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The purpose of a power supply is to take electrical energy in one form and convert it into another. There are many types of power supply. Most are designed to convert high voltage AC mains electricity to a suitable low voltage supply for electronic circuits and other devices such as computers, fax machines and telecommunication equipment. In Singapore, supply from 230V, 50Hz AC mains is converted into smooth DC using AC-DC power supply.

A power supply can by broken down into a series of blocks, each of which performs a particular function. A transformer first steps down high voltage AC to low voltage AC. A rectifier circuit is then used to convert AC to DC. This DC, however, contains ripples, which can be smoothened by a filter circuit. Power supplies can be ‘regulated’ or ‘unregulated’. A regulated power supply maintains a constant DC output voltage through ‘feedback action’. The output voltage of an unregulated supply, on the other hand, will not remain constant. It will vary depending on varying operating conditions, for example when the magnitude of input AC voltage changes.

Power supplies are designed to produce as little ripple voltage as possible, as the ripple can cause several problems. For Example

- In audio amplifiers, too much ripple shows up as an annoying 50 Hz or 100 Hz audible hum.
- In video circuits, excessive ripple shows up as “hum” bars in the picture.
- In digital circuits it can cause erroneous outputs from logic circuits.

**Half-wave Rectifier with Capacitor Filter**

The capacitor is the most basic filter type and is the most commonly used. The half-wave rectifier for power supply application is shown below. A capacitor filter is connected in parallel with the load. The rectifier circuit is supplied from a transformer.

*Circuit operation*

The operation of this circuit during positive half cycle of the source voltage is shown in figure 8. During the positive half cycle, diode D1 will conduct, and the capacitor charges rapidly. As the input starts to go negative, D1 turns off, and the capacitor will slowly discharge through the load.

*Figure 1: Half wave rectifier with capacitor filter – positive half cycle*

(a) Unfiltered output from the half wave Rectifier

(b) When the next pulse does arrive, it charges the capacitor back to full charge as shown on the right. The thick line shows the charge – discharge waveform at the capacitor.

(c) The load sees a reasonably constant DC voltage now, with a ripple voltage on top of it.

During each positive half cycle, the capacitor charges during the interval *t1 *to *t2*. During this interval, the diode will be forward biased. Due to this charging, the voltage across the capacitor *vo *will be equal to the AC peak voltage *Vm *on the secondary side of the transformer at *t2 *(assuming diode forward voltage drop is zero).The capacitor will supply current to load resistor *RL *during time interval *t2 *to *t3*. During this interval, diode will be reverse biased since the AC voltage is less than the output voltage *vo*. Due to the large energy stored in the capacitor, the capacitor voltage will not reduce much during *t2 *to *t3*, and the voltage *vo *will remain close to the peak value. As can be seen, addition of the capacitor results in much better quality output voltage.

*Average load voltage*

In practical applications, a very large capacitor is used so that the output voltage is close to the peak value. The average voltage (also called DC output voltage) across the load can therefore be approximated to:

V_{avg} = V_{m}

*Calculation of capacitance*

The voltage waveforms show a small AC component called “ripple” present in the output voltage. This ripple can be minimized by choosing the largest capacitance value that is practical. The capacitor is typically “electrolytic” type, and is very large (several hundreds or even thousands of microfarads). We can calculate the required value of the filter capacitor as follows.

The charge removed from the capacitor during the discharge cycle (i.e., t2 to t3) is:

*Δ _{Q} =I_{L }T*

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Where IL is the average load current and T is the period of the AC voltage. As the interval *t1 *to t2 is very small, the discharge time can be approximated to T.

If Vp-p is the peak-to-peak ripple voltage, and C is the capacitance, the charge removed from the capacitor can also be expressed as:

*Δ _{Q}* = V

_{p-p }C

From these two equations, capacitance C can be calculated as

C = *I _{L }T/* V

_{p-p}

The amount of ripple voltage left by a given filter depends on the three things:

Type of rectifier (half or full wave)

The capacity of the filter capacitor

The load resistance

**The Full-Wave Rectifier**

The full wave rectifier consists of two diodes and a resister as shown in Figure.

The transformer has a centre-tapped secondary winding. This secondary winding has a lead attached to the centre of the winding. The voltage from the centre tap to either end terminal on this winding is equal to one half of the total voltage measured end-to-end.

*Circuit Operation*

Figure shows the operation during the positive half cycle of the full wave rectifier. Note that diode D1 is forward biased and diode D2 is reverse biased. Note the direction of the current through the load.

*Calculating Load Voltage and Currents*

Using the ideal diode model, the peak load voltage for the full wave rectifier is *m V *. The full wave rectifier produces twice as many output pulses as the half wave rectifier. This is the same as saying that the full wave rectifier has twice the output frequency of a half wave rectifier. For this reason, the average load voltage (i.e DC output voltage) is found as

V_{avg} = 2V_{m}/Π

*Figure 14: Average DC Voltage for a Full Wave Rectifier*

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*Peak Inverse Voltage*

When one of the diodes in a full-wave rectifier is reverse biased, the peak voltage across that diode will be approximately equal to *Vm*. This point is illustrated in figure 13. With the polarities shown, D1 is conducting and D2 is reverse biased. Thus the cathode of D1 will be at *Vm*. Since this point is connected directly to the cathode of D2, its cathode will also be *Vm*. With *–Vm *applied to the anode of D2, the total voltage across the diode D2 is *2Vm*. Therefore, the maximum reverse voltage across either diode will be twice the peak load voltage.

*PIV =* 2*Vm*

**Full-Wave Rectifier with Capacitor filter**

Similar to the half-wave rectifier, smoothing is performed by a large value capacitor connected across the load resistance to act as a reservoir, supplying current to the output when the varying DC voltage from the rectifier is falling. Note that smoothing significantly increases the average DC voltage to almost the peak value. However,

Smoothing is not perfect due to the capacitor voltage falling a little as it discharges, giving a small **ripple voltage**.

For many circuits a ripple which is 10% of the supply voltage is satisfactory and the equation below** **gives the required value for the smoothing capacitor. In the full-wave circuit, the capacitor discharges for only a** **half-cycle before being recharged. Hence the capacitance required is only half as much in the full-wave circuit** **as for the half-wave circuit.

C = *I _{L }T/* 2V

_{p-p}

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**The Full Wave Bridge Rectifier**

In many power supply circuits, the bridge rectifier is used. The bridge rectifier produces almost double the output voltage as a full wave centre-tapped transformer rectifier using the same secondary voltage.The advantage of using this circuit is that no center-tapped transformer is required.

*Basic Circuit Operation*

During the positive half cycle, both D3 and D1 are forward biased. At the same time, both D2 and D4 are reverse biased. Note the direction of current flow through the load. During the negative half cycle D2 and D4 are forward biased and D1 and D3 are reverse biased. Again note that current through the load is in the same direction although the secondary winding polarity has reversed.

*Peak Inverse Voltage*

In order to understand the Peak Inverse Voltage across each diode, look at figure below. It is a simplified version of figure 17 showing the circuit conditions during the positive half cycle. The load and ground connections are removed because we are concerned with the diode conditions only. In this circuit, diodes D1 and D3 are forward biased and act like closed switches. They can be replaced with wires. Diodes D2 and D4 are reverse biased and act like open switches.

Peak inverse voltage = *V _{m}*

**Full Wave Bridge Rectifier with Capacitor Filter**

The voltage obtained across the load resistor of the full-wave bridge rectifier described above has a large amount of ripple. A capacitor filter may be added to smoothen the ripple in the output, as shown below.

The rectifier circuits discussed above can be used to charge batteries and to convert AC voltages into constant DC voltages. Full-wave and bridge rectifier are more commonly used than half-wave rectifier.