Question 1

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

Accepted Answer

A) True 
 B)False

Question 2

In Figure 7-18,the negative supply

Accepted Answer

A)  forward-biases the emitter diode. 
B)  forward-biases the collector diode. 
C)  reverse-biases the emitter diode. 
D)  reverse-biases the collector diode. 
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 3

In Figure 7-5,if the fuse blows open

Accepted Answer

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. 
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 4

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

Accepted Answer

A) True 
 B)False

Question 5

The schematic diagram of a pnp transistor shows

Accepted Answer

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. 
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 6

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

Accepted Answer

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 RC by the sum of RE and RC and then multiply by 10 V. 
D)  divide RE by the sum of RE and RC and then multiply by 10 V. 
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 RC by the sum of RE and RC and then multiply by 10 V. 
D)  divide RE by the sum of RE and RC and then multiply by 10 V.

Question 7

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

Accepted Answer

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. 
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 8

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

Accepted Answer

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

Question 9

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

Accepted Answer

A) True 
 B)False

Question 10

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

Accepted Answer

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

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

In Figure 7-18,the negative supply

In Figure 7-5,if the fuse blows open

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

The schematic diagram of a pnp transistor shows

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

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

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

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

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

Introduction

Semoconductors

Diode Theory Key

Diode Circuits

Special-Purpose Diodes Key

BJT Fumdamentals

Basic BJT Amplifiers

Multistage,CC,and CB Amplifiers

Power Amplifiers

JFETS

Mosfets

Thyristors

Frequency Effects

Differential Amplifiers

Operational Amplifiers

Negative Feedback

Linear OP AMP Circuits

Active Filters Key

Nonlinear Op-Amp Circuit Applications Key

Oscillators Key

Regulated Power Supplies Key

Filters

Exam 7: BJT Biasing

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

In Figure 7-18,the negative supply

In Figure 7-5,if the fuse blows open

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

The schematic diagram of a pnp transistor shows

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

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

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

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

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

Introduction

Semoconductors

Diode Theory Key

Diode Circuits

Special-Purpose Diodes Key

BJT Fumdamentals

Basic BJT Amplifiers

Multistage,CC,and CB Amplifiers

Power Amplifiers

JFETS

Mosfets

Thyristors

Frequency Effects

Differential Amplifiers

Operational Amplifiers

Negative Feedback

Linear OP AMP Circuits

Active Filters Key

Nonlinear Op-Amp Circuit Applications Key

Oscillators Key

Regulated Power Supplies Key

Filters

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