Diode Applications Practice Exam: 40 Solved Problems

DaysHoursMinSec

This offer has expired!

SubscribePREMIUM PROMO: Remove Ads!! Get Freebies!!! and Get EXTRA MONTHS before Time EXPIRED!

You finished the 40-item Diode Applications Practice Exam. Now check your work. Every item below follows the same Given, Find, Solution format used throughout this series, with no step skipped. If you got an item wrong, do not just read the answer. Find the exact step where your process broke down, then solve a similar problem from scratch before you move on.

Part A — Rectification (Items 1 to 5)

Full lesson on this topic is in Part 1 — Rectification.

Item 1. The DC (average) output voltage of a half-wave rectifier is given by what formula?

Choices: (A) $V_m/\pi$ (B) $2V_m/\pi$ (C) $V_m/2$ (D) $0.636V_m$

Given: Half-wave rectifier

Find: $V_{dc}$ formula

Solution:

Step 1: Recall the two standard DC output formulas by rectifier type.

$V_{dc}(\text{half-wave}) = \dfrac{V_m}{\pi} \qquad V_{dc}(\text{full-wave}) = \dfrac{2V_m}{\pi}$

Step 2: The question asks specifically for half-wave, which uses only one term in the numerator.

✓ ANSWER: A — $V_m/\pi$

Examiner note: Choice (B), $2V_m/\pi$ , is the full-wave formula — the most common distractor for students who mix up the two rectifier types. Choice (D), $0.636V_m$ , is numerically equal to $2V_m/\pi$ expressed as a decimal, so it is really the same trap in a different form.

Item 2. In a full-wave bridge rectifier, the PIV rating required for each diode is what?

Choices: (A) $V_m$ (B) $2V_m$ (C) $V_m/2$ (D) $4V_m$

Given: Full-wave bridge rectifier

Find: PIV rating per diode

Solution:

Step 1: Recall the PIV rule for each rectifier configuration.

$PIV(\text{bridge}) = V_m \qquad PIV(\text{center-tap}) = 2V_m$

Step 2: A bridge rectifier splits the reverse voltage across two non-conducting diodes at any instant, halving the requirement compared to a center-tap design.

✓ ANSWER: A — $V_m$

Examiner note: Choice (B), $2V_m$ , is the center-tap PIV requirement — swapping these two is the single most common rectifier PIV error on the board exam.

Item 3. A full-wave rectifier is fed from a 60 Hz AC source. What is the output ripple frequency?

Choices: (A) 60 Hz (B) 120 Hz (C) 180 Hz (D) 240 Hz

Given: Full-wave rectifier, $f_{in} = 60$ Hz

Find: $f_{ripple}$

Solution:

Step 1: Recall that full-wave rectification produces one output pulse per half-cycle of input.

Step 2: Apply the doubling rule.

$f_{ripple} = 2 \times f_{in} = 2 \times 60 = 120 \text{ Hz}$

✓ ANSWER: B — 120 Hz

Examiner note: Choice (A), 60 Hz, is the half-wave rectifier’s ripple frequency — a full-wave rectifier always doubles the input frequency at the output.

Item 4. What is the number of diodes required to build a standard full-wave bridge rectifier?

Choices: (A) 1 (B) 2 (C) 4 (D) 6

Given: Full-wave bridge rectifier

Find: Diode count

Solution:

Step 1: Recall the bridge topology: four diodes arranged so two conduct on each half-cycle.

Step 2: Confirm against the center-tap design, which uses only 2 diodes but requires a center-tapped transformer instead.

✓ ANSWER: C — 4

Examiner note: Choice (B), 2, describes the center-tap rectifier, not the bridge — read the question carefully to see which topology is actually named.

Item 5. In a full-wave center-tapped rectifier, the PIV rating required for each diode is what?

Choices: (A) $V_m$ (B) $2V_m$ (C) $3V_m$ (D) $4V_m$

Given: Full-wave center-tapped rectifier

Find: PIV rating per diode

Solution:

Step 1: In a center-tap design, the non-conducting diode sees the full secondary winding voltage in reverse — both halves of the transformer add together across it.

Step 2: Apply the formula.

$PIV = 2V_m$

✓ ANSWER: B — $2V_m$

Examiner note: This is the opposite pairing of Item 2 — center-tap needs the higher PIV, bridge needs the lower one. Keep the two straight by remembering that the bridge splits reverse voltage across two diodes in series, while the center-tap does not.

Part B — Diode Configurations (Items 6 to 9)

Full lesson on this topic is in Part 2 — Diode Configurations.

Item 6. Three silicon diodes are connected in series, all forward biased. Using the practical model, what is the total voltage drop across all three?

Choices: (A) 0.7 V (B) 1.4 V (C) 2.1 V (D) 2.8 V

Given: 3 series Si diodes, all forward biased, practical model

Find: Total voltage drop

Solution:

Step 1: Each forward biased silicon diode drops 0.7 V under the practical model.

Step 2: In series, voltage drops add.

$V_{total} = 3 \times 0.7 = 2.1 \text{ V}$

✓ ANSWER: C — 2.1 V

Examiner note: Choice (D), 2.8 V, would be the answer for four diodes, not three — a simple miscount trap. Choice (A) is the drop for a single diode only.

Item 7. When using the ON/OFF assumption method to analyze a multi-diode circuit, a diode assumed OFF is verified correct if what condition holds?

Choices: (A) It shows forward current (B) It shows reverse or sub-threshold voltage across it (C) It dissipates maximum power (D) It has zero resistance

Given: ON/OFF assumption method

Find: Verification condition for an OFF assumption

Solution:

Step 1: Recall the method: assume a state for each diode, solve the circuit as if that state were fixed, then check consistency.

Step 2: An OFF assumption is only valid if the resulting voltage across that diode is reverse biased or below its threshold — otherwise the assumption contradicts the solved circuit.

✓ ANSWER: B — It shows reverse or sub-threshold voltage across it

Examiner note: Choice (A), forward current, is actually the check used to verify an ON assumption — the reverse pairing of what this question asks about.

Item 8. Two diodes with different threshold voltages are connected in parallel across the same source. Which diode conducts first?

Choices: (A) The one with higher $V_D$ (B) The one with lower $V_D$ (C) Both conduct simultaneously and equally (D) Neither conducts

Given: Two parallel diodes, different threshold voltages

Find: Which conducts first

Solution:

Step 1: As the source voltage rises from zero, the diode requiring less voltage to forward bias reaches its threshold first.

Step 2: Once that lower-threshold diode conducts, it clamps the shared node voltage near its own $V_D$ , which can keep the higher-threshold diode from ever reaching its own turn-on point.

✓ ANSWER: B — The one with lower $V_D$

Examiner note: This is the same first-principles reasoning used throughout diode circuit analysis: the path of least resistance to conduction wins, and here that means the lowest threshold voltage.

Item 9. In a diode logic gate, a diode conducts only under which condition?

Choices: (A) Its input condition matches the gate’s forward-bias requirement (B) It is always ON regardless of input (C) It is always OFF regardless of input (D) The gate has no power supply

Given: Diode logic gate

Find: Condition for conduction

Solution:

Step 1: A diode logic gate uses each diode’s forward-bias condition as the switching mechanism for that input line.

Step 2: A diode conducts only when its specific input state satisfies its own forward-bias requirement relative to the rest of the circuit.

✓ ANSWER: A — Its input condition matches the gate’s forward-bias requirement

Examiner note: This is a conceptual question tying diode configurations directly to basic digital logic — a favorite way board exams cross reference topics.

Part C — DC Load Line Analysis and Q-Point (Items 10 to 14)

Full lesson on this topic is in Part 3 — DC Load Line Analysis and Q-Point.

Item 10. What is the DC load line equation for a series diode circuit with source $V_{DD}$ and resistor $R$ ?

Choices: (A) $I_D = (V_{DD} - V_D)/R$ (B) $I_D = V_{DD}/V_D$ (C) $I_D = V_D/R$ (D) $I_D = V_{DD} \times R$

Given: Series diode circuit, source $V_{DD}$ , resistor $R$

Find: Load line equation

Solution:

Step 1: Apply KVL around the loop: $V_{DD} = I_D R + V_D$ .

Step 2: Solve for $I_D$ .

$I_D = \dfrac{V_{DD} - V_D}{R}$

✓ ANSWER: A — $I_D = (V_{DD} - V_D)/R$

Examiner note: This equation is derived directly from KVL, not memorized separately — if you forget it, rederive it from the loop in ten seconds rather than guessing.

Item 11. What is the slope of the DC load line on the diode I-V characteristic curve?

Choices: (A) $-1/R$ (B) $1/R$ (C) $-R$ (D) $R$

Given: DC load line equation

Find: Slope

Solution:

Step 1: Rewrite the load line equation in slope-intercept form with $I_D$ as the vertical axis and $V_D$ as the horizontal axis.

$I_D = -\dfrac{1}{R}V_D + \dfrac{V_{DD}}{R}$

Step 2: The coefficient of $V_D$ is the slope.

✓ ANSWER: A — $-1/R$

Examiner note: The negative sign matters — it reflects that $I_D$ decreases as $V_D$ increases along the load line, opposite to the diode’s own forward characteristic curve, which rises.

Item 12. What does the Q-point (quiescent point) of a diode circuit represent?

Choices: (A) The diode’s maximum rated current (B) The intersection of the load line and the diode’s characteristic curve (C) The reverse breakdown voltage (D) The AC resistance point

Given: DC load line and diode characteristic curve

Find: Definition of Q-point

Solution:

Step 1: The load line represents every possible operating combination allowed by the external circuit ( $V_{DD}$ and $R$ ).

Step 2: The diode’s own characteristic curve represents every combination the diode itself can physically produce. The one point satisfying both simultaneously is the Q-point.

✓ ANSWER: B — The intersection of the load line and the diode’s characteristic curve

Examiner note: This graphical definition is tested constantly. Do not confuse it with a numeric rating like maximum current or breakdown voltage — the Q-point is a specific operating condition, not a component spec.

Item 13. If the source voltage $V_{DD}$ increases while $R$ stays constant, what happens to the load line?

Choices: (A) It shifts left (B) It shifts right, x-intercept increases (C) It stays the same (D) It becomes vertical

Given: Increasing $V_{DD}$ , constant $R$

Find: Effect on the load line

Solution:

Step 1: The x-intercept of the load line (where $I_D = 0$ ) occurs at $V_D = V_{DD}$ .

Step 2: A larger $V_{DD}$ pushes this x-intercept further right, while the slope $-1/R$ stays fixed since $R$ has not changed.

✓ ANSWER: B — It shifts right, x-intercept increases

Examiner note: The load line shifts as a parallel line to itself — same slope, new position — because only the intercept changes, not the slope, when $V_{DD}$ alone changes.

Item 14. If $R$ increases while $V_{DD}$ stays constant, what happens to the slope of the load line?

Choices: (A) It increases in magnitude (steeper) (B) It decreases in magnitude (flatter) (C) It stays the same (D) It becomes zero

Given: Increasing $R$ , constant $V_{DD}$

Find: Effect on load line slope

Solution:

Step 1: The slope magnitude is $1/R$ .

Step 2: As $R$ gets larger, $1/R$ gets smaller, so the slope magnitude decreases and the line flattens.

✓ ANSWER: B — It decreases in magnitude (flatter)

Examiner note: This is the opposite effect of Item 13 — that item changes the intercept only, this one changes the slope only. Keep the two variables’ effects separate in your memory rather than blending them into one vague rule.

Part D — Diode Clippers (Items 15 to 19)

Full lesson on this topic is in Part 4 — Diode Clippers.

Item 15. In a series positive clipper, what is the output during the positive half-cycle?

Choices: (A) $V_{in} - V_D$ (B) 0 V (C) $V_D$ (D) $V_{in}$

Given: Series positive clipper

Find: $V_{out}$ during positive half-cycle

Solution:

Step 1: In a series positive clipper, the diode is oriented to block during the positive half-cycle.

Step 2: A blocked series diode carries no current, so no voltage develops across the load.

✓ ANSWER: B — 0 V

Examiner note: Choice (C), $V_D$ , is the parallel clipper answer for the clipped half-cycle, not the series clipper answer — these two circuit types give different outputs during clipping and must not be swapped.

Item 16. In a parallel positive clipper, what is the output clamped to during the clipped (positive) half-cycle?

Choices: (A) 0 V (B) $V_D$ (0.7 V for Si) (C) $V_{in}$ (D) $2V_D$

Given: Parallel positive clipper

Find: $V_{out}$ during clipping

Solution:

Step 1: In a parallel clipper, the diode conducts during the clipped half-cycle rather than blocking.

Step 2: A conducting diode has its own forward voltage across it, and since the diode is directly across the output node, the output sits at that same voltage.

✓ ANSWER: B — $V_D$

(0.7 V for Si)

Examiner note: Choice (A), 0 V, is the series clipper answer — this is the direct reverse pairing of Item 15, and confusing the two is the most common clipper error on the board exam.

Item 17. In a positive biased clipper, the clipping threshold voltage equals what?

Choices: (A) $V$ (B) $V_D$ (C) $V + V_D$ (D) $V - V_D$

Given: Positive biased clipper, bias voltage $V$

Find: $V_{clip}$

Solution:

Step 1: The diode must overcome both the bias voltage and its own forward drop before it conducts.

Step 2: Add both quantities together.

$V_{clip} = V + V_D$

✓ ANSWER: C — $V + V_D$

Examiner note: Choice (A), just $V$ alone, is the most common wrong answer — it forgets the diode’s own forward drop entirely, treating the diode as ideal when the problem calls for the practical model.

Item 18. The presence of which component signals a clamper rather than a clipper?

Choices: (A) Resistor (B) Capacitor (C) Inductor (D) Transistor

Given: Diode waveform-shaping circuit identification

Find: Distinguishing component

Solution:

Step 1: A clamper needs a way to store a DC shift voltage between cycles.

Step 2: Only a capacitor can hold that stored charge — a clipper has no such element.

✓ ANSWER: B — Capacitor

Examiner note: This single-component check is faster than analyzing the full output waveform description on an identification question — use it first before doing any other analysis.

Item 19. How is the current through $R$ during clipping in a biased clipper computed?

Choices: (A) $V_{in}/R$ (B) $(V_{in} - V_{clip})/R$ (C) $V_{clip}/R$ (D) $(V_{in} + V_{clip})/R$

Given: Biased clipper, diode conducting (clipping active)

Find: $I_R$ formula

Solution:

Step 1: The clip voltage appears across the diode-bias branch, not across $R$ .

Step 2: The remaining voltage, which does drop across $R$ , is the difference between the input and the clipped output.

$I_R = \dfrac{V_{in} - V_{clip}}{R}$

✓ ANSWER: B — $(V_{in} - V_{clip})/R$

Examiner note: Choice (A) is the single most common arithmetic error on clipper current problems — dividing the full input voltage by $R$ without first subtracting the clipped output voltage.

Part E — Diode Clampers (Items 20 to 23)

Full lesson on this topic is in Part 5 — Diode Clampers.

Item 20. Every clamper circuit requires which three elements?

Choices: (A) Diode, resistor, inductor (B) Diode, capacitor, resistive load (C) Two diodes only (D) Capacitor and inductor only

Given: Clamper circuit requirements

Find: The three required elements

Solution:

Step 1: A clamper needs a diode to control charging direction, a capacitor to store the DC shift, and a resistive load to complete the circuit.

Step 2: Confirm this matches the minimum working configuration described throughout the post — an independent DC source is optional (for a biased clamper), but these three are not.

✓ ANSWER: B — Diode, capacitor, resistive load

Examiner note: Choice (A) swaps in an inductor, which plays no role in a standard diode clamper — inductors belong to filter circuits, not clamping circuits.

Item 21. For a negative clamper (unbiased, silicon diode), what is the maximum output voltage approximately equal to?

Choices: (A) 0 V (B) +0.7 V (C) −0.7 V (D) $V_m$

Given: Negative clamper, unbiased, Si diode

Find: $V_{out(max)}$

Solution:

Step 1: Recall the negative clamper output formula.

$V_{out(max)} = +V_D$

Step 2: For a silicon diode, $V_D = 0.7$ V.

✓ ANSWER: B — +0.7 V

Examiner note: Choice (A), exactly 0 V, ignores the diode drop entirely — a trap for students treating the diode as ideal when the problem specifies silicon.

Item 22. What is the RC time constant design guideline for a clamper?

Choices: (A) $RC \geq 10T$ (B) $RC = T$ (C) $RC \leq T/10$ (D) $RC = 0$

Given: Clamper RC design rule

Find: The correct guideline

Solution:

Step 1: Recall that the capacitor must hold its charge steady across the non-conducting portion of each cycle.

Step 2: This requires the discharge path to be much slower than the signal itself.

$RC \geq 10T$

✓ ANSWER: A — $RC \geq 10T$

Examiner note: Choice (C) inverts the direction of the inequality entirely — a design with $RC \leq T/10$ would discharge far too quickly and produce heavy droop, the opposite of the design goal.

Item 23. If $RC$ is too small relative to the input period, what happens to the clamped output?

Choices: (A) It remains perfectly flat (B) It droops and distorts (C) Its amplitude increases (D) It becomes pure DC

Given: Undersized $RC$ relative to signal period

Find: Effect on the output

Solution:

Step 1: A small $RC$ means the capacitor discharges significantly through the resistor between charging pulses.

Step 2: This discharge causes the clamped voltage level to sag between cycles instead of staying flat.

✓ ANSWER: B — It droops and distorts

Examiner note: This is the standard qualitative follow-up to any RC sizing problem — know the failure mode in words, not just the design formula in symbols.

Part F — Zener Diodes and Voltage Regulation (Items 24 to 30)

Full lesson on this topic is in Part 6 — Zener Diodes and Voltage Regulation.

Item 24. In a zener regulator, how is the current through the series resistor calculated?

Choices: (A) $I_R = (V_{in} - V_Z)/R$ (B) $I_R = V_Z/R$ (C) $I_R = V_{in}/R$ (D) $I_R = (V_{in} + V_Z)/R$

Given: Zener regulator circuit

Find: $I_R$ formula

Solution:

Step 1: Apply KVL: the source voltage drops across the resistor and the zener in series.

$V_{in} = I_R R + V_Z$

Step 2: Solve for $I_R$ .

$I_R = \dfrac{V_{in} - V_Z}{R}$

✓ ANSWER: A — $I_R = (V_{in} - V_Z)/R$

Examiner note: Choice (C) ignores the zener’s own voltage drop entirely, treating the full source voltage as if it appeared across $R$ alone — an easy but incorrect shortcut.

Item 25. How is the zener current $I_Z$ found?

Choices: (A) $I_Z = I_R + I_L$ (B) $I_Z = I_R - I_L$ (C) $I_Z = I_L - I_R$ (D) $I_Z = I_R \times I_L$

Given: Zener regulator current distribution

Find: $I_Z$ formula

Solution:

Step 1: At the node where $R$ , the zener, and the load meet, apply KCL.

$I_R = I_Z + I_L$

Step 2: Solve for $I_Z$ .

$I_Z = I_R - I_L$

✓ ANSWER: B — $I_Z = I_R - I_L$

Examiner note: This is a direct KCL application — the zener carries whatever current the load does not use, never more than $I_R$ supplies in total.

Item 26. At the no-load condition (load removed), what is the zener current?

Choices: (A) Zero (B) Equal to $I_R$ (C) Equal to $I_L$ (D) Undefined

Given: Zener regulator, $R_L$ removed

Find: $I_Z$ at no-load

Solution:

Step 1: With no load connected, $I_L = 0$ .

Step 2: Apply $I_Z = I_R - I_L$ with $I_L = 0$ .

$I_Z = I_R - 0 = I_R$

✓ ANSWER: B — Equal to $I_R$

Examiner note: This is why no-load is always the worst case for zener current — every bit of $I_R$ that would normally split off to the load instead flows entirely through the zener.

Item 27. The no-load condition should be checked against which zener current limit?

Choices: (A) $I_{ZK}$ (B) $I_{ZM}$ (C) Average $I_Z$ (D) Maximum $I_L$

Given: No-load design check

Find: Correct limit to check against

Solution:

Step 1: No-load produces the highest possible $I_Z$ for a given design, as shown in Item 26.

Step 2: The highest current must be checked against the maximum rated current, $I_{ZM}$ , to avoid exceeding the diode’s power rating.

✓ ANSWER: B — $I_{ZM}$

Examiner note: Choice (A), $I_{ZK}$ , is the limit used for the opposite extreme condition (full-load) — see Item 28. Mixing up which limit applies to which condition is the most common zener design error.

Item 28. The full-load (minimum $R_L$ ) condition should be checked against which limit?

Choices: (A) $I_{ZK}$ (B) $I_{ZM}$ (C) $I_R$ (D) $V_Z$

Given: Full-load design check

Find: Correct limit to check against

Solution:

Step 1: Full-load pulls the maximum load current, leaving the least amount of current for the zener.

Step 2: This lowest zener current must stay above the knee current $I_{ZK}$ , or the zener drops out of breakdown and stops regulating.

✓ ANSWER: A — $I_{ZK}$

Examiner note: No-load checks the maximum limit ( $I_{ZM}$ ); full-load checks the minimum limit ( $I_{ZK}$ ). These two items are a matched pair — learn them together, not separately.

Item 29. How is zener power dissipation calculated?

Choices: (A) $P_Z = V_Z \times I_Z$ (B) $P_Z = V_Z/I_Z$ (C) $P_Z = I_Z^2 \times R$ (D) $P_Z = V_Z^2/I_Z$

Given: Zener diode power dissipation

Find: $P_Z$ formula

Solution:

Step 1: Power dissipated by any two-terminal device equals the voltage across it times the current through it.

Step 2: Apply this directly to the zener.

$P_Z = V_Z \times I_Z$

✓ ANSWER: A — $P_Z = V_Z \times I_Z$

Examiner note: Choice (C), $I_Z^2 \times R$ , is the power formula for a resistor, not a zener diode — the zener does not have a fixed resistance value to plug into that formula.

Item 30. How is load regulation calculated?

Choices: (A) $(V_{NL} - V_{FL})/V_{NL} \times 100\%$ (B) $(V_{NL} - V_{FL})/V_{FL} \times 100\%$ (C) $(V_{FL} - V_{NL})/V_{NL} \times 100\%$ (D) $V_{FL}/V_{NL} \times 100\%$

Given: No-load and full-load output voltages

Find: Load regulation formula

Solution:

Step 1: Recall the load regulation formula.

$\text{Load Regulation} = \dfrac{V_{NL} - V_{FL}}{V_{FL}} \times 100\%$

Step 2: Note that the denominator is specifically the full-load value, not the no-load value.

✓ ANSWER: B — $(V_{NL} - V_{FL})/V_{FL} \times 100\%$

Examiner note: Choice (A) uses $V_{NL}$ as the denominator instead of $V_{FL}$ — the numerator and denominator terms look similar enough that examinees often place the wrong value on the bottom without noticing.

Part G — Voltage Multiplier Circuits (Items 31 to 34)

Full lesson on this topic is in Part 7 — Voltage Multiplier Circuits.

Item 31. What is the ideal output voltage of a half-wave voltage doubler?

Choices: (A) $V_m$ (B) $2V_m$ (C) $3V_m$ (D) $4V_m$

Given: Half-wave voltage doubler

Find: $V_{out}$

Solution:

Step 1: One capacitor clamps the input near its peak; the second capacitor then charges to the sum of the clamped level and the incoming peak.

Step 2: This produces an ideal output of twice the peak input voltage.

$V_{out} \approx 2V_m$

✓ ANSWER: B — $2V_m$

Examiner note: Choice (A), $V_m$ , would only apply to a plain rectifier stage with no doubling action at all.

Item 32. Compared to a half-wave doubler, what does a full-wave doubler offer?

Choices: (A) A higher output voltage (B) Lower ripple and better regulation for the same output (C) A lower output voltage (D) No PIV requirement

Given: Full-wave doubler vs half-wave doubler

Find: The advantage of full-wave

Solution:

Step 1: Both configurations produce the same ideal output, $2V_m$ , so the difference is not in output magnitude.

Step 2: The full-wave doubler recharges its capacitors twice per input cycle instead of once, doubling the ripple frequency and improving regulation.

✓ ANSWER: B — Lower ripple and better regulation for the same output

Examiner note: Choice (A) incorrectly assumes a higher output voltage — both doubler types share the same $2V_m$ formula, so output magnitude is not the distinguishing factor at all.

Item 33. What is the minimum PIV rating required for each diode in a voltage multiplier?

Choices: (A) $V_m$ (B) $2V_m$ (C) $3V_m$ (D) $4V_m$

Given: Voltage multiplier diode PIV requirement

Find: Minimum PIV

Solution:

Step 1: Each diode in a multiplier blocks the sum of the AC input swing and a charged capacitor voltage simultaneously.

Step 2: This combination requires a PIV rating of at least twice the peak input voltage.

$PIV \geq 2V_m$

✓ ANSWER: B — $2V_m$

Examiner note: This requirement does not scale up with the number of stages in the multiplier — it stays fixed at $2V_m$ per diode whether the circuit is a doubler or a quadrupler.

Item 34. A voltage quadrupler produces an ideal output of approximately what?

Choices: (A) $2V_m$ (B) $3V_m$ (C) $4V_m$ (D) $5V_m$

Given: Voltage quadrupler

Find: $V_{out}$

Solution:

Step 1: Recall the general multiplier pattern for $n$ stages.

$V_{out} \approx nV_m$

Step 2: A quadrupler has $n = 4$ stages.

$V_{out} \approx 4V_m$

✓ ANSWER: C — $4V_m$

Examiner note: Match the multiplier’s name directly to its stage count — doubler = 2, tripler = 3, quadrupler = 4 — and the corresponding multiple of $V_m$ follows immediately.

Part H — Special Purpose Diodes (Items 35 to 40)

Full lesson on this topic is in Part 8 — Special Purpose Diodes.

Item 35. Which diode emits light when forward biased?

Choices: (A) Zener (B) LED (C) Schottky (D) Varactor

Given: Five special purpose diode types

Find: The one that emits light

Solution:

Step 1: Recall each diode’s defining characteristic from the identification reference table.

Step 2: Light emission on forward bias is the LED’s single defining property, distinct from all the others.

✓ ANSWER: B — LED

Examiner note: This is direct recall once the five diode types and their one-line characteristics are memorized as a set.

Item 36. What are a Schottky diode’s main advantages?

Choices: (A) High PIV and slow switching (B) Low forward drop and fast switching (C) Negative resistance (D) Light detection

Given: Schottky diode

Find: Its main advantages

Solution:

Step 1: A Schottky diode uses a metal-semiconductor junction instead of a standard p-n junction.

Step 2: This junction type produces a lower forward voltage drop and eliminates minority carrier storage delay, giving faster switching.

✓ ANSWER: B — Low forward drop and fast switching

Examiner note: Choice (C), negative resistance, belongs to the tunnel diode, and choice (D), light detection, belongs to the photodiode — both are distinct diode types covered later in this same post.

Item 37. Which diode’s junction capacitance varies with reverse voltage, making it useful in electronic tuning?

Choices: (A) Photodiode (B) Varactor (C) Tunnel (D) LED

Given: Voltage-controlled capacitance application

Find: Correct diode type

Solution:

Step 1: Recall the varactor’s defining relationship.

$C_j \propto \dfrac{1}{\sqrt{V_R}}$

Step 2: This voltage-controlled capacitance is exploited specifically in tuning circuits, where a DC control voltage adjusts an LC tank’s resonant frequency electronically.

✓ ANSWER: B — Varactor

Examiner note: “Voltage-controlled capacitance” and “electronic tuning” are the standard signal phrases pointing directly to the varactor on identification and application-matching items.

Item 38. A photodiode used for light detection is typically operated in which mode?

Choices: (A) Forward bias (B) Reverse bias (C) No bias (D) Avalanche only

Given: Photodiode detection application

Find: Typical operating mode

Solution:

Step 1: Reverse bias widens the photodiode’s depletion region.

Step 2: A wider depletion region improves both sensitivity and response speed, which is why detection applications use reverse bias rather than forward bias.

✓ ANSWER: B — Reverse bias

Examiner note: Do not confuse this with a solar cell, which is used unbiased or under load to generate power rather than to sense light quickly and accurately.

Item 39. Which diode type exhibits a negative resistance region in its forward characteristic curve?

Choices: (A) Schottky (B) Varactor (C) Tunnel (D) Photodiode

Given: Negative resistance property

Find: Correct diode type

Solution:

Step 1: Recall that heavy doping in the tunnel diode produces a region where forward current decreases as forward voltage increases.

Step 2: No other diode type covered in this series shares this property.

✓ ANSWER: C — Tunnel

Examiner note: Any board exam item describing “current decreasing as voltage increases” in a forward characteristic curve points directly to the tunnel diode — no other option needs serious consideration.

Item 40. Why do blue LEDs have a higher forward voltage than red LEDs?

Choices: (A) Blue light has lower photon energy (B) Blue light has higher photon energy (C) Blue LEDs use a different current rating (D) There is no actual difference

Given: Blue LED vs red LED forward voltage comparison

Find: Explanation for the difference

Solution:

Step 1: Blue light has a higher frequency and shorter wavelength than red light.

Step 2: Higher frequency light carries higher photon energy, which requires a larger junction energy gap — and therefore a higher forward voltage — to produce.

✓ ANSWER: B — Blue light has higher photon energy

Examiner note: This item links LED behavior directly to basic physics rather than circuit math — know the direction of the relationship: higher frequency light means higher forward voltage, not lower.

What to Do With Your Score

Go back to the Diode Applications Series Index and check your raw score against the score interpretation table there. If you missed three or more items in any single part, that part is your weak spot. Go back to that part’s full post, not just this solutions page, and work through one additional similar problem from scratch, not from memory and not by rereading the answer.

If you are reviewing for ECE, give extra weight to Part F. Zener diodes and voltage regulation carry the most points of any single topic in this series, and the no-load versus full-load current check is the single most commonly confused pair of concepts on the actual board exam.

This completes both the Semiconductor Diode Fundamentals and Diode Applications series — together, the full diode syllabus for ELEX1. Follow PinoyBIX on Facebook to get notified when new posts and series go live.

Published by PinoyBIX.org — Engineering Education for Every Filipino Student. Electronics · Mathematics · Board Exam Review · Premium Content.

Please do Subscribe on YouTube!

P inoyBIX educates thousands of reviewers and students a day in preparation for their board examinations. Also provides professionals with materials for their lectures and practice exams. Help me go forward with the same spirit.

“Will you subscribe today via YOUTUBE?”

What You Also Get: FREE ACCESS & DOWNLOAD via GDRIVE

TIRED OF ADS?

Become Premium Member and experienced complete ads-free content browsing.
Full Content Access to Premium Solutions Exclusive for Premium members
Access to PINOYBIX FREEBIES folder
Download Reviewers and Learning Materials Free
Download Content: You can see download/print button at the bottom of each post.

PINOYBIX FREEBIES FOR PREMIUM MEMBERSHIP:

CIVIL ENGINEERING REVIEWER
CIVIL SERVICE EXAM REVIEWER
CRIMINOLOGY REVIEWER
ELECTRONICS ENGINEERING REVIEWER (ECE/ECT)
ELECTRICAL ENGINEERING & RME REVIEWER
FIRE OFFICER EXAMINATION REVIEWER
LET REVIEWER
MASTER PLUMBER REVIEWER
MECHANICAL ENGINEERING REVIEWER
NAPOLCOM REVIEWER
Additional upload reviewers and learning materials are also FREE

FOR A LIMITED TIME

If you subscribe for PREMIUM today!

You will receive an additional 1 month of Premium Membership FREE.

For Bronze Membership an additional 2 months of Premium Membership FREE.

For Silver Membership an additional 3 months of Premium Membership FREE.

For Gold Membership an additional 5 months of Premium Membership FREE.

Join the PinoyBIX community.

DaysHoursMinSec

This offer has expired!

SubscribeMembership Plan

Part A — Rectification (Items 1 to 5)

Item 1. The DC (average) output voltage of a half-wave rectifier is given by what formula?

Item 2. In a full-wave bridge rectifier, the PIV rating required for each diode is what?

Item 3. A full-wave rectifier is fed from a 60 Hz AC source. What is the output ripple frequency?

Item 4. What is the number of diodes required to build a standard full-wave bridge rectifier?

Item 5. In a full-wave center-tapped rectifier, the PIV rating required for each diode is what?

Part B — Diode Configurations (Items 6 to 9)

Item 6. Three silicon diodes are connected in series, all forward biased. Using the practical model, what is the total voltage drop across all three?

Item 7. When using the ON/OFF assumption method to analyze a multi-diode circuit, a diode assumed OFF is verified correct if what condition holds?

Item 8. Two diodes with different threshold voltages are connected in parallel across the same source. Which diode conducts first?

Item 9. In a diode logic gate, a diode conducts only under which condition?

Part C — DC Load Line Analysis and Q-Point (Items 10 to 14)

Item 11. What is the slope of the DC load line on the diode I-V characteristic curve?

Item 12. What does the Q-point (quiescent point) of a diode circuit represent?

Part D — Diode Clippers (Items 15 to 19)

Item 15. In a series positive clipper, what is the output during the positive half-cycle?

Item 16. In a parallel positive clipper, what is the output clamped to during the clipped (positive) half-cycle?

Item 17. In a positive biased clipper, the clipping threshold voltage equals what?

Item 18. The presence of which component signals a clamper rather than a clipper?

Part E — Diode Clampers (Items 20 to 23)

Item 20. Every clamper circuit requires which three elements?

Item 21. For a negative clamper (unbiased, silicon diode), what is the maximum output voltage approximately equal to?

Item 22. What is the RC time constant design guideline for a clamper?

Part F — Zener Diodes and Voltage Regulation (Items 24 to 30)

Item 24. In a zener regulator, how is the current through the series resistor calculated?

Item 26. At the no-load condition (load removed), what is the zener current?

Item 27. The no-load condition should be checked against which zener current limit?

Item 29. How is zener power dissipation calculated?

Item 30. How is load regulation calculated?

Part G — Voltage Multiplier Circuits (Items 31 to 34)

Item 31. What is the ideal output voltage of a half-wave voltage doubler?

Item 32. Compared to a half-wave doubler, what does a full-wave doubler offer?

Item 33. What is the minimum PIV rating required for each diode in a voltage multiplier?

Item 34. A voltage quadrupler produces an ideal output of approximately what?

Part H — Special Purpose Diodes (Items 35 to 40)

Item 35. Which diode emits light when forward biased?

Item 36. What are a Schottky diode’s main advantages?

Item 37. Which diode’s junction capacitance varies with reverse voltage, making it useful in electronic tuning?

Item 38. A photodiode used for light detection is typically operated in which mode?

Item 39. Which diode type exhibits a negative resistance region in its forward characteristic curve?

Item 40. Why do blue LEDs have a higher forward voltage than red LEDs?

What to Do With Your Score

Related Content

201 Terms and Definitions in Algebra for the Mathematics Engineering Board Exam

201 Plane Trigonometry Terms and Definitions | Mathematics Board Exam Review

Lecture 11: Advanced Applied Optimization Problems in Differential Calculus: Business, Engineering, and Real-World Applications

Lecture 8: Related Rates Problems in Calculus – Complete Guide to Dynamic Rate-of-Change Solutions

TIRED OF ADS?

PINOYBIX FREEBIES FOR PREMIUM MEMBERSHIP:

FOR A LIMITED TIME

Add Comment Cancel Reply