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3.4 Electronics & Devices

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FE Electrical and Computer Exam Exam

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1 FE Electrical & Computer: Exam Map, Handbook & Strategy

1.1 Current Exam Facts 1.2 CBT Format & Scaled Scoring 1.3 The FE Reference Handbook 1.4 Approved Calculators & Test-Day Logistics 1.5 Study Plan & Time Strategy

2 Mathematics, Economics & Ethics

2.1 Algebra, Calculus & Differential Equations 2.2 Complex Numbers & Laplace Transforms 2.3 Probability & Statistics 2.4 Engineering Economics 2.5 Ethics & Professional Practice

3 Circuits, Electronics & Electromagnetics

3.1 DC Circuit Analysis 3.2 AC Steady-State & Phasors 3.3 Transient Response (RL, RC, RLC) 3.4 Electronics & Devices 3.5 Electromagnetics

4 Signals, Systems, Controls & Communications

4.1 Linear Systems & Transfer Functions 4.2 Signal Processing & Sampling 4.3 Control Systems 4.4 Communications

5 Digital Systems, Computers, Networks & Software

5.1 Number Systems & Boolean Algebra 5.2 Combinational & Sequential Logic 5.3 Computer Architecture & Microprocessors 5.4 Computer Networks 5.5 Software Engineering & Algorithms

6 Power, Materials & Final Readiness

6.1 Power Fundamentals & Three-Phase 6.2 Transformers & Rotating Machines 6.3 Power Electronics & Transmission 6.4 Properties of Electrical Materials & Engineering Sciences 6.5 Integrated Practice & Final Readiness

3.4 Electronics & Devices

Key Takeaways Explain This Topic AI

A silicon diode conducts when forward-biased above roughly 0.7 V and blocks current when reverse-biased.

A BJT in active mode has collector current I_C = beta * I_B; a MOSFET in saturation has drain current proportional to (V_GS - V_th)^2.

Ideal op-amp golden rules: no current flows into the inputs, and negative feedback forces V+ = V- (the virtual short).

An inverting amplifier has gain -R_f/R_in; a non-inverting amplifier has gain 1 + R_f/R_in.

An op-amp integrator output is the time integral of the input scaled by -1/(R*C).

Last updated: June 2026

Diodes and rectification

A diode conducts current in one direction. In the common constant-voltage-drop model, a silicon diode is ON (a fixed 0.7 V forward drop) or OFF (open, reverse-biased). To analyze, assume a state, solve, and confirm it is consistent (a forward-ON diode must carry positive current; a reverse-OFF diode must have negative terminal voltage). If the check fails, flip the assumption and re-solve.

Key applications:

Half-wave rectifier: passes one polarity of an AC waveform.

Full-wave bridge rectifier: four diodes convert both half-cycles to one polarity; output ripple is smoothed by a filter capacitor.

Zener diode: operated in reverse breakdown at a fixed V_Z to regulate voltage.

For the ideal diode model the forward drop is neglected (0 V). FE problems usually specify which model to use, so read carefully — using 0.7 V when the problem says "ideal" is a needless error.

Worked diode example: a 10 V source drives a series diode and a 1 kohm resistor. Assume the diode is ON with a 0.7 V drop; then the resistor sees 10 - 0.7 = 9.3 V and the current is 9.3/1000 = 9.3 mA, which is positive, so the ON assumption is consistent. If the source were reversed (-10 V), the assumed-ON current would come out negative, contradicting the assumption, so the diode is actually OFF and no current flows.

The full diode equation I = I_s*(e^(V/(nV_T)) - 1) describes the exponential I-V curve, where V_T = kT/q is about 26 mV at room temperature, but FE problems almost always use the simpler constant-drop or ideal models.

Bipolar junction transistor (BJT)

A bipolar junction transistor (BJT) has three regions of operation:

Cutoff: both junctions reverse-biased, transistor off (I_C = 0); acts as an open switch.

Active (forward-active): base-emitter forward, base-collector reverse; acts as a linear amplifier with I_C = beta * I_B and I_E = I_C + I_B = (beta + 1) * I_B.

Saturation: both junctions forward-biased; acts as a closed switch with a small V_CE(sat) around 0.2 V.

The current gain beta (also written h_FE) typically ranges from about 50 to 250. The base-emitter junction drop V_BE is about 0.7 V when conducting. Worked bias check: if a base resistor sets I_B = 20 microamps and beta = 100, then in the active region I_C = 100 * 20e-6 = 2 mA — but only if the collector supply and load resistor allow that current without driving V_CE below ~0.2 V; otherwise the device saturates and I_C is limited by the external circuit, not by beta*I_B.

MOSFET regions

A metal-oxide-semiconductor field-effect transistor (MOSFET) is voltage-controlled — the insulated gate draws essentially no DC current. For an n-channel enhancement MOSFET with threshold voltage V_th:

Cutoff: V_GS < V_th, no drain current (I_D = 0).

Triode (ohmic): V_DS < V_GS - V_th; behaves as a voltage-controlled resistor.

Saturation (active): V_DS >= V_GS - V_th; I_D = (k/2) (V_GS - V_th)^2, the square-law region used for amplification, where k is the transconductance parameter (k = mu C_ox*W/L).

The quantity (V_GS - V_th) is the overdrive voltage . Because the gate is insulated, MOSFETs have very high input impedance, which is why CMOS dominates digital logic. Note the naming inversion versus the BJT: a BJT amplifies in active , and a MOSFET amplifies in saturation — "saturation" means opposite things in the two devices, a frequent FE trap.

Worked MOSFET example: an n-channel device with V_th = 1 V and k = 0.5 mA/V^2 is biased at V_GS = 3 V with V_DS = 5 V. First confirm the region: V_GS - V_th = 2 V overdrive, and V_DS (5 V) >= 2 V, so it is in saturation. Then I_D = (k/2) (V_GS - V_th)^2 = (0.5e-3/2) (2)^2 = 0.25e-3*4 = 1 mA. Doubling the overdrive to 4 V would quadruple I_D to 4 mA because of the square law. For amplifier biasing the device must stay in saturation across the signal swing; for switching (logic) the device toggles between cutoff (open) and deep triode (low-resistance on).

Ideal op-amp golden rules and worked gain

An operational amplifier (op-amp) has very high open-loop gain and high input impedance. With negative feedback, two golden rules apply:

No current flows into either input (infinite input impedance).

The inputs are forced to the same voltage, V+ = V- (the virtual short ; a virtual ground when V+ is grounded).

Apply these with KCL at the inverting node to derive every standard circuit.

  • Configuration Output / Gain
  • Inverting V_out = -(R_f/R_in) * V_in
  • Non-inverting V_out = (1 + R_f/R_in) * V_in
  • Voltage follower (buffer) V_out = V_in (gain = 1)
  • Summing (inverting) V_out = -R_f*(V_1/R_1 + V_2/R_2 + ...)
  • Difference (subtractor) V_out = (R_f/R_in)*(V_2 - V_1), matched resistors
  • Integrator V_out = -(1/(R*C)) * integral of V_in dt
  • Differentiator V_out = -R*C * dV_in/dt

Worked op-amp example: an inverting amplifier with R_in = 2 kohm and R_f = 20 kohm has gain -R_f/R_in = -10. The inverting node is a virtual ground (0 V) because V+ is grounded; the input current V_in/R_in flows entirely through R_f (golden rule 1), so V_out = -(R_f/R_in)*V_in. For V_in = +0.5 V, V_out = -10 * 0.5 = -5 V. The non-inverting gain can never be less than 1; the inverting gain is negative (polarity inverts).