A: one nucleus and one or more electrons; protons, electrons and neutrons
Q: The nucleus of an atom is made up of
Q: The atomic number of silicon is
Q: The atomic number of germanium is
Q: The valence shell in a silicon
A: in the most distant orbit from the nucleus
A: a valence electron breaks away from the atom
Q: The most widely used
semiconductive material in electronic devices is
Q: The energy bond in which free electrons exist is the
Q: Electron-hole pairs are produced by
Q: In a semiconductor crystal, the atoms are held together by
A: the interaction of valence
electrons, forces of attraction and covalent bonds
Q: Each atom in a silicon crystal has
A: eight valence electrons, four of its own and four shared
Q: The current in a
A: the free electrons are
thermally produced and there are as many electrons as there are
Q: The difference between
A: a wider energy gap between the valence bond and the conductive bond, the number of free
Q: The process of adding an impurity to an intrinsic
Q: A trivalent impurity is added to silicon to create
Q: The purpose of a pentavalent impurity is to
A: increase the number of free electrons
Q: The majority carriers in an n-type semiconductor are
Q: Holes in an n-type semiconductor are
A: minority carriers that are thermally produced
A: the boundary of a p-type and an n-type material
Q: The depletion region is created by
A: ionization, diffusion and recombination
Q: The depletion region consists of
A: positive and negative ions and no majority carriers
A: a dc voltage is applied to control the operation of the device
A: an external voltage is applied that is positive at the anode and negative at the cathode or an
external voltage is applied that is positive at the p region and
Q: When a pn junction is forward bias,
A: the current is produced by both holes and electrons
Q: Although current is blocked in reverse bias,
A: there is very small current due to minority carriers
Q: For a silicon diode, the value of the forward bias voltage is
Q: When a voltmeter is placed
across a forward-biased diode, it will read a voltage approximately equal to
Q: A silicon diode is in series with a 1 k resistor and a 5 V battery. If the anode is connected to the
positive battery terminal, the cathode voltage with respect to the negative battery terminal is
Q: The positive lead of the
ohmmeter is connected to the
anode of a diode and the negative lead is connected to the cathode. The diode is
Q: The average value of the
half-wave rectified voltage with a peak value of 200 V
Q: When a 60 Hz sinusoidal
voltage is applied to the input of a half-wave rectifier, the output
Q: The peak value of the input to half-wave rectifier is 10 V. The
approximate peak value of the output is
Q: When a 60 Hz sinusoidal
voltage is applied to the input of a full-wave rectifier the output
Q: The total secondary voltage in a center-tapped full-wave rectifier is 125 V rms. Neglecting the diode drop, the rms voltage output is
A: 62.5 V
Q: When the peak output voltage is 100 V, the PIV for each diode in a center-tapped full-wave rectifier is (neglecting the diode drop)
A: 200 V
Q: When the rms output voltage of a full-wave bridge rectifier is 20 V, the peak inverse voltage
across the diode is (neglecting the diode drop)
A: 28.3 V PIV = Vp
Q: The ideal dc output voltage of a capacitor filter is equal to
A: the peak value of the rectified voltage
Q: A certain power supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor is
A: 0.005
r = V r(p-p) / V (dc) r = 100 mV / 20 V
Q: A 60 V peak full-wave rectified voltage is applied to a capacitor
filter. If f = 120 Hz, RL = 10 k and C = 10 F, the ripple voltage is
A: 5.0 V
V r(p-p) = Vm / (fRLC)
= 60 / (120 x 10k x 10) = 5 V
Q: If the load resistance of a capacitor-filtered full-wave
rectifier is reduced, the ripple voltages
A: increases
Q: A 10 V(p-p) sinusoidal voltage is applied across a silicon diode
and series resistor. The maximum voltage across the diode is
Q: If the input voltage to a tripler has an rms value of 12 V, the dc
A: 32.4 V
Vdc = 22 V rms / or use
Q: If one of the diodes in a full wave bridge rectifier opens, the output is
Q: What happens to one of the diodes in a full-wave bridge
rectifier if it is observed that the output has a 60 Hz ripple
Q: The cathode of a Zener diode in a voltage regulator is normally
Q: If a Zener diode has a Zener voltage of 3.6 V, it operates in
Q: For a certain 12 V Zener diode, a 10 mA change in Zener current produces a 0.1 V change in Zener voltage. The Zener impedance for this current change is
A: 10
Zz = Vz / Iz = 0.1 V / 10 mA = 10
Q: The data sheet for a particular Zener gives Vz = 10 V and Izt =
A: 20
Zz = Vz / Iz = 10 V / 500 mA = 20
Q: Line regulation is determined by
A: change in output voltage and input voltage
Q: Load regulation is determined by
A: changes in load current and output voltage
Load Reg. = (Vnl – Vfl)/Vfl x 100% Therefore, load regulation is the same as voltage regulation.
A: the load has infinite resistance or the load has zero resistance
A: a variable capacitance that depends on the reverse voltage
A: emits light when forward-biased
Q: Compared with a visible red LED, an infrared LED
A: produces light with longer wavelengths
Q: The internal resistance of a photodiode
A: decreases with light intensity when reversed -biased
Q: A diode that has a negative
Q: An infrared LED is optically coupled to a photodiode. When
the LED is turned off, the reading on an ammeter in series with a
Q: In order for a system to
function properly, the various
types of circuits that make up the system must
A: properly biased, properly connected, and properly
Q: The three terminals of a
bipolar junction transistor are called
Q: In a pnp transistor, the p regions are
Q: For operation as an amplifier, the base of an npn transistor must be
A: positive with respect to the emitter
A: greater than the base current and the collector current
Q: If Ic is 50 times larger than Ib, then dc is
Q: The approximate voltage
across the forward-biased base-emitter junction of a silicon BJT is
Q: The bias condition for a
transistor to be used as a linear amplifier is
Q: If the output of a transistor
amplifier is 5 V rms and the input is 100 mV rms, the voltage gain is
Q: When operated in cutoff and
Q: Once in saturation, a further increase in base current will
Q: If the base-emitter junction is open, the collector voltage is
Q: The maximum value of a collector current in a biased transistor is
Q: Ideally, a dc load line is straight line drawn on the
collector characteristic curves between
Q: 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 is
A: being driven into saturation and operating nonlinearly
Q: the dc beta h(FE) for a given type of transistor
A: varies with temperature and from device to device
Q: The disadvantage of base bias is that
A: essentially dependent of dc and provide a stable bias point
Q: In an emitter bias circuit R(B) = 2.7 k and V(EE) = 15 V. The
A: cannot be determined unless Vcc is given
Q: The input resistance at the base of the biased transistor depends mainly on
Q: In a certain voltage divider biased npn transistor, V(B) is
2.95 V. The dc emitter voltage is approximately
A: 2.25 V
V(B) = V(BE) + V(EE)
Q: Voltage divider bias can be essentially independent of __
Q: Collector feedback is based on the principle of
Q: In a voltage-divider biased npn transistor, if the upper
voltage-divider resistor (the one
Q: In a voltage-divider biased npn transistor, if the lower
voltage-divider resistor (the one
A: the transistor may be driven into saturation
Q: A small-signal amplifier
A: uses only a small portion of its load line
Q: The parameter h(FE) corresponds to
Q: If the dc emitter current in a certain transistor amplifier is 3 mA, the approximate value of r’e is
A: 8.33
r’e = 25 mV / I(E) = 25 mV / 3 mA = 8.33
Q: For a common-collector
amplifier, R(E) = 100 , r’e = 10 , and ac = 150. The input
A: 16.5 k
Rin(base) = [R(E) + r’e] = 150(100 + 10) = 16.5 k
Q: A certain common emitter amplifier has a voltage gain of 100. If the emitter bypass
A: the voltage gain will decrease with bypass capacitor
Av = Rc / r’e
without bypass capacitor Av = Rc / [r’e + R(E)]
Q: For a common-emitter
amplifier, Rc = 1 k, R(E) = 390, r’e = 15 , and ac = 15075.
Assuming that R(E) is completely bypassed at the operating
A: 66.7
Q: In a certain common-collector circuit, the current gain is 50. The power gain is approximately
Q: In a darlington configuration, each transistor has an ac beta of 125. If R(E) is 560 , the input
A: 8.75 M
Zi = R(B) //[ri + (D)R(E)]
Zi (D)R(E) (125)2 x 560