text{V}\), \(R_F = 250~\text{k}\Omega\), \(R_C = 4.7~\text{k}\Omega\), \(R_E = 1.2~\text{k}\Omega\)
\(I_B \ll I_C\), so the current in \(R_C\) is taken as \(I_C\) and \(I_E \approx I_C\).

Part (a) — base loop

Start at ground, up through \(R_E\), across the junction, back through \(R_F\), through \(R_C\) to the rail. The current in \(R_C\) is \(I_C + I_B \approx I_C\):

\[\begin{aligned} V_{CC} &= I_C R_C + I_B R_F + V_{BE} + I_E R_E \\ &\approx \beta I_B R_C + I_B R_F + V_{BE} + \beta I_B R_E \\ \Rightarrow\; I_B &= \frac{V_{CC} - V_{BE}}{R_F + \beta(R_C + R_E)} \end{aligned}\]

\[\begin{aligned} I_B &= \frac{9.3~\text{V}}{250~\text{k} + 90(5.9~\text{k})} = \frac{9.3}{781.00~\text{k}\Omega} = 11.91~\mu\text{A} \\ I_C &= \beta I_B = 1.072~\text{mA} \\ V_{CE} &= V_{CC} - I_C(R_C + R_E) = 10 - (1.072~\text{mA})(5.9~\text{k}\Omega) = 3.677~\text{V} \end{aligned}\]

\(I_B = 11.9~\mu\text{A}\), \(I_C = 1.07~\text{mA}\), \(V_{CE} = 3.68~\text{V}\)

Part (b) — \(\beta = 180\)

\[\begin{aligned} I_B &= \frac{9.3}{250~\text{k} + 180(5.9~\text{k})} = 7.088~\mu\text{A} \\ I_C &= 1.276~\text{mA}, \qquad V_{CE} = 2.472~\text{V} \end{aligned}\]

\(I_C\) rises by only 19.1\% when \(\beta\) doubles.

Part (c) — why

The base resistor is fed from the collector, not the supply. If \(I_C\) tries to rise, \(V_C\) falls, which reduces the voltage across \(R_F\) and therefore reduces \(I_B\) — negative voltage feedback around the device.

Sanity check on the denominator: \(\beta(R_C+R_E) = 531.0~\text{k}\Omega\) is already comparable with \(R_F = 250~\text{k}\Omega\), so \(\beta\) largely cancels between numerator and denominator. Make \(R_F\) small relative to \(\beta(R_C+R_E)\) and the bias becomes almost \(\beta\)-independent.

PROBLEM 05

Stability Factor and Leakage-Driven Q-Point Drift

Problem Statement

The stability factor \(S(I_{CO}) = \partial I_C / \partial I_{CO}\) measures how strongly the reverse saturation current pushes the Q-point around. Compare two silicon stages, both with \(\beta = 100\):

Circuit A — fixed bias, base resistor returned to \(V_{CC}\), emitter grounded.
Circuit B — voltage-divider bias with \(R_{Th} = 3.545~\text{k}\Omega\) and \(R_E = 1.5~\text{k}\Omega\).

Evaluate \(S(I_{CO})\) for each circuit.
\(I_{CO} = 0.5~\mu\text{A}\) at 25\(^\circ\)C and doubles for every 10\(^\circ\)C rise. Find \(\Delta I_C\) in each circuit when the junction reaches 65\(^\circ\)C.
State which circuit you would put into production and why.

Solution

\(\beta = 100\); Circuit B: \(R_{Th} = 3.545~\text{k}\Omega\), \(R_E = 1.5~\text{k}\Omega\)
\(I_{CO}(25^\circ\text{C}) = 0.5~\mu\text{A}\), doubling every 10\(^\circ\)C
Constant \(\beta\) and constant \(V_{BE}\); only \(I_{CO}\) is allowed to vary.

Part (a) — the two stability factors

The general result for the emitter-biased/divider topology is

\[S(I_{CO}) = (\beta+1)\,\frac{1 + R_{Th}/R_E}{(\beta+1) + R_{Th}/R_E}\]

Fixed bias is the limit \(R_E \to 0\), i.e. \(R_{Th}/R_E \to \infty\), which gives \(S = \beta+1\):

\[\begin{aligned} S_A &= \beta + 1 = 101 \\ \frac{R_{Th}}{R_E} &= \frac{3.545}{1.5} = 2.364 \\ S_B &= 101 \times \frac{1 + 2.364}{101 + 2.364} = 101 \times \frac{3.364}{103.36} = 3.287 \end{aligned}\]

\(S_A = 101\) (fixed bias), \(S_B = 3.29\) (voltage divider).

Part (b) — leakage rise and the resulting collector-current drift

\[\begin{aligned} I_{CO}(65^\circ\text{C}) &= I_{CO}(25^\circ\text{C}) \times 2^{(T_2-T_1)/10} = 0.5~\mu\text{A} \times 2^{4} = 8.00~\mu\text{A} \\ \Delta I_{CO} &= 8.00 - 0.5 = 7.50~\mu\text{A} \end{aligned}\]

\[\begin{aligned} \Delta I_{C,A} &= S_A\,\Delta I_{CO} = 101 \times 7.50~\mu\text{A} = 0.7575~\text{mA} \\ \Delta I_{C,B} &= S_B\,\Delta I_{CO} = 3.287 \times 7.50~\mu\text{A} = 24.7~\mu\text{A} \end{aligned}\]

\(\Delta I_C = 0.757~\text{mA}\) for fixed bias against only \(24.7~\mu\text{A}\) for the divider — a factor of 30.7 better.

Part (c) — verdict

Circuit B. A drift of 0.757 mA is the same order as a typical quiescent current and would move the Q-point by volts; 24.7 \(\mu\)A is negligible. Note that \(S_B\) is set almost entirely by the ratio \(R_{Th}/R_E\), not by \(\beta\): keeping \(R_{Th}\) small (a stiff divider) and \(R_E\) large is what buys the stability, and it is exactly the same design choice that made the approximate analysis valid.

PROBLEM 06

Design of a Voltage-Divider Bias Network

Problem Statement

Design. From a \(V_{CC} = 12~\text{V}\) supply, bias a silicon transistor of \(\beta = 100\) at \(I_{CQ} = 2~\text{mA}\), \(V_{CEQ} = 6~\text{V}\) using a four-resistor voltage-divider network. Use the standard rules of thumb \(V_E \approx 0.1V_{CC}\) and \(R_2 \le \beta R_E/10\), and round every resistor to the nearest E24 value. Then verify the design exactly and quote the stability factor you have bought.

Solution

Target: \(I_{CQ} = 2~\text{mA}\), \(V_{CEQ} = 6~\text{V}\), \(V_{CC} = 12~\text{V}\), \(\beta = 100\), \(V_{BE} = 0.7~\text{V}\)
Rules: \(V_E = 0.1V_{CC}\); divider bleeder current \(\ge 10 I_B\), enforced as \(R_2 \le \beta R_E/10\).

Step 1 — emitter resistor from the 10\% rule

\[\begin{aligned} V_E &= 0.1 V_{CC} = 1.20~\text{V} \\ I_E &= \frac{\beta+1}{\beta} I_C = \frac{101}{100}(2~\text{mA}) = 2.020~\text{mA} \\ R_E &= \frac{V_E}{I_E} = \frac{1.20~\text{V}}{2.020~\text{mA}} = 594.1~\Omega \;\Rightarrow\; R_E = 620~\Omega\ \text{(E24)} \end{aligned}\]

Step 2 — collector resistor from the collector loop

\[\begin{aligned} V_{CC} &= I_C R_C + V_{CE} + V_E \\ R_C &= \frac{V_{CC} - V_{CE} - V_E}{I_C} = \frac{12 - 6 - 1.20}{2~\text{mA}} = 2400~\Omega \;\Rightarrow\; R_C = 2.4~\text{k}\Omega\ \text{(E24)} \end{aligned}\]

Step 3 — the divider

The base must sit at \(V_B = V_E + V_{BE} = 1.90~\text{V}\). The 1/10 rule caps \(R_2\):

\[\begin{aligned} R_2 &\le \frac{\beta R_E}{10} = \frac{100 \times 620~\Omega}{10} = 6.200~\text{k}\Omega \;\Rightarrow\; R_2 = 6.2~\text{k}\Omega \\ I_2 &= \frac{V_B}{R_2} = \frac{1.90~\text{V}}{6.2~\text{k}\Omega} = 306.5~\mu\text{A} \\ I_B &= \frac{I_C}{\beta} = 20.0~\mu\text{A}, \qquad I_1 = I_2 + I_B = 326.5~\mu\text{A} \\ R_1 &= \frac{V_{CC} - V_B}{I_1} = \frac{10.1~\text{V}}{326.5~\mu\text{A}} = 30.94~\text{k}\Omega \;\Rightarrow\; R_1 = 30~\text{k}\Omega\ \text{(E24)} \end{aligned}\]

\(R_1 = 30~\text{k}\Omega\), \(R_2 = 6.2~\text{k}\Omega\), \(R_C = 2.4~\text{k}\Omega\), \(R_E = 620~\Omega\)

Step 4 — exact verification with the chosen E24 values

\[\begin{aligned} R_{Th} &= 30~\text{k} \parallel 6.2~\text{k} = 5.1381~\text{k}\Omega, \qquad E_{Th} = 2.055~\text{V} \\ I_B &= \frac{1.355}{5.138~\text{k} + 101(0.62~\text{k})} = 20.00~\mu\text{A} \\ I_C &= 2.000~\text{mA}, \qquad V_{CE} = 12 - 4.800 - 1.252 = 5.947~\text{V} \end{aligned}\]

Quantity	Specified	Achieved	Error
\(I_{CQ}\) (mA)	2.000	2.000	\(0.0064\%\)
\(V_{CEQ}\) (V)	6.000	5.947	\(-0.880\%\)

\[S(I_{CO}) = 101\,\frac{1 + 8.287}{101 + 8.287} = 8.583\]

Design meets \(I_C\) to 0.0064\% and \(V_{CE}\) to 0.880\%, with \(S(I_{CO}) = 8.58\).

The E24 rounding of \(R_E\) from 594.1 \(\Omega\) to 620 \(\Omega\) is absorbed almost perfectly because \(R_1\) was then computed from the actual \(V_B\) — always propagate the rounded value forward rather than restarting from the ideal one.