A sinusoidal alternating voltage is to be rectified. (a) Suggest one advantage of full-wave rectification as compared with half-wave rectification.

Question 5

A sinusoidal alternating voltage is to be rectified.

(a) Suggest one advantage of full-wave rectification as compared with half-wave rectification. [1]

(b) The rectification is produced using the circuit of Fig. 7.1.

Fig. 7.1

All the diodes may be considered to be ideal.

The variation with time t of the alternating voltage applied to the circuit is shown in Fig. 7.2 and in Fig. 7.3.

(i) On the axes of Fig. 7.2, draw a graph to show the variation with time t of the potential difference across diode A. [1]

(ii) On the axes of Fig. 7.3, draw a graph to show the variation with time t of the potential difference across diode B. [1]

(c)

(i) On Fig. 7.1, draw the symbol for a capacitor, connected into the circuit so as to provide smoothing. [1]

(ii) Fig. 7.4 shows the variation with time t of the smoothed potential difference across the resistor R in Fig. 7.1.

1. State how the amount of smoothing may be increased. [1]

2. On Fig. 7.4, draw the variation with time t of the potential difference across resistor R for increased smoothing. [2]

Reference: Past Exam Paper – November 2009 Paper 41 Q7

Solution:

(a) Example:

More (output) power is available

There is less ripple for same smoothing capacitor

(any sensible suggestion)

(b)

(i) The curve should be drawn showing half-wave rectification

{Diode A conducts only when the upper terminal of the supply is positive. That is, when the line of the alternating voltage applied is positive, current flows and the p.d. across the resistor is positive. This means that the p.d. across the diode is ZERO. Here, we want to plot the graph of p.d. across the diode, not the resistor.}

(ii) The curve is similar to (i) but with a phase shift of 180°

{Diode B conducts only when the lower terminal of the supply is positive. That is, when the line of the alternating voltage applied is negative, current flows through the resistor and the p.d. across the resistor is positive. This means that the p.d. across the diode is ZERO. Again, here want to plot the graph of p.d. across the diode. So, the p.d. across the diode is positive when the applied alternating voltage is positive (that is, when the diode is reverse-biased).}

(c)

(i) A correct symbol, connected in parallel with R    

(ii)

1. By using a larger capacitor / second capacitor in parallel with R

(not increase R)

2. The curve has the same peak values, with a correct shape giving less ripple

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}