A bridge rectifier contains four ideal diodes A, B, C and D, as shown in Fig. 11.1.

Question 4

A bridge rectifier contains four ideal diodes A, B, C and D, as shown in Fig. 11.1.

Fig. 11.1

The output of the rectifier is connected to a load L of resistance 2.4 kΩ.

(a) On Fig. 11.1, mark with the letter P the positive terminal of the load. [1]

(b) The variation with time t of the potential difference V across the input to the rectifier is shown in Fig. 11. 2.

Fig. 11.2

Calculate the root-mean-square (r.m.s.) current in the load L. [2]

(c) The potential difference across the load L is to be smoothed using a capacitor.

(i) On Fig. 11.1, draw the symbol for a capacitor, connected to produce smoothing. [1]

(ii) The minimum potential difference across the load L with the smoothing capacitor

connected is 3.0 V.

On Fig. 11.2, sketch the variation with time t of the potential difference across the load L.

[3]

[Total: 7]

Reference: Past Exam Paper – June 2016 Paper 42 Q11

Solution:

(a) point P shown at ‘lower end’ of load

{A diode would only allow the current to flow if they point in the same direction.

When current flows from the upper terminal of the supply (upper terminal is positive), currents flows through A (not B), then to the load (not D).

When current flows from the lower terminal of the supply (lower terminal is positive), currents flows through D (not C), then to the load (not A).}

(b)

{ Vr.m.s. = V_max / √2

From the graph, V_max = 6.0 V}

Vr.m.s. = 6.0 / √2 = 4.24 V

{From Ohm’s law: Ir.m.s. = Vr.m.s. / R }

Ir.m.s. = 4.24 / (2.4×10³)

Ir.m.s. = 1.8×10^-3 A

(c)

(i) capacitor in parallel with load

(ii)

line from peak to curve at 3.0 V for either half- or full-wave rectified

correct curvature on line (gradient becoming more shallow)

line drawn as for full-wave rectified

{The diodes are arranged in such a way to produce a full-wave rectifier. That is, the p.d. across the load is always positive.

The capacitor produces a smoothing effect – that is, the positive (varying) p.d. will not return to 0 V.

The question tells us that the minimum p.d. is 3.0 V – the curve cannot go below 3.0 V}