Electronics practice questions

Answer: B) 200

• AIncorrect. That adds instead.
• BCorrect. \(|A_v| \approx g_m R_C\).
• CIncorrect. That divides instead.
• DIncorrect. That's off by a factor of 10.

Answer: B) How fast its output voltage can change

• AIncorrect. It limits rate of change.
• BCorrect. Usually given in volts per microsecond.
• CIncorrect. It limits rate of change.
• DIncorrect. It limits rate of change.

Answer: B) An open circuit carrying zero current

In the ideal-switch model, a forward-biased diode is a short circuit and a reverse-biased diode is an open circuit, so no reverse current flows.

• AIncorrect. In the ideal-switch model, a forward-biased (conducting) diode is the one shown as a short circuit.
• BCorrect. The ideal-switch model treats a reverse-biased diode as an open switch, so no current flows through it whatever the reverse voltage is.
• CIncorrect. A 0.7 V source belongs to the constant-voltage-drop model, and only for a forward-biased silicon diode.
• DIncorrect. The dynamic (small-signal) resistance describes a forward-biased diode around its operating point. It is not used for reverse bias in the ideal model.

Answer: B) \(CV^2f\)

• AIncorrect. Power rises with frequency.
• BCorrect. Lowering supply voltage greatly cuts power.
• CIncorrect. It goes with voltage squared.
• DIncorrect. Switching dominates in active logic.

Answer: C) 4

• AIncorrect. That's half-wave.
• BIncorrect. That's a center-tapped rectifier.
• CCorrect. Two conduct on each half-cycle.
• DIncorrect. That's a three-phase bridge.

Answer: C) \((V_{GS} - V_t)^2\)

• AIncorrect. Current rises with \(V_{GS}\).
• BIncorrect. In saturation, \(V_{DS}\) has little effect.
• CCorrect. This is the square-law model.
• DIncorrect. The gate draws almost no current.

Answer: A) Pass a control signal between circuits with no galvanic connection

An optocoupler sends a signal optically from an internal LED to a photodetector. This keeps the two circuits galvanically isolated, for example to protect low-voltage logic from high-voltage or noisy circuits.

• ACorrect. The LED's light carries the signal across an insulating gap to the photodetector, so the input and output circuits stay electrically isolated.
• BIncorrect. That job uses an exposed photodiode or photoresistor. An optocoupler's detector is sealed and responds only to its own internal LED.
• CIncorrect. That describes a Zener or TVS diode used for surge protection, not an optocoupler.
• DIncorrect. Scaling by a turns ratio is what a transformer does. An optocoupler couples through light and has no turns ratio.

Answer: A) The peak-to-peak ripple voltage across the load decreases

Ripple in a capacitor-input filter is about I_load/(f·C), so a larger capacitor reduces ripple. It also shortens diode conduction, which raises the peak diode current.

• ACorrect. Ripple is about I_load/(f·C), so a larger C discharges less between charging peaks and the ripple shrinks.
• BIncorrect. With a larger C the diodes conduct for a shorter part of each cycle, so they must deliver the same charge in a shorter pulse. The peak diode current goes up.
• CIncorrect. The ripple frequency is set by the rectifier type. A full-wave rectifier gives twice the line frequency whatever the capacitor size.
• DIncorrect. Less discharge between peaks keeps the output closer to the peak value, so the average DC output rises slightly.

Answer: C) Drain-source on-resistance, R_DS(on)

In the fully-on (ohmic) state, a MOSFET switch dissipates about I_D² × R_DS(on), so a low on-resistance minimizes conduction loss. Capacitance and threshold mainly affect switching behavior.

• AIncorrect. V_GS(th) sets where the device starts to conduct. Once it is fully enhanced, conduction loss depends on on-resistance, not threshold. A very low threshold can also cause false turn-on from noise.
• BIncorrect. Lower C_iss reduces gate-drive energy and switching time, so it affects switching loss. It has essentially no effect on loss while the device is steadily on.
• CCorrect. When fully on, the MOSFET acts like a resistor, and conduction loss is about I_D² × R_DS(on). A smaller R_DS(on) directly lowers that loss.
• DIncorrect. g_fs describes how drain current responds to gate voltage in the saturation region. It does not set conduction loss in the fully-on (ohmic) state, and a low value is not a goal.

Answer: B) Reproduce a reference current in another branch so it can bias a stage or serve as an active load

A current mirror uses matched transistors to copy (or scale) a reference current into one or more output branches. It is a core building block for IC biasing and active loads.

• AIncorrect. That describes a voltage reference, such as a bandgap or Zener reference. A current mirror copies a current, not a voltage.
• BCorrect. Two matched transistors with equal V_BE carry nearly equal collector currents. The output branch therefore copies (or scales) the reference current, which is widely used for biasing and for active loads.
• CIncorrect. That is a transimpedance amplifier (current-to-voltage converter). A current mirror's output is a current.
• DIncorrect. That is a level-shifter stage. A current source may bias a level shifter, but copying a current is a different function.