explain the use of four diodes (bridge rectifier) for the full-wave rectification of an alternating current

Rectification and Smoothing – Bridge (Full‑Wave) Rectifier

Objective

• Distinguish graphically between half‑wave and full‑wave rectification.
• Explain why a single diode gives half‑wave rectification while four diodes in a bridge give full‑wave rectification.
• Analyse the effect of a single smoothing capacitor on the ripple voltage.

1. Half‑Wave vs. Full‑Wave Rectification

Figure 1 – Graphical comparison of the two rectification types

2. Bridge Rectifier Circuit

The bridge rectifier consists of four identical diodes (D1–D4) arranged in a diamond shape. The AC source is connected to the two opposite corners (labelled AC terminals). The load is connected across the remaining two corners (labelled + and –). The direction of current for each half‑cycle is shown in the schematic below.

3. Operation During One AC Cycle

1. Positive half‑cycle (\(v_{\text{in}} > 0\))

   • D1 and D2 are forward‑biased (conduct); D3 and D4 are reverse‑biased (off).
   • Current path: AC + → D1 → + load → – load → D2 → AC –.
   • Output voltage across the load:

     \[

     v{\text{out}} = v{\text{in}} - 2V_D

     \]

     where \(V_D\) is the forward voltage drop of a silicon diode (≈ 0.7 V).

2. Negative half‑cycle (\(v_{\text{in}} < 0\))

   • D3 and D4 become forward‑biased; D1 and D2 reverse‑biased.
   • Current path: AC + → D3 → + load → – load → D4 → AC –.
   • The polarity of the load voltage is unchanged; its magnitude is the same as in the positive half‑cycle:

     \[

     v{\text{out}} = |v{\text{in}}| - 2V_D .

     \]

4. Resulting Output Waveform

Because a conducting diode pair is present during both halves of the AC cycle, the output consists of a series of positive half‑sine pulses, each lasting half a mains period. The ripple (pulsating‑DC) frequency is therefore:

\[

f{\text{ripple}} = 2f{\text{mains}} .

\]

For a 50 Hz mains supply, the ripple frequency is 100 Hz.

5. Smoothing with a Single Capacitor

5.1 Role of the Filter Capacitor

A capacitor \(C\) placed directly across the load stores charge when the diode pair conducts (the peaks of the waveform) and releases it when the diodes are off (the valleys). This “fills in” the valleys, reducing the ripple voltage \(V_{\text{ripple}}\).

5.2 Derivation of the Ripple‑Voltage Approximation (including load resistance)

During the interval \(\Delta t\) when the diodes are non‑conducting, the capacitor discharges through the load resistance \(R_L\). Assuming the ripple is small, the discharge can be approximated as linear:

\[

\Delta V = \frac{I_{\text{load}}\;\Delta t}{C},

\qquad I{\text{load}} = \frac{V{\text{DC}}}{R_L}.

\]

For a full‑wave rectifier the discharge interval is half a mains period:

\[

\Delta t = \frac{1}{2f_{\text{mains}}}.

\]

Substituting \(I_{\text{load}}\) and \(\Delta t\) gives the peak‑to‑peak ripple:

\[

V{\text{ripple}} \approx \frac{V{\text{DC}}}{RL}\;\frac{1}{2f{\text{mains}}C}

= \frac{I{\text{load}}}{2f{\text{mains}}C}.

\]

5.3 Ripple Factor

The ripple factor \(r\) expresses the ripple as a fraction of the average (DC) output:

\[

r = \frac{V{\text{ripple}}}{V{\text{DC}}}

= \frac{1}{2f{\text{mains}}RL C}.

\]

A smaller \(r\) means a smoother DC supply.

5.4 Influence of Diode Forward Drop

The two forward‑biased diodes in each half‑cycle reduce the maximum voltage to which the capacitor can charge:

\[

V{\text{peak,\,cap}} = V{\text{peak,\,AC}} - 2V_D .

\]

Consequently the average DC output is also reduced by roughly \(2V_D\). The ripple formula above is unchanged because it depends on the load current, not directly on the peak voltage.

6. Example Calculation

1. DC (average) output voltage

   \[

   V{\text{DC}} \approx V{\text{peak}} - 2V_D

   = 14.1 - 1.4 = 12.7\ \text{V}

   \]

2. Load current

   \[

   I{\text{load}} = \frac{V{\text{DC}}}{R_L}

   = \frac{12.7\ \text{V}}{1.0\ \text{kΩ}}

   = 12.7\ \text{mA}

   \]

3. Peak‑to‑peak ripple

   \[

   V{\text{ripple}} \approx \frac{I{\text{load}}}{2fC}

   = \frac{12.7\times10^{-3}}{2\times50\times1000\times10^{-6}}

   \approx 0.127\ \text{V}

   \]

4. Ripple factor

   \[

   r = \frac{V{\text{ripple}}}{V{\text{DC}}}

   = \frac{0.127}{12.7}

   \approx 0.01\;(1\%)

   \]

The ripple is only about 0.13 V (≈ 1 % of the DC level), which is sufficiently smooth for most low‑power electronics used in Cambridge AS & A‑Level experiments.

7. Advantages of the Bridge (Full‑Wave) Rectifier

• Both halves of the AC cycle are utilised → higher average output voltage and better utilisation of the transformer.
• No centre‑tapped transformer is required, reducing cost, size and weight.
• Ripple frequency is doubled (2f), making filtering with a given capacitor more effective.
• When combined with a single smoothing capacitor, a low ripple factor can be achieved with modest component values.

8. Summary

A bridge rectifier uses four diodes to convert an AC input into a full‑wave rectified output. The output consists of positive half‑sine pulses occurring twice per mains cycle, giving a ripple frequency of \(2f{\text{mains}}\). Each conducting diode pair introduces a total forward drop of \(2VD\), reducing the peak and average output voltage accordingly. Adding a filter capacitor across the load smooths the waveform; the peak‑to‑peak ripple is approximated by

\[

V{\text{ripple}} \approx \frac{I{\text{load}}}{2f_{\text{mains}}C},

\]

and the ripple factor by

\[

r = \frac{1}{2f{\text{mains}}RL C}.

\]

These relationships enable the design of reliable low‑voltage DC supplies for Cambridge IGCSE and A‑Level physics investigations.