‫الکترونیک صنعتی‬

Controlled rectifier

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Thyristor characteristics
A thyristor is a four-layer semiconductor device of PNPN
structure with three pn junctions. It has three terminals:
anode, cathode, and gate. Figure shows the thyristor symbol
and the sectional view of three pn-junctions.

When the anode voltage is made positive with respect to the cathode,
the junctions J1 and J3 are forward biased. The junction J2 is reverse
biased, and only a small leakage current flows from anode to cathode.

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The thyristor is then said to be in the forward blocking, or off-state, condition
and the leakage current is known as off-state current 𝐼𝐷 .
If the anode-to-cathode voltage 𝑉𝐴𝐾 is increased to a sufficiently large value, the
reverse-biased junction J2 breaks. This is known as avalanche breakdown and
the corresponding voltage is called forward breakdown voltage 𝑉𝐵𝑂 .
The device is then in a conducting state, or on-state. The voltage drop would be
due to the ohmic drop is small, typically, 1 V. In the on-state, the anode current
is limited by an external impedance or a resistance, RL, as shown in Figure.

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The anode current must be more than a value known as latching current
𝐼𝐿 to maintain the required amount of carrier flow across the junction;
otherwise, the device reverts to the blocking condition as the anode-to-
cathode voltage is reduced.

Latching current 𝐼𝐿 is the minimum anode current required to maintain

the thyristor in the on-state immediately after a thyristor has been turned
on and the gate signal has been removed.

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When the cathode voltage is positive with respect to the
anode, the junction J2 is forward biased but junctions J1
and J3 are reverse biased. This is like two series-connected
diodes with reverse voltage across them. The thyristor is in
the reverse blocking state and a reverse leakage current,
known as reverse current 𝐼𝑅 , flows through the device.

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Once a thyristor conducts, it behaves like a conducting diode and there
is no control over the device. However, if the forward anode current is
reduced below a level known as the holding current 𝐼𝐻 , the thyristor is
in the blocking state.

The holding current is on the order of milli-amperes and is less than

the latching current 𝐼𝐿 ,. That is, 𝐼𝐿 > 𝐼𝐻 .

Holding current 𝐼𝐻 is the minimum anode current to maintain the

thyristor in the on-state.

The holding current is less than the latching current.

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A thyristor can be turned on by increasing the forward voltage 𝑉𝐴𝐾
beyond 𝑉𝐵𝑂 , but such a turn-on could be destructive.

In practice, the forward voltage is maintained below 𝑉𝐵𝑂 and the

thyristor is turned on by applying a positive voltage between its gate
and cathode. This is shown in Figure by dashed lines.

Once a thyristor is turned on by a gating signal and its anode current is

greater than the holding current, the device continues to conduct due to
positive feedback, even if the gating signal is removed.

Gate current. If a thyristor is forward biased, the injection of gate

current by applying positive gate voltage between the gate and
cathode terminals turns on the thyristor. As the gate current is
increased, the forward blocking voltage is decreased. 7
Controlled Half- wave rectifier
To control the output of a half-wave rectifier is to use an 𝑆𝐶𝑅1 instead
of a diode. Figure a shows a basic controlled half-wave rectifier with a
resistive load. Two conditions must be met before the SCR can conduct:

1. The SCR must be forward-biased (𝑉𝑆𝐶𝑅 > 0).

2. A current must be applied to the gate of the SCR.

Unlike the diode, the SCR will not begin to conduct as soon as the source becomes
positive. Conduction is delayed until a gate current is applied, which is the basis
for using the SCR as a means of control. Once the SCR is conducting, the gate
current can be removed and the SCR remains on until the current goes to zero.

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Resistive Load
Figure b shows the voltage waveforms for a controlled half-wave rectifier
with a resistive load. A gate signal is applied to the SCR at 𝜔t=𝛼, where 𝛼 is
the delay angle. The average (dc) voltage across the load resistor in Fig. a is:

2 /R, where the rms voltage

The power absorbed by the resistor is 𝑉𝑟𝑚𝑠
across the resistor is computed from:

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Example: Controlled Half-Wave Rectifier with Resistive Load
Design a circuit to produce an average voltage of 40 V across a 100Ω
load resistor from a 120-V rms 60-Hz ac source. Determine the power
absorbed by the resistance and the power factor.

■ Solution

If an uncontrolled half-wave rectifier is used, the average voltage will be 𝑉𝑚 /𝜋 =(120 2)/𝜋=54 V.
Some means of reducing the average resistor voltage to the design specification of 40 V must be
found.

A series resistance or inductance could be added to an uncontrolled rectifier, or a controlled rectifier

could be used. The controlled rectifier of Fig. a has the advantage of not altering the load or
introducing losses, so it is selected for this application.

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Equation is rearranged to determine the required delay angle:

𝑉𝑟𝑚𝑠 is:

Load power is:

The power factor of the circuit is:

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 RL Load
The analysis of this circuit is similar to that of the uncontrolled rectifier. The
current is the sum of the forced and natural responses:

The constant A is determined from the initial condition i(𝛼)= 0:

Substituting for A and simplifying,

The extinction angle 𝛽 is defined as the angle

at which the current returns to zero, as in the 12

case of the uncontrolled rectifier. When 𝜔t=𝛽

which must be solved numerically for 𝛽. The angle 𝛽 − 𝛼 is called
the conduction angle 𝛾 Figure b shows the voltage waveforms.

The average (dc) output voltage is:

The average current is computed from

2 R, where the rms current is computed from:

Power absorbed by the load is 𝐼𝑟𝑚𝑠

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Example: Controlled Half-Wave Rectifier with RL Load

For the circuit of Fig. a, the source is 120 V rms at 60 Hz, R=20 Ω, L=0.04 H,
and the delay angle is 450 . Determine (a) an expression for i(𝜔t), (b) the
average current, (c) the power absorbed by the load, and (d) the power factor.

■ Solution
(a) From the parameters given,

Substituting the preceding quantities into Eq., current is expressed as

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The preceding equation is valid from 𝛼 to 𝛽, where 𝛽 is found numerically by setting
the equation to zero and solving for 𝜔t, with the result 𝛽 =3.79 rad (2170 ).
The conduction angle is 𝛾 = 𝛽 − 𝛼=3.79-0.785=3.01 rad =1720 .

(b) Average current is:

2 R, where:
(c) The power absorbed by the load is computed from 𝐼𝑟𝑚𝑠

(d) The power factor is:

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RL-Source Load
The analysis of this circuit is very similar to that of the uncontrolled half-
wave rectifier. The major difference is that for the uncontrolled rectifier,
conduction begins as soon as the source voltage reaches the level of the dc
voltage. For the controlled rectifier, conduction begins when a gate signal
is applied to the SCR, provided the SCR is forward-biased.
Thus, the gate signal may be applied at any time that the ac source is larger
than the dc source:

Current is expressed as in Eq., with 𝛼

specified within the allowable range:

where A is determined from: 16

 Example: Controlled Rectifier with RL-Source Load
The controlled half-wave rectifier has an ac input of 120 V rms at 60
Hz, R=2Ω L=20 mH, and 𝑉𝑑𝑐 =100 V. The delay angle 𝛼 is 450 .

Determine (a) an expression for the current, (b) the power absorbed by
the resistor, and (c) the power absorbed by the dc source in the load.

■ Solution:
From the parameters given,

(a) The minimum delay angle is:

which indicates that 45 is allowable.

Current is:

where the extinction angle 𝛽 is found numerically to be 3.37 rad from the equation i(𝛽)=0.
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 2
(b) Power absorbed by the resistor is 𝐼𝑟𝑚𝑠 R, where 𝐼𝑟𝑚𝑠 is:

(c) Power absorbed by the dc source is𝐼0 𝑉𝑑𝑐 , where Io is:

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 Controlled full- wave rectifiers
Controlled full-wave rectifiers are shown in Figs a, b. For the bridge
rectifier, SCRs S1 and S2 will become forward-biased when the source
becomes positive but will not conduct until gate signals are applied.
Similarly, S3 and S4 will become forward-biased when the source
becomes negative but will not conduct until they receive gate signals.

For the center-tapped transformer rectifier, S1 is forward-biased when

vs is positive, and S2 is forward-biased when vs is negative, but each
will not conduct until it receives a gate signal.

The delay angle is the angle interval between the forward biasing of the SCR and the gate signal application.
If the delay angle is zero, the rectifiers behave exactly as uncontrolled rectifiers with diodes. The discussion
that follows generally applies to both bridge and center-tapped rectifiers.
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Resistive Load
The output voltage waveform for a controlled full-wave rectifier with a
resistive load is shown in Fig. c. The average component of this waveform is
determined from:

Average output current is then:

The power delivered to the load is a function of the input voltage, the
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delay angle, and the load components; P=𝐼𝑟𝑚𝑠 R is used to determine
the power in a resistive load, where:

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The rms current in the source is the same as the rms current in the load.
Example: Controlled Full-Wave Rectifier with Resistive Load
The full-wave controlled bridge rectifier of Fig. a has an ac input of 120 V rms at
60 Hz and a 20 Ω load resistor. The delay angle is 400 Determine the average
current in the load, the power absorbed by the load, and the source volt-amperes.
■ Solution
The average output voltage is:

Average load current is:

Power absorbed by the load is

Determined from the rms current:

The rms current in the source is also 5.80 A,

and the apparent power of the source is:

Power factor is: 21

RL Load, Discontinuous Current
Load current for a controlled full-wave rectifier with an RL load
(Fig. a) can be either continuous or discontinuous, and a separate
analysis is required for each.

Starting the analysis at 𝜔t=0 with zero load current, SCRs S1 and
S2 in the bridge rectifier will be forward-biased and S3 and S4 will
be reverse-biased as the source voltage becomes positive.

Gate signals are applied to S1 and S2 at 𝜔t=𝛼, turning S1 and S2

on. With S1 and S2 on, the load voltage is equal to the source
voltage. For this condition, the circuit is identical to that of the
controlled half-wave rectifier.

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The above current function becomes zero at 𝜔t=𝛽 If 𝛽 < 𝜋 + 𝛼, the
current remains at zero until 𝜔t= 𝜋 + 𝛼 when gate signals are applied to
S3 and S4 which are then forward-biased and begin to conduct. This mode
of operation is called discontinuous current, which is illustrated in Fig. b.

Analysis of the controlled full-wave rectifier operating in the discontinuous

current mode is identical to that of the controlled half-wave rectifier except
that the period for the output current is 𝜋 rather than 2𝜋 rad.

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 Example: Controlled Full-Wave Rectifier, Discontinuous Current
A controlled full-wave bridge rectifier has a source of 120 V
rms at 60 Hz, R=10 Ω, L=20 mH, and 𝛼 = 60𝑜 Determine
(a) an expression for load current, (b) the average load
current, and (c) the power absorbed by the load.

■ Solution
From the parameters given: 𝑉𝑚 =120 2 =169.7 V

Solving io(𝛽)=0 numerically for 𝛽, 𝛽 =3.78 rad (216𝑜 ). Since 𝜋 + 𝛼 =4.19> 𝛽, the
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current is discontinuous, and the above expression for current is valid.
(b) Average load current is determined from the numerical integration of :

2 R, where:
(c) Power absorbed by the load occurs in the resistor and is computed from 𝐼𝑟𝑚𝑠

1 𝛽 2
𝐼𝑟𝑚𝑠 = 𝑖 (𝜔𝑡)𝑑 𝜔𝑡 = 8.35 𝐴
𝜋 𝛼 𝑜

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RL Load, Continuous Current
If the load current is still positive at 𝜔t=𝜋 + 𝛼 when gate signals are
applied to S3 and S4 in the above analysis, S3 and S4 are turned on
and S1 and S2 are forced off. Since the initial condition for current in
the second half-cycle is not zero, the current function does not repeat.
Equation (P. 23) is not valid in the steady state for continuous current.
For an RL load with continuous current, the steady-state current and
voltage waveforms are generally as shown in Fig. c.

The boundary between continuous and discontinuous current occurs

when 𝛽 for Eq. (P. 23) is 𝜋 + 𝛼 (𝛽=𝜋 + 𝛼). The current at 𝜔t=𝜋 + 𝛼
must be greater than zero for continuous-current operation.

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Using: Solving for 𝛼:

Using:

for continuous current

Both Eqs. can be used to check whether the load current is continuous or discontinuous.
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A method for determining the output voltage and current for the continuous current case is to use the
Fourier series. The Fourier series for the voltage waveform for continuous-current case shown in Fig. c
is expressed in general form as:

The dc (average) value is:

The amplitudes of the ac terms are calculated from:

Where:

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Figure shows the relationship between normalized harmonic content of the
output voltage and delay angle.
The Fourier series for current is determined by superposition as was done
for the uncontrolled rectifier earlier. The rms current is determined by
combining the rms currents at each frequency.

As the harmonic number increases, the impedance for the inductance

increases. Therefore, it may be necessary to solve for only a few terms
of the series to be able to calculate the rms current. If the inductor is
large, the ac terms will become small, and the current is essentially dc. 29
Example: Controlled Full-Wave Rectifier with RL Load, Continuous Current
A controlled full-wave bridge rectifier of Fig. a has a source of
120 V rms at 60 Hz, an RL load where R=10 Ω and L=100 mH.
The delay angle 𝛼 = 600 . (a) Verify that the load current is
continuous. (b) Determine the dc (average) component of the
current. (c) Determine the power absorbed by the load.

■ Solution
(a) Equation is used to verify that the current is continuous:

continuous current

(b) The voltage across the load is expressed in terms of the Fourier series of Eq. The dc term is computed:
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(c) The amplitudes of the ac terms are computed
and are summarized in the following table where,

The rms current is:

2 R.
Power is computed from 𝐼𝑟𝑚𝑠

Note that the rms current could be approximated accurately from the dc term and one ac term
(n=2). Higher-frequency terms are very small and contribute little to the power in the load.
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Single-phase full Converters with a highly inductive load
The load current is continuous and ripple free. T1 and T2
are turned off due to line or natural commutation and the
load current is transferred from T1 and T2 to T3 and T4.
Figure b shows the regions of converter operation.

During the period from α to π, the input voltage vs and input current is are positive,
and the power flows from the supply to the load. The converter is said to be
operated in rectification mode.

During the period from π to π + α, the input voltage vs is negative and the input
current is is positive, and reverse power flows from the load to the supply. The
converter is said to be operated in inversion mode.

Depending on the value of α, the average output voltage could be either positive or
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negative and it provides two-quadrant operation.
The average output voltage can be found from

and 𝑉𝑑𝑐 can be varied from 2𝑉𝑚 /π to -2𝑉𝑚 /π by varying α from 0 to π. The maximum
average output voltage is 𝑉𝑑𝑚 = 2𝑉𝑚 /π and the normalized average output voltage is

The rms value of the output voltage is given by

With a purely resistive load, thyristors T1 and T2 can conduct from

α to π, and thyristors T3 and T4 can conduct from α + π to 2π.
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 Example: Input power Factor of a Single-phase Full Converter
The full converter in Figure is connected to a 120-V, 60-Hz supply. The load current Ia is
continuous and its ripple content is negligible. The turns ratio of the transformer is unity.
(a) Express the input current in a Fourier series; determine the HF of the input current, DF,
and input PF. (b) If the delay angle is α = π/3, calculate 𝑉𝑑𝑐 , 𝑉𝑛 , 𝑉𝑟𝑚𝑠 , HF, DF, and PF.

Solution
a. The instantaneous input current can be expressed in a Fourier series as:

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 Because 𝑎0 = 0, the input current can be written as:

𝜑𝑛 ,the displacement angle of the nth harmonic current is:

The rms value of the nth harmonic input current is:

and the rms value of the fundamental current is: 35

The rms value of the input current can be calculated from:

Is can also be determined directly from:

HF is found as:

DF and PF are:

Note: The fundamental component of input current is always 90.03% of Ia and the HF
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remains constant at 48.34%.

 Single-phase dual Converters
We have seen that single-phase full converters with
inductive loads allow only a two-quadrant operation. If
two of these full converters are connected back to back,
as shown in Figure a, both the output voltage and the
load current flow can be reversed. The system provides
a four-quadrant operation and is called a dual converter.

Dual converters are normally used in high-power variable-speed drives. If α1 and α2 are the
delay angles of converters 1 and 2, respectively, the corresponding average output voltages are
Vdc1 and Vdc2. The delay angles are controlled such that one converter operates as a rectifier
and the other converter operates as an inverter; but both converters produce the same average
output voltage. Figures b–f show the output waveforms for two converters, where the two
average output voltages are the same. Figure b shows the v-i characteristics of a dual converter.
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average output voltages are

Because one converter is rectifying and the other one is inverting,

Therefore,

Because the instantaneous output voltages of the two converters are out of phase,
there can be an instantaneous voltage difference and this can result in circulating
current between the two converters. This circulating current cannot flow through the
load and is normally limited by a circulating current reactor Lr, as shown in Figure a.

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The instantaneous voltage difference starting from ωt = π - α1.
Because the two average output voltages during the interval ωt =
π + α1 to 2π - α1 are equal and oppositive, their contributions to
the instantaneous circulating current ir is zero.

For α1 = 0, only the converter 1 operates; for α1 = π, only the

converter 2 operates. For 0 … α1 6 π/2, the converter 1
supplies a positive load current +io and thus the circulating
current can only be positive. For π/2 6 α1 … π, the converter
2 supplies a negative load current -io and thus only a negative
circulating current can flow. At α1 = π/2, the converter 1
supplies positive circulating during the first half-cycle, and
the converter 2 supplies negative circulating during the
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second half-cycle.
its magnitude becomes minimum when ωt = nπ, n = 0, 2, 4, c, and maximum
when ωt = nπ, n = 1, 3, 5, c. If the peak load current is Ip, one of the converters
that controls the power flow may carry a peak current of 1Ip + 4Vm/ωLr2.
The dual converters can be operated with or without a circulating current. In case
of operation without circulating current, only one converter operates at a time
and carries the load current, and the other converter is completely blocked by
inhibiting gate pulses.

Gating sequence. The gating sequence is as follows:

1. Gate the positive converter with a delay angle of α1 = α. 40

2. Gate the negative converter with a delay angle of α2 = π - α through gateisolating circuits.
Single-phase Series Converters

For high-voltage applications, two or more converters can be connected

in series to share the voltage and also to improve the PF. In figure a the
turns ratio between the primary and secondary is Np/Ns = 2. Due to the
fact that there are no freewheeling diodes, one of the converters can not
be bypassed and both converters must operate at the same time.
In rectification mode, one converter is fully advanced (α1= 0) and the
delay angle of the other converter, α2, is varied from 0 to π to control the
dc output voltage.

In the inversion mode, one converter is fully retarded, α2 = π, and the

delay angle of the other converter, α1, is varied from 0 to π to control the
average output voltage.
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In rectification mode, one converter is fully
advanced (α1 = 0) and the delay angle of the
other converter, α2, is varied from 0 to π to
control the dc output voltage.
In the inversion mode, one converter is fully
retarded, α2 = π, and the delay angle of the
other converter, α1, is varied from 0 to π to
control the average output voltage.
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The average output voltages of two full converters are:

The resultant average output voltage is:

The maximum average output voltage for α1 = α2 = 0 is Vdm = 4Vm/π.

 In the rectification mode: α1 = 0 and 0<α2< π; then:

and the normalized dc output voltage is:

 In the inversion mode: 0<α1<π and α2 = π; then:

and the normalized average output voltage is:

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Example: Finding the input power factor of a series single-phase full converter
The load current (with an average value of Ia) of series-full converters in is
continuous and the ripple content is negligible. The turns ratio of the transformer is
Np/Ns = 2. The converters operate in rectification mode such that α1 = 0 and α2
varies from 0 to π. (a) Express the input supply current in Fourier series, and
determine the input current HF, DF, and input PF. (b) If the delay angle is α2 = π/2
and the peak input voltage is Vm = 162 V, calculate Vdc, Vn, Vrms, HF, DF, and PF.

Solution
a. The waveform for input current is shown in Figure 10.11b and the
instantaneous input supply current can be expressed in a Fourier series as:

where ϕn = -nα2/2

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The rms value of the nth harmonic input current is:

The rms value of fundamental current is:

The rms input current is found as:

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 b. α1 = 0 and α2 = π/2.

Note: The performance of series-full converters is

similar to that of single-phase semiconverters.
Key Point:
• Semiconverters and full converters can be connected
in series to share the voltage and also to improve the
input PF.

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