Full-wave and bridge rectifiers

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WHAT is a full-wave rectifier?

WHY two topologies? Both achieve the same goal (flip the negative half up). They differ in how they route the current so that current always flows the same direction through the load.

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HOW the center-tapped rectifier works

Picture a transformer secondary split at the middle into a top half and a bottom half, sharing a common center point (the tap = 0 V reference).

• On the positive half-cycle: the top winding is positive, so diode $D_1$ conducts, current flows through $R_L$ top→bottom.
• On the negative half-cycle: the bottom winding is positive, so diode $D_2$ conducts, current flows through $R_L$ in the same direction.

Only one diode conducts at a time, so you lose one diode drop ($0.7$ V for silicon).

$V_{out,peak} = \frac{V_{secondary,peak}}{2} - 0.7\text{ V}$

(the "/2" because each half of the secondary only carries half the total secondary voltage.)

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HOW the bridge rectifier works

Four diodes arranged in a diamond. AC comes in on two corners; the load sits across the other two.

• On the positive half-cycle: diodes $D_1$ and $D_2$ (a diagonal pair) conduct.
• On the negative half-cycle: the other diagonal pair $D_3$ and $D_4$ conduct.

In both cases current is steered through $R_L$ in the same direction.

Two diodes conduct in series each half-cycle, so you lose two diode drops:

$V_{out,peak} = V_{secondary,peak} - 2(0.7\text{ V})$

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DERIVATION: Average (DC) value from first principles

We want the average height of the rectified waveform. The full-wave output is $v(t)=V_m|\sin\omega t|$.

Average over one full period $T=\frac{2\pi}{\omega}$. Because the two humps are identical, average over just half a period $[0,\pi/\omega]$:

$V_{dc}=\frac{1}{\pi/\omega}\int_0^{\pi/\omega} V_m\sin(\omega t)\,dt$

Let $\theta=\omega t$, $d\theta=\omega\,dt$:

$V_{dc}=\frac{V_m}{\pi}\int_0^{\pi}\sin\theta\,d\theta=\frac{V_m}{\pi}\big[-\cos\theta\big]_0^{\pi}=\frac{V_m}{\pi}\big[-(-1)-(-1)\big]$

$\boxed{V_{dc}=\frac{2V_m}{\pi}\approx 0.637\,V_m}$

Why this step? The integral is the "total area" under one hump; dividing by the width gives the average height — that is literally what a DC meter reads.

Compare: half-wave gives $V_{dc}=V_m/\pi\approx0.318V_m$. Full-wave is exactly double — makes sense, we kept both humps.

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DERIVATION: RMS value

$V_{rms}=\sqrt{\frac{1}{\pi}\int_0^{\pi}(V_m\sin\theta)^2\,d\theta}=V_m\sqrt{\frac{1}{\pi}\int_0^{\pi}\sin^2\theta\,d\theta}$

Use $\sin^2\theta=\tfrac{1}{2}(1-\cos2\theta)$:

$\int_0^\pi\sin^2\theta\,d\theta=\frac{1}{2}\Big[\theta-\frac{\sin2\theta}{2}\Big]_0^\pi=\frac{\pi}{2}$

$\boxed{V_{rms}=\frac{V_m}{\sqrt2}\approx0.707\,V_m}$

Why this matters: rectifying doesn't change RMS vs. the full sine, because squaring erases the sign difference.

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DERIVATION: Ripple factor & efficiency

Why derive $r$ this way? Any waveform = DC part + AC (ripple) part. $V_{rms}^2 = V_{dc}^2 + V_{ac,rms}^2$ (they're orthogonal). Rearranging gives $V_{ac,rms}=\sqrt{V_{rms}^2-V_{dc}^2}$, then divide by $V_{dc}$.

Ripple frequency $=2f$ (two humps per input cycle) — key advantage: a smoothing capacitor of the same size filters full-wave ripple far better than half-wave.

Rectification efficiency: $\eta=\dfrac{P_{dc}}{P_{ac}}=\dfrac{(2V_m/\pi)^2}{(V_m/\sqrt2)^2}=\dfrac{4/\pi^2}{1/2}=\dfrac{8}{\pi^2}\approx81.2\%$ (double the half-wave's 40.6%).

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Worked Examples

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Active Recall

Recall Cover and answer

1. Why is full-wave $V_{dc}$ exactly double the half-wave value?
2. How many diodes conduct at once in a bridge? In a center-tapped?
3. Which topology needs the higher PIV diodes, and by what factor?
4. What is the ripple frequency for a 60 Hz supply?
5. Derive $V_{dc}=2V_m/\pi$ in two lines.

Recall Explain to a 12-year-old (Feynman)

Imagine a swing that goes forward and backward. A half-wave rectifier only lets you push it when it swings forward — half the pushes are wasted. A full-wave rectifier is a smart friend who catches the swing on the backswing too and turns that into a forward push. So the swing gets pushed twice as often and moves much more smoothly. The "bridge" version does this with four one-way gates so the swing always ends up going the same way, no matter which direction it started.

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Flashcards

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Connections

• Half-wave rectifiers — the 50%-wasteful predecessor
• Capacitor filter / smoothing — exploits the $2f$ ripple for smaller caps
• Peak Inverse Voltage (PIV) — selects diode ratings
• Diode I-V characteristics — origin of the $0.7$ V drop
• Transformers — center-tap vs. full secondary
• Zener regulator — what comes after smoothing
• RMS and average values — the integral machinery reused here

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, half-wave rectifier AC ka sirf ek half (positive) leta hai aur negative half ko phenk deta hai — matlab aadhi power waste. Full-wave rectifier smart hai: wo negative half ko delete karne ke bajaye usse ulta karke upar flip kar deta hai. Isse output mein ek cycle mein do humps aate hain, isliye ripple frequency $2f$ ho jaati hai aur DC zyada smooth milta hai. Yehi reason hai ki phir chhota capacitor bhi easily filter kar deta hai.

Do designs hote hain. Center-tapped mein transformer ke beech mein ek tap hota hai; jo half positive ho jaaye, uska diode current ko load ke through same direction mein bhej deta hai. Yahan ek time par sirf ek diode conduct karta hai, isliye sirf $0.7$ V drop, lekin har diode ko $2V_m$ ka PIV jhelna padta hai. Bridge design mein 4 diodes ek diamond shape mein lage hote hain; do-do diode ka diagonal pair alternate half-cycles mein on hota hai. Yahan do diode series mein conduct karte hain ($1.4$ V drop) par PIV sirf $V_m$ — isliye high voltage supplies mein bridge sasta padta hai.

Important formulas yaad rakho: $V_{dc}=2V_m/\pi\approx0.637V_m$ (half-wave ka bilkul double), $V_{rms}=V_m/\sqrt2\approx0.707V_m$, aur ripple factor $\approx0.48$ (half-wave ka $1.21$ se kaafi kam). Efficiency $8/\pi^2\approx81\%$. Exam mein trap yeh hota hai ki log same output voltage maan lete hain dono designs ka — nahi! Pehle effective $V_m$ nikaalo, phir sahi number of diode drops subtract karo (center-tap = 1, bridge = 2). Bas yeh dhyan rakhoge to sab sawaal ho jaayenge.