Half Wave Rectifier Circuit

Half Wave Rectifier Circuit

Q. What is a half wave rectifier circuit?

Q. How does a half wave rectifier circuit work?

Q. What is use of half wave rectifier?

Q. How is a diode used as a half wave rectifier?

• The simplest circuit for generation of high direct voltage is the half wave rectifier shown in figure A.

Figure A

Figure B

Figure C

R_L – Load resistance,

C – Capacitance to smoothen the dc output voltage.

• If the capacitor is not connected, pulsating dc voltage is obtained at the output terminals whereas with the capacitance C the pulsation is reduced at the output.

Working of Half Wave Rectifier Circuit :

1. Assuming an ideal transformer and small internal resistance of the diode during conduction the capacitance C is charged to the maximum voltage V_max during conduction of diode D.
2. Assuming that there is No Load (NL) connected, the dc voltage across capacitance remains constant at V_max whereas the supply voltage oscillates between ±  and during negative half cycle the potential of point A becomes – V_max  and hence diode must be rated for 2 V_max. (This would also be the case if the transformer is grounded at A instead of B as shown in Figure A.
3. Such a circuit is known as Villard’s voltage Doubler circuit in which the output voltage would be taken across D, as shown in Figure D.
4. However, this dc voltage oscillates between zero and 2 V_max and is needed for cascade circuit.

Figure D

1. If the circuit is loaded, the output does not remain constant at V_max After point E (refer Figure C, the supply voltage becomes less than the capacitor voltage, diode stops conducting. The capacitor cannot discharge back into the ac system because of one way action of diode.
2. Hence the current flows out of C, to the load (R_L) as current i_L. While giving up this energy, the capacitor voltage also decreases at a rate depending on the time constant C.R. of the circuit and reaches point F corresponding to V_min. Beyond F, the supply voltage is greater than the capacitor voltage and hence the diode D startsconducting and charging the capacitor C to V_min again and it also supplies current to the load.
3. This second pulse is of shorter duration than the initial charging pulse as it serves mainly to restore the energy given out by the capacitance to the load. Thus, while each pulse of diode current lasts much less than a half cycle, the load receives current more continuously from capacitor C.

Mathematical Analysis of Half Wave Rectifier Circuit :

• Suppose at any time the voltage of the capacitor is V and it decreases by an amount dV over the time dt then charge delivered by the capacitor during this time is

dθ = C dV

• Therefore, if voltage changes from V_max to V_min the charge delivered by capacitor is

• Equation C shows that the ripple in a rectifier output depends upon the load current and circuit parameter like f and C.
• The product f.C is an important factor for design of rectifiers.
• The higher the frequency of supply and larger the value of filtering capacitors, the smaller will be the ripple in d.c. output.