7.2.3.1 Voltage Relationships

The average value of the load voltage v_L is V_dc and it is defined as

$V_{\mathit{dc}}=\frac{1}{T}{\displaystyle {\int }_0^T}{v}_L\left(t\right)dt$

In the case of a half-wave rectifier, Fig. 7.2 indicates that load voltage $v_L\left(t\right)=0$ for the negative half cycle. Note that the angular frequency of the source $\omega =2\pi /T$, and Eq. (7.1) can be rewritten as

$V_{\mathit{dc}}=\frac{1}{2\pi }{\displaystyle {\int }_0^{\pi }}{V}_{m}sin\omega td\left(\omega t\right)$

Therefore,

$\mathrm{Half}-\mathrm{wave}{V}_{\mathit{dc}}=\frac{V_m}{\pi }=0.318V_m$

In the case of a full-wave rectifier, Figs. 7.4 and 7.6 indicate that $v_L\left(t\right)=V_m|sin\omega t|$ for both the positive and negative half cycles. Hence, Eq. (7.1) can be rewritten as

$V_{\mathit{dc}}=\frac{1}{\pi }{\displaystyle {\int }_0^{\pi }}{V}_{m}sin\omega td\left(\omega t\right)$

Therefore,

$\mathrm{Full}-\mathrm{wave}{V}_{\mathit{dc}}=\frac{2V_m}{\pi }=0.636V_m$

The root-mean-square (rms) value of load voltage v_L is V_L, which is defined as

$V_L={\left[\frac{1}{T}{\displaystyle {\int }_0^T}{v}_L^2\left(t\right)dt\right]}^{1/2}$

In the case of a half-wave rectifier, $v_L\left(t\right)=0$ for the negative half cycle; therefore, Eq. (7.6) can be rewritten as

$V_L=\sqrt{\frac{1}{2\pi }{\displaystyle {\int }_0^{\pi }}{\left(V_{m}sin\omega t\right)}^{2}d\left(\omega t\right)}$

or

$\mathrm{Half}-\mathrm{wave}{V}_L=\frac{V_m}{2}=0.5V_m$

In the case of a full-wave rectifier, $v_L\left(t\right)=V_m|sin\omega t|$ for both the positive and negative half cycles. Hence, Eq. (7.6) can be rewritten as

$V_L=\sqrt{\frac{1}{\pi }{\displaystyle {\int }_0^{\pi }}{\left(V_{m}sin\omega t\right)}^{2}d\left(\omega t\right)}$

or

$\mathrm{Full}-\mathrm{wave}{V}_L=\frac{V_m}{\sqrt{2}}=0.707V_m$

The result of Eq. (7.10) is as expected because the rms value of a full-wave rectified voltage should be equal to that of the original ac voltage.

7.2.3.2 Current Relationships

The average value of load current i_L is I_dc, and because load R is purely resistive, it can be found as

$I_{\mathit{dc}}=\frac{V_{\mathit{dc}}}{R}$

The rms value of load current i_L is I_L, and it can be found as

$I_L=\frac{V_L}{R}$

In the case of a half-wave rectifier, from Eq. (7.3),

$\mathrm{Half}-\mathrm{wave}{I}_{\mathit{dc}}=\frac{0.318V_m}{R}$

and from Eq. (7.8),

$\mathrm{Half}-\mathrm{wave}{I}_L=\frac{0.5V_m}{R}$

In the case of a full-wave rectifier, from Eq. (7.5),

$\mathrm{Full}-\mathrm{wave}{I}_{\mathit{dc}}=\frac{0.636V_m}{R}$

and from Eq. (7.10),

$\mathrm{Full}-\mathrm{wave}{I}_L=\frac{0.707V_m}{R}$

7.2.3.3 Rectification Ratio

The rectification ratio, which is a figure of merit for comparing the effectiveness of rectification, is defined as

$\sigma =\frac{P_{\mathit{dc}}}{P_L}=\frac{V_{\mathit{dc}}{T}_{\mathit{dc}}}{V_L{I}_L}$

In the case of a half-wave diode rectifier, the rectification ratio can be determined by substituting Eqs. (7.3), (7.13), (7.8), and (7.14) into Eq. (7.17):

$\mathrm{Half}-\mathrm{wave}\sigma =\frac{{\left(0.318V_m\right)}^2}{{\left(0.5V_m\right)}^2}=40.5\%$

In the case of a full-wave rectifier, the rectification ratio is obtained by substituting Eqs. (7.5), (7.15), (7.10), and (7.16) into Eq. (7.17):

$\mathrm{Full}-\mathrm{wave}\sigma =\frac{{\left(0.636V_m\right)}^2}{{\left(0.707V_m\right)}^2}=81\%$

7.2.3.4 Form Factor

The form factor (FF) is defined as the ratio of the root-mean-square value (heating component) of a voltage or current to its average value:

$\mathrm{F}\mathrm{F}=\frac{V_L}{V_{\mathit{dc}}}\mathrm{or}\frac{I_L}{I_{\mathit{dc}}}$

$\mathrm{Half}-\mathrm{wave}\mathrm{F}\mathrm{F}=\frac{0.5V_m}{0.318V_m}=1.57$

In the case of a full-wave rectifier, the FF can be found by substituting Eqs. (7.16) and (7.15) into Eq. (7.20):

$\mathrm{Full}-\mathrm{wave}\mathrm{F}\mathrm{F}=\frac{0.707V_m}{0.636V_m}=1.11$

7.2.3.5 Ripple Factor

The ripple factor (RF), which is a measure of the ripple content, is defined as

$\mathrm{R}\mathrm{F}=\frac{V_{ac}}{V_{\mathit{dc}}}$

where V_ac is the effective (rms) value of the ac component of load voltage v_L:

$V_{ac}=\sqrt{V_L^2-V_{\mathit{dc}}^2}$

Substituting Eq. (7.24) into Eq. (7.23), the RF can be expressed as

$\mathrm{R}\mathrm{F}=\sqrt{{\left(\frac{V_L}{V_{\mathit{dc}}}\right)}^2-1}=\sqrt{{\mathrm{F}\mathrm{F}}^2-1}$

In the case of a half-wave rectifier,

$\mathrm{Half}-\mathrm{wave}\mathrm{R}\mathrm{F}=\sqrt{{1.57}^2-1}=1.21$

In the case of a full-wave rectifier,

$\mathrm{Full}-\mathrm{wave}\mathrm{R}\mathrm{F}=\sqrt{{1.11}^2-1}=0.482$

7.2.3.6 Transformer Utilization Factor

The transformer utilization factor (TUF), which is a measure of the merit of a rectifier circuit, is defined as the ratio of the dc output power to the transformer volt-ampere (VA) rating required by the secondary winding:

$\mathrm{T}\mathrm{U}\mathrm{F}=\frac{P_{\mathit{dc}}}{V_s{I}_s}=\frac{V_{\mathit{dc}}{I}_{\mathit{dc}}}{V_s{I}_s}$

where V_s and I_s are the rms voltage and rms current ratings of the secondary transformer:

$V_s=\frac{V_m}{\sqrt{2}}=0.707V_m$

The rms value of the transformer secondary current I_s is the same as that of the load current I_L. For a half-wave rectifier, I_s can be found from Eq. (7.14):

$\mathrm{Half}-\mathrm{wave}{I}_s=\frac{0.5V_m}{R}$

For a full-wave rectifier, I_s is found from Eq. (7.16):

$\mathrm{Full}-\mathrm{wave}{I}_s=\frac{0.707V_m}{R}$

Therefore, the TUF of a half-wave rectifier can be obtained by substituting Eqs. (7.3), (7.13), (7.29), and (7.30) into Eq. (7.28):

$\mathrm{Half}-\mathrm{wave}\mathrm{T}\mathrm{U}\mathrm{F}=\frac{{0.318}^2}{0.707\times 0.5}=0.286$

The poor TUF of a half-wave rectifier signifies that the transformer employed must have a 3.496 (1/0.286) VA rating in order to deliver 1 W dc output power to the load. In addition, the transformer secondary winding has to carry a dc current that may cause magnetic core saturation. As a result, half-wave rectifiers are used only when the current requirement is small.

In the case of a full-wave rectifier with center-tapped transformer, the circuit can be treated as two half-wave rectifiers operating together. Therefore, the transformer secondary VA rating, V_sI_s, is double that of a half-wave rectifier, but the output dc power is increased by a factor of four due to higher rectification ratio as indicated by Eqs. (7.5) and (7.15). Therefore, the TUF of a full-wave rectifier with center-tapped transformer can be found from Eq. (7.32):

$\mathrm{Full}-\mathrm{wave}\mathrm{T}\mathrm{U}\mathrm{F}=\frac{4\times {0.318}^2}{2\times 0.707\times 0.5}=0.572$

In the case of a bridge rectifier, it has the highest TUF in single-phase rectifier circuits because the currents flowing in both the primary and secondary windings are continuous sinewaves. By substituting Eqs. (7.5), (7.15), (7.29), and (7.31) into Eq. (7.28), the TUF of a bridge rectifier can be found:

$\mathrm{Bridge}\mathrm{T}\mathrm{U}\mathrm{F}=\frac{{0.636}^2}{{\left(0.707\right)}^2}=0.81$

The transformer primary VA rating of a full-wave rectifier is equal to that of a bridge rectifier since the current flowing in the primary winding is also a continuous sinewave.

7.2.3.7 Harmonics

Full-wave rectifier circuits with resistive load do not produce harmonic currents in their transformers. In half-wave rectifiers, harmonic currents are generated. The amplitudes of the harmonic currents of a half-wave rectifier with resistive load, relative to the fundamental, are given in Table 7.1. The extra loss caused by the harmonics in the resistive loaded rectifier circuits is often neglected because it is not high compared with other losses. However, with nonlinear loads, harmonics can cause appreciable loss and other problems such as poor power factor and interference.

A basic three-phase star rectifier circuit is shown in Fig. 7.7. This circuit can be considered as three single-phase half-wave rectifiers combined together. Therefore, it is sometimes referred to as a three-phase half-wave rectifier. The diode in a particular phase conducts during the period when the voltage on that phase is higher than that on the other two phases. The voltage waveforms of each phase and the load are shown in Fig. 7.8. It is clear that, unlike the single-phase rectifier circuit, the conduction angle of each diode is 2π/3, instead of π. This circuit finds uses where the required dc output voltage is relatively low, and the required output current is too large for a practical single-phase system.

Taking phase R as an example, diode D conducts from π/6 to 5π/6. Therefore, using Eq. (7.1), the average value of the output can be found as

$V_{\mathit{dc}}=\frac{3}{2\pi }{\displaystyle {\int }_{\pi /6}^{5\pi /6}}{V}_{m}sin\theta d\theta$

or

$V_{\mathit{dc}}=V_m\frac{3}{\pi }\frac{\sqrt{3}}{2}=0.827V_m$

Similarly, using Eq. (7.6), the rms value of the output voltage can be found as

$V_L=\sqrt{\frac{3}{2\pi }{\displaystyle {\int }_{\pi /6}^{5\pi /6}}{\left(V_{m}sin\theta \right)}^{2}d\theta }$

or

$V_L=V_m\sqrt{\frac{3}{2\pi }\left(\frac{\pi }{3}+\frac{\sqrt{3}}{4}\right)}=0.84V_m$

In addition, the rms current in each transformer secondary winding can also be found as

$I_s=I_m\sqrt{\frac{1}{2\pi }\left(\frac{\pi }{3}+\frac{\sqrt{3}}{4}\right)}=0.485I_m$

where $I_m=V_m/R$.

Based on the relationships stated in Eqs. (7.43), (7.45), (7.46), all the important design parameters of the three-phase star rectifier can be evaluated, as listed in Table 7.3, which is given at the end of Section 7.3.2. Note that, as with a single-phase half-wave rectifier, the three-phase star rectifier shown in Fig. 7.7 has direct currents in the secondary windings that can cause a transformer core saturation problem. In addition, the currents in the primary do not sum to zero. Therefore, it is preferable not to have star-connected primary windings.

7.3.1.2 Three-Phase Interstar Rectifier Circuit

The transformer core saturation problem in the three-phase star rectifier can be avoided by a special arrangement in its secondary windings, known as zigzag connection. The modified circuit is called the three-phase interstar or zigzag rectifier circuit, as shown in Fig. 7.9. Each secondary-phase voltage is obtained from two equal-voltage secondary windings (with a phase displacement of π/3) connected in series so that the dc magnetizing forces due to the two secondary windings on any limb are equal and opposite. At the expense of extra secondary windings (increasing the transformer secondary rating factor from 1.51 to 1.74 VA/W), this circuit connection eliminates the effects of core saturation and reduces the transformer primary rating factor to the minimum of 1.05 VA/W. Apart from transformer ratings, all the design parameters of this circuit are the same as those of a three-phase star rectifier (therefore not separately listed in Table 7.3). Furthermore, a star-connected primary winding with no neutral connection is equally permissible because the sum of all primary phase currents is zero at all times.

7.3.1.3 Three-Phase Double-Star Rectifier With Interphase Transformer

This circuit consists essentially of two three-phase star rectifiers with their neutral points interconnected through an interphase transformer or reactor (Fig. 7.10). The polarities of the corresponding secondary windings in the two interconnected systems are reversed with respect to each other, so that the rectifier output voltage of one three-phase unit is at a minimum when the rectifier output voltage of the other unit is at a maximum as shown in Fig. 7.11. The function of the interphase transformer is to cause the output voltage v_L to be the average of the rectified voltages v1 and v2 as shown in Fig. 7.11. In addition, the ripple frequency of the output voltage is now six times that of the mains, and therefore, the component size of the filter (if there is any) becomes smaller. In a balanced circuit, the output currents of two three-phase units flowing in opposite directions in the interphase transformer winding will produce no dc magnetization current. Similarly, the dc magnetization currents in the secondary windings of two three-phase units cancel each other out.

By virtue of the symmetry of the secondary circuits, the three primary currents add up to zero at all times. Therefore, a star primary winding with no neutral connection would be equally permissible.

7.5.1.1 Voltage and Current Waveforms of Full-Wave Rectifier With Inductor-Input DC Filter

Fig. 7.22 shows a single-phase full-wave rectifier with an inductor-input dc filter. The voltage and current waveforms are illustrated in Fig. 7.23.

When the inductance of L_f is infinite, the current through the inductor and the output voltage are constant. When inductor L_f is finite, the current through the inductor has a ripple component, as shown by the dotted lines in Fig. 7.23. If the input inductance is too small, the current decreases to zero (becoming discontinuous) during a portion of the time between the peaks of the rectifier output voltage. The minimum value of inductance required to maintain a continuous current is known as the critical inductance L_C.

7.5.1.2 Critical Inductance L_C

In the case of single-phase full-wave rectifiers, the critical inductance can be found as

$\mathrm{Full}-\mathrm{wave}{L}_C=\frac{R}{6\pi f_i}$

where f_i is the input mains frequency.

In the case of polyphase rectifiers, the critical inductance can be found as

$\mathrm{Poly}-\mathrm{phase}{L}_C=\frac{R}{3\pi m\left(m^2-1\right)f_i}$

where m is ratio of the lowest ripple frequency to the input frequency, for example, $m=6$ for a three-phase bridge rectifier.

7.5.1.3 Determining the Input Inductance for a Given Ripple Factor

In practice, the choice of the input inductance depends on the required ripple factor of the output voltage. The ripple voltage of a rectifier without filtering can be found by means of Fourier analysis. For example, the coefficient of the nth harmonic component of the rectified voltage v_L shown in Fig. 7.22 can be expressed as

$v_{L_n}=\frac{-4V_m}{\pi \left(n^2-1\right)}$

where $n=2,4,8,\dots$, etc.

The dc component of the rectifier voltage is given by Eq. (7.5). Therefore, in addition to Eq. (7.27), the ripple factor can also be expressed as

$\mathrm{R}\mathrm{F}=\sqrt{2{\displaystyle \sum _{n=2,4,8,}}{\left(\frac{1}{n^2-1}\right)}^2}$

Considering only the lowest-order harmonic $\left(n=2\right)$, the output ripple factor of a simple inductor-input dc filter (without C_f) can be found, from Eqs. (7.64) and (7.69), as

$\mathrm{Filtered}\mathrm{R}\mathrm{F}=\frac{0.4714}{\sqrt{1+{\left(4\pi f_i{L}_f/R\right)}^2}}$

7.5.1.4 Harmonics of the Input Current

In general, the total harmonic distortion (THD) of an input current is defined as

$\mathrm{T}\mathrm{H}\mathrm{D}=\sqrt{{\left(\frac{I_s}{I_{s1}}\right)}^2-1}$

where I_s is the rms value of the input current and I_s1 and the rms value of the fundamental component of the input current. The THD can also be expressed as

$\mathrm{T}\mathrm{H}\mathrm{D}=\sqrt{{\displaystyle \sum _{n=2,3,4,}}{\left(\frac{I_{sn}}{I_{s1}}\right)}^2}$

where I_sn is the rms value of the nth harmonic component of the input current.

Moreover, the input power factor is defined as

$\mathrm{P}\mathrm{F}=\frac{I_{s1}}{I_s}cos\varphi$

where ϕ is the displacement angle between the fundamental components of the input current and voltage.

Assume that inductor L_f of the circuit shown in Fig. 7.22 has an infinitely large inductance. The input current is then a square wave. This input current contains undesirable higher harmonics that reduce the input power factor of the system. The input current can be easily expressed as

$i_s=\frac{4I_m}{\pi }{\displaystyle \sum _{n=1,3,5,}}\frac{1}{n}sin2n\pi f_{i}t$

The rms values of the input current and its fundamental component are I_m and $4I_m/\left(\pi \surd 2\right)$, respectively. Therefore, the THD of the input current of this circuit is 0.484. Since the displacement angle is $\varphi =0$, the power factor is $4/\left(\pi \surd 2\right)=0.9\text{.}$

The power factor of the circuit shown in Fig. 7.22 can be improved by installing an ac filter between the source and the rectifier, as shown in Fig. 7.24.

Considering only the harmonic components, the equivalent circuit of the rectifier given in Fig. 7.24 can be found as shown in Fig. 7.25. The rms value of the nth harmonic current appearing in the supply can then be obtained using the current divider rule:

$I_{sn}=\left|\frac{1}{1-{\left(2n\pi f_i\right)}^2{L}_i{C}_i}\right|I_{rn}$

where I_rn is the rms value of the nth harmonic current of the rectifier.

Applying Eq. (7.73) and knowing $I_{rn}/I_{r1}=1/n$ from Eq. (7.74), the THD of the rectifier with input filter shown in Fig. 7.24 can be found as

$\mathrm{Filtered}\mathrm{T}\mathrm{H}\mathrm{D}=\sqrt{{\displaystyle \sum _{n=3,5}\frac{1}{n^2}}{\left|\frac{1}{1-{\left(2n\pi f_i\right)}^2{L}_i{C}_i}\right|}^2}$

The important design parameters of typical single-phase and three-phase rectifiers with inductor-input dc filter are listed in Table 7.5. Note that, in a single-phase half-wave rectifier, a freewheeling diode is required to be connected across the input of the dc filters such that the flow of load current can be maintained during the negative half cycle of the supply voltage.

7.5.2.1 Inrush Current

The resistor R_inrush in Fig. 7.26 is used to limit the inrush current imposed on the diodes during the instant when the rectifier is being connected to the supply. The inrush current can be very large because capacitor C has zero charge initially. The worst case occurs when the rectifier is connected to the supply at its maximum voltage. The worst-case inrush current can be estimated from

$I_{\mathit{inrush}}=\frac{V_m}{R_{\mathit{sec}}+R_{ESR}}$

where R_sec is the equivalent resistance looking from the secondary transformer and R_ESR is the equivalent series resistance (ESR) of the filtering capacitor. Hence, the employed diode should be able to withstand the inrush current for a half cycle of the input voltage. In other words, the maximum allowable surge current (I_FSM) rating of the employed diodes must be higher than the inrush current. The equivalent resistance associated with the transformer windings and the filtering capacitor is usually sufficient to limit the inrush current to an acceptable level. However, in cases where the transformer is omitted, for example, the rectifier of an off-line switch-mode supply, resistor R_inrush must be added for controlling the inrush current.

Consider as an example, a single-phase bridge rectifier, which is to be connected to a 120 V to 60 Hz source (without transformer). Assume that the I_FSM rating of the diodes is 150 A for an interval of 8.3 ms. If the ESR of the filtering capacitor is zero, the value of the resistor for limiting inrush current resistance can be estimated to be 1.13 Ω using Eq. (7.81).

Fig. 7.28 shows the basic circuit of a forward converter. Fig. 7.29 shows the idealized steady-state waveforms for continuous-mode operation (the current in L₁ being continuous). These waveforms are obtained from PSpice simulations, based on the following assumptions:

• Rectifier diode D_R, flywheel diode D_F, and magnetic-reset clamping diode D_M are ideal diodes with infinitely fast switching speed.

• Electronic switch M1 is an idealized MOS switch with infinitely fast switching speed and

$\begin{array}{c}\mathrm{On}-\mathrm{state}\mathrm{resistance}=0.067\mathrm{\Omega }\\ \mathrm{Off}-\mathrm{state}\mathrm{resistance}=1\mathrm{M}\mathrm{\Omega }\end{array}$

It should be noted that PSpice does not allow a switch to have zero on-state resistance and infinite off-state resistance.

• Transformer T₁ has a coupling coefficient of 0.99999999. PSpice does not accept a coupling coefficient of 1.

• The switching operation of the converter has reached a steady state.

Referring to the circuit shown in Fig. 7.28 and the waveforms shown in Fig. 7.29, the operation of the converter can be explained as follows:

1. For $0<t<DT$ (D is the duty cycle of the MOS switch M₁ and T is the switching period of the converter. M₁ is turned on when V1 (VPULSE) is 15 V and turned off when V1(VPULSE) is 0 V).
The switch M₁ is turned on at $t=0$.
The voltage at node 3, denoted as V(3), is

$V\left(3\right)=0\mathrm{f}\mathrm{o}\mathrm{r}0<t<DT$

The voltage induced at node 6 of the secondary winding L_S is

$V\left(6\right)=V_{\mathit{IN}}\left(N_s/N_p\right)$

This voltage drives a current I(DR) (current through rectifier diode D_R) into the output circuit to produce the output voltage V_o. The rate of increase of I(DR) is given by

$\frac{dI\left(DR\right)}{dt}=\left[V_{\mathit{IN}}\frac{N_S}{N_P}-V_o\right]\frac{1}{L_1}$

where V_o is the dc output voltage of the converter.
The flywheel diode D_F is reversely biased by V(9), the voltage at node 9:

$V\left(9\right)=V_{\mathit{IN}}\left(N_S/N_P\right)\mathrm{f}\mathrm{o}\mathrm{r}0<t<DT$

The magnetic-reset clamping diode D_M is reversely biased by the negative voltage at node 100. Assuming that L_M and L_P have the same number of turns, we have

$V\left(100\right)=-V_{\mathit{IN}}\mathrm{f}\mathrm{o}\mathrm{r}0<t<DT$

A magnetizing current builds up linearly in L_P. This magnetizing current reaches the maximum value of (V_INDT)/L_P at $t=DT$.

2. For $DT<t<2DT$
The switch M₁ is turned off at $t=DT$.
The collapse of magnetic flux induces a back emf in L_M, which is equal to L_P, to turn-on the clamping diode D_M. The magnetizing current in L_M drops (from the maximum value of (V_IN DT)/L_P, as mentioned above) at the rate of V_IN/L_P. It reaches zero at $t=2DT$.
The back emf induced across L_P is equal to V_IN. The voltage at node 3 is

$V\left(3\right)=2V_{\mathit{IN}}\mathrm{f}\mathrm{o}\mathrm{r}DT<t<2DT$

The back emf across L_S forces D_R to stop conducting.
The inductive current in L₁ forces the flywheel diode D_F to conduct. I(L1) (current through L₁) falls at the rate of

$\frac{dI\left(L1\right)}{dt}=\frac{-V_o}{L_1}$

The voltage across D_R, denoted as V(6,9) (the voltage at node 6 with respect to node 9), is

$\begin{array}{c}V\left(DR\right)=V\left(6,9\right)\\ =-V_{\mathit{IN}}\left(N_S/N_P\right)\mathrm{f}\mathrm{o}\mathrm{r}DT<t<2DT\end{array}$

3. For $2DT<t<T$
D_M stops conducting at $t=2DT$. The voltage across L_M then falls to zero.
The voltage across L_P is zero:

$V\left(3\right)=V_{\mathit{IN}}$

The voltage across L_S is also zero:

$V\left(6\right)=0$

Inductive current I(L1) continues to fall at the rate of

$\frac{dI\left(L1\right)}{dt}=\frac{-V_o}{L_1}$

The switching cycle restarts when the switch M₁ is turned on again at $t=T$.

From the waveforms shown in Fig. 7.29, the following useful information (for continuous-mode operation) can be found:

• The output voltage V_o is equal to the average value of V(9):

$V_o=\mathrm{D}\frac{N_S}{N_P}{V}_{\mathit{IN}}$

• The maximum current in the forward rectifying diode D_R and flywheel diode D_F is

$\begin{array}{c}I{\left(DR\right)}_{\mathit{max}}=I{\left(DF\right)}_{\mathit{max}}\\ =I_o+\frac{1}{2}\frac{V_o}{L_1}\left(1-D\right)T\end{array}$

where $V_o=\mathrm{D}{V}_{\mathit{IN}}\left(N_S/N_P\right)$ and I_o is the output loading current.

• The maximum reverse voltage of D_R and D_F is

$\begin{array}{c}V{\left(DR\right)}_{\mathit{max}}=V{\left(DF\right)}_{\mathit{max}}\\ =V{\left(6,9\right)}_{\mathit{max}}=V_{\mathit{IN}}\frac{N_S}{N_P}\end{array}$

• The maximum reverse voltage of D_M is

$V{\left(DM\right)}_{\mathit{max}}=V_{\mathit{IN}}$

• The maximum current in D_M is

$I{\left(DM\right)}_{\mathit{max}}=DT\frac{V_{\mathit{IN}}}{L_P}$

• The maximum current in the switch M1, denoted as ID(M1), is

$\begin{array}{c}\mathit{ID}{\left(M1\right)}_{\mathit{max}}=\frac{N_S}{N_P}I{\left(DR\right)}_{\mathit{max}}+I{\left(DM\right)}_{\mathit{max}}\\ =\frac{N_S}{N_P}\left[I_o+\frac{1}{2}\frac{V_o}{L_1}\left(1-D\right)T\right]+DT\frac{V_{\mathit{IN}}}{L_P}\end{array}$

It should, however, be understood that, due to the nonideal characteristics of practical components, the idealized waveforms shown in Fig. 7.29 cannot actually be achieved in the real world. In the following, the effects of nonideal diodes and transformers will be examined.

7.6.1.2 Circuit Using Ultra-Fast Diodes

Fig. 7.30 shows the waveforms of the forward converter (circuit given in Fig. 7.28) when ultrafast diodes are used as D_M, D_R, and D_F. (Note that ultrafast diodes are actually much slower than Schottky diodes.) The waveforms are obtained by PSpice simulations, based on the following assumptions:

• D_M is an MUR460 ultrafast diode. D_R and D_F are MUR1560 ultrafast diodes.

• M₁ is an IRF640 MOS transistor.

• Transformer T₁ has a coupling coefficient of 0.99999999 (which may be assumed to be 1).

• The switching operation of the converter has reached a steady state.

It is observed that a large spike appears in the current waveforms of diodes D_R and D_F (denoted as I(DR) and I(DF) in Fig. 7.30) whenever the MOS transistor M₁ is turned on. This is due to the relatively slow reverse recovery of the flywheel diode D_F. During the reverse recovery time, the positive voltage suddenly appearing across L _S (which is equal to V_IN (N_S/N_P)) drives a large transient current through D_R and D_F. This current spike results in large current stress and power dissipation in D_R, D_F, and M₁.

A method of reducing the current spikes is to use Schottky diodes as D_R and D_F, as described below.

7.6.1.3 Circuit Using Schottky Diodes

In order to reduce the current spikes caused by the slow reverse recovery of rectifiers, Schottky diodes are now used as D_R and D_F. The assumptions made here are (referring to the circuit shown in Fig. 7.28) the following:

• D_R and D_F are MBR2540 Schottky diodes.

• D_M is an MUR460 ultrafast diode.

• M₁ is an IRF640 MOS transistor.

• Transformer T₁ has a coupling coefficient of 0.99999999.

• The switching operation of the converter has reached a steady state.

The new simulated waveforms are given in Fig. 7.31. It is found that, by employing Schottky diodes as D_R and D_F, the amplitudes of the current spikes in ID(M1), I(DR), and I(DF) can be reduced to practically zero. This solves the slow-speed problem of ultrafast diodes.

7.6.1.4 Circuit With Practical Transformer

The simulation results given above in Figs. 7.29–7.31 (for the forward converter circuit shown in Fig. 7.28) are based on the assumption that transformer T1 has effectively no leakage inductance (with coupling coefficient $K=0.99999999$). It is, however, found that when a practical transformer (having a slightly lower K) is used, severe ringings occur. Fig. 7.32 shows some simulation results to demonstrate this phenomenon, where the following assumptions are made:

• D_R and D_F are MBR2540 Schottky diodes. D_M is an MUR460 ultrafast diode.

• M₁ is an IRF640 MOS transistor.

• Transformer T₁ has a practical coupling coefficient of 0.996.

• The effective winding resistance of L_P is 0.1 Ω. The effective winding resistance of L_M is 0.4 Ω. The effective winding resistance of L_S is 0.01 Ω.

• The effective series resistance of the output filtering capacitor is 0.05 Ω.

• The switching operation of the converter has reached a steady state.

The resultant waveforms shown in Fig. 7.32 indicate that there are large voltage and current ringings in the circuit. These ringings are caused by the resonant circuits formed by the leakage inductance of the transformer and the parasitic capacitances of diodes and transistor.

A practical converter may therefore need snubber circuits to damp these ringings, as described below.

7.6.1.5 Circuit With Snubber Across The Transformer

In order to suppress the ringing voltage caused by the resonant circuit formed by transformer leakage inductance and the parasitic capacitance of the MOS switch, a snubber circuit, shown as R₁ and C₁ in Fig. 7.33, is now connected across the primary winding of transformer T₁. The new waveforms are shown in Fig. 7.34. Here, the drain-to-source voltage waveform of the MOS transistor, V(3), is found to be acceptable. However, there are still large ringing voltages across the output rectifiers (V(6,9) and V(9)).

In order to damp the ringing voltages across the output rectifiers, additional snubber circuits across the rectifiers may therefore also be required in a practical circuit, as described below.

7.6.1.6 Practical Circuit

Fig. 7.35 shows a practical forward converter with snubber circuits added also to rectifiers (R₂C₂ for D_R and R₃C₃ for D_F) to reduce the voltage ringing. Figs. 7.36 and 7.37 show the resultant voltage and current waveforms. Fig. 7.36 is for continuous-mode operation $\left(R_L=0.35\mathrm{\Omega }\right)$, where I(L1) (current in L₁) is continuous. Fig. 7.37 is for discontinuous-mode operation $\left(R_L=10\mathrm{\Omega }\right)$, where I(L1) becomes discontinuous due to an increased value of R_L. These waveforms are considered to be acceptable.

The design considerations of diode rectifier circuits in high-frequency converters will be discussed later in Section 7.6.3.

Fig. 7.38 shows the basic circuit of a flyback converter. Due to its simple circuit, this type of converter is widely used in low-cost low-power applications. Discontinuous-mode operation (meaning that the magnetizing current in the transformer falls to zero before the end of each switching cycle) is often used because it offers the advantages of easy control and low diode reverse recovery loss. Fig. 7.39 shows the idealized steady-state waveforms for discontinuous-mode operation. These waveforms are obtained from PSpice simulations, based on the following assumptions:

• D_R is an idealized rectifier diode with infinitely fast switching speed.

• M₁ is an idealized MOS switch with infinitely fast switching speed and

$\begin{array}{c}\mathrm{On}-\mathrm{stateresistance}=0.067\mathrm{\Omega }\\ \mathrm{Off}-\mathrm{stateresistance}=1\mathrm{M}\mathrm{\Omega }\end{array}$

• Transformer T₁ has a coupling coefficient of 0.99999999.

• The switching operation of the converter has reached a steady state.

Referring to the circuit shown in Fig. 7.38 and the waveforms shown in Fig. 7.39, the operation of the converter can be explained as follows:

1. For $0<t<DT$
The switch M₁ is turned on at $t=0$:

$V\left(3\right)=0\mathrm{f}\mathrm{o}\mathrm{r}0<t<DT$

The current in M₁, denoted as ID(M1), increases at the rate of

$\frac{d\mathit{ID}\left(M1\right)}{dt}=\frac{V_{\mathit{IN}}}{L_P}$

The output rectifier D_R is reversely biased.

2. For $DT<t<\left(D+D_2\right)T$
The switching M₁ is turned off at $t=DT$.
The collapse of magnetic flux induces a back emf in L_S to turn-on the output rectifier D_R. The initial amplitude of the rectifier current I(DR), which is also denoted as I(LS), can be found by equating the energy stored in the primary winding current I(LP) just before $t=DT$ to the energy stored in the secondary winding current I(LS) just after $t=DT$:

$\frac{1}{2}{L}_P{\left[I\left(LP\right)\right]}^2=\frac{1}{2}{L}_S{\left[I\left(LS\right)\right]}^2$

$\frac{1}{2}{L}_P{\left[\frac{V_{\mathit{IN}}}{L_P}DT\right]}^2=\frac{1}{2}{L}_S{\left[I\left(LS\right)\right]}^2$

$I\left(LS\right)=\sqrt{\frac{L_P}{L_S}}\frac{V_{\mathit{IN}}}{L_P}DT$

$I\left(LS\right)=\frac{N_P}{N_S}\frac{V_{\mathit{IN}}}{L_P}DT$

The amplitude of I(LS) falls at the rate of

$\frac{dI\left(LS\right)}{dt}=\frac{-V_o}{L_S}$

and I(LS) falls to zero at t=(D+D₂)T. Since $D_2{V}_o=V_{\mathit{IN}}\left(N_S/N_P\right)D$,

$D_2=\frac{V_{\mathit{IN}}}{V_o}\frac{N_S}{N_P}D$

D₂ is effectively the duty cycle of the output rectifier D_R.

3. For $\left(D+D_2\right)T<t<T$
The output rectifier D_R is off.
The output capacitor C_L provides the output current to the load R_L.
The switching cycle restarts when the switch M₁ is turned on again at $t=T$.

From the waveforms shown in Fig. 7.39, the following information (for discontinuous-mode operation) can be obtained:

• The maximum value of the current in the switch M₁ is

$\mathit{ID}{\left(M1\right)}_{\mathit{max}}=\frac{V_{\mathit{IN}}}{L_P}DT$

• The maximum value of the current in the output rectifier D_R is

$I{\left(DR\right)}_{\mathit{max}}=\frac{N_P}{N_S}\frac{V_{\mathit{IN}}}{L_P}DT$

• The output voltage V_o can be found by equating the input energy to the output energy within a switching cycle.

$V_{\mathit{IN}}\times \left[\mathrm{Charge}\mathrm{taken}\mathrm{from}{V}_{\mathit{IN}}\mathrm{in}\mathrm{a}\mathrm{switching}\mathrm{cycle}\right]=\frac{V_o^2}{R_L}T$

$V_{\mathit{IN}}\left[\frac{1}{2}DT\frac{DT}{L_P}{V}_{\mathit{IN}}\right]=\frac{V_o^2}{R_L}T$

$V_o=\sqrt{\frac{R_{L}T}{2L_P}}D{V}_{\mathit{IN}}$

• The maximum reverse voltage of D_R, V(6,9) (which is the voltage at node 6 with respect to node 9), is

$V{\left(DR\right)}_{\mathit{max}}=V{\left(6,9\right)}_{\mathit{max}}=V_{\mathit{IN}}\frac{N_S}{N_P}+V_o$

7.6.2.2 Practical Circuit

When a practical transformer (with leakage inductance) is used in the flyback converter circuit shown in Fig. 7.38, there will be large ringings. In order to reduce these ringings to practically acceptable levels, snubber and clamping circuits have to be added. Fig. 7.40 shows a practical flyback converter circuit where a resistor-capacitor snubber (R₂C₂) is used to damp the ringing voltage across the output rectifier D_R and a resistor-capacitor-diode clamping (R₁C₁D_S) is used to clamp the ringing voltage across the switch M₁. What the diode D_S does here is to allow the energy stored by the current in the leakage inductance to be converted to the form of a dc voltage across the clamping capacitor C₁. The energy transferred to C₁ is then dissipated slowly in the parallel resistor R₁, without ringing problems.

The simulated waveforms of the flyback converter (circuit given in Fig. 7.40) for discontinuous-mode operation are shown in Fig. 7.41, where the following assumptions are made:

• D_R and D_S are MUR460 ultrafast diodes.

• M₁ is an IRF640 MOS transistor.

• Transformer T₁ has a practical coupling coefficient of 0.992.

• The effective winding resistance of L_P is 0.025 Ω. The effective winding resistance of L_S is 0.1 Ω.

• The effective series resistance of the output filtering capacitor C_L is 0.05 Ω.

• The switching operation of the converter has reached a steady state.

The waveforms shown in Fig. 7.40 are considered to be acceptable.