BJT Biasing and Stability Questions and Answers

Answer: A

The ratio fixes their proportion and the parallel value fixes their size:

\[ \begin{aligned} \frac{R_{2}}{R_{1}+R_{2}} &= \frac{2}{12} \\ R_{1} \| R_{2} &= 15 \ \mathrm{k\Omega} \end{aligned} \]

which gives about 90 k and 18 k, the larger arm to the supply. Reversing them would put the base at ten volts. Two conditions determine two unknowns, and setting both up before solving avoids the guesswork.

Answer: B

A stiff divider is a voltage source, so the base voltage is what it holds constant, and the emitter resistor then converts that into a current. The base current is whatever the device happens to draw and is exactly the beta-dependent quantity the arrangement is avoiding. Specifying the quantity a circuit actually controls, rather than one that follows from it, is what keeps a design's assumptions visible.

Answer: C

The emitter current is set by the base voltage and the emitter resistor, so almost nothing changes - only a very slight rise as the reduced base current loads the divider a little less. Doubling the collector current is what fixed bias would do. The stage certainly does not saturate, and the base voltage rise is real but tiny. That a design's behaviour survives a twofold change in its least reliable parameter is the practical test of whether it is beta-independent.

Answer: D

The emitter voltage divided by the emitter resistor gives the collector current directly, which is the quantity the design specifies. The base-emitter drop is nearly always about 0.7 V and says little, in-circuit resistance measurements are unreliable, and total supply current includes the divider. Choosing the measurement that converts in one step into the designed quantity is worth a moment's thought before any probe is placed.

Answer: B

Use the Thevenin equivalent of the base divider, with \(V_{CC} = 15\) V, \(R_1 = 6\ \text{k}\Omega\) and \(R_2 = 3\ \text{k}\Omega\).

\[ \begin{aligned} V_{TH} = 15 \times \frac{3}{6 + 3} \\ &= 5\ \text{V},\qquad R_{TH} = 6 \parallel 3 \\ &= 2\ \text{k}\Omega \end{aligned} \]

The base loop equation with \(I_E = (\beta + 1) I_B\) is

\[ 5 = I_B R_{TH} + 0.7 + (\beta + 1) I_B R_E \]

\[ \begin{aligned} I_B = \frac{4.3}{2 + 101 \times 1}\ \text{mA} \\ &= 0.0417\ \text{mA} \end{aligned} \]

\[ \begin{aligned} I_C = \beta I_B = 100 \times 0.0417 \\ &= 4.17 \approx 4.2\ \text{mA} \end{aligned} \]

The answer is 4.2 mA, option B. The other values are off by a factor of ten.