Diode Models and Rectifier Circuits

Learning Objectives

By the end of this lesson you should be able to:

• Pick the right diode model (ideal, constant voltage drop, piecewise linear, exponential) for a given exam question.
• Test whether a diode is forward or reverse biased using a quick open-circuit check.
• Compute peak current, DC output voltage, and peak inverse voltage (PIV) for half-wave and full-wave rectifiers.
• Compare the half-wave, center-tapped, and bridge rectifier topologies on diode count, output, and ripple frequency.
• Explain why a capacitor filter reduces ripple and why the bridge wins in practice.

Quick Review

In Module 1 you learned that any linear network collapses to a Thevenin source $V_{th}$ in series with $R_{th}$. That is exactly the tool we use here: replace everything the diode sees with a Thevenin equivalent, then test the bias. Keep Ohm's law and KVL warm, you will use them in every example below.

Theory

What a diode does

A diode is a two-terminal device that conducts easily in one direction (anode to cathode, forward bias) and blocks in the other (reverse bias). For a silicon diode, conduction starts at roughly 0.6 to 0.7 V across the junction. That single number drives almost all hand analysis in this course.

The four models, from crudest to finest

[!TIP] Exams almost always want the CVD model unless they explicitly say "ideal" or hand you $r_d$. State your assumption in one line and move on. Graders award method marks for that line.

The bias test that never fails

1. Guess a state (ON or OFF) for each diode.
2. Replace ON diodes with a 0.7 V source, OFF diodes with an open circuit.
3. Solve the linear circuit for the diode current or voltage.
4. Check consistency: an ON diode must have $I > 0$; an OFF diode must have $V < 0.7\ \text{V}$. If the check fails, flip the guess and redo.

flowchart TD
    A["🔌<br/>Circuit with diodes"] --> B["Guess ON or OFF"]
    B --> C["Replace diodes<br/>with model"]
    C --> D["Solve with KVL<br/>and Ohm's law"]
    D --> E{"Result<br/>consistent?"}
    E -->|Yes| F["✅<br/>Answer"]
    E -->|No| B

Half-wave rectifier

One diode, one load resistor. It passes the positive half-cycle and blocks the negative one.

• Ideal peak load voltage: $V_m$
• CVD peak load voltage: $V_m - 0.7$
• DC (average) output: $V_{dc} = \dfrac{V_m - 0.7}{\pi}$
• Ripple frequency: $f$ (same as the mains)
• PIV across the diode: $V_m$

Full-wave rectifiers

Two families, same output shape:

• Center-tapped: 2 diodes, needs a center-tapped transformer, PIV is $2V_m - 0.7$ (stressful on the diode).
• Bridge: 4 diodes, two conduct per half-cycle, PIV is about $V_m$, standard in real supplies.

For both, the load sees two humps per mains cycle:

$$ V_{dc} = \frac{2(V_m - V_{drop,total})}{\pi} $$

The ripple frequency doubles to $2f$ (120 Hz on a 60 Hz mains), which is why a bridge is easier to filter: the capacitor refills twice as often and the ripple is far smaller.

Worked Examples

Example 1: Half-wave with CVD model

A 12 V rms transformer secondary feeds one diode and a 1 kΩ load.

1. Peak secondary voltage: $V_m = 12\sqrt{2} = 16.97\ \text{V}$
2. Peak load voltage: $16.97 - 0.7 = 16.27\ \text{V}$
3. Peak load current: $16.27 / 1000 = 16.3\ \text{mA}$
4. DC output: $16.27 / \pi = 5.18\ \text{V}$, so $I_{dc} = 5.18\ \text{mA}$
5. PIV: 16.97 V, so specify a diode rated at least 50 V

Example 2: Bridge rectifier, same transformer

1. Two diodes conduct, total drop $1.4\ \text{V}$
2. Peak load voltage: $16.97 - 1.4 = 15.57\ \text{V}$
3. DC output: $2(15.57)/\pi = 9.91\ \text{V}$, nearly double the half-wave result
4. Ripple frequency: 120 Hz, not 60 Hz

Example 3: Bias check in a DC circuit

A 5 V source drives a 1 kΩ resistor and a diode in series.

Guess ON: $I = (5 - 0.7)/1000 = 4.3\ \text{mA}$. Positive, so the guess holds. Reverse the diode and $I = 0$, an OFF state consistent with $V = -5\ \text{V} < 0.7\ \text{V}$.

==State your diode model first, then solve.== That habit prevents 90 percent of diode mistakes.

Guided Practice

Work each one on paper before opening the answer.

1. Ideal model sanity check. A 9 V source, a 470 Ω resistor, and an ideal forward-biased diode. Find the current.

2. Same circuit, CVD model. Recompute.

3. Half-wave design. $V_m = 20\ \text{V}$, $R = 500\ \Omega$, CVD model. Find $V_{dc}$, $I_{dc}$, and the minimum PIV rating.

4. Bridge comparison. $V_m = 20\ \text{V}$, $R = 500\ \Omega$, CVD model, bridge topology.

5. Spot the error. A student claims the bridge output is about 3.7 V from a 20 V peak input because "four diodes drop 2.8 V."

6. Reasoning question. Why does a bridge rectifier need a larger transformer for the same diode count? Answer in one sentence.

A silicon diode in the constant voltage drop model is replaced by a source of {{c1::0.7 V}} when forward biased.
For a half-wave rectifier, the DC output voltage equals {{c2::V_m}} divided by pi when the diode is ideal.
A bridge rectifier uses {{c3::four}} diodes and conducts through {{c4::two}} of them per half-cycle.
The ripple frequency of a full-wave rectifier is {{c5::twice}} the mains frequency.
The peak inverse voltage seen by the single diode in a half-wave rectifier is approximately {{c6::V_m}}.

[
  { "front": "What is the constant voltage drop for a forward-biased silicon diode?", "back": "About 0.7 V." },
  { "front": "When is the ideal diode model acceptable?", "back": "For quick bias checks or when the 0.7 V drop is negligible next to the supply." },
  { "front": "Which model includes a series resistance r_d?", "back": "The piecewise linear model." },
  { "front": "State the Shockley diode equation.", "back": "I = I_S (e^(V/(nV_T)) - 1), with V_T about 26 mV at room temperature." },
  { "front": "How many diodes conduct at once in a bridge rectifier?", "back": "Two, in series with the load, alternating pairs each half-cycle." },
  { "front": "DC output of an ideal half-wave rectifier?", "back": "V_m divided by pi." },
  { "front": "DC output of an ideal full-wave rectifier?", "back": "2 V_m divided by pi." },
  { "front": "What is the ripple frequency of a half-wave rectifier?", "back": "f, the same as the mains frequency." },
  { "front": "What is the ripple frequency of a full-wave rectifier?", "back": "2f, twice the mains frequency." },
  { "front": "PIV of a half-wave rectifier diode?", "back": "Approximately V_m." },
  { "front": "PIV of a center-tapped full-wave rectifier diode?", "back": "About 2 V_m, which is its main disadvantage." },
  { "front": "Which rectifier needs a center-tapped transformer?", "back": "The two-diode full-wave rectifier." },
  { "front": "What does the bias test require after solving?", "back": "A consistency check: ON diodes need positive current, OFF diodes need less than 0.7 V." },
  { "front": "Why does a capacitor filter work better after a bridge rectifier?", "back": "The capacitor recharges at 2f, so the discharge interval and the ripple are smaller." }
]

Expert Corner

• Always write down the model you are using. Half the marks lost on diode questions come from mixing ideal and CVD results in the same problem.
• Reverse bias means an open circuit, not a short. Students who draw a wire through a reverse-biased diode get spectacularly wrong answers.
• PIV is a rating, not an operating point. Choose at least twice the calculated value for real hardware.
• Sanity-check magnitude: a full-wave output is about double the half-wave output for the same transformer. If your numbers do not show that, recheck the $\pi$ factor.
• The 0.7 V is temperature dependent (about -2 mV per °C). Ignore it for exams, remember it exists in the lab.

Summary & Next Steps

You now have a repeatable routine: replace the diode with a model, test the bias, solve with KVL, and check consistency. For rectifiers, remember the three numbers that carry most exam marks: DC output ($V_m/\pi$ or $2V_m/\pi$), ripple frequency ($f$ or $2f$), and PIV ($V_m$, $2V_m$, or $V_m$ for bridge).

Next lesson moves from two-terminal devices to three: BJT and MOSFET Switching Basics. The bias-testing habit you just built carries straight over, except the control terminal decides the state instead of the polarity across the device.