Deck 7: BJT Biasing

Question 1

Emitter bias is used to stabilize the Q point on the load line against current gain changes.

Answer 1

True

Question 2

Collector-feedback bias is still sensitive to changes in current gain,but it is used because of its

A) small size.
B) low cost.
C) simplicity.
D) complexity.

Answer 2

C

Question 3

A firm voltage divider means that the collector current will be approximately 10 percent lower than the stiff value.

Answer 3

True

Question 4

With the switch in the position as shown in Figure 7-4 (a),the LED is on.

Question 5

If R_E is adjusted to its maximum value in Figure 7-16,the Q point moves toward saturation.

Question 6

Emitter-feedback bias uses a resistor that provides negative feedback in attempt to stabilize the operating point.

Question 7

Emitter-feedback bias is also called self-bias.

Question 8

Emitter-feedback and collector-feedback are typically not adequate because

A) there is too much negative feedback.
B) there is not enough negative feedback.
C) circuitry is too complex.
D) there is not enough positive feedback.

Question 9

The emitter-base junction voltage of the silicon transistor shown in Figure 7-12 is approximated to be 0.3 V.

Question 10

In the circuit shown in Figure 7-26,if the base voltage measured 0 V and the emitter voltage measured 0 V,what should the troubleshooter do?

A) replace the transistor
B) replace R_E
C) check R₁
D) check voltage source and leads

Question 11

In the circuit shown in Figure 7-26,if R₁ opens

A) the collector voltage will be 10 V.
B) the collector voltage will be 0 V.
C) the base voltage will be 10 V.
D) the emitter voltage will be 10 V.

Question 12

In the circuit shown in Figure 7-12,the first step in finding the emitter current is to determine the voltage across R₂.

Question 13

Voltage-divider bias is really emitter bias in disguise.

Question 14

Collector-feedback bias is also called

A) self-bias.
B) collector follower.
C) negative collector.
D) voltage-divider.

Question 15

A well-designed voltage-divider bias circuit is one in which the voltage divider appears stiff to the input resistance of the base.

Question 16

Since voltage-divider bias is derived from emitter bias,the Q point is virtually immune to changes in current gain.

Question 17

Some electronic equipment has a power supply that produces both positive and negative supply voltages.

Question 18

The emitter current in the circuit shown in Figure 7-12 can be calculated by dividing +10 V by 1 kW.

Question 19

If R_E is adjusted it its minimum value in Figure 7-16,the Q point moves toward cut-off.

Question 20

With the values shown at the operating point labeled Q,the voltage across R_C would be 10 V.

Question 21

A well-designed voltage-divider bias circuit is one in which the voltage divider appears

A) stiff to the input resistance of the base.
B) soft to the input resistance of the base.
C) stiff to the output resistance of the collector.
D) stiff to the input resistance of the emitter.

Question 22

To prevent meter loading,a voltmeter used to make transistor circuit voltage measurements should have an input impedance of ________.

A) 10 kW
B) 100 kW
C) 1 MW
D) 10 MW

Question 23

Collector-feedback bias is also called

A) collector-follower bias.
B) emitter-feedback bias.
C) voltage-divider bias.
D) self-bias.

Question 24

What type transistor is used when the electronics equipment has a negative power supply?

A) field effect
B) small-signal
C) npn
D) pnp

Question 25

In Figure 7-18,the positive supply

A) forward-biases the collector diode.
B) reverse-biases the collector diode.
C) forward-biases the emitter diode.
D) reverse-biases the emitter diode.

Question 26

In Figure 7-18,the negative supply

A) forward-biases the emitter diode.
B) forward-biases the collector diode.
C) reverse-biases the emitter diode.
D) reverse-biases the collector diode.

Question 27

The intent of emitter-feedback bias is to

A) reduce the source voltage.
B) swamp out the variations in current gain.
C) swamp out the base current.
D) increase current gain.

Question 28

While troubleshooting the circuit shown in Figure 7-26,the base voltage measures 1.8 V and the collector voltage measures 10 V,what should the technician do?

A) replace the transistor
B) replace R_E
C) check R₁
D) check voltage source and leads

Question 29

By exposing a transistor's collector junction to light,a manufacturer can produce

A) an LED.
B) a seven-segment display.
C) a common-emitter circuit.
D) a phototransistor.

Question 30

Using a firm voltage divider means that the collector current will be approximately

A) 5 percent lower than the stiff value.
B) 10 percent lower than the stiff value.
C) 50 percent lower than the stiff value.
D) 10 percent higher than the stiff value.

Question 31

The most widely used transistor biasing circuit is called

A) negative bias.
B) voltage-multiplier bias.
C) voltage-divider bias.
D) full-wave bridge bias.

Question 32

Emitter-feedback and collector-feedback bias circuits are inadequate because

A) there is not enough negative feedback.
B) there is too much negative feedback.
C) circuits are too complex.
D) there is not enough positive feedback.

Question 33

Emitter-feedback bias has never become popular because the movement of the Q point is

A) stationary at cutoff.
B) stationary at saturation.
C) too large for applications that must be mass-produced.
D) too small for applications that must be mass-produced.

Question 34

In Figure 7-5,if the fuse blows open

A) the transistor is biased on and the green LED illuminates.
B) the transistor is biased off and the green LED illuminates.
C) the transistor is biased on and the red LED illuminates.
D) the transistor is biased off and the red LED illuminates.

Question 35

The biasing configuration shown in Figure 7-18 is derived from emitter bias and is referred to as

A) emitter derived bias (EDB).
B) bi-polar emitter bias (BPEB).
C) two-supply emitter bias (TSEB).
D) one-supply emitter bias (OSEB).

Question 36

In Figure 7-5,when the fuse is not blown

A) the transistor is biased on and the green LED illuminates.
B) the transistor is biased off and the green LED illuminates.
C) the transistor is biased on and the red LED illuminates.
D) the transistor is biased off and the red LED illuminates.

Question 37

The main advantage of an optocoupler is

A) electrical isolation between input and output circuits.
B) electrical connection between input and output circuits.
C) inexpensive manufacturing costs.
D) miniature size.

Question 38

Voltage-divider bias produces a fixed value of emitter current that results in a stable operating point that is

A) always located at cutoff.
B) always located at saturation.
C) dependent on the current gain.
D) independent of the current gain.

Question 39

With a pnp transistor,the base-emitter junction will be forward biased

A) when the emitter voltage is 0.7 V above the base voltage.
B) when the emitter voltage is 0.7 V below the base voltage.
C) when the transistor is in cut-off.
D) when the collector voltage is 0.7 V above the base voltage.

Question 40

Collector-feedback bias is more effective than emitter-feedback bias and is used in practice because of its

A) ability to change current gain.
B) low cost.
C) simplicity.
D) small size.

Question 41

The voltage measured across the base-emitter junction of the silicon transistor shown in Figure 7-29 (a)is ________.

A) 0.3 V
B) 1.3 V
C) 0.7 V
D) 1.7 V

Question 42

In order to determine the emitter current for the circuit shown in Figure 7-21,you must

A) subtract 0.7 V from 15 V and then divide by 20 kW.
B) subtract 0.7 V from 15 V and then divide by 10 kW.
C) subtract 0.7 V from 15 V and then divide by 33 kW.
D) add 0.7 V to 15 V and then divide by 20 kW.

Question 43

The schematic diagram of a pnp transistor shows

A) the emitter as an arrow pointing in.
B) the emitter as an arrow pointing out.
C) the collector as an arrow pointing in.
D) the collector as an arrow pointing out.

Question 44

In Figure 7-21,what effect do the capacitors have on the bias voltage in each stage?

A) increase
B) decrease
C) causes a cutoff condition
D) no effect

Question 45

In order to calculate the base voltage for the circuit shown in Figure 7-29 (b)

A) divide R₂ by the sum of R₁ and R₂ and then multiply by 10 V.
B) divide R₁ by the sum of R₁ and R₂ and then multiply by 10 V.
C) divide R_C by the sum of R_E and R_C and then multiply by 10 V.
D) divide R_E by the sum of R_E and R_C and then multiply by 10 V.

Question 46

With ambiguous troubles,the troubleshooter very often must

A) replace all suspected components.
B) memorize the circuit diagram.
C) disconnect one suspected component and test it.
D) disconnect all suspected components and apply power.

Question 47

What should a troubleshooter do if the base,collector,and emitter voltages in a transistor circuit all measure 0 V?

A) replace the transistor
B) check supply voltage and leads
C) disconnect base resistors
D) check collector resistor

Question 48

Once the base voltage has been calculated for the circuit shown in Figure 7-29 (b),emitter voltage can be calculated by

A) adding 0.7 V to the base voltage.
B) subtracting 0.7 V from the base voltage.
C) adding 0.3 V to the base voltage.
D) subtracting 0.3 V from the base voltage.

Question 49

In the circuit shown in Figure 7-29 (a),the base voltage can be calculated by

A) using the voltage divider rule using R₁ and R₂ and the supply voltage.
B) using the voltage divider rule using R_C and R_E and the supply voltage.
C) multiplying collector current by the collector resistor.
D) multiplying emitter current by the emitter resistor.

Question 50

Whenever you have a circuit with npn transistors,you can often use the same circuit with pnp transistors and

A) a negative power supply.
B) and an ac source.
C) a positive power supply.
D) a floating ground.