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Thyristor and Single-Phase Half-Wave Controlled Rectifier

Abstract

This chapter introduces the thyristor semiconductor device together with its respective I–V characteristics, operation modes, and dynamic switching characteristics. In this chapter, the half-wave thyristor rectifier topology is analyzed for different types of loads and a block diagram of a thyristor gate drive circuit is presented. Solved problems and simulation results are presented to help the reader understand the theory presented in the chapter.

Keywords

Half-wave thyristor rectifier; Simulation; Single-phase thyristor rectifier; Thyristor; Thyristor power circuit; Thyristor states of operation; Thyristor switching angles

3.0. Introduction

In this chapter the single-phase half-wave thyristor rectifier topologies will be analyzed. The thyristor in contrast with a power diode is a semiconductor device whereby its turn-on time can be controlled through its gate. Therefore, if the rectifier topologies are implemented using thyristor devices, then their output voltage can be controlled.

To simplify the analysis of the rectifier topologies of this chapter, the following assumptions have been made:

• For the circuit analysis, only the steady-state operation of the circuit will be considered.

• The semiconductor devices are ideal which means that their transient times, internal resistances, and conduction voltages are neglected.

• The switches are ideal, which means they exhibit either zero or infinite resistance to current.

• The internal resistances of the passive elements are neglected.

• The wire resistances, self-inductances, and capacitances are considered negligible.

• The input source is ideal and does not have internal resistance.

3.1. Thyristor–Silicon Controlled Rectifier

The thyristor is a silicon-controlled rectifier (SCR) and a four layer pnpn structure as shown in Fig. 3.1. The thyristor can be considered as a combination of two transistors as shown in Fig. 3.1(c). The thyristor unlike a diode has an additional terminal called gate. This terminal gives the ability for a thyristor, when it is forward biased, to be turned on at specific time instants. If the thyristor is forward biased (i.e., V_AK > 0) by applying a gating voltage pulse between the gate and the cathode, then the thyristor turns on and current flows from the anode to the cathode. The gating pulse triggers the thyristor only during the positive step of the pulse and due to its internal construction latches to the on state. When the thyristor is in the on state latches to this state and, consequently, there is no need any more for the gating pulse. The conduction state of the thyristor will end when the thyristor becomes reverse biased (i.e., V_AK < 0), as happens with a diode.

According to Fig. 3.1(c), when the thyristor is forward biased and a gating pulse is applied to the thyristor gate, transistor T₂ starts conducting (see Fig. 3.2(a)) allowing current to flow through the base of transistor. Therefore, a current i_A starts flowing from the external circuit to the device T₁ (see Fig. 3.2(b)). The collector current of transistor T₁ is injected into the base of transistor T₂ (i.e., i_B2 = i_C1) resulting in the conduction of T₂ (see Fig. 3.2(c)). These operation stages create the latching process of the transistors, which cause the device to remain in conduction state, even if the gating pulse is removed from the gate. For the thyristor to go to the blocking state, the anode current i_SCR must be reduced to zero. In reality, the turn off of a thyristor is achieved at a current value slightly greater than zero and this current is called holding current and is symbolized by i_H. This happens because the latching process cannot be effective when the anode current becomes less than the holding current. As the thyristor moves from forward-blocking state to the conduction state, the external circuit must allow sufficient anode current to flow to keep the device latched. The minimum anode current that will cause the device to remain in forward conduction as it switches from forward-blocking is called the latching current and is symbolized by i_L. The holding current value is lower than the latching current and both are given by the manufacturer's specifications. The i_A–v_AK characteristic curve of a typical thyristor device is presented in Fig. 3.3.

Figure 3.1 Operation of a typical thyristor.
(a) Thyristor electrical schematic symbol; (b) thyristor power circuit; (c) two transistor behavioral model of a thyristor; (d) thyristor structure.

Figure 3.2 Thyristor behavioral model turn-on stages.
(a) Stage 1 (T₂ is turned on); (b) stage 2 (T₁ is turned on); (c) stage 3 (current flows from anode to cathode). The arrows show the current flow.

Figure 3.3 i_A–v_AK characteristic curve of a typical thyristor.

The gate voltage and current levels which are required to turn on a thyristor depend on the junction temperature of the device. When the junction temperature increases the gate gets more sensitive. The opposite is valid for temperature values below the nominal as shown in Fig. 3.4. For reliable operation in the case of low-temperature applications, adequate current should be supplied to the gate of the device.

As can be seen from the i_A–v_AK characteristic of Fig. 3.3, a thyristor operates in one of the following three states:

a) Conduction state: When the thyristor is forward biased (i.e., v_AK > 0) and a gating signal is applied between its gate and cathode, the thyristor turns on and current starts flowing from the anode to the cathode.

b) Forward-blocking state: During this state although the thyristor is forward biased (i.e., v_AK > 0), it is not conducting but it is in forward-blocking state. If a gating pulse is applied, then the thyristor will be switched into the conduction state.

c) Reverse-blocking state: During this state the thyristor is reverse biased (i.e., v_AK ≤ 0) and, consequently, it is not conducting even if a gating signal is applied to its gate terminal. If the thyristor is in the conduction state and a negative voltage is applied between the anode and cathode, it moves into the reverse-blocking state and, consequently, turns off.

If the forward-blocking voltage exceeds the manufacturer's specification, namely the forward critical blocking voltage (Fig. 3.3), the thyristor switches from the forward-blocking state to the conduction state without a gating pulse. If the reverse-blocking voltage exceeds the manufacturer's specification, the thyristor loses the blocking capability and current I_Α takes high values and the thyristor will be destroyed. This reverse critical voltage is called breakdown voltage.

Figure 3.4 Critical gate currents and voltages required for a typical thyristor to be turned on.

Fig. 3.5 presents the data sheet of a typical thyristor. As Fig. 3.5 indicates, the gate trigger current requirement of the specific thyristor is I_GT = 100 mA for a junction temperature of 25°C.

3.1.1. Thyristor Dynamic Behavior

Fig. 3.6(a) shows a thyristor circuit which will be used to study the switching characteristics of a thyristor. Moreover, Fig. 3.6(b)–(e) present the different waveforms which will be used to explain the dynamic behavior of a thyristor. As Fig. 3.6(a) indicates, initially the input voltage v_A is positive and, consequently, the thyristor is in the forward-blocking state. If a gating pulse is applied during this state, there is a delay time t_d before the anode current i_A starts rising. This delay time, shown in Fig. 3.6(d), is approximately some microseconds and it is due to the charge movement inside the thyristor device. At the end of time t_d the current starts flowing through the thyristor. When the thyristor is in the on state the current starts increasing until it reaches its final value I_T, with a rate of rise di/dt (Fig. 3.6(d)). The maximum value of di/dt is given by the manufacturer and has to be respected by the designer because beyond this value the thyristor will be destroyed due to local overheating (hot spot). This hot spot is due to the fact that during the thyristor turn on the anode current i_A does not have the appropriate time to be uniformly distributed within the thyristor filaments and there is one section of a filament through which the excessive current flows. This phenomenon is shown in Fig. 3.7. To avoid thyristor hot spot phenomenon, an inductor of few μH could be connected in series to the thyristor. The same care should be taken with the rate of rise of the voltage dv_AK/dt (see Fig. 3.6(e)) when the thyristor goes from the on state to the off state. If the dv_AK/dt exceeds the manufacturer's specifications, then the thyristor will turn on randomly without a gating pulse. The time required for current i_Α to increase from 10% to 90% of its nominal value is called rise time (see Fig. 3.6(d), t_R). The sum of time intervals t_d and t_r is the turn on time (t_on) of the thyristor (see Fig. 3.6(d)).

Figure 3.5 Data sheet of a typical thyristor.
Courtesy of SEMIKRON.

The thyristor turn off starts at the instant t₀ where negative voltage is applied across the anode and cathode terminals (Fig. 3.6(b) and (e)). During thyristor turn off, a switching time delay from the on state to the off state occurs. As Fig. 3.6(b) indicates, the anode current from the instant t₀ decreases with a rate of rise di/dt and at the instant t₁ becomes zero. After time instant t_l, anode current becomes negative with the same di/dt slope as before. The reason for the reversal of anode current is due to the presence of carriers stored in the four layers. The reverse recovery current removes excess carriers from the end junctions J₁ and J₃ between the instants t_l and t₃. At instant t₂, when about 60% of the stored charges are removed from the two outer layers, carrier density across J₁ and J₃ begins to decrease and with this reverse recovery current also starts decaying. The reverse current decay is fast in the beginning but gradual thereafter. When the process has ended, the junctions J₁ and J₃ return to blocking state and the anode current at the instant t₃ is approximately zero. The thyristor is now able to block negative voltages, because the junctions J₁ and J₃ are reverse biased. However, junction J₂ remains forward biased due to its trapped charges. The trapped charges around J₂, cannot flow to the external circuit, therefore, these trapped charges must decay only by recombination. This recombination is possible if a reverse voltage is maintained across thyristor, though the magnitude of this voltage is not important. The phenomenon related to the negative current flow in the thyristor is called reverse recovery and the time reverse recovery time t_rr. The time for the recombination of charges between t₃ and t₄ is called gate recovery time t_gr. As Fig. 3.6(e) indicates, t_q = t_rr + t_gr is the time required by the thyristor to obtain again its forward-blocking capability. It is worth noticing that, if the thyristor voltage V_AK takes positive values before obtaining the forward-blocking capability (i.e., before the end of the time determined by the manufacturer t_q), then the thyristor will be triggered into conduction state without the need of a switch-on pulse to its gate. The turn-off time provided to the thyristor by the power circuit is called circuit turn-off time t_c. It is defined as the time between the instant time the thyristor current becomes zero and the instant time of the reverse voltage due to power circuit reaches zero (see Fig. 3.6(d)). Time t_c must be greater than t_q for reliable turn off, otherwise the device may turn on at an undesired instant, a process called commutation failure.

Figure 3.6 Dynamic switching characteristics of a typical thyristor.
(a) A thyristor circuit; (b) circuit input voltage; (c) thyristor gating pulse; (d) thyristor current; (e) thyristor voltage v_AK.

Figure 3.7 Thyristor overheating phenomenon due to di/dt.

The turn-off process of a thyristor is called commutation. The commutation of a thyristor can be achieved using the following techniques:

a) Natural commutation, which can be performed by:

i) Load commutation, where the thyristor is turned off due to the nature of the load and

ii) Line commutation, where the thyristor is turned off due to the nature of the input ac voltage source. For example, when an ac voltage is applied to the anode of a thyristor, at a certain time, for an instant, the voltage across the thyristor or the current that flows through the anode goes to zero and, consequently, the thyristor will turn off. This type of commutation is used in thyristor rectifiers.

b) Forced commutation, which can be performed by:

i) Forced impulse current commutation, where an auxiliary current pulse forces the thyristor into the blocking state. This type of commutation is used in inverters and dc–dc converters that are implemented with thyristors.

Figure 3.8 Thyristor behavioral model turn-off stages.
(a) Stage 1 (T₁ and T₂ are on and silicon-controlled rectifier is conducting); (b) stage 2 (T₁ is turned off); (c) stage 3 (T₂ is turned off and current stops flowing from anode to cathode).

ii) Forced impulse voltage commutation, where an auxiliary voltage pulse forces the thyristor transition into blocking state. Such a commutation is applied to thyristor-based dc–dc converters.

The turn-off stages are shown in Fig. 3.8.

The creation of the thyristor commutation voltages and currents are achieved by applying auxiliary circuits, called commutation circuits. The commutation circuits are part of any power converter, which is implemented with thyristors, and the only way to turn them off is by using force commutation techniques.

3.2. Single-Phase Half-Wave Thyristor Rectifier With Resistive Load

Fig. 3.9 presents the power circuit of a single-phase half-wave controlled rectifier. When during the positive half-cycle a gating pulse i_G is applied to thyristor Q, the thyristor will go to conduction state connecting the input ac voltage across the load. At the angle ωt = π the output voltage and current become zero and, consequently, thyristor will go to the blocking state disconnecting the input ac voltage from the load. Fig. 3.9(b) shows the waveforms of the rectifier. Using the waveforms of Fig. 3.9(b), the following equations are obtained:

$v_{o} = \sqrt{2} {\tilde{V}}_{i} \sin ω t = Ri for α < ω t < π$ $v_{o} = \sqrt{2} {\tilde{V}}_{i} \sin ω t = Ri for α < ω t < π$

(3.1)

$i_{o} = i = \frac{\sqrt{2} {\tilde{V}}_{i}}{R} \sin ω t for α < ω t < π$ $i_{o} = i = \frac{\sqrt{2} {\tilde{V}}_{i}}{R} \sin ω t for α < ω t < π$

(3.2)

According to Fig. 3.9(b) the average output voltage is:

${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{π} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) d (ω t) = {\frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (- \cos ω t) |}_{α}^{π} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (1 + \cos α)$ ${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{π} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) d (ω t) = {\frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (- \cos ω t) |}_{α}^{π} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (1 + \cos α)$

(3.3)

Figure 3.9 Single-phase half-wave thyristor rectifier with resistive load.
(a) Power circuit; (b) key waveforms.

Therefore, the average output current is:

${\bar{I}}_{o} = \frac{{\bar{V}}_{o}}{R}$ ${\bar{I}}_{o} = \frac{{\bar{V}}_{o}}{R}$

(3.4)

Moreover, the rms output voltage is:

${\tilde{V}}_{o} = output rms voltage = {[\frac{1}{2 π} \int_{α}^{π} {[\sqrt{2} {\tilde{V}}_{i} \sin (ω t)]}^{2} d ω t]}^{1 / 2} = \frac{{\tilde{V}}_{i}}{\sqrt{2}} {[1 - \frac{α}{π} + \frac{\sin 2 α}{2 π}]}^{1 / 2}$ ${\tilde{V}}_{o} = output rms voltage = {[\frac{1}{2 π} \int_{α}^{π} {[\sqrt{2} {\tilde{V}}_{i} \sin (ω t)]}^{2} d ω t]}^{1 / 2} = \frac{{\tilde{V}}_{i}}{\sqrt{2}} {[1 - \frac{α}{π} + \frac{\sin 2 α}{2 π}]}^{1 / 2}$

(3.5)

where α is the firing or the delay angle of the thyristor. The firing angle is the angle at which the gate signal is applied to the thyristor and is measured with respect to the beginning of the ac input source.

Figure 3.10 Control characteristic of the rectifier of Fig. 3.10.

Finally, the output active power, the dc output power, and the power factor of the rectifier are respectively given by:

$P_{o} = output active power = {\tilde{I}}_{o}^{2} R = \frac{{\tilde{V}}_{o}^{2}}{R}$ $P_{o} = output active power = {\tilde{I}}_{o}^{2} R = \frac{{\tilde{V}}_{o}^{2}}{R}$

(3.6)

${\bar{P}}_{o} = output dc power = {\bar{V}}_{o} {\bar{I}}_{o}$ ${\bar{P}}_{o} = output dc power = {\bar{V}}_{o} {\bar{I}}_{o}$

(3.7)

$λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = \frac{input active power}{input apparent power} = \frac{P_{o}}{{\tilde{V}}_{i} {\tilde{I}}_{i}} = \frac{{\tilde{I}}_{o}^{2} R}{{\tilde{V}}_{i} {\tilde{I}}_{i}} = \frac{{\tilde{I}}_{o} R}{{\tilde{V}}_{i}} (Assuming that the rectifier is ideal)$ $λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = \frac{input active power}{input apparent power} = \frac{P_{o}}{{\tilde{V}}_{i} {\tilde{I}}_{i}} = \frac{{\tilde{I}}_{o}^{2} R}{{\tilde{V}}_{i} {\tilde{I}}_{i}} = \frac{{\tilde{I}}_{o} R}{{\tilde{V}}_{i}} (Assuming that the rectifier is ideal)$

(3.8)

where

{\tilde{V}}_{i}

${\tilde{V}}_{i}$

= rms input voltage;

{\tilde{I}}_{i}

${\tilde{I}}_{i}$

= rms input current;

{\tilde{I}}_{o}

${\tilde{I}}_{o}$

= rms output current; ω = angular frequency of the input voltage.

Fig. 3.10 presents the control characteristic of the rectifier where

{\bar{V}}_{o (normalized)} = \frac{{\bar{V}}_{o} (α)}{{\bar{V}}_{o (α = 0 °)}}

${\bar{V}}_{o (normalized)} = \frac{{\bar{V}}_{o} (α)}{{\bar{V}}_{o (α = 0 °)}}$

3.3. Single-Phase Half-Wave Thyristor Rectifier With Resistive–Inductive Load

If during the first half-cycle of the input voltage a gating pulse i_G is applied to the thyristor of the rectifier of Fig. 3.11(a), then the thyristor will go to the conduction state connecting the input ac voltage source across the R–L load. The waveforms of the rectifier during the conduction of the thyristor are shown in Fig. 3.11(b). Moreover, when the thyristor is conducting, the following equations hold:

Figure 3.11 Single-phase half-wave thyristor rectifier with R–L load.
(a) Power circuit; (b) key waveforms.

$v_{L} + v_{R} = v_{o} = v$ $v_{L} + v_{R} = v_{o} = v$

(3.9)

$L \frac{{di}_{o}}{dt} + {Ri}_{o} = \sqrt{2} {\tilde{V}}_{i} \sin ω t for α \leq ω t \leq β$ $L \frac{{di}_{o}}{dt} + {Ri}_{o} = \sqrt{2} {\tilde{V}}_{i} \sin ω t for α \leq ω t \leq β$

(3.10)

Solving Eq. (3.10) the circuit current is obtained:

$i_{o} = i_{i} = i_{F} + i_{N} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + {Ae}^{- (\frac{ω t}{τ})} for α \leq ω t \leq β$ $i_{o} = i_{i} = i_{F} + i_{N} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + {Ae}^{- (\frac{ω t}{τ})} for α \leq ω t \leq β$

(3.11)

where

$φ = \tan^{- 1} (\frac{ω L}{R}), | Z_{o} | = {[R^{2} + {(ω L)}^{2}]}^{1 / 2} and τ = load time constant = \frac{L}{R}$ $φ = \tan^{- 1} (\frac{ω L}{R}), | Z_{o} | = {[R^{2} + {(ω L)}^{2}]}^{1 / 2} and τ = load time constant = \frac{L}{R}$

(3.12)

Using the initial condition i(ωt = a) = 0 in Eq. (3.11), the value of the constant A can be calculated and is given by

A = - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) \cdot e^{(\frac{α}{ωτ})}

$A = - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) \cdot e^{(\frac{α}{ωτ})}$

. Substituting A into Eq. (3.11) yields:

$i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) \cdot e^{(\frac{α}{ωτ})} \cdot e^{- (\frac{ω t}{ωτ})} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\sin (ω t - φ) - \sin (α - φ) \cdot e^{(\frac{α - ω t}{ωτ})}] for α \leq ω t \leq β$ $i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) \cdot e^{(\frac{α}{ωτ})} \cdot e^{- (\frac{ω t}{ωτ})} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\sin (ω t - φ) - \sin (α - φ) \cdot e^{(\frac{α - ω t}{ωτ})}] for α \leq ω t \leq β$

(3.13)

Moreover, using the second initial condition of the circuit i(ωt = β) = 0 in Eq. (3.13), the following equation is obtained:

$\sin (β - φ) = \sin (α - φ) \cdot e^{(\frac{α - β}{ωτ})} where τ = load time constant = \frac{L}{R}$ $\sin (β - φ) = \sin (α - φ) \cdot e^{(\frac{α - β}{ωτ})} where τ = load time constant = \frac{L}{R}$

(3.14)

Therefore, knowing the values of α, φ, R, L, and ω, then using Eq. (3.14) and numerical analysis methods the value of β can be calculated. Next, when the value of angle β is known, the conduction angle γ can be found as follows:

$γ = β - α$ $γ = β - α$

(3.15)

At this point it should be noted, that after angle π the output current has a negative slope di/dt resulting in a negative voltage across the inductor terminals making the thyristor forward biased after angle π. Although the voltage in the thyristor anode after π takes negative values the voltage of the cathode terminal due to inductor is more negative. This lasts until turn-off angle β.

Fig. 3.12 presents a family of curves, through which the value of γ can be found when the values of α and φ are known.

Using Eq. (3.13), the average and rms output current can be calculated as follows:

${\bar{I}}_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \frac{1}{2 π} \int_{α}^{α + γ} [\sin (ω t - φ) - \sin (α - φ) \cdot e^{(\frac{α - ω t}{ωτ})}] d (ω t)$ ${\bar{I}}_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \frac{1}{2 π} \int_{α}^{α + γ} [\sin (ω t - φ) - \sin (α - φ) \cdot e^{(\frac{α - ω t}{ωτ})}] d (ω t)$

(3.16)

${\tilde{I}}_{o} = {\tilde{I}}_{i} = {[\frac{1}{2 π} \int_{α}^{α + γ} {[\frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\sin (ω t - φ) - \sin (α - φ) \cdot e^{(\frac{α - ω t}{ωτ})}]]}^{2} d (ω t)]}^{1 / 2}$ ${\tilde{I}}_{o} = {\tilde{I}}_{i} = {[\frac{1}{2 π} \int_{α}^{α + γ} {[\frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\sin (ω t - φ) - \sin (α - φ) \cdot e^{(\frac{α - ω t}{ωτ})}]]}^{2} d (ω t)]}^{1 / 2}$

(3.17)

Figure 3.12 γ versus α with φ as parameter.

Fig. 3.13 presents a family of curves which can be used to find the value of

{\bar{I}}_{o}

${\bar{I}}_{o}$

when the values α and φ are known. Also, Fig. 3.14 presents a family of curves which can be used to find the value of

{\tilde{I}}_{o}

${\tilde{I}}_{o}$

when the values α and φ are known.

According to Fig. 3.8(b) the average output voltage is:

${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) (d ω t) = {\frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (- \cos ω t) |}_{α}^{β} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (\cos α - \cos β)$ ${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) (d ω t) = {\frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (- \cos ω t) |}_{α}^{β} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (\cos α - \cos β)$

(3.18)

Figure 3.13 Normalized average output current, ${\bar{I}}_{oN}$ ${\bar{I}}_{oN}$ , versus α with φ as parameter.

Figure 3.14 Normalized rms output current, ${\tilde{I}}_{oN}$ ${\tilde{I}}_{oN}$ , versus α with φ as parameter.

Knowing that the magnitude of the load impedance in zero or dc frequency is

| Z_{o, 0} | = \sqrt{R^{2} + {(0 \times L)}^{2}} = R

$| Z_{o, 0} | = \sqrt{R^{2} + {(0 \times L)}^{2}} = R$

, then the average output current is:

${\bar{I}}_{o} = \frac{{\bar{V}}_{o}}{R}$ ${\bar{I}}_{o} = \frac{{\bar{V}}_{o}}{R}$

(3.19)

If the rectifier is ideal, then the output active power and the power factor of the rectifier are respectively given by:

$P_{o} = output active power = \frac{{\tilde{V}}_{o}^{2}}{R} = input active power = P_{i}$ $P_{o} = output active power = \frac{{\tilde{V}}_{o}^{2}}{R} = input active power = P_{i}$

(3.20)

$λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = \frac{P_{o}}{S_{i}} = \frac{output active power}{input apparent power} = \frac{P_{o}}{{\tilde{V}}_{i} {\tilde{I}}_{i}}$ $λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = \frac{P_{o}}{S_{i}} = \frac{output active power}{input apparent power} = \frac{P_{o}}{{\tilde{V}}_{i} {\tilde{I}}_{i}}$

(3.21)

It should be mentioned, that since the voltage across the resistive–inductive load is not a sinusoidal waveform, the rms output current

{\tilde{I}}_{o} \neq {\tilde{V}}_{o} / | Z_{o} |

${\tilde{I}}_{o} \neq {\tilde{V}}_{o} / | Z_{o} |$

(where Z_o is the load impedance). The rms output current can be found using Eq. (3.17).

3.4. Single-Phase Half-Wave Thyristor Rectifier With Inductive Load

Fig. 3.15 shows the power circuit and the key waveforms of a single-phase half-wave controlled rectifier with pure inductive load. Since the load is purely inductive, substituting R = 0 into Eq. (3.13), the circuit current for the interval α ≤ ωt ≤ β is:

Figure 3.15 Single-phase half-wave thyristor rectifier with inductive load.
(a) Power circuit; (b) key waveforms.

$i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} (\cos α - \cos ω t) for α \leq ω t \leq β$ $i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} (\cos α - \cos ω t) for α \leq ω t \leq β$

(3.22)

Eq. (3.22) indicates that the circuit current consists of a dc component

\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} \cos α

$\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} \cos α$

and a fundamental harmonic component

- \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} \cos ω t = \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} \sin (ω t - \frac{π}{2})

$- \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} \cos ω t = \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} \sin (ω t - \frac{π}{2})$

, which has a phase difference of

\frac{π}{2}

$\frac{π}{2}$

with respect to the ac input source voltage. Regarding the waveforms of Fig. 3.15(b), the turn-off angle β of the thyristor is given by:

$\begin{matrix} i_{o} (ω t = β) = 0 = \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} (\cos α - \cos β) \\ or & \cos α = \cos β or β = 2 π - α \end{matrix}$ $\begin{matrix} i_{o} (ω t = β) = 0 = \frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} (\cos α - \cos β) \\ or & \cos α = \cos β or β = 2 π - α \end{matrix}$

(3.23)

Furthermore it should be noted that according to Eq. (3.22) the maximum value of the output current appears at ωt = π where the component (cosα – cosωt) takes its maximum value.

Using Eq. (3.22) the average output current is:

${\bar{I}}_{o} = (\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L}) (\frac{1}{2 π}) \int_{α}^{2 π - α} (\cos α - \cos ω t) d ω t = (\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L}) (\frac{1}{π}) [(π - α) \cos α + \sin α]$ ${\bar{I}}_{o} = (\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L}) (\frac{1}{2 π}) \int_{α}^{2 π - α} (\cos α - \cos ω t) d ω t = (\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L}) (\frac{1}{π}) [(π - α) \cos α + \sin α]$

(3.24)

Similarly, from Eq. (3.22), the rms output current is:

${\tilde{I}}_{o} = {[\frac{1}{2 π} \int_{α}^{2 π - α} {[\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} (\cos α - \cos ω t)]}^{2} d (ω t)]}^{1 / 2}$ ${\tilde{I}}_{o} = {[\frac{1}{2 π} \int_{α}^{2 π - α} {[\frac{\sqrt{2} {\tilde{V}}_{i}}{ω L} (\cos α - \cos ω t)]}^{2} d (ω t)]}^{1 / 2}$

(3.25)

Moreover, using Fig. 3.15(b) the rms and average output voltages are:

${\tilde{V}}_{o} = {[\frac{1}{2 π} \int_{α}^{2 π - α} {(\sqrt{2} {\tilde{V}}_{i} \sin ω t)}^{2} d (ω t)]}^{1 / 2} = {\tilde{V}}_{i} {[\frac{π - α + (\frac{1}{2}) \sin 2 α}{π}]}^{1 / 2}$ ${\tilde{V}}_{o} = {[\frac{1}{2 π} \int_{α}^{2 π - α} {(\sqrt{2} {\tilde{V}}_{i} \sin ω t)}^{2} d (ω t)]}^{1 / 2} = {\tilde{V}}_{i} {[\frac{π - α + (\frac{1}{2}) \sin 2 α}{π}]}^{1 / 2}$

(3.26)

${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{2 π - α} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} {[- \cos ω t]}_{α}^{2 π - α} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [- \cos (2 π - α) + \cos α] = 0$ ${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{2 π - α} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} {[- \cos ω t]}_{α}^{2 π - α} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [- \cos (2 π - α) + \cos α] = 0$

(3.27)

where

{\tilde{V}}_{i}

${\tilde{V}}_{i}$

= rms input voltage.

The input active power, the input reactive power, and the power factor of the rectifier are given by the following equations:

$P_{i} = P_{o} = {\bar{V}}_{o} {\bar{I}}_{o} = 0 \cdot {\bar{I}}_{o} = 0 W$ $P_{i} = P_{o} = {\bar{V}}_{o} {\bar{I}}_{o} = 0 \cdot {\bar{I}}_{o} = 0 W$

(3.28(a))

$Q_{i} = {\tilde{V}}_{i} {\tilde{I}}_{i} = {\tilde{V}}_{i} {\tilde{I}}_{o} VAR and S_{i} = Q_{i} VA$ $Q_{i} = {\tilde{V}}_{i} {\tilde{I}}_{i} = {\tilde{V}}_{i} {\tilde{I}}_{o} VAR and S_{i} = Q_{i} VA$

(3.28(b))

$λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = 0$ $λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = 0$

(3.29)

3.5. Single-Phase Half-Wave Thyristor Rectifier With R–L–E Load

Fig. 3.16 presents the power circuit and the key waveforms of a single-phase half-wave thyristor rectifier with R–L–E load. The dc load source E can be a battery bank or the counter EMF of a dc motor. As Fig. 3.16(b) indicates, special attention shall be given so that the position of the firing angle α must be after the intersection angle η. At the intersection angle η the input voltage is equal to dc voltage E. If the position of the firing angle α is before the intersection angle η, then the thyristor will be unable to go to the conduction state because it is reverse biased (v_AK = v_i − E). However, if the gating signal is applied after the angle η, where v_AK > 0 the thyristor is forward biased and therefore can be switched to the conduction state. Using Fig. 3.16(b) the intersection angle η is given by:

Figure 3.16 Single-phase half-wave thyristor rectifier with R–L–E load.
(a) Power circuit; (b) key waveforms.

$η = \sin^{- 1} (\frac{E}{\sqrt{2} {\tilde{V}}_{i}}) = \sin^{- 1} (m)$ $η = \sin^{- 1} (\frac{E}{\sqrt{2} {\tilde{V}}_{i}}) = \sin^{- 1} (m)$

(3.30)

Moreover, the circuit current, when the thyristor conducts, is given by the following equation:

$\begin{matrix} i_{o} = steady - state component + transient - state component + dc component \\ = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + {Ae}^{- (\frac{ω t}{ωτ})} - \frac{E}{R} for α \leq ω t \leq β \end{matrix}$ $\begin{matrix} i_{o} = steady - state component + transient - state component + dc component \\ = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + {Ae}^{- (\frac{ω t}{ωτ})} - \frac{E}{R} for α \leq ω t \leq β \end{matrix}$

(3.31)

Using the initial condition i_o(ωt = 0°) = 0, then from Eq. (3.31) the value of constant A is found:

$A = [- \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) + \frac{E}{R}] e^{(\frac{α}{ωτ})}$ $A = [- \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) + \frac{E}{R}] e^{(\frac{α}{ωτ})}$

Substituting the value of A into Eq. (3.31) the final form of the circuit current will be:

$i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + [- \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) + \frac{E}{R}] e^{(\frac{α}{ωτ})} \cdot e^{- (\frac{ω t}{ωτ})} - \frac{E}{R} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + [- \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) + \frac{E}{R}] e^{(\frac{α - ω t}{ωτ})} - \frac{E}{R} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\frac{m}{\cos φ} - {Be}^{(\frac{α - ω t}{\tan φ})}] for α \leq ω t \leq β$ $i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + [- \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) + \frac{E}{R}] e^{(\frac{α}{ωτ})} \cdot e^{- (\frac{ω t}{ωτ})} - \frac{E}{R} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + [- \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ) + \frac{E}{R}] e^{(\frac{α - ω t}{ωτ})} - \frac{E}{R} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\frac{m}{\cos φ} - {Be}^{(\frac{α - ω t}{\tan φ})}] for α \leq ω t \leq β$

(3.32)

where

$B = \frac{m}{\cos φ} - \sin (α - φ) \cos φ = \frac{R}{| Z_{o} |} \tan φ = \frac{ω L}{R} = ωτ$ $B = \frac{m}{\cos φ} - \sin (α - φ) \cos φ = \frac{R}{| Z_{o} |} \tan φ = \frac{ω L}{R} = ωτ$

(3.33)

$m = load dc source factor = \frac{E}{\sqrt{2} {\tilde{V}}_{i}}$ $m = load dc source factor = \frac{E}{\sqrt{2} {\tilde{V}}_{i}}$

Using the initial conditions of the circuit (i(ωt = α) = 0 and i(ωt = β) = 0) into Eq. (3.31), the following equation is obtained:

$\frac{(m / \cos φ) - \sin (α + γ - φ)}{(m / \cos φ) - \sin (α - φ)} = e^{- γ / \tan φ}$ $\frac{(m / \cos φ) - \sin (α + γ - φ)}{(m / \cos φ) - \sin (α - φ)} = e^{- γ / \tan φ}$

(3.34)

Using Eq. (3.34) the value of the thyristor conduction angle γ can be calculated using numerical analysis methods when the values of α, φ, and m are known.

From Fig. 3.16(b) the average output current of the half-wave rectifier is:

${\bar{I}}_{o} = \frac{1}{2 π} \int_{α}^{β} i_{o} (ω t) d (ω t) = \frac{{\bar{V}}_{o} - E}{R}$ ${\bar{I}}_{o} = \frac{1}{2 π} \int_{α}^{β} i_{o} (ω t) d (ω t) = \frac{{\bar{V}}_{o} - E}{R}$

(3.35)

When

{\bar{I}}_{o}

${\bar{I}}_{o}$

is known, the average output voltage is

{\bar{V}}_{o} = {\bar{I}}_{o} R + E

${\bar{V}}_{o} = {\bar{I}}_{o} R + E$

. Also, from Fig. 3.16(b) the rms output current is:

${\tilde{I}}_{o} = {\tilde{I}}_{i} = {[\frac{1}{2 π} \int_{α}^{β} {(i_{o})}^{2} d (ω t)]}^{1 / 2}$ ${\tilde{I}}_{o} = {\tilde{I}}_{i} = {[\frac{1}{2 π} \int_{α}^{β} {(i_{o})}^{2} d (ω t)]}^{1 / 2}$

(3.36)

where i_o is given by Eq. (3.31).

Finally, the active power delivered to R and E, the input active power, and the power factor of the rectifier are given by:

$P_{R} = power consumed by the load resistance = {\tilde{I}}_{o}^{2} R$ $P_{R} = power consumed by the load resistance = {\tilde{I}}_{o}^{2} R$

(3.37)

$P_{E} = power consumed by the load dc source E = {\bar{I}}_{o} E$ $P_{E} = power consumed by the load dc source E = {\bar{I}}_{o} E$

(3.38)

$P_{i} = input active power = P_{R} + P_{E}$ $P_{i} = input active power = P_{R} + P_{E}$

(3.39)

$λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = \frac{P_{R} + P_{E}}{{\tilde{V}}_{i} {\tilde{I}}_{i}}$ $λ = power factor at the ac side = \frac{P_{i}}{S_{i}} = \frac{P_{R} + P_{E}}{{\tilde{V}}_{i} {\tilde{I}}_{i}}$

(3.40)

3.6. Gate Drive Circuits for Thyristors

The gating pulses of a thyristor are generated by the control circuit. These pulses before applied to the gate of the thyristor are going through the so-called gate drive circuit that provides the following features:

1) Converts the logic-level control signals into the appropriate voltage and current for efficient, reliable, turn on of the thyristor module.

2) Provides ohmic isolation so that the logic signals are not connected to the high voltage present in the power circuit.

3) Discharges the parasitic capacitors of the semiconductor device.

Figure 3.17 Block diagram of a thyristor gate drive circuit.

Fig. 3.17 shows the block diagram of a gate drive circuit that drives one thyristor of a rectifier bridge. Fig. 3.18 shows four different thyristor gate drive circuits, three with ohmic isolation and one without.

Figure 3.18 Thyristor gate drive circuits.
(a) With transformer ohmic isolation; (b) with opto-coupler isolation; (c) without ohmic isolation. SCR, silicon-controlled rectifier.

Example 3.1

The specifications for the circuit shown below include the following:

v_{i} = 222 \sqrt{2} \sin 314 t

$v_{i} = 222 \sqrt{2} \sin 314 t$

, α = 45°, R = 20 Ω, and L = 10 mH.

Find the relation of the circuit current and calculate the average output voltage and current.

Solution

Using Eq. (3.11) the circuit current is given by:

$i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\sin (ω t - φ) - \sin (α - φ) e^{(\frac{α - ω t}{ωτ})}]$ $i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} [\sin (ω t - φ) - \sin (α - φ) e^{(\frac{α - ω t}{ωτ})}]$

(1)

where

$φ = \tan^{- 1} (\frac{ω L}{R}) = \tan^{- 1} (\frac{314 \times 10 \times 10^{- 3}}{20}) = 9 ° ω = 314 rads / s α = 45 ° or 0.785 rads | Z_{o} | = {[R^{2} + {(ω L)}^{2}]}^{1 / 2} = {[20^{2} + {(314 \times 10 \times 10^{- 3})}^{2}]}^{1 / 2} = 20.24 Ω τ = \frac{L}{R} = \frac{10 \times 10^{- 3}}{20} = 0.0005 s ωτ = 314 \times 0.0005 = 0.157 rads {\tilde{V}}_{i} = 222 V$ $φ = \tan^{- 1} (\frac{ω L}{R}) = \tan^{- 1} (\frac{314 \times 10 \times 10^{- 3}}{20}) = 9 ° ω = 314 rads / s α = 45 ° or 0.785 rads | Z_{o} | = {[R^{2} + {(ω L)}^{2}]}^{1 / 2} = {[20^{2} + {(314 \times 10 \times 10^{- 3})}^{2}]}^{1 / 2} = 20.24 Ω τ = \frac{L}{R} = \frac{10 \times 10^{- 3}}{20} = 0.0005 s ωτ = 314 \times 0.0005 = 0.157 rads {\tilde{V}}_{i} = 222 V$

Therefore, Eq. (1) becomes:

$i_{o} = 15.51 [\sin (ω t - 9 °) - \sin (α - 9 °) e^{(\frac{α - ω t}{ωτ})}]$ $i_{o} = 15.51 [\sin (ω t - 9 °) - \sin (α - 9 °) e^{(\frac{α - ω t}{ωτ})}]$

(2)

Using the initial condition i_o(ωt = β) = 0, then from Eq. (2) the following equation is found:

$\sin (β - 9 °) = \sin (α - φ) e^{(\frac{α - β}{ωτ})}$ $\sin (β - 9 °) = \sin (α - φ) e^{(\frac{α - β}{ωτ})}$

(3)

Solving Eq. (3) with numerical methods, it is found that β = 189°.

The average output voltage and current are given by:

${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) (d ω t) = {\frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (- \cos ω t) |}_{α}^{β} = \frac{222 \sqrt{2}}{2 π} (\cos 45 ° - \cos 189 °) = 84.7 V {\bar{I}}_{o} = \frac{{\bar{V}}_{o}}{R} = \frac{84.7}{20} = 4.23 A$ ${\bar{V}}_{o} = \frac{1}{2 π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) (d ω t) = {\frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (- \cos ω t) |}_{α}^{β} = \frac{222 \sqrt{2}}{2 π} (\cos 45 ° - \cos 189 °) = 84.7 V {\bar{I}}_{o} = \frac{{\bar{V}}_{o}}{R} = \frac{84.7}{20} = 4.23 A$

Example 3.2

For the circuit shown below the input voltage is

v_{i} = \sqrt{2} {\tilde{V}}_{i} \sin ω t

$v_{i} = \sqrt{2} {\tilde{V}}_{i} \sin ω t$

. If the firing angle α is different than 0° and 180°, draw the output waveforms and calculate the average and rms output voltage.

Solution

When a gating pulse is applied to thyristor and the thyristor is forward biased, then the thyristor will conduct connecting the ac input source to the load. As can be seen from the above circuit for the thyristor to become forward biased the anode voltage, which is supplied by the input ac source, must be higher than the cathode voltage which is supplied by the dc load source E. Therefore, the firing angle must be applied to the thyristor after the intersection point between the input voltage and E. The thyristor will remain in conduction state until angle β where the voltage across its terminals is reduced to zero. Afterward, V_AK takes negative values and hence the thyristor becomes reverse biased and, consequently, turns off (locking state) disconnecting the ac supply from the load. When the thyristor is in blocking state, the output voltage is equal to dc load source E. The voltage and current waveforms of the given circuit are shown in Fig. 3.19:

The intersection point angle η that indicates the minimum value of the firing angle α is given by the following equation:

$η = α_{min} = \sin^{- 1} (\frac{E}{\sqrt{2} {\tilde{V}}_{i}}) = \sin^{- 1} m$ $η = α_{min} = \sin^{- 1} (\frac{E}{\sqrt{2} {\tilde{V}}_{i}}) = \sin^{- 1} m$

(1)

The differential equation of the circuit and the current function during the thyristor conduction state are given by the following equations:

$\sqrt{2} {\tilde{V}}_{i} \sin ω t = i_{o} R + L (\frac{{di}_{o}}{dt}) + E for α \leq ω t \leq β$ $\sqrt{2} {\tilde{V}}_{i} \sin ω t = i_{o} R + L (\frac{{di}_{o}}{dt}) + E for α \leq ω t \leq β$

(2)

$\begin{matrix} i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + {Ae}^{\frac{- t}{τ}} - \frac{E}{R} \\ where : | Z_{o} | = \sqrt{R^{2} + {(ω L)}^{2}}, φ = \tan^{- 1} (\frac{ω L}{R}) and τ = \frac{L}{R} \end{matrix}$ $\begin{matrix} i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) + {Ae}^{\frac{- t}{τ}} - \frac{E}{R} \\ where : | Z_{o} | = \sqrt{R^{2} + {(ω L)}^{2}}, φ = \tan^{- 1} (\frac{ω L}{R}) and τ = \frac{L}{R} \end{matrix}$

(3)

Using the circuit initial condition i_ο(ωt = α) = 0 the constant A is found and Eq. (3) becomes:

$i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) - \frac{E}{R} + [\frac{E}{R} - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ)] e^{\frac{- R}{ω L} (ω t - α)}$ $i_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (ω t - φ) - \frac{E}{R} + [\frac{E}{R} - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ)] e^{\frac{- R}{ω L} (ω t - α)}$

(4)

Moreover, substituting the circuit initial condition i_ο(ωt = β) = 0 into Eq. (4) the turn-off angle β is found and is given by:

$i_{o} = 0 = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (β - φ) - \frac{E}{R} + [\frac{E}{R} - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ)] e^{\frac{- R}{ω L} (β - α)}$ $i_{o} = 0 = \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (β - φ) - \frac{E}{R} + [\frac{E}{R} - \frac{\sqrt{2} {\tilde{V}}_{i}}{| Z_{o} |} \sin (α - φ)] e^{\frac{- R}{ω L} (β - α)}$

(5)

$or \frac{(m / \cos φ) - \sin (α + γ - φ)}{(m / \cos φ) - \sin (α - φ)} = e^{- γ / \tan φ}$ $or \frac{(m / \cos φ) - \sin (α + γ - φ)}{(m / \cos φ) - \sin (α - φ)} = e^{- γ / \tan φ}$

(6)

Using the above output voltage waveform, the average output voltage can be calculated as follows:

${\bar{V}}_{o} = \frac{1}{2 π} \int_{0}^{2 π} v_{o} (ω t) d (ω t) = \frac{1}{2 π} [\int_{0}^{α} v_{o} (ω t) d (ω t) + \int_{α}^{β} v_{o} (ω t) d (ω t) + \int_{β}^{2 π} v_{o} (ω t) d (ω t)] where v_{o} = \sqrt{2} {\tilde{V}}_{i} \sin ω t for α \leq ω t \leq β$ ${\bar{V}}_{o} = \frac{1}{2 π} \int_{0}^{2 π} v_{o} (ω t) d (ω t) = \frac{1}{2 π} [\int_{0}^{α} v_{o} (ω t) d (ω t) + \int_{α}^{β} v_{o} (ω t) d (ω t) + \int_{β}^{2 π} v_{o} (ω t) d (ω t)] where v_{o} = \sqrt{2} {\tilde{V}}_{i} \sin ω t for α \leq ω t \leq β$

(7)

Knowing that from 0° to α and β to 2π, the output voltage v_o = E, the average output voltage will be:

${\bar{V}}_{o} = \frac{1}{2 π} [\int_{0}^{α} Ed (ω t) + \sqrt{2} \int_{α}^{β} {\tilde{V}}_{i} \sin (ω t) d (ω t) + \int_{β}^{2 π} Ed (ω t)] = \frac{1}{2 π} [{E (ω t) |}_{0}^{α} +] {\sqrt{2} {\tilde{V}}_{i} (- \cos ω t) |}_{α}^{β} + {E (ω t) |}_{β}^{2 π} = \frac{1}{2 π} [E (α - 0 °) - \sqrt{2} {\tilde{V}}_{i} (\cos β - \cos α) + E (2 π - β)] = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (\cos α - \cos β) + [\frac{2 π - (β - α)}{2 π}] E$ ${\bar{V}}_{o} = \frac{1}{2 π} [\int_{0}^{α} Ed (ω t) + \sqrt{2} \int_{α}^{β} {\tilde{V}}_{i} \sin (ω t) d (ω t) + \int_{β}^{2 π} Ed (ω t)] = \frac{1}{2 π} [{E (ω t) |}_{0}^{α} +] {\sqrt{2} {\tilde{V}}_{i} (- \cos ω t) |}_{α}^{β} + {E (ω t) |}_{β}^{2 π} = \frac{1}{2 π} [E (α - 0 °) - \sqrt{2} {\tilde{V}}_{i} (\cos β - \cos α) + E (2 π - β)] = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (\cos α - \cos β) + [\frac{2 π - (β - α)}{2 π}] E$

(8)

Finally, the rms output voltage can be calculated using the following equation:

${\tilde{V}}_{o} = \sqrt{\frac{1}{2 π} [\int_{0}^{2 π} v_{o}^{2} (ω t) d (ω t)]}$ ${\tilde{V}}_{o} = \sqrt{\frac{1}{2 π} [\int_{0}^{2 π} v_{o}^{2} (ω t) d (ω t)]}$

(9)

Example 3.3

For the half-wave controlled rectifier shown below calculate the average output voltage and current.

Solution

The key waveforms of the above rectifier are shown in Fig. 3.20. As can be seen from the waveforms of Fig. 3.20 the intersection angle η is given by:

$η = \sin^{- 1} (\frac{48}{220 \sqrt{2}}) = 8.87 °$ $η = \sin^{- 1} (\frac{48}{220 \sqrt{2}}) = 8.87 °$

Since the firing angle α = 10° is higher than 8.87°, the thyristor is forward biased and, consequently, will conduct connecting the ac source across the load. This mode of operation will end when the thyristor voltage V_AK will become equal to zero and from there on the thyristor becomes reverse biased and turns off. The voltage V_AK becomes zero when for once more the ac source becomes equal to the dc load source E and that happens at the turn-off angle β = 180°−8.87°=171.13°. Therefore, the thyristor conduction angle for one cycle will be:

$γ = β - α = 171.13 ° - 10 ° = 161.13 ° .$ $γ = β - α = 171.13 ° - 10 ° = 161.13 ° .$

From the above circuit, when the thyristor is conducting, the output voltage and current are respectively given by:

$v_{o} = i_{o} R - 48$ $v_{o} = i_{o} R - 48$

$i_{o} = \frac{v_{o} - 48}{R} = \frac{v_{o} - 48}{1} = v_{o} - 48$ $i_{o} = \frac{v_{o} - 48}{R} = \frac{v_{o} - 48}{1} = v_{o} - 48$

The current waveform will have a maximum value of

\frac{220 \sqrt{2} - 48}{1} = 263.13 A

$\frac{220 \sqrt{2} - 48}{1} = 263.13 A$

Using the output voltage waveform, the average output voltage of the rectifier is:

${\bar{V}}_{o} = \frac{1}{T} \int_{0}^{T} v_{o} (t) dt = \frac{1}{2 π} \int_{0}^{2 π} v_{o} (ω t) d (ω t) = \frac{1}{2 π} [\int_{0}^{10 °} 48 (d ω t) + \int_{10 °}^{171.13 °} 220 \sqrt{2} \sin ω t d (ω t) + \int_{171.13 °}^{360 °} 48 d (ω t)] = 124.23 V$ ${\bar{V}}_{o} = \frac{1}{T} \int_{0}^{T} v_{o} (t) dt = \frac{1}{2 π} \int_{0}^{2 π} v_{o} (ω t) d (ω t) = \frac{1}{2 π} [\int_{0}^{10 °} 48 (d ω t) + \int_{10 °}^{171.13 °} 220 \sqrt{2} \sin ω t d (ω t) + \int_{171.13 °}^{360 °} 48 d (ω t)] = 124.23 V$

Therefore, the average output current is:

${\bar{I}}_{o} = \frac{{\bar{V}}_{o} - 48}{R} = \frac{124.23 - 48}{1} = 76.23 A$ ${\bar{I}}_{o} = \frac{{\bar{V}}_{o} - 48}{R} = \frac{124.23 - 48}{1} = 76.23 A$

Example 3.4

For a single-phase half-wave thyristor rectifier with resistive load, express its output voltage in terms of Fourier series. The input voltage is

v_{i} = \sqrt{2} {\tilde{V}}_{i} \sin (ω t)

$v_{i} = \sqrt{2} {\tilde{V}}_{i} \sin (ω t)$

Solution

The output voltage of the rectifier of Fig. 3.9(a) can be represented by the following Fourier series:

$v_{o} = {\bar{V}}_{o} + \sum_{n = 1,2,3,}^{\infty} {\hat{V}}_{o 1, n} \sin (n ω t) + \sum_{n = 1,2,3,}^{\infty} {\hat{V}}_{o 2, n} \cos (n ω t) = {\bar{V}}_{o} + {\hat{V}}_{o 1,1} \sin (ω t) + {\hat{V}}_{o 1,2} \sin (2 ω t) + {\hat{V}}_{o 1,3} \sin (3 ω t) \dots + {\hat{V}}_{o 2,1} \cos (ω t) + {\hat{V}}_{o 2,2} \cos (2 ω t) + {\hat{V}}_{o 2,3} \cos (3 ω t) \dots$ $v_{o} = {\bar{V}}_{o} + \sum_{n = 1,2,3,}^{\infty} {\hat{V}}_{o 1, n} \sin (n ω t) + \sum_{n = 1,2,3,}^{\infty} {\hat{V}}_{o 2, n} \cos (n ω t) = {\bar{V}}_{o} + {\hat{V}}_{o 1,1} \sin (ω t) + {\hat{V}}_{o 1,2} \sin (2 ω t) + {\hat{V}}_{o 1,3} \sin (3 ω t) \dots + {\hat{V}}_{o 2,1} \cos (ω t) + {\hat{V}}_{o 2,2} \cos (2 ω t) + {\hat{V}}_{o 2,3} \cos (3 ω t) \dots$

(1)

where n = harmonic order.

The output voltage v_o of the single-phase half-wave thyristor rectifier with R–L load, shown in Fig. 3.9(b), does not present any symmetry and, consequently, the average output voltage and the Fourier coefficients of the output voltage are given by the following equations:

${\bar{V}}_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (\cos α - \cos β)$ ${\bar{V}}_{o} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (\cos α - \cos β)$

(2)

and

${\hat{V}}_{o 1, n} = \frac{1}{π} \int_{0}^{2 π} v_{o} \sin (n ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) \sin (n ω t) d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} \int_{α}^{β} [\cos (1 - n) ω t - \cos (1 + n) ω t] d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [{\frac{\sin (1 - n) ω t}{1 - n} |}_{α}^{β} - {\frac{\sin (1 + n) ω t}{1 + n} |}_{α}^{β}] = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [\frac{1}{1 - n} (\sin (1 - n) β - \sin (1 - n) α) - \frac{1}{1 + n} (\sin (1 + n) β - \sin (1 + n) α)]$ ${\hat{V}}_{o 1, n} = \frac{1}{π} \int_{0}^{2 π} v_{o} \sin (n ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) \sin (n ω t) d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} \int_{α}^{β} [\cos (1 - n) ω t - \cos (1 + n) ω t] d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [{\frac{\sin (1 - n) ω t}{1 - n} |}_{α}^{β} - {\frac{\sin (1 + n) ω t}{1 + n} |}_{α}^{β}] = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [\frac{1}{1 - n} (\sin (1 - n) β - \sin (1 - n) α) - \frac{1}{1 + n} (\sin (1 + n) β - \sin (1 + n) α)]$

(3)

${\hat{V}}_{o 2, n} = \frac{1}{π} \int_{0}^{2 π} v_{o} \cos (n ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) \cos (n ω t) d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} \int_{α}^{β} [\sin (1 + n) ω t + \sin (1 - n) ω t] d (ω t) = \frac{\sqrt{2} V_{i}}{2 π} [{- \frac{\cos (1 + n) ω t}{1 + n} |}_{α}^{β} - {\frac{\cos (1 - n) ω t}{1 - n} |}_{α}^{β}] = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [\frac{1}{1 + n} (\cos (1 + n) α - \cos (1 + n) β) + \frac{1}{1 - n} (\cos (1 - n) α - \cos (1 - n) β)]$ ${\hat{V}}_{o 2, n} = \frac{1}{π} \int_{0}^{2 π} v_{o} \cos (n ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) \cos (n ω t) d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} \int_{α}^{β} [\sin (1 + n) ω t + \sin (1 - n) ω t] d (ω t) = \frac{\sqrt{2} V_{i}}{2 π} [{- \frac{\cos (1 + n) ω t}{1 + n} |}_{α}^{β} - {\frac{\cos (1 - n) ω t}{1 - n} |}_{α}^{β}] = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} [\frac{1}{1 + n} (\cos (1 + n) α - \cos (1 + n) β) + \frac{1}{1 - n} (\cos (1 - n) α - \cos (1 - n) β)]$

(4)

The amplitude of the fundamental component cannot be calculated from Eqs. (3) and (4) and for this reason the following equation is used:

${\hat{V}}_{o 1,1} = \frac{1}{π} \int_{α}^{β} v_{o} \sin (ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) \sin (ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \frac{(1 - \cos 2 ω t)}{2} d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} {(ω t - \frac{\sin 2 ω t}{2}) |}_{α}^{β} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (β - α + \frac{\sin 2 α}{2} - \frac{\sin 2 β}{2})$ ${\hat{V}}_{o 1,1} = \frac{1}{π} \int_{α}^{β} v_{o} \sin (ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \sin (ω t) \sin (ω t) d (ω t) = \frac{1}{π} \int_{α}^{β} \sqrt{2} {\tilde{V}}_{i} \frac{(1 - \cos 2 ω t)}{2} d (ω t) = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} {(ω t - \frac{\sin 2 ω t}{2}) |}_{α}^{β} = \frac{\sqrt{2} {\tilde{V}}_{i}}{2 π} (β - α + \frac{\sin 2 α}{2} - \frac{\sin 2 β}{2})$

(5)

3.7. Simulation Examples Using the Power Simulation Software Program

Figs. 3.21 and 3.22 present the simulation results of the single-phase half-wave thyristor rectifier using the power simulation software program for two different loads.

Example 3.5

For the circuit shown below, draw the time variations of the variables i, v_L, and v_C when the thyristor is triggered to conduction and the capacitor is initially charged with a voltage of v_C(0) = −50 V.

Solution

At t = 0 when the thyristor is triggered to conduction, the following expression is valid:

$V - v_{L} - v_{C} - V_{C} (0) = 0 or v_{L} = V - v_{C} - v_{C} (0)$ $V - v_{L} - v_{C} - V_{C} (0) = 0 or v_{L} = V - v_{C} - v_{C} (0)$

(1)

Moreover, the voltage across the inductor at t = 0 is:

$v_{L} (0) = {L \frac{di}{dt} |}_{t = 0} = 150 V$ $v_{L} (0) = {L \frac{di}{dt} |}_{t = 0} = 150 V$

${\frac{di}{dt} |}_{t = 0} = \frac{150}{100 \times 10^{- 6}} = 1.5 \times 10^{6} \frac{A}{s}$ ${\frac{di}{dt} |}_{t = 0} = \frac{150}{100 \times 10^{- 6}} = 1.5 \times 10^{6} \frac{A}{s}$

(2)

Figure 3.21 Power simulation results of the single-phase half-wave thyristor rectifier with ${\tilde{V}}_{i} = 222 V$ ${\tilde{V}}_{i} = 222 V$ 50 Hz, R = 20 Ω, L = 10 mH, and α = 45^ο.

Figure 3.22 Power simulation results of the single-phase half-wave thyristor rectifier with ${\tilde{V}}_{i} = 222 V$ ${\tilde{V}}_{i} = 222 V$ 50 Hz, R = 20 Ω, and L = 150 mH.

The current of the resonant circuit is given by the following equation:

$i = e^{- ξ t} (B_{1} \cos ω_{R} t + B_{2} \sin ω_{R} t) where ξ = \frac{R}{2 L} = 0 ω_{R} = resonant frequency = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{100 \times 10^{- 6} \times 25 \times 10^{- 6}}} = 2 \times 10^{4} \frac{rad}{s}$ $i = e^{- ξ t} (B_{1} \cos ω_{R} t + B_{2} \sin ω_{R} t) where ξ = \frac{R}{2 L} = 0 ω_{R} = resonant frequency = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{100 \times 10^{- 6} \times 25 \times 10^{- 6}}} = 2 \times 10^{4} \frac{rad}{s}$

(3)

Differentiating Eq. (3) and using Eq. (2) for t = 0 the value of B₂ is obtained:

$\begin{matrix} {\frac{di}{dt} |}_{t = 0} = e^{0} (- B_{1} ω_{R} \sin 0 + B_{2} ω_{R} \cos 0) = 1.5 \times 10^{6} (\frac{A}{s}) \\ or B_{2} = \frac{1.5 \times 10^{6}}{ω_{R}} = \frac{1.5 \times 10^{6}}{2 \cdot 10^{4}} = 75 \end{matrix}$ $\begin{matrix} {\frac{di}{dt} |}_{t = 0} = e^{0} (- B_{1} ω_{R} \sin 0 + B_{2} ω_{R} \cos 0) = 1.5 \times 10^{6} (\frac{A}{s}) \\ or B_{2} = \frac{1.5 \times 10^{6}}{ω_{R}} = \frac{1.5 \times 10^{6}}{2 \cdot 10^{4}} = 75 \end{matrix}$

Next, using the initial condition, where at ωt = 0 and the current i = 0 the value of B₁ is obtained:

$i = 0 = e^{0} (B_{1} \cos 0 + B_{2} \sin 0) or B_{1} = 0$ $i = 0 = e^{0} (B_{1} \cos 0 + B_{2} \sin 0) or B_{1} = 0$

Substituting the values of B₁ and B₂ into Eq. (3), the circuit current is given by:

$i = 75 \sin ω_{R} t$ $i = 75 \sin ω_{R} t$

(4)

Moreover, the voltages across the inductor L and capacitor C are respectively given by:

$v_{L} = L \frac{di}{dt} = L \frac{d}{dt} (75 \sin ω_{R} t) = 75 ω_{R} L \cos ω_{R} t = 75 \times 2 \times 10^{4} \times 100 \times 10^{- 6} \cos ω_{R} t = 150 \cos ω_{R} t$ $v_{L} = L \frac{di}{dt} = L \frac{d}{dt} (75 \sin ω_{R} t) = 75 ω_{R} L \cos ω_{R} t = 75 \times 2 \times 10^{4} \times 100 \times 10^{- 6} \cos ω_{R} t = 150 \cos ω_{R} t$

(5)

$v_{C} = 100 - v_{L} = 100 - 150 \cos ω_{R} t$ $v_{C} = 100 - v_{L} = 100 - 150 \cos ω_{R} t$

(6)

As can be seen from Eqs. (5) and (6) when ωt = π/2 the inductor voltage becomes zero and the capacitor voltage becomes equal to 100 V. Therefore, at ωt = π/2 the voltage across the thyristor becomes:

$v_{AK} = 100 - v_{L} - v_{C} = 100 - 0 - 100 = 0 V$ $v_{AK} = 100 - v_{L} - v_{C} = 100 - 0 - 100 = 0 V$

and, consequently, the thyristor is naturally commutated (i.e., turned off) and the diode becomes forward biased and starts conducting.

Therefore, based on the above equations and results, the time variations of i, v_L, and v_C are presented in Fig. 3.23.

Example 3.6

For the circuit shown below, draw the time variations of i, v_L, and v_C when the thyristor is triggered at t = 0.

Solution

At t = 0, as soon as the thyristor Q is switched into the conduction state, the diode D is reverse biased and, consequently, the equivalent circuit during this mode is:

Using the equivalent circuit the following equation is obtained:

$100 - v_{L} - v_{C} + 75 = 0 or L \frac{di}{dt} + \frac{1}{C} \int_{0}^{t} i \cdot dt = 175 or L \frac{d^{2} i}{{dt}^{2}} + \frac{i}{C} = 0$ $100 - v_{L} - v_{C} + 75 = 0 or L \frac{di}{dt} + \frac{1}{C} \int_{0}^{t} i \cdot dt = 175 or L \frac{d^{2} i}{{dt}^{2}} + \frac{i}{C} = 0$

(1)

Therefore, using Eq. (1) the circuit characteristic equation will be:

${LS}^{2} + \frac{1}{C} = 0$ ${LS}^{2} + \frac{1}{C} = 0$

(2)

The roots of the above equation are given by:

$S_{1,2} = \pm j \sqrt{\frac{1}{LC}} = \pm j ω_{R}$ $S_{1,2} = \pm j \sqrt{\frac{1}{LC}} = \pm j ω_{R}$

Therefore, the equivalent circuit current is given by:

$i = A_{1} e^{s_{1} t} + A_{2} e^{s_{2} t} = A_{1} e^{j ω_{R} t} + A_{2} e^{- j ω_{R} t}$ $i = A_{1} e^{s_{1} t} + A_{2} e^{s_{2} t} = A_{1} e^{j ω_{R} t} + A_{2} e^{- j ω_{R} t}$

(3)

It is well known that:

$\begin{matrix} e^{{jω}_{R} t} = {cosω}_{R} t + {jsinω}_{R} t \\ e^{- {jω}_{R} t} = {cosω}_{R} t - {jsinω}_{R} t \end{matrix}$ $\begin{matrix} e^{{jω}_{R} t} = {cosω}_{R} t + {jsinω}_{R} t \\ e^{- {jω}_{R} t} = {cosω}_{R} t - {jsinω}_{R} t \end{matrix}$

Therefore, Eq. (3) becomes:

$i = A_{1} (\cos ω_{R} t + j \sin ω_{R} t) + A_{2} (\cos ω_{R} t - j \sin ω_{R} t) = A_{1} \cos ω_{R} t + {jA}_{1} \sin ω_{R} t + A_{2} \cos ω_{R} t - {jA}_{2} \sin ω_{R} t = (A_{1} + A_{2}) \cos ω_{R} t + ({jA}_{1} - {jA}_{2}) \sin ω_{R} t = B_{1} \cos ω_{R} t + B_{2} \sin ω_{R} t$ $i = A_{1} (\cos ω_{R} t + j \sin ω_{R} t) + A_{2} (\cos ω_{R} t - j \sin ω_{R} t) = A_{1} \cos ω_{R} t + {jA}_{1} \sin ω_{R} t + A_{2} \cos ω_{R} t - {jA}_{2} \sin ω_{R} t = (A_{1} + A_{2}) \cos ω_{R} t + ({jA}_{1} - {jA}_{2}) \sin ω_{R} t = B_{1} \cos ω_{R} t + B_{2} \sin ω_{R} t$

(4)

Using the circuit initial condition the constants Β₁ and Β₂ can be calculated as follows:

At t = 0, current i = 0 and, therefore i = 0 = B₁cos0 + B₂sin0 or B₁ = 0.

Moreover, differentiating the current expression, the following equation results:

$\begin{matrix} \frac{di}{dt} = - B_{1} ω_{R} \sin ω_{R} t + B_{2} ω_{R} \cos ω_{R} t = \frac{v_{L}}{L} \\ where v_{L} = L \frac{di}{dt} = 100 - v_{C} + v_{C} (0) \end{matrix}$ $\begin{matrix} \frac{di}{dt} = - B_{1} ω_{R} \sin ω_{R} t + B_{2} ω_{R} \cos ω_{R} t = \frac{v_{L}}{L} \\ where v_{L} = L \frac{di}{dt} = 100 - v_{C} + v_{C} (0) \end{matrix}$

(5)

At t = 0 the following equation holds:

$v_{L} (0) = 100 + v_{C} (0) = 100 + 75 = 175 V$ $v_{L} (0) = 100 + v_{C} (0) = 100 + 75 = 175 V$

Therefore

${\frac{di}{dt} |}_{t = 0} = - B_{1} ω_{R} \sin 0 + B_{2} ω_{R} \cos 0 = \frac{175}{L} or B_{2} = \frac{175}{ω_{R} L} = \frac{175}{\sqrt{L / C}}$ ${\frac{di}{dt} |}_{t = 0} = - B_{1} ω_{R} \sin 0 + B_{2} ω_{R} \cos 0 = \frac{175}{L} or B_{2} = \frac{175}{ω_{R} L} = \frac{175}{\sqrt{L / C}}$

Substituting the values of B₁ and B₂ into Eq. (4) yields:

$i = \frac{175}{\sqrt{L / C}} \sin ω_{R} t = 350 \sin \frac{10^{6}}{60} t$ $i = \frac{175}{\sqrt{L / C}} \sin ω_{R} t = 350 \sin \frac{10^{6}}{60} t$

(6)

Moreover, using Eq. (6) the inductor voltage is:

$v_{L} = L \frac{di}{dt} = 30 \cdot 10^{- 6} \frac{10^{6}}{60} 350 \cos \frac{10^{6}}{60} t = 175 \cos \frac{10^{6}}{60} t = 175 \cos ω_{R} t$ $v_{L} = L \frac{di}{dt} = 30 \cdot 10^{- 6} \frac{10^{6}}{60} 350 \cos \frac{10^{6}}{60} t = 175 \cos \frac{10^{6}}{60} t = 175 \cos ω_{R} t$

(7)

The period of the current i and voltage v_L is:

$T = \frac{1}{f} = \frac{1}{ω_{R} / 2 π} = \frac{2 π}{ω_{R}} = \frac{2 π}{10^{6} / 60} = \frac{120 π}{10^{6}} = 377 μ s$ $T = \frac{1}{f} = \frac{1}{ω_{R} / 2 π} = \frac{2 π}{ω_{R}} = \frac{2 π}{10^{6} / 60} = \frac{120 π}{10^{6}} = 377 μ s$

As can be seen from the given circuit at the instant the inductor voltage becomes equal to v_L = −75 V, the diode D becomes forward biased entering the conduction state. Also, at that instant the voltage across the capacitor becomes v_C = V − v_L = 100 − (−75) = 175 V making the thyristor reverse biased and, consequently, turns off. Therefore, the equivalent circuit during this mode becomes:

The time instant at which the diode is turned on can be calculated using the following equation:

$v_{L} = 175 \cos ω_{R} t = - 75 V \Rightarrow ω_{R} t = \cos^{- 1} (\frac{- 75}{175}) = 115 °$ $v_{L} = 175 \cos ω_{R} t = - 75 V \Rightarrow ω_{R} t = \cos^{- 1} (\frac{- 75}{175}) = 115 °$

$Therefore, t = \frac{115}{ω_{R}} \frac{π}{180} = \frac{115}{10^{6} / 60} \frac{π}{180} = 120.8 μ s$ $Therefore, t = \frac{115}{ω_{R}} \frac{π}{180} = \frac{115}{10^{6} / 60} \frac{π}{180} = 120.8 μ s$

Thus, the current i at t = 120.8 μs (i.e., at angle 115^ο) will be:

$i = 350 \sin 115 ° = 316 A$ $i = 350 \sin 115 ° = 316 A$

Furthermore, for the circuit path L-D-75 V, the following equation holds:

$v_{L} = L \frac{{di}_{D}}{dt} or \frac{{di}_{D}}{dt} = \frac{- 75}{L} = \frac{- 75}{30 \times 10^{- 6}} = - 2.5 \times 10^{6} \frac{A}{s}$ $v_{L} = L \frac{{di}_{D}}{dt} or \frac{{di}_{D}}{dt} = \frac{- 75}{L} = \frac{- 75}{30 \times 10^{- 6}} = - 2.5 \times 10^{6} \frac{A}{s}$

Therefore the time required for the current i_D to be reduced to zero is given by:

$\frac{316}{t_{x}} = 2.5 \times 10^{6} or t_{x} = 126.5 μ s$ $\frac{316}{t_{x}} = 2.5 \times 10^{6} or t_{x} = 126.5 μ s$

Using the above results, the time variations of i, v_L, v_C, and i_D are presented in Fig. 3.24.

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Table of Contents for 3. Thyristor and Single-Phase Half-Wave Controlled Rectifier

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Abstract

Keywords

3.0. Introduction

3.1. Thyristor–Silicon Controlled Rectifier

3.1.1. Thyristor Dynamic Behavior

3.2. Single-Phase Half-Wave Thyristor Rectifier With Resistive Load

3.3. Single-Phase Half-Wave Thyristor Rectifier With Resistive–Inductive Load

3.4. Single-Phase Half-Wave Thyristor Rectifier With Inductive Load

3.5. Single-Phase Half-Wave Thyristor Rectifier With R–L–E Load

3.6. Gate Drive Circuits for Thyristors

Solution

Solution

Solution

Solution

3.7. Simulation Examples Using the Power Simulation Software Program

Solution

Solution

Table of Contents for
3. Thyristor and Single-Phase Half-Wave Controlled Rectifier