Electronics is 7–11 items on the E&C specification, and Properties of Electrical Materials adds 4–6 items. They share a page because the semiconductor subtopic of the materials area is the ground floor of every device question.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Electrical and Computer
items · Electronics7–11
items · Properties of Electrical Materials4–6
problems on this page18
Checked
What Electronics covers
Five subtopics in the July 2020 specification:
Discrete devices: models, biasing and performance of diodes, transistors and thyristors.
Amplifiers: single-stage (e.g., common emitter), differential, biasing.
Operational amplifiers: ideal and nonideal behavior.
Instrumentation: measurements, data acquisition and transducers.
Power electronics: rectifiers, inverters and converters.
18 electronics and materials problems
0 / 18 answered · 0 correct
Question 1 of 18 · Properties of Electrical Materials · 4–6 items
An n-type silicon sample is doped with Nd = 1017 cm−3 (ni = 1010 cm−3). The electron mobility is 1350 cm2/V·s and the hole mobility is 480 cm2/V·s. Assuming complete ionization and q = 1.6 × 10−19 C, what is the conductivity?
This converts n to m−3 but leaves the mobility in cm2/V·s.
Question 2 of 18 · Properties of Electrical Materials · 4–6 items
A slab of n-type silicon doped to Nd = 1017 cm−3 carries I = 50 mA along its length in a perpendicular magnetic field B = 0.8 T. The slab thickness along the field is t = 0.5 mm. With q = 1.6 × 10−19 C, what Hall voltage develops?
Given
Donor concentration Nd = 1017 cm−3
Current I = 50 mA
Magnetic field B = 0.8 T
Thickness t = 0.5 mm
Elementary charge q = 1.6 × 10−19 C
Find
Hall voltage VH (V)
Solution
n = 1017 cm−3 = 1023 m−3
VH = IB/(qnt) = (0.050 A × 0.8 T)/(1.6 × 10−19 C × 1023 m−3 × 0.5 × 10−3 m) = 5.0 mV
This converts 1017 cm−3 with 103 instead of 106, giving n = 1020 m−3.
This puts the thickness in the numerator; VH = IB/(qnt) divides by t.
Question 3 of 18 · Properties of Electrical Materials · 4–6 items
A silicon sample is doped with 1016 phosphorus atoms/cm3. Given that the intrinsic carrier concentration ni = 1010 cm−3, what is the approximate hole concentration at room temperature?
Given
Donor concentration ND = 1016 cm−3
Intrinsic concentration ni = 1010 cm−3
Find
Hole concentration p (cm−3)
Solution
n · p = ni2 → p = ni2/ND
p = (1010)2/1016 = 104 cm−3
n ≈ ND = 1016 cm−3, so p = ni2/n = 1020/1016 = 104 cm−3.
This is ni, which holds only in intrinsic material; doping suppresses the minority carriers.
This is the electron (majority) concentration, not the hole concentration.
This is ni2, which has units of cm−6; it must still be divided by n.
Question 4 of 18 · Properties of Electrical Materials · 4–6 items
Silicon dioxide has a relative permittivity of about 3.9. A parallel-plate capacitor with an SiO2 dielectric has a plate area of 1 mm2 and a separation of 10 nm. What is its capacitance? Use ε0 = 8.85 × 10−12 F/m.
Given
Relative permittivity εr = 3.9
Plate area A = 1 mm2
Separation d = 10 nm
Permittivity ε0 = 8.85 × 10−12 F/m
Find
Capacitance C (nF)
Solution
C = εrε0A/d
C = 3.9 × 8.85 × 10−12 F/m × 1 × 10−6 m2/(10 × 10−9 m) = 3.45 nF
This takes 1 mm2 as 10−3 m2; 1 mm2 = 10−6 m2.
This converts 1 mm2 as 10−4 m2 (the cm2 factor), making C 100 times too large: 345 nF.
This takes 10 nm as 10−9 m; 10 nm = 10−8 m.
C = εrε0A/d = 3.9 × 8.85 × 10−12 × 10−6/10−8 = 3.45 × 10−9 F.
Question 5 of 18 · Electronics · 7–11 items
An ideal op amp is configured as an inverting amplifier with input resistor Rin = 10 kΩ and feedback resistor Rf = 100 kΩ. What is the voltage gain?
Given
Input resistor Rin = 10 kΩ
Feedback resistor Rf = 100 kΩ
Find
Voltage gain Av
Solution
Av = −Rf/Rin = −100 kΩ/10 kΩ = −10
This mixes the noninverting form 1 + Rf/Rin with the inverting sign.
Av = −Rf/Rin = −100 kΩ/10 kΩ = −10.
The magnitude is right, but an inverting amplifier's output is 180° out of phase, so the gain is negative.
This is the noninverting gain 1 + Rf/Rin; here the input drives the inverting terminal.
Question 6 of 18 · Electronics · 7–11 items
A full-wave bridge rectifier uses four identical diodes, each with a forward voltage drop of 0.7 V. If the transformer secondary provides 24 Vrms, what is the peak DC output voltage?
Given
Secondary voltage V = 24 V rms
Diode forward drop VD = 0.7 V each
Find
Peak DC output voltage (V)
Solution
Vp,out = √2·Vrms − 2VD
Vp,out = 1.414 × 24 V − 2 × 0.7 V = 32.5 V
Vp = √2 × 24 − 2 × 0.7 = 33.9 − 1.4 = 32.5 V: two diodes conduct in series on each half-cycle.
This subtracts only one diode drop; a bridge always has two diodes in the conducting path.
This is the secondary peak with no diode drops at all.
This adds the two 0.7 V drops instead of subtracting them: 33.9 + 1.4 = 35.3 V.
Question 7 of 18 · Electronics · 7–11 items
An enhancement MOSFET with VGS > Vth operates in the triode (linear) region. Which condition must also be satisfied?
Solution
triode: VGS > Vth and VDS < VGS − Vth
With the channel on (VGS > Vth), a small VDS below the overdrive keeps the channel continuous: triode region.
This is the diode-connected case; with VDS = VGS > VGS − Vth the device is in saturation.
Below threshold the MOSFET is cut off and no channel exists.
This is the saturation (active) condition, where the channel is pinched off at the drain.
Question 8 of 18 · Electronics · 7–11 items
A buck dc-dc converter operates with an input voltage of 12 V and a duty cycle of 0.5. Assuming ideal components, what is the output voltage?
Given
Input voltage Vin = 12 V
Duty cycle D = 0.5
Find
Output voltage Vout (V)
Solution
Vout = D · Vin = 0.5 × 12 V = 6 V
This is the boost relation Vin/(1 − D).
This is the input voltage; a buck converter steps down.
An ideal buck converter gives Vout = D·Vin = 0.5 × 12 = 6 V.
This is the buck-boost relation −D·Vin/(1 − D).
Question 9 of 18 · Electronics · 7–11 items
A noninverting amplifier has R1 = 2.00 kΩ from the inverting input to ground and R2 = 18.0 kΩ from the output to the inverting input. The op amp has open-loop DC gain AOL = 1.00 × 105 and a single dominant pole at fp = 10.0 Hz. What are the closed-loop gain and the approximate closed-loop −3 dB bandwidth?
This uses R2/R1 = 9, the inverting gain magnitude; the noninverting gain is 1 + R2/R1.
This keeps the open-loop pole; negative feedback widens the bandwidth by the loop gain.
This is the gain-bandwidth product itself, the bandwidth only at unity gain.
Question 10 of 18 · Electronics · 7–11 items
A power semiconductor dissipates 12.0 W in steady state. The maximum junction temperature is 150 °C and the ambient is 40.0 °C. The junction-to-case resistance is 0.800 °C/W and the case-to-sink contact resistance is 0.200 °C/W. What is the maximum allowable sink-to-ambient thermal resistance RθSA?
A silicon controlled rectifier (SCR) has been triggered into conduction. Which condition turns it OFF?
Solution
SCR turn-off: IA < IH (natural or forced commutation); the gate can turn it on but not off.
Raising the forward anode voltage keeps the device conducting; at breakover it can even turn an off SCR on.
Once latched, an SCR stays on until its anode current drops below the holding current, for example at a natural current zero.
After latching the gate loses control, so removing the gate signal leaves the SCR conducting.
More gate current only helps turn-on; a conducting SCR ignores the gate.
Question 13 of 18 · Electronics · 7–11 items
A BJT common-emitter amplifier operating in the active region has a collector resistor RC = 2 kΩ and current gain β = 100. If the base current is 20 µA, what is the voltage drop across the collector resistor?
Given
Collector resistor RC = 2 kΩ
Current gain β = 100
Base current IB = 20 µA
Find
Voltage drop across RC (V)
Solution
IC = β · IB = 100 × 20 µA = 2 mA
VRC = IC · RC = 2 mA × 2 kΩ = 4 V
This multiplies 2 mA by 2 Ω, losing the kilo prefix on RC.
This uses the base current, 20 µA × 2 kΩ; the collector carries β times more.
IC = βIB = 100 × 20 µA = 2 mA, and VRC = ICRC = 2 mA × 2 kΩ = 4 V.
This reads 20 µA as 20 mA, a factor of 1000 too large.
Question 14 of 18 · Electronics · 7–11 items
An op-amp has an open-loop gain of 100,000 and is used in a non-inverting configuration with a closed-loop gain of 10. What is the approximate loop gain?
Given
Open-loop gain AOL = 100,000
Closed-loop gain = 10
Find
Loop gain
Solution
β ≈ 1/ACL = 1/10 = 0.1
T = AOL · β = 100,000 × 0.1 = 10,000
This is the open-loop gain; the feedback network attenuates it by β = 0.1 around the loop.
β ≈ 1/ACL = 0.1, so the loop gain is AOLβ = 100,000 × 0.1 = 10,000.
This is the closed-loop gain, 1/β, not the loop gain.
This is the feedback factor β = 1/10 by itself; the loop gain is AOL·β.
Question 15 of 18 · Electronics · 7–11 items
A noninverting amplifier has a closed-loop gain of 20.0, and its op amp has a slew rate of 5.00 V/µs. At 200 kHz, what are the largest sinusoidal output peak that avoids slew-rate distortion and the corresponding input peak?
Given
Closed-loop gain Acl = 20.0
Slew rate SR = 5.00 V/µs
Frequency f = 200 kHz
Find
Maximum output and input peak amplitudes (V)
Solution
SR = 2πf · Vout,pk → Vout,pk = SR/(2πf)
Vout,pk = (5.00 V/µs)/(2π × 200 kHz) = 3.98 V
Vin,pk = 3.98 V/20.0 = 0.199 V
This is SR/f with no 2π; a sine of peak V has a maximum slope of 2πfV.
This applies the slew limit to the input and multiplies by the gain; the limit belongs to the output.
An ideal buck converter runs from Vin = 24 V at fs = 200 kHz with duty cycle D = 0.25 and an inductor L = 100 µH, supplying 2.0 A in continuous conduction. What are the peak-to-peak inductor ripple current and the minimum output capacitance that keeps the peak-to-peak output ripple below 30 mV (ESR negligible)?
Given
Input voltage Vin = 24 V
Switching frequency fs = 200 kHz
Duty cycle D = 0.25
Inductance L = 100 µH
Load current Iload = 2.0 A
Maximum output ripple ΔV = 30 mV
Find
Inductor ripple ΔIL (A) and minimum capacitance C (µF)
The ripple current is right, but C = ΔIL/(fsΔV) drops the 8 of the triangular-ripple charge.
Question 17 of 18 · Electronics · 7–11 items
An ideal op-amp integrator has an input resistor R = 100 kΩ and a feedback capacitor C = 1 µF. What is the integrator time constant?
Given
Input resistor R = 100 kΩ
Feedback capacitor C = 1 µF
Find
Time constant τ (ms)
Solution
τ = R · C = 100 kΩ × 1 µF = 0.1 s = 100 ms
This reads 1 µF as 1 nF, a factor of 1000 too small.
This is RC/(2π), mixing the time constant with the 1/(2πRC) corner frequency.
τ = RC = 100 × 103 Ω × 1 × 10−6 F = 0.1 s = 100 ms.
This multiplies RC by 2π; the time constant is simply RC.
Question 18 of 18 · Electronics · 7–11 items
A noninverting amplifier has R1 = 10 kΩ from the inverting input to ground and R2 = 90 kΩ from the output to the inverting input. The op amp has open-loop gain AOL = 50,000 and open-loop output resistance Ro = 50 Ω; the load is 2.0 kΩ. What are the closed-loop voltage gain (to four significant figures) and the approximate closed-loop output resistance?
Given
Resistor R1 = 10 kΩ (to ground)
Feedback resistor R2 = 90 kΩ
Open-loop gain AOL = 50,000
Open-loop output resistance Ro = 50 Ω
Load RL = 2.0 kΩ
Find
Closed-loop gain Acl and output resistance (Ω)
Solution
β = R1/(R1 + R2) = 10/(10 + 90) = 0.1
Acl = AOL/(1 + AOLβ) = 50,000/5001 = 9.998
Rout = Ro/(1 + AOLβ) = 50 Ω/5001 = 0.0100 Ω
This takes β = R1/R2 = 1/9; for a noninverting stage β = R1/(R1 + R2) = 0.1.
This is the ideal gain with the output resistance left unchanged; feedback divides Ro by 1 + AOLβ.
The gain is right, but Ro was multiplied by the loop gain; series-shunt feedback lowers the output resistance.
With negative feedback, an ideal op amp has no input current and no voltage between its inputs. Those two rules solve an ideal op-amp circuit in a few lines:
inverting: Av = −Rf/Rin noninverting: Av = 1 + Rf/Rin
Nonideal questions bring back the limits: finite open-loop gain, gain-bandwidth product, slew rate, output resistance. For discrete devices, the stem tells you the model to use, ideal or constant-drop diode, and which region a transistor is in; decide the region first, then pick the equation. Power electronics items are averages over a switching cycle: a converter’s output follows its duty cycle.
Properties of electrical materials
A separate area of the specification, 4–6 items, with three groups of subtopics:
Semiconductors: energy bands, doping, drift and diffusion current, tunneling, p-n junction theory.
Electrical properties: conductivity and resistivity, permittivity, magnetic permeability, noise.
Thermal properties: thermal conductivity and expansion.
Three relations carry most of it:
R = ρL/A σ = q(nμn + pμp) n·p = ni2
In a doped sample the majority carrier density is close to the dopant density, and the mass-action law gives the minority one. Mechanical properties of materials (strength, hardness, processing) belong to FE Mechanical.
Common traps in electronics
Diode model. A silicon diode drops about 0.7 V in the constant-drop model and nothing in the ideal model. Use the model the stem names.
α or β. β = α/(1 − α). Using one for the other is off by a factor of about β.
Golden rules without feedback. An op amp with no negative feedback, a comparator, saturates at a rail; the virtual short does not apply.
Ripple frequency. A full-wave rectifier ripples at twice the line frequency; a half-wave one at the line frequency.
Signal or bias. Small-signal gain uses the incremental model; a DC operating point uses the large-signal one. One stem may ask for both.
Sources: Sedra and Smith, Microelectronic Circuits; Streetman and Banerjee, Solid State Electronic Devices.
How it connects to other areas
Bridge circuits, ADC resolution and op-amp signal conditioning count as electrical instrumentation and stay here; selecting a thermocouple or RTD for a process is FE Mechanical measurements. Device circuits rest on Circuit Analysis; amplifier frequency response uses the transfer functions of Linear Systems; inverters and rectifiers feeding motors meet Power Systems.