Half-Wave and Full-Wave Rectifiers

About this problem set

Worked problems on half-wave, centre-tapped and bridge rectifiers: DC and rms output, ripple factor, form factor, rectification efficiency, PIV, transformer utilisation factor, the cost of the silicon diode drop, a transformer/diode design, and regulation when the winding and diode resistances are not negligible.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

PROBLEM 01

Half-Wave Rectifier Figures of Merit

Problem Statement

A half-wave rectifier is fed from a transformer whose secondary delivers \( 24~\text{V} \) rms at \( 50~\text{Hz} \). The load is a pure resistance \( R_L = 500~\Omega \) and the diode may be treated as ideal.

Find the peak, DC and rms load voltages.
Find \( I_{dc} \) and \( I_{rms} \).
Find the form factor and the ripple factor.
Find the rectification efficiency and the peak inverse voltage.

Solution

Secondary voltage \( V_s = 24~\text{V} \) (rms), \( f = 50~\text{Hz} \)
\( R_L = 500~\Omega \), ideal diode \( (V_D = 0,\; r_f = 0) \)

Step 1 — Peak values

The secondary rms value is converted to a peak, and the peak load current follows from Ohm's law.

\[\begin{aligned} V_m &= \sqrt{2}\,V_s = \sqrt{2}\times 24 = 33.9~\text{V} \\ I_m &= \frac{V_m}{R_L} = \frac{33.9}{500} = 67.9~\text{mA} \end{aligned}\]

Step 2 — DC and rms values

A half-wave waveform is one sine loop per period, so the average is \( V_m/\pi \) and the rms is \( V_m/2 \) (not \( V_m/\sqrt{2} \) — half the period is zero).

\[\begin{aligned} V_{dc} &= \frac{V_m}{\pi} = \frac{33.9}{\pi} = 10.8~\text{V} \\ I_{dc} &= \frac{I_m}{\pi} = 21.6~\text{mA} \\ V_{rms} &= \frac{V_m}{2} = 17~\text{V},\qquad I_{rms} = \frac{I_m}{2} = 33.9~\text{mA} \end{aligned}\]

(a) \( V_m = 33.9~\text{V} \), \( V_{dc} = 10.8~\text{V} \), \( V_{rms} = 17~\text{V} \)

(b) \( I_{dc} = 21.6~\text{mA} \), \( I_{rms} = 33.9~\text{mA} \)

Step 3 — Form factor and ripple factor

\[\begin{aligned} \text{FF} &= \frac{I_{rms}}{I_{dc}} = \frac{I_m/2}{I_m/\pi} = \frac{\pi}{2} = 1.571 \\ r &= \sqrt{\text{FF}^2 - 1} = \sqrt{2.467 - 1} = 1.211 \end{aligned}\]

Step 4 — Efficiency and PIV

Efficiency is the ratio of DC power in the load to the total power delivered to it.

\[\begin{aligned} \eta &= \frac{P_{dc}}{P_{ac}} = \frac{I_{dc}^2 R_L}{I_{rms}^2 R_L} = \frac{(I_m/\pi)^2}{(I_m/2)^2} = \frac{4}{\pi^2} = 40.53\% \\ P_{dc} &= I_{dc}^2 R_L = 233~\text{mW},\qquad P_{ac} = I_{rms}^2 R_L = 576~\text{mW} \end{aligned}\]

When the diode is off, the whole secondary peak appears across it while the load sits at 0 V, so \( \text{PIV} = V_m \).

(d) \( \eta = 40.5\% \), \( \text{PIV} = 33.9~\text{V} \)

A ripple factor of 1.21 means the ripple is larger than the DC it rides on — a half-wave rectifier is useless without a filter, and its \( 40.5\% \) ceiling on efficiency is a property of the waveform, not of the diode.

PROBLEM 02

Effect of the 0.7 V Diode Drop

Problem Statement

A half-wave rectifier uses a silicon diode \( (V_D = 0.7~\text{V}) \) fed from a \( 12~\text{V} \) rms secondary into \( R_L = 1~\text{k}\Omega \).

Find \( V_{dc} \) using the usual approximation \( V_{dc} = (V_m - V_D)/\pi \), and compare with the ideal-diode answer.
The diode does not conduct for the full half cycle. Find the conduction angle and the exact \( V_{dc} \) obtained by averaging only over the conduction interval.
Is the approximation of part (a) safe here?

Solution

\( V_s = 12~\text{V} \) rms \( \Rightarrow V_m = \sqrt{2}\times 12 = 17~\text{V} \)
\( V_D = 0.7~\text{V} \) (constant-voltage-drop model), \( R_L = 1~\text{k}\Omega \)

Step 1 — Approximate DC output

Subtract the drop from the peak, then average as if the loop were still a full half sine.

\[\begin{aligned} v_{L,\text{peak}} &= V_m - V_D = 17 - 0.7 = 16.3~\text{V} \\ V_{dc} &\approx \frac{V_m - V_D}{\pi} = \frac{16.3}{\pi} = 5.18~\text{V} \end{aligned}\]

With an ideal diode the same circuit would give \( V_m/\pi = 5.4~\text{V} \), so the drop costs \( 4.12\% \) of the output.

(a) \( V_{dc} \approx 5.18~\text{V} \) versus \( 5.4~\text{V} \) for an ideal diode

Step 2 — Conduction angle

The diode turns on only once \( V_m\sin\theta \) exceeds \( V_D \).

\[\begin{aligned} \theta_1 &= \sin^{-1}\!\left(\frac{V_D}{V_m}\right) = \sin^{-1}\!\left(\frac{0.7}{17}\right) = 2.36^\circ \\ \text{conduction} &= \pi - 2\theta_1 = 180^\circ - 2(2.36^\circ) = 175.3^\circ \end{aligned}\]

Step 3 — Exact average

Average \( (V_m\sin\theta - V_D) \) over \( \theta_1 \to \pi-\theta_1 \) and divide by the full period \( 2\pi \).

\[\begin{aligned} V_{dc} &= \frac{1}{2\pi}\int_{\theta_1}^{\pi-\theta_1}\left(V_m\sin\theta - V_D\right)d\theta \\ &= \frac{1}{2\pi}\Big[\,2V_m\cos\theta_1 - V_D(\pi - 2\theta_1)\,\Big] \\ &= \frac{1}{2\pi}\Big[\,2(17)(0.99915) - 0.7(3.059)\Big] = 5.06~\text{V} \\ I_{dc} &= \frac{V_{dc}}{R_L} = 5.06~\text{mA} \end{aligned}\]

(b) Conduction angle \( = 175.3^\circ \), exact \( V_{dc} = 5.06~\text{V} \), \( I_{dc} = 5.06~\text{mA} \)

Step 4 — Verdict

The two answers differ by \( 0.123~\text{V} \), i.e. the simple formula runs \( 2.42\% \) high. The reason is visible in the algebra: the approximation removes only \( V_D/\pi = 0.223~\text{V} \) from the average, whereas the diode actually removes \( V_D(\pi - 2\theta_1)/2\pi = 0.341~\text{V} \) — close to \( V_D/2 \), because the drop is present only while the diode conducts.

(c) Acceptable for engineering work — the error is \( 2.42\% \) here and shrinks as \( V_m/V_D \) grows, but it becomes serious for a low-voltage secondary of only a few volts.

Note the PIV is unchanged at \( V_m = 17~\text{V} \): on the negative half cycle the load carries no current, so the entire secondary voltage stands across the diode.

PROBLEM 03

Centre-Tapped Full-Wave Rectifier and Its PIV

Problem Statement

A centre-tapped full-wave rectifier uses a transformer with \( 18\text{-}0\text{-}18~\text{V} \) rms secondary (\( 18~\text{V} \) rms from the tap to each end) and two silicon diodes. \( R_L = 100~\Omega \), \( f = 50~\text{Hz} \).

Find \( V_{dc} \), \( I_{dc} \) and \( I_{rms} \).
Find the ripple factor and the rectification efficiency.
Find the PIV of each diode and explain why it is \( 2V_m \) and not \( V_m \).

Solution

Each half secondary: \( 18~\text{V} \) rms \( \Rightarrow V_m = \sqrt{2}\times 18 = 25.5~\text{V} \)
One diode in the conduction path at a time \( \Rightarrow \) one drop, \( V_D = 0.7~\text{V} \)
\( R_L = 100~\Omega \), ripple frequency \( = 2f = 100~\text{Hz} \)

Step 1 — Peak load quantities

\[\begin{aligned} v_{L,\text{peak}} &= V_m - V_D = 25.5 - 0.7 = 24.8~\text{V} \\ I_m &= \frac{24.8}{100} = 248~\text{mA} \end{aligned}\]

Step 2 — DC and rms values

Both half cycles now reach the load, so the average doubles relative to half-wave and the rms becomes the full-sine value \( I_m/\sqrt{2} \).

\[\begin{aligned} V_{dc} &= \frac{2(V_m - V_D)}{\pi} = \frac{2(24.8)}{\pi} = 15.8~\text{V} \\ I_{dc} &= \frac{2I_m}{\pi} = 158~\text{mA} \\ I_{rms} &= \frac{I_m}{\sqrt{2}} = 175~\text{mA} \end{aligned}\]

(a) \( V_{dc} = 15.8~\text{V} \), \( I_{dc} = 158~\text{mA} \), \( I_{rms} = 175~\text{mA} \)

Step 3 — Ripple factor and efficiency

\[\begin{aligned} \text{FF} &= \frac{I_{rms}}{I_{dc}} = \frac{I_m/\sqrt{2}}{2I_m/\pi} = \frac{\pi}{2\sqrt{2}} = 1.111 \\ r &= \sqrt{\text{FF}^2 - 1} = 0.4834 \\ \eta &= \frac{I_{dc}^2 R_L}{I_{rms}^2 R_L} = \frac{8}{\pi^2} = 81.06\% \end{aligned}\]

(b) \( r = 0.483 \; (48.3\%) \), \( \eta = 81.1\% \)

Step 4 — Peak inverse voltage

Take the instant when the upper half of the secondary is at \( +V_m \). Diode \( D_1 \) conducts, so the cathode of the off diode \( D_2 \) sits at \( V_m - V_D \) (the load voltage). But \( D_2 \)'s anode is tied to the lower end of the secondary, which is at \( -V_m \) with respect to the tap. The reverse voltage across \( D_2 \) is therefore the whole secondary, end to end.

\[\begin{aligned} \text{PIV} &= 2V_m - V_D = 2(25.5) - 0.7 = 50.2~\text{V} \end{aligned}\]

This doubled PIV is the price of the centre tap: each half winding is idle for half the cycle (rms current only \( I_m/2 = 124~\text{mA} \) per half), and the diodes must be rated for twice the peak the bridge demands.

PROBLEM 04

Bridge Versus Centre-Tap: TUF and PIV

Problem Statement

The same \( 18~\text{V} \) rms secondary is used in two ways: (i) as a single winding feeding a silicon bridge rectifier, and (ii) as an \( 18\text{-}0\text{-}18~\text{V} \) centre-tapped winding feeding two diodes. \( R_L = 100~\Omega \) in both cases.

Find \( V_{dc} \) and the PIV for the bridge.
Derive the transformer utilisation factor for the bridge and for the centre-tapped secondary, and state the form factor of a full-wave output.
Tabulate the comparison and say when each topology is preferred.

Solution

\( V_m = \sqrt{2}\times 18 = 25.5~\text{V} \) (peak of the winding used)
Bridge: two diodes conduct in series each half cycle \( \Rightarrow 2V_D = 1.4~\text{V} \)
Centre tap: one diode conducts \( \Rightarrow V_D = 0.7~\text{V} \), but only half the winding is used

Step 1 — Bridge output

\[\begin{aligned} v_{L,\text{peak}} &= V_m - 2V_D = 25.5 - 1.4 = 24.1~\text{V} \\ V_{dc} &= \frac{2(V_m - 2V_D)}{\pi} = 15.3~\text{V} \\ I_{dc} &= \frac{V_{dc}}{R_L} = 153~\text{mA},\qquad I_{rms} = \frac{I_m}{\sqrt{2}} = 170~\text{mA} \end{aligned}\]

When a bridge diode is reverse biased, its two conducting partners clamp it across the load, so it sees only the peak load voltage plus one forward drop:

\[\begin{aligned} \text{PIV} &= V_m - V_D = 24.8~\text{V} \;\approx\; V_m \end{aligned}\]

(a) Bridge: \( V_{dc} = 15.3~\text{V} \), \( \text{PIV} \approx V_m = 25.5~\text{V} \)

Step 2 — Transformer utilisation factor

\( \text{TUF} = P_{dc}/(\text{VA rating of the winding}) \). For the bridge the secondary carries a full sine, \( I_{rms} = I_m/\sqrt{2} \):

\[\begin{aligned} \text{TUF}_{\text{bridge}} &= \frac{(2I_m/\pi)^2 R_L}{\left(\dfrac{V_m}{\sqrt2}\right)\left(\dfrac{I_m}{\sqrt2}\right)} = \frac{4I_m^2R_L/\pi^2}{I_m^2R_L/2} = \frac{8}{\pi^2} = 0.8106 \end{aligned}\]

For the centre tap each half winding conducts only alternate half cycles, so its rms current is \( I_m/2 \) and the total secondary VA is \( 2\times(V_m/\sqrt2)(I_m/2) \):

\[\begin{aligned} \text{TUF}_{\text{sec}} &= \frac{4I_m^2R_L/\pi^2}{V_mI_m/\sqrt2} = \frac{4\sqrt2}{\pi^2} = 0.5732 \\ \text{TUF}_{\text{pri}} &= \frac{8}{\pi^2} = 0.8106 \\ \overline{\text{TUF}} &= \tfrac12\left(0.5732 + 0.8106\right) = 0.6919 \end{aligned}\]

Form factor of any full-wave sine output:

\[\begin{aligned} \text{FF} &= \frac{V_m/\sqrt2}{2V_m/\pi} = \frac{\pi}{2\sqrt2} = 1.111\quad\text{(half-wave: } \pi/2 = 1.571\text{)} \end{aligned}\]

(b) \( \text{TUF}_{\text{bridge}} = 0.811 \), \( \text{TUF}_{\text{centre-tap}} = 0.692 \), \( \text{FF} = 1.111 \)

Step 3 — Comparison

Quantity	Centre-tapped FW	Bridge
Diodes	2	4
Winding needed	18-0-18 V	18 V
Drops in series	1 \( (0.7~\text{V}) \)	2 \( (1.4~\text{V}) \)
\( V_{dc} \) here	\( 15.8~\text{V} \)	\( 15.3~\text{V} \)
PIV	\( 2V_m = 50.9~\text{V} \)	\( V_m = 25.5~\text{V} \)
TUF	\( 0.692 \)	\( 0.811 \)
Ripple factor	\( 0.483 \) (same waveform)
\( \eta_{max} \)	\( 81.1\% \)

(c) The bridge gives \( 1.17\times \) better copper utilisation and half the PIV, at the cost of \( 0.446~\text{V} \) more DC output lost in the extra diode drop.

Rule of thumb: use the bridge for anything above about 12 V DC, where an extra 0.7 V is negligible; use the centre tap for low-voltage supplies where that 0.7 V is a large fraction of the output.

PROBLEM 05

Design of a 12 V Bridge Rectifier

Problem Statement

Design. A silicon bridge rectifier working from the \( 230~\text{V} \), \( 50~\text{Hz} \) mains must deliver \( 12~\text{V} \) DC into a resistive load drawing \( 500~\text{mA} \). No filter is used.

Choose the transformer secondary voltage and the turns ratio.
Specify the average forward current, rms current and PIV each diode must withstand, and pick a standard rectifier from the 1N400x family.
Find the power dissipated in each diode and the secondary VA rating required.

Solution

Required \( V_{dc} = 12~\text{V} \) at \( I_{dc} = 500~\text{mA} \Rightarrow R_L = 24~\Omega \)
Bridge \( \Rightarrow \) two silicon drops in the conduction path, \( 2V_D = 1.4~\text{V} \)
Mains \( 230~\text{V} \) rms primary, \( f = 50~\text{Hz} \)

Step 1 — Work backwards from the DC specification

Invert \( V_{dc} = 2(V_m - 2V_D)/\pi \) to find the peak the load must see, then add the diode drops back to get the secondary peak.

\[\begin{aligned} v_{L,\text{peak}} &= \frac{\pi V_{dc}}{2} = \frac{\pi (12)}{2} = 18.8~\text{V} \\ V_m &= v_{L,\text{peak}} + 2V_D = 18.8 + 1.4 = 20.2~\text{V} \\ V_s &= \frac{V_m}{\sqrt2} = 14.3~\text{V (rms)} \end{aligned}\]

\[\begin{aligned} n &= \frac{N_p}{N_s} = \frac{230}{14.3} = 16.06\;\approx\; 16:1 \end{aligned}\]

The nearest standard secondary is \( 15~\text{V} \) rms, for which \( n = 230/15 = 15.3:1 \) and the delivered output becomes

\[\begin{aligned} V_{dc} &= \frac{2(15\sqrt2 - 1.4)}{\pi} = 12.6~\text{V}\quad(5.1\%\text{ high — a useful margin}) \end{aligned}\]

(a) Ideal secondary \( = 14.3~\text{V} \) rms (\( n = 16.1:1 \)); specify the standard \( 15~\text{V} \) secondary, \( n = 15.3:1 \), giving \( 12.6~\text{V} \) DC.

Step 2 — Diode ratings

Each diode conducts on alternate half cycles, so it carries half the load's average current; the rms current in a diode is the half-wave value \( I_m/2 \).

\[\begin{aligned} I_m &= \frac{v_{L,\text{peak}}}{R_L} = \frac{18.8}{24} = 785~\text{mA} \\ I_{F(av)} &= \frac{I_{dc}}{2} = 250~\text{mA} \\ I_{D,rms} &= \frac{I_m}{2} = 393~\text{mA} \\ \text{PIV} &= V_m - V_D = 19.5~\text{V} \end{aligned}\]

Applying the usual \( 2\times \) safety margin on both current and voltage gives \( \ge 0.5~\text{A} \) and \( \ge 40.5~\text{V} \).

(b) \( I_{F(av)} = 250~\text{mA} \), \( \text{PIV} = 19.5~\text{V} \) → a 1N4001 (\( 1~\text{A} \), \( 50~\text{V} \)) is comfortably adequate.

Step 3 — Dissipation and transformer VA

\[\begin{aligned} P_D &= V_D \times I_{F(av)} = 0.7 \times 0.25 = 0.175~\text{W per diode} \\ I_{s,rms} &= \frac{I_m}{\sqrt2} = 555~\text{mA} \\ \text{VA}_{sec} &= V_s I_{s,rms} = 14.3 \times 0.5554 = 7.95~\text{VA} \end{aligned}\]

(c) \( P_D = 175~\text{mW} \) per diode; secondary rating \( \approx 7.95~\text{VA} \) — specify a \( 15~\text{V} \), \( 1~\text{A} \) transformer.

Sanity check on the utilisation: the winding is rated \( 7.95~\text{VA} \) to deliver \( V_{dc}I_{dc} = 6~\text{W} \), a ratio of \( 0.755 \). That sits a little under the ideal bridge TUF of \( 0.811 \), as it must once the \( 2V_D \times I_{dc} = 0.7~\text{W} \) dissipated in the diode drops is taken out of the winding's output.

PROBLEM 06

Voltage Regulation with Finite Source Resistance

Problem Statement

A centre-tapped full-wave rectifier has \( V_m = 30~\text{V} \) peak per half secondary. The half-winding resistance is \( r_s = 2~\Omega \) and each diode has a forward (bulk) resistance \( r_f = 1.5~\Omega \). Model the diodes as ideal switches in series with \( r_f \) — the \( 0.7~\text{V} \) offset is neglected so the regulation caused by the resistances alone is isolated. \( R_L = 50~\Omega \).

Find \( I_{dc} \) and the full-load DC output voltage.
Find the percentage voltage regulation and confirm it against \( \%\text{Reg} = (r_f + r_s)/R_L \).
Find the rectification efficiency, and recompute the regulation if the load is changed to \( 25~\Omega \).

Solution

\( V_m = 30~\text{V} \), \( r_s = 2~\Omega \), \( r_f = 1.5~\Omega \), \( R_L = 50~\Omega \)
Full wave \( \Rightarrow I_{dc} = 2I_m/\pi \) with \( I_m = V_m/(R_L + r_f + r_s) \)

Step 1 — Peak and DC current

The winding and diode resistances are in series with the load throughout conduction, so they simply add to \( R_L \).

\[\begin{aligned} R_{total} &= R_L + r_f + r_s = 50 + 1.5 + 2 = 53.5~\Omega \\ I_m &= \frac{V_m}{R_{total}} = \frac{30}{53.5} = 0.5607~\text{A} \\ I_{dc} &= \frac{2I_m}{\pi} = 0.357~\text{A} = 357~\text{mA} \\ V_{dc(FL)} &= I_{dc}R_L = 0.357 \times 50 = 17.8~\text{V} \end{aligned}\]

(a) \( I_{dc} = 357~\text{mA} \), \( V_{dc} = 17.8~\text{V} \)

Step 2 — Regulation

At no load no current flows, so nothing is dropped in \( r_f + r_s \) and the output rises to the ideal value.

\[\begin{aligned} V_{dc(NL)} &= \frac{2V_m}{\pi} = \frac{60}{\pi} = 19.1~\text{V} \\ \%\text{Reg} &= \frac{V_{dc(NL)} - V_{dc(FL)}}{V_{dc(FL)}}\times 100 = \frac{19.1 - 17.8}{17.8}\times 100 = 7\% \end{aligned}\]

Check with the closed form. Since \( V_{dc} = 2V_m/\pi - I_{dc}(r_f+r_s) \), the rectifier behaves as an ideal \( 2V_m/\pi \) source behind \( (r_f + r_s) \):

\[\begin{aligned} \%\text{Reg} &= \frac{r_f + r_s}{R_L}\times 100 = \frac{3.5}{50}\times 100 = 7\% \end{aligned}\]

(b) \( \%\text{Reg} = 7\% \) — both routes agree.

Step 3 — Efficiency, and a heavier load

\[\begin{aligned} I_{rms} &= \frac{I_m}{\sqrt2} = 0.3965~\text{A} \\ P_{dc} &= I_{dc}^2R_L = 6.372~\text{W},\qquad P_{loss} = I_{rms}^2(r_f + r_s) = 0.5503~\text{W} \\ \eta &= \frac{P_{dc}}{I_{rms}^2R_{total}} = \frac{8/\pi^2}{1 + (r_f+r_s)/R_L} = 75.8\% \end{aligned}\]

\[\begin{aligned} R_L = 25~\Omega:\quad I_{dc} &= \frac{2(30)}{\pi(25 + 3.5)} = 0.6701~\text{A} \\ V_{dc} &= 16.8~\text{V},\qquad \%\text{Reg} = \frac{3.5}{25}\times 100 = 14\% \end{aligned}\]

Regulation is inversely proportional to \( R_L \): the heavier the load, the worse the droop. Good regulation therefore demands \( r_f + r_s \ll R_L \), which is why mains supplies use fat secondary wire and low-\( r_f \) rectifiers.