BJT Biasing and Stability Questions and Answers

Q: A divider must be stiff against a base impedance of \( 150 \ \mathrm{k\Omega} \). Its Thevenin resistance should be no more than about

The usual criterion is a tenth of the reflected impedance:\[ \begin{aligned} R_{th} &\le \frac{150 \ \mathrm{k}}{10} \\ &= 15 \ \mathrm{k\Omega} \end{aligned} \]giving about 15 kilohms. Matching the two would leave the base voltage moving noticeably with beta, and a larger value would be worse still. A factor of ten is the conventional margin for treating one impedance as negligible beside another, and it recurs throughout circuit design.

Q: Why is the base voltage rather than the base current specified in a divider design?

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.

Q: Which single measurement best confirms a divider-biased stage is working as designed?

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.

111.

A divider must be stiff against a base impedance of \( 150 \ \mathrm{k\Omega} \). Its Thevenin resistance should be no more than about

A.150 kilohms, matching the impedance it works into
B.75 kilohms, half the impedance looking into the base
C.1.5 megohms
D.15 kilohms

Answer: D

The usual criterion is a tenth of the reflected impedance:

\[ \begin{aligned} R_{th} &\le \frac{150 \ \mathrm{k}}{10} \\ &= 15 \ \mathrm{k\Omega} \end{aligned} \]

giving about 15 kilohms. Matching the two would leave the base voltage moving noticeably with beta, and a larger value would be worse still. A factor of ten is the conventional margin for treating one impedance as negligible beside another, and it recurs throughout circuit design.

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112.

With that Thevenin resistance and a base voltage of \( 2 \ \mathrm{V} \) from a \( 12 \ \mathrm{V} \) supply, the two divider resistors are about

A.90 k and 18 k
B.18 k and 90 k, with the larger arm to ground
C.15 k and 15 k
D.60 k and 12 k, which give the right ratio but the wrong parallel value

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.

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113.

Why is the base voltage rather than the base current specified in a divider design?

A.Voltage is the quantity the emitter resistor responds to directly
B.Voltage is the quantity the divider actually fixes
C.Base current cannot be measured without breaking the circuit open
D.Base current varies with the collector-emitter voltage applied

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.

Discuss in Forum

114.

What happens to a divider-biased stage if the transistor is replaced by one of twice the beta?

A.The stage saturates because the base current has halved
B.The base voltage rises as the divider is loaded less heavily
C.Almost nothing changes
D.The collector current doubles in proportion to the gain

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.

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115.

Which single measurement best confirms a divider-biased stage is working as designed?

A.The voltage between base and emitter at the device
B.The resistance of the two divider arms in circuit
C.The current drawn by the whole stage from the supply
D.The voltage across the emitter resistor

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.

Discuss in Forum

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