# 💡 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
| Model | Forward behavior | Use it when |
| --- | --- | --- |
| Ideal switch | 0 V drop, perfect conductor | quick bias checks, "does it conduct at all" |
| Constant voltage drop (CVD) | Fixed 0.7 V drop | default for hand analysis and exams |
| Piecewise linear | 0.7 V plus a small series resistance $r_d$ | the problem gives you $r_d$ |
| Exponential (Shockley) | $I = I_S\left(e^{V/(nV_T)} - 1\right)$ | simulation and small-signal work, $V_T \approx 26\ \text{mV}$ |
> [!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.
```mermaid
flowchart TD
A["🔌
Circuit with diodes"] --> B["Guess ON or OFF"]
B --> C["Replace diodes
with model"]
C --> D["Solve with KVL
and Ohm's law"]
D --> E{"Result
consistent?"}
E -->|Yes| F["✅
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.
Click to reveal answer
$I = 9/470 = 19.1\ \text{mA}$
Click to reveal answer
$I = (9 - 0.7)/470 = 17.7\ \text{mA}$
Click to reveal answer
Peak load $= 19.3\ \text{V}$; $V_{dc} = 19.3/\pi = 6.14\ \text{V}$; $I_{dc} = 12.3\ \text{mA}$; PIV $\approx 20\ \text{V}$, pick a 40 V part for margin.
Click to reveal answer
Peak load $= 20 - 1.4 = 18.6\ \text{V}$; $V_{dc} = 2(18.6)/\pi = 11.8\ \text{V}$; ripple frequency $= 2f$.
Click to reveal answer
Only two diodes conduct per half-cycle, so the total drop is 1.4 V, and the DC formula still divides by $\pi$, not by 4.
Click to reveal answer
Because two diode drops are in series in the conduction path, the peak load voltage is lower than a center-tapped design for the same secondary voltage.