Topic 3 -  Rectification

In the exam you are expected to know about:

• Half wave rectification;

• Full wave rectification;

• The bridge rectifier;

• Choice of suitable diodes from specifications;

• Effect of a capacitor on output waveform;

• How ripple voltage and current are affected by the value of a capacitor.

 Rectification is the process by which alternating current, usually from the mains supply, is turned into direct current.  Alternating Current can be easily converted to direct current using a diode as a rectifier. 

Half Wave Rectification

The incorporation of a single diode is the simplest way of rectifying AC.  Only the forward half-cycle is passed.  The reverse is blocked.  The rectified current is a series of pulses.  This is adequate for a crude circuit, for example the low voltage fan motor for a hair dryer.

Question 1 What is the advantage of half wave rectification? ANSWER

Full wave Rectification

In the case above we see that both forward and reverse half cycles are rectified. 

·        Two half-wave rectifiers are placed back to back.

·        The load is connected to a centre tapping of the transformer.

·        This is called a centre-tap full-wave rectifier.

·        It always needs a transformer with a centre tap.

A centre tapping  is where a connection is made to the secondary coil half way between the outside terminals.  This allows for an arbitrary zero point for the voltage.  During the forward half cycle, the top terminal will be positive relative to the centre tap, while the bottom terminal will be negative relative to the centre tap, and vice versa for the reverse half cycle.

A more common method to get full wave rectification is to use four diodes in a bridge rectifier circuit, two allowing current from each half cycle to pass.

Question 2 Give one advantage and one disadvantage of using the method above to rectify an alternating current to a direct current, compared to half wave rectification.

ANSWER

Bridge Rectifier

The arrows show the forward and reverse half cycles:

In this circuit we have four diodes, A, B, C, and D. 

·        In the forward half-cycle, shown by the black arrows, we can see that both B and D conduct. The rectified current goes clockwise through the load

·        In the reverse half-cycle, shown by the grey arrows, we can see that both A and C conduct.  Because the two diodes are conducting on the reverse half cycle, the current then passes through the load clockwise.

·        The load current goes clockwise through the load regardless of whether it is being driven by the forward or reverse half cycle.

The two photographs show a bridge rectifier using LED’s.  Note how two LED’s on

the opposite sides of the bridge are glowing:  

On this diagram the input AC waveform is shown for comparison.  The CRO would show the rectified AC as:

If we looked at the CRO trace in reality, we would not be able to distinguish the forward or backward half-cycles from the trace. 

The rectified rms AC voltage will be about 1.2 V less than the input rms voltage as there is a 0.6 V voltage drop across each diode.

Since this is still a sinusoidal waveform, the DC equivalent voltage would be the rms voltage, not the peak voltage.

Bridge rectifiers usually come in a ready made pack, rather than having to solder four diodes together.  On the pack the AC inputs and the positive and negative terminals are marked. 

When selecting a component, we need to assess:

·        Working voltages.

·        Current that will be passed.

·        What kind of cooling will be needed.

Heavy-duty rectifiers have heat sinks, which are fins to conduct excess heat to the air.  Sometimes in high power applications, a fan may be used.  We need to check the specifications for the diodes when we plan the circuits.  Any electronics catalogue will provide these.

Question 3 What physics processes do heat sinks use?  Explain how their structure is suitable for these processes. ANSWER

The table shows some data for diodes:

If we wanted to use a diode in a circuit that contains a 50 V motor that takes a current of 4.0 A, we would need to chose the P600D diode.  The 1N5401 has a sufficient working voltage, but can only take a current of 3 A.  If we tried to use the 1N5401 diode with the motor above, it would get very hot and would burn out quickly.

Question 4 A diode is required to work at a voltage of 600 V, and needs to supply a current of 0.35 A.  Which one would you use?  Explain your answer.  ANSWER

 Now go on to Capacitive Smoothing

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