Answer MET Question 23

Question: By means of a schematic circuit diagram illustrate the peak rectifier. If the supply voltage is v(t) =Vm Sin wt, what is the voltage across the load resistor? 

Answer: Peak Rectifiers.
The output of the rectifier circuits is pulsating significantly with time. Hence, it’s not useful as the output from a DC power supply. One way to reduce this ripple is to use a  filtering capacitor.

Rectification
Where direct current is required from an a.c. installation to provide power for d.c. equipment, then a.c. has to be rectified. The process of rectification is one where the flow of current in one direction is permitted to pass, but flow in the reverse direction is resisted, or channelled in the other direction. The modern rectifying device is a semi-conductor junction rectifier. These are pn diodes which, like the thermionic valves and metal rectifiers formerly used, act as electrical non-retun valves when connected to an alternating current circuit. That is, they permit current flow in one direction but resist a reverse flow.

Half-wave rectification
a.c. supply connected to a load with a rectifier in the circuit. When terminal T is positive relative to terminal B, conventional current flows in a direction that agrees with that of the arrow symbol representing the rectifying diode.

Current passes through the rectifier to the load and the rectifier is said to be forward biased. When the situation changes and B is positive relative to T, then current flow in the circuit would tend to be the other way. This flow is resisted by the rectifier. The effect of the single rectifier is to produce half-wave rectification and, as with  alternating current. The half sine waves indicate unidirectional although not continuous flow of current through the load as a result of the pattern of voltage developed. To obtain a d.c. supply with less ripple, the pulsations can be reduced by a capacitor smoothing circuit.

Full-wave rectification
Both half-cycles of the alternating current input can be applied to the load with an arrangement of two diodes and a transformer having a centre tap.

Each pn diode conducts in turn when the end of the secondary winding which supplies it has full potential relative to the centre tap. A high-voltage transformer is needed for this method of full-wave rectification. The double winding is more expensive than the cost of extra rectifiers for a bridge rectifier.

Bridge full-wave rectifier
Four pn diodes in a bridge circuit between the transformer secondary and the load will give full-wave rectification without the need loss centre tap. The diodes work in series pairs to complete a circuit carrying current through the load.

When terminal T of the transformer secondary has higher potential than B, then current follows a path from T through diode D1, to the load and completes its travel through D2, back to terminal B of the secondary. Current flows in the opposite direction when potential B is higher than that of T. The path taken is then from B through D3, to the load and returning via D4. to terminal T. A unidirection current flow is provided for the load and smoothing can be applied to reduce ripple.