ELEG 3143 Power Electronics Converter and Rectifiers

Outline Half-Wave Rectifiers Center-Tap Full-Wave Rectifier Full-Wave Bridge Rectifier

Half-Wave Rectifiers Figure 3.1 shows a simple half-wave rectifier with a single pn junction diode connected in series to the load resistor. It is given an alternating current as input. Input voltage is given to a step down transformer and the resulting reduced output of transformer is given to the diode, D and load resistor, RL. The output voltage is measured across load resistor RL.

Figure 3.1 Half wave rectifier circuit. Half-Wave Rectifiers Figure 3.1 Half wave rectifier circuit.

Half-Wave Rectifiers During the positive half cycles of the input wave, the diode, D will be forward biased and during the negative half cycles of input wave, the diode, D will be reverse biased. Since the diode passes current only during one half cycle of the input wave, the output waveform is shown in Figure 3.2. The output is positive or significant during the positive cycles of input wave. At the same time, output is zero or insignificant during negative half cycles of input wave.

Figure 3.2 Half wave rectifier output. Half-Wave Rectifiers Figure 3.2 Half wave rectifier output.

Power Supply Specifications of a rectifier The most important characteristics which are required to be specified for a power supply are the required output dc voltage the average and peak currents in the diode, the peak inverse voltage (PIV) of diode, the regulation and the ripple factor.

Half Wave Rectifier Analysis The following parameters will be explained for the analysis of Half Wave Rectifier: Peak Inverse Voltage (PIV) DC or average output voltage and current Root Mean Square (RMS) output voltage and current Rectification Efficiency Ripple Factor

Peak Inverse Voltage (PIV) PIV rating of a diode is important in its design stages. It is the maximum voltage that the rectifying diode has to withstand, during the reverse biased period. When the diode is reverse biased, during the negative half cycle, there will be no current flow through the load resistor RL. Hence, there will be no voltage drop through the load resistance RL which causes the entire input voltage to appear across the diode. Thus, Vs(max), the peak secondary voltage appears across the voltage. PIV = Vs(max) ------ (Eq27)

DC or average output voltage and current Average or DC Value 𝐴 𝑚𝑎𝑥 𝜋 Assuming Vs = Vs(max) sin (ωt) and diode has a forward resistance of Rf ohms and VF = 0 Average or DC voltage 𝑉 𝑎𝑣𝑒 = 𝑉 𝑠(max) 𝜋 Average or DC current 𝐼 𝑎𝑣𝑒 = 𝑉 𝑠(max) 𝜋( 𝑅 𝐹 + 𝑅 𝐿 )

Root Mean Square (RMS) output voltage and current RMS Value 𝐴 𝑚𝑎𝑥 2 Assuming Vs = Vs(max) sin (ωt) and diode has a forward resistance of Rf ohms and VF = 0 RMS voltage 𝑉 𝑟𝑚𝑠 = 𝑉 𝑠(max) 2 RMS current 𝐼 𝑟𝑚𝑠 = 𝑉 𝑠(max) 2 ( 𝑅 𝐹 + 𝑅 𝐿 )

Rectification Efficiency Rectification efficiency is defined as the ratio between the output power to the ac input power. Efficiency, η = 𝑃 𝑑𝑐 𝑃 𝑎𝑐 DC power delivered to the load 𝑃 𝑑𝑐 = 𝐼 𝑑𝑐 2 𝑅 𝐿 = 𝐼 𝑚𝑎𝑥 𝜋 2 𝑅 𝐿

Rectification Efficiency AC power input to the transformer, Pac = power dissipated in diode junction + power dissipated in load resistance, RL. 𝑃 𝑎𝑐 = 𝐼 𝑟𝑚𝑠 2 𝑅 𝐹 + 𝐼 𝑟𝑚𝑠 2 𝑅 𝐿 = 𝐼 𝑚𝑎𝑥 2 4 𝑅 𝐹 + 𝑅 𝐿 Rectification efficiency 𝜂= 𝑃 𝑑𝑐 𝑃 𝑎𝑐 = 4 𝜋 2 𝑅 𝐿 𝑅 𝐿 + 𝑅 𝐹 The maximum efficiency that can be obtained by the half wave rectifier is 40.6 %. This is obtained if RF is neglected.

𝑅𝐹,𝛾= 𝑉 𝑎𝑐−𝑟𝑖𝑝𝑝𝑙𝑒(𝑟𝑚𝑠) 𝑉 𝐷𝐶 Ripple Factor Ripple factor is defined as the ratio of rms value of ac component to the dc component in the output. 𝑅𝐹,𝛾= 𝑉 𝑎𝑐−𝑟𝑖𝑝𝑝𝑙𝑒(𝑟𝑚𝑠) 𝑉 𝐷𝐶 𝑅𝐹,𝛾= 1 2 3 𝑓 𝑅 𝐿 𝐶

Advantages and Disadvantages of Half Wave Rectifier A half wave rectifier is rarely used in practice. It is never preferred as the power supply of an audio circuit because of the very high ripple factor. High ripple factor will result in noises in input audio signal, which in turn will affect audio quality. Advantage of a half wave rectifier is only that its cheap, simple and easy to construct. It is cheap because of the low number of components involved. Simple is because of the straight forwardness in circuit design.

Center-Tap Full-Wave Rectifier In the case of center-tap full wave rectifier, only two diodes are used, and are connected to the opposite ends of a center-tapped secondary transformer as shown in Figure 3.3. The center-tap is usually considered as the ground point or the zero voltage reference point.

Center-Tap Full-Wave Rectifier Figure 3.3 Center-tap full wave rectifier.

Center-Tap Full-Wave Rectifier An AC input is applied to the primary coils of the transformer, this input makes the secondary ends P1 and P2 become positive and negative alternately. For the positive half of the ac signal, the secondary point D1 is positive, GND point will have zero volt and P2 will be negative. At this instant, diode, D1 will be forward biased and diode, D2 will be reverse biased. Thus the current flow will be in the direction P1 – D1 – C – A – B – GND. Thus, the positive half cycle appears across the load resistance, RLoad.

Center-Tap Full-Wave Rectifier During the negative half cycle, the secondary ends P1 becomes negative and P2 becomes positive. At this instant, the diode, D1 will be negative and D2 will be positive with the zero reference point being the GND. Thus, the diode, D2 is forward biased and D1 is reverse biased. The current flow will be in the direction P2 – D2 – C – A – B – GND.

Center-Tap Full-Wave Rectifier Figure 3.4 Center-tap full-wave rectifier waveform.

Center-Tap Full-Wave Rectifier When comparing the current flow in the positive and negative half cycles, we can conclude that the direction of the current flow is the same (through load resistance, RLoad). When compared to the Half-wave rectifier, both the half cycles are used to produce the corresponding output. The frequency of the rectified output voltage is twice the input frequency.

Center-tap Full-Wave Rectifier Analysis The following parameters will be explained for the analysis of Center-tap Full-Wave Rectifier: Peak Inverse Voltage (PIV) DC or average output voltage and current Root Mean Square (RMS) output voltage and current Rectification Efficiency Ripple Factor

Peak Inverse Voltage (PIV) 𝑃𝐼𝑉 𝑜𝑓 𝐷 1 = 𝑉 𝑠 max + 𝑉 𝑠 max =2 𝑉 𝑠 max 𝑃𝐼𝑉 𝑜𝑓 𝐷 2 = 2 𝑉 𝑠 max

DC or average output voltage and current Average or DC Value 2𝐴 𝑚𝑎𝑥 𝜋 Assuming Vs = Vs(max) sin (ωt) and diode has a forward resistance of Rf ohms and VF = 0 Average or DC voltage 𝑉 𝑎𝑣𝑒 = 2𝑉 𝑠(max) 𝜋 Average or DC current 𝐼 𝑎𝑣𝑒 = 2𝑉 𝑠(max) 𝜋( 𝑅 𝐹 + 𝑅 𝐿 )

Root Mean Square (RMS) output voltage and current RMS Value 𝐴 𝑚𝑎𝑥 2 Assuming Vs = Vs(max) sin (ωt) and diode has a forward resistance of Rf ohms and VF = 0 RMS voltage 𝑉 𝑟𝑚𝑠 = 𝑉 𝑠(max) 2 RMS current 𝐼 𝑟𝑚𝑠 = 𝑉 𝑠(max) 2 ( 𝑅 𝐹 + 𝑅 𝐿 )

Rectification Efficiency Rectification efficiency is defined as the ratio between the output power to the ac input power. Efficiency, η = 𝑃 𝑑𝑐 𝑃 𝑎𝑐 DC power delivered to the load 𝑃 𝑑𝑐 = 𝐼 𝑑𝑐 2 𝑅 𝐿 = 2𝐼 𝑚𝑎𝑥 𝜋 2 𝑅 𝐿

Rectification Efficiency AC power input to the transformer, Pac = power dissipated in diode junction + power dissipated in load resistance, RL. 𝑃 𝑎𝑐 = 𝐼 𝑟𝑚𝑠 2 𝑅 𝐹 + 𝐼 𝑟𝑚𝑠 2 𝑅 𝐿 = 𝐼 𝑚𝑎𝑥 2 2 𝑅 𝐹 + 𝑅 𝐿 Rectification efficiency 𝜂= 𝑃 𝑑𝑐 𝑃 𝑎𝑐 = 8 𝜋 2 𝑅 𝐿 𝑅 𝐿 + 𝑅 𝐹 The maximum efficiency that can be obtained by the half wave rectifier is about 81%. This is obtained if RF is neglected.

Full-Wave Bridge Rectifier A full wave rectifier is a circuit arrangement which makes use of both half cycles of input alternating current (AC) and converts them to direct current (DC). A full wave rectifier is much more efficient than a half wave rectifier. Figure 3.5 shows a of a bridge. The transformer secondary is connected to two diametrically opposite points of the bridge at points A & C. The load resistance, RL is connected to bridge through points B and D.

Full-Wave Bridge Rectifier Figure 3.5 Full-wave bridge rectifier

Full-Wave Bridge Rectifier During the first half cycle of the input voltage, the upper end of the transformer secondary winding is positive with respect to the lower end. Thus during the first half cycle diodes, D1 and D3 are forward biased and current flows through arm AB, enter the load resistance, RL, and returns back flowing through arm DC. During this half of each input cycle, the diodes, D2 and D4 are reverse biased and current is not allowed to flow in arms AD and BC. Figure 3.6 indicates the current flow from the source to the load resistance, then returns to the source, thus completing the circuit.

Full-Wave Bridge Rectifier Figure 3.6 Full-wave bridge waveform.

Full-Wave Bridge Rectifier During second half cycle of the input voltage, the lower end of the transformer secondary winding is positive with respect to the upper end. Thus diodes, D2 and D4 become forward biased and current flows through arm CB, enters the load resistance, RL and returns back to the source flowing through arm DA.

Full-Wave Bridge Rectifier Figure 3.7 Current flow during the first half cycle of the input voltage.

Full-Wave Bridge Rectifier Figure 3.8 Current flow during the second half cycle of the input voltage.

Peak Current Peak value of the current flowing through the load resistance, RL, is given as 𝐼 𝑚𝑎𝑥 = 𝑉 𝑠 max 2 𝑅 𝐹 + 𝑅 𝐿

DC or average output voltage and current Average or DC voltage 𝑉 𝑎𝑣𝑒 = 2 𝐼 𝑚𝑎𝑥 𝑅 𝐿 𝜋 Average or DC current 𝐼 𝑎𝑣𝑒 = 2 𝐼 𝑚𝑎𝑥 𝜋 = 2𝑉 𝑠(max) 𝜋(2 𝑅 𝐹 + 𝑅 𝐿 )

Root Mean Square (RMS) output voltage and current RMS Value 𝐴 𝑚𝑎𝑥 2 Assuming Vs = Vs(max) sin (ωt) and diode has a forward resistance of Rf ohms and VF = 0 RMS voltage 𝑉 𝐿𝑜𝑎𝑑 𝑟𝑚𝑠 = 𝐼 𝑟𝑚𝑠 𝑅 𝐿𝑜𝑎𝑑 = 𝐼 𝑚𝑎𝑥 𝑅 𝐿𝑜𝑎𝑑 2 RMS current 𝐼 𝑟𝑚𝑠 = 𝐼 𝑚𝑎𝑥 2

Rectification Efficiency 𝑃 𝑑𝑐 = 4 𝜋 2 𝐼 𝑚𝑎𝑥 2 𝑅 𝐿 𝑃 𝑎𝑐 = 𝐼 𝑚𝑎𝑥 2 2 2 𝑅 𝐹 + 𝑅 𝐿𝑜𝑎𝑑 𝜂= 8 𝑅 𝐿 𝜋 2 2 𝑅 𝐹 + 𝑅 𝐿𝑜𝑎𝑑