IFC page 8 and 9

lebelehi's version from 2017-01-13 14:38


Question Answer
Ideally, a dc load line is a straight line drawn on the collector characteristic curves betweenVCE (cutoff) and IC (sat)
If a sinusoidal voltage is applied to the base of a biased npn transistor and the resulting sinusoidal collector voltage is clipped near zero volts, the transistor isBeing driven into saturation. Operating nonlinearly
The input resistance at the base of a biased transistor depends mainly onβDC and RE
In a voltage-divider biased transistor circuit such as in fig 5-13, R IN(BASE) can generally be neglected in calculations whenR IN(BASE) > 10 R2
In a certain voltage-divider biased npn transistor, VB is 2.95 V. The dc emitter voltage is approximately2.25 V
Voltage - divider bias Can be essentially independent of βDC
Emitter bias is essentially independent of βDC Provide a stable bias point
The disadvantage of base bias is thatIt is too beta dependent
Collector-feedback bias isBased on the principle of negative feedback
In a voltage-divider biased npn transistor, if the upper voltage-divider resistor (the one connected to VCC) opens,The transistor goes into cutoff
In a voltage-divider biased npn transistor, if the lower voltage-divider resistor (the one connected to ground) opens,The transistor may be driven into saturation
In a voltage-divider biased npn transistor, there is no base current, but the base voltage is approximately correct. The most likely problem(s) isThe base-emitter junction is open. The emitter resistor is open
A small-signal amplifierUses only a small portion of its load line
The parameter hfe corresponds toβac
If the dc emitter current in a certain transistor amplifier is 3 mA, the approximate value of r’e is8.33 Ω
A certain common-emitter amplifier has a voltage gain of 100. If the emitter bypass capacitor is removed,The voltage gain will decrease
The transistors in a class B amplifier are biasedat midpoint of the load line
For a common-emitter amplifier, RC = 1.0 kΩ, RE = 390Ω, r’e = 15Ω, and βac = 75. Assuming that RE is completely bypassed at the operating frequency, the voltage gain is66.7
In the circuit of Question 118, if the frequency is reduced to the point where XC(bypass) = RE, the voltage gainis less
In a common-emitter amplifier with voltage- divider bias, Rin(base) = 68 kΩ, R1 = 33 kΩ, and R2 = 15kΩ. The total input resistance is8.95 kΩ
A CE amplifier is driving a 10 kΩ load. If RC = 2.2 kΩ and r’e = 10Ω, the voltage gain is approximately180
For a common-collector amplifier, RE = 100 Ω, r’e = 10 Ω, and βac = 150. The ac input resistance at the base is16.5 kΩ
If a 10 mV signal is applied to the base of the emitter-follower circuit in Question 122, the output signal is approximately10 mV
In a certain emitter-follower circuit, the current gain is 50. The power gain is approximately50Av
In a Darlington pair configuration, each transistor has an ac beta of 125. If RE is 560Ω, the input resistance is8.75 MΩ
The input resistance of a common-base amplifier isvery low
Each stage of a four-stage amplifier has a voltage gain of 15. The overall voltage gain is50,625
The overall gain found in Question 127 can be expressed in decibels as94.1 dB
A differential amplifieris used in op-amps and has one input and one output
When a differential amplifier is operated single- ended,one input is grounded and a signal is applied to the other
In the double-ended differential mode,opposite polarity signals are applied to the inputs
In the common mode,an identical signal appears on both inputs *
An amplifier that operates in the linear region at all times isClass A
A certain class A power amplifier delivers 5 W to a load with an input signal power of 100 mW. The power gain is50
The peak current a class A power amplifier can deliver to a load depends on thequiescent current
For maximum output, a class A power amplifier must maintain a value of quiescent current that isone-half the peak load current
A certain class A power amplifier is the ratio of the power delivered to the load to thepower from the dc power supply
The maximum efficiency of a class A power amplifier is25%
Crossover distortion is a problem forclass B amplifiers
A BJT class B push-pull amplifier with no transformer coupling usescomplementary symmetry transistors
A current mirror in a push-pull amplifier should give an ICQ that isequal to the current in the bias resistors and diodes
The maximum efficiency of a class B push-pull amplifier is79%
The output of a certain two-supply class B push-pull amplifier has a VCC of 20V. If the load resistance is 50Ω, the value of IC(sat) is0.4 A
The maximum efficiency of a class AB amplifier isslightly less than a class B
The power dissipation of a class C amplifier is normallyvery low
The efficiency of a class C amplifier isgreater than classes A, B, or AB
The transistor in a class C amplifier conducts fora very small percentage of the input cycle
The JFET isa unipolar device. a voltage-controlled device
The channel of a JFET is between thedrain and source
A JFET always operates withthe gate-to-source pn junction reverse- biased
For VGS = 0 V, the drain current becomes constant when VDS exceedsVP
The constant-current region of a FET lies betweenpinch-off and breakdown
IDSS isthe maximum possible drain current
Drain current in the constant-current region increases when thegate-to-source bias voltage decreases
In a certain FET circuit, VGS = 0 V, VDD = 15 V, IDSS = 15 mA, and RD = 470 Ω. If RD is decreased to 330 Ω, IDSS is15 mA
At cutoff, the JFET channel iscompletely closed by the depletion region
A certain JFET datasheet VGS(off) = - 4V. The pinch-off voltage, VP, is +4V
The JFET in Question an n-channel
For a certain JFET, IGSS = 10 nA at VGS = 10 V. The input resistance is1000 MΩ
For a certain p-channel JFET, VGS(off) = 8V. The value of VGS for an approximate midpoint bias is2.34 V