Lecture Note 7 - · PDF file · 2016-12-23Lecture Note 7 Uncontrolled and Controlled Rectifiers ... The main difference between ... The type of commutation used in controlled rectifier

Lecture Note

7

Uncontrolled and Controlled Rectifiers

Prepared by Dr. Oday A Ahmed Website: https://odayahmeduot.wordpress.com Email: [email protected]

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Lecture Note 7: Uncontrolled and Controlled Rectifier

Instructure: Dr. Oday A Ahmed

1

single-phase diode and SCR rectifiers

The diode rectifiers are referred to as uncontrolled rectifiers, which make use of

power semiconductor diodes to carry the load current. The diode rectifiers give a

fixed dc output voltage (fixed average output voltage) and each diode rectifying

element conducts for one half cycle duration (T/2 seconds), that is the diode

conduction angle = 1800 or π radians.

We cannot control (we cannot vary) the dc output voltage or the average

dc load current in a diode rectifier circuit.

Controlled SCR rectifiers are line-commutated ac to dc power converters that are

used to convert a fixed voltage, fixed frequency ac power supply into variable dc

output voltage.

Important NOTE:

The thyristor current and the load current begin to flow once the thyristors are

triggered (turned ON) at angle called firing angle α. The main difference between

the diode rectifier and SCR rectifier is the required trigger for firing the SCR.

Hence, for ωt= α=00 the behaviour of SCR rectifier is look like the diode

rectifier. Thus, the controlled SCR rectifier will only be explained here.

The thyristor remains reverse biased during the negative half cycle of input

supply. The type of commutation used in controlled rectifier circuits is referred

to AC line commutation or Natural commutation or AC phase commutation.

SCR Rectifier can be controlled by α from 00 (operate like a diode rectifier)

to 1800 (provided zero DC voltage for R-load). Where the SCR conduction

angle δ =π –α

Single quadrant ac-dc converters where the output voltage is only positive

and cannot be made negative for a given polarity of output current.

Single quadrant converters can also be designed to provide only negative dc

output voltage.

two quadrant converters so that the output voltage can be made either

positive or negative for a given polarity of output load current can be

achieved by using fully controlled bridge converter circuit.



2

Note: the operation of the rectifier has different effects on the power quality

when it is work. Also its operation is effected by the loads type. These effects can

be summarized below:

A. The effect of RL load with low inductance on the output DC voltage and

on the input current.

B. The effect of RL load with heavy inductive on the input current power

factor and current waveform shape. In addition, the effect on the average

power that transferred from the source to load, as stated in Lec.7.

C. The effect of output C filter on the input current shape and Peak value and

RMS current.

D. The effect of reverse recovery current on the output DC voltage of the

rectifier.

E. The effect of leakage inductance of the input transformer on the output

voltage of the rectifier.

F. The effect of RLE load on the output voltage and operation of the

converter.

I. Single Phase Half Wave Phase Controlled Rectifier



3

A. R-Load

o V1= Vmsin α (hence this voltage depend on

α, if α=0 V1= 0V)

o vT1 = (0.7+Vf) for real SCR for ωt = α →π

o peak reverse voltage PIV = Vm

o iT1 = reverese leakge current when SCR OFF for real SCR

o vp represents the primary supply voltage = Vmsinωt.

o vS represents the secondary supply voltage = Vmsinωt for turns ratio n=1.

The maximum value (peak value) of that flow via the RL, T1, and secondary

winding is calculated as:

SCR naturally turns off when the current flowing through it falls to zero at

ωt =π.

State T1 vo vT1 iT1 iO iS

iO(max)

For ωt =

0 → α OFF 0 V1 0 0 0 0

For ωt =

α →π ON vS 0

𝑣𝑜𝑅𝐿

𝑣𝑜𝑅𝐿

𝑣𝑜𝑅𝐿

𝑉𝑚𝑅𝐿

For ωt =

π →2π OFF 0 Vm 0 0 0 0



4

Since conduction angle δ =π –α, hence maximum conduction angle

when α=00

RMS value of input ac supply voltage across transformer

secondary.

Drive an Expression for output DC Voltage:

For single-phase diode HWR, α=00 →

The control characteristic of SCR HWR is

shown aside:

DC Load current is equal to:

Drive an Expression for output RMS Voltage:

RMS Load current is equal to:



5

For diode Rectifier (α=00)

Input power factor pf

Since the shape input current is still sinusoidal, DF=1 and DPF=pf (Can you

guess the value of fundamental components of input current, FF, RF, and

CF?).

B. RL-Load

In practice, most of the loads are of RL type.

Note: Load current increases slowly due the

inductance since the inductance in the load forced the load current to lag load

voltage.

The load current flowing through T1would not fall to zero at ωt=π, when vs

starts to become negative (due to magnetic field in the inductor).

T1 will continue to conduct io until all the inductive energy stored in L is

completely utilized and current via T1 falls to zero at ωt=β.

β is referred to as the Extinction angle or advance angle.

β is measured from the point of the beginning of the positive half cycle of input

supply to the point where the load current falls to zero.

conduction angle of SCR δ = β – α

δ depends on the firing angle and the load impedance angle ϕ

T1 ON For ωt = α → β

ωt = β→2π

ON OFF

T1 Off



6

At ωt =β, io = 0, VL=0 and VR =0 and vs appears as reverse bias across SCR.

then SCR turned off

peak reverse voltage PIV = Vm

negative in output voltage reduces Vo(dc) when compared to a purely R-Load.

Derive an expression for inductive load current, during α <ωt< β

During ωt = α → β, the load current is determined as following:

io=is+it where, is steady state load current, it: transient load current

Thus,

A1 is the initial transient current which is changing with α which found as

explained in Lecture09.

Based on the above equation the steady-state component of load current in related

to source voltage and transient components at different firing angles are plotted

below:

is i

t

Source voltage Vmsinωt

The current via L is produce from

sinusoidal source but the current via it is

rising slowly exponentially. Hence, the

original of current is sinusoidal entered

with it exponential current component.



7

λ

Max I at di/dt=0, which occurs at vs=vR

βππ/2

vR

vL

Vm

Voltage

current

io

it

is

λ βφ

Ito

Iso

Imax

it(β)

is(β)

io(β)=0

Voltage decreasing, L realising energyVoltage increasing, L stored energy

Steady state load current

φ

If α<Φ : δ>π, as for α1 and α2

If α=Φ : δ=π, as for α3

If α>Φ : δ<π, as for α4

If α= π : δ=0

Ito measure

from the

beginning

of SCR

triggered

not at zero-

crossing

voltage.

(i.e. at ωt =

α

The Figure

shows the load

impedance

phase angle

related to io at

α=0:

Φ=900



8

The load current is the same one for the current via T1 and source (with no

FWD) is:

TO CALCULATE EXTINCTION ANGLE β

As shown in Lecture09:

Iterative solutions to

find β

The exact value for β

exist only for the purely

resistive load, Φ= 0,

and the purely

inductive load, Φ = ½π.



9

TO CALCULATE MAX LOAD CURRENT AT ωt=λ

To find the maximum value of the load current which should be occurred at ωt=λ, 𝑑𝑖

𝑑𝜔𝑡= 0 =

𝑑

𝑑𝜔𝑡(𝑉𝑚𝑍[sin(𝜔𝑡 − 𝜙) − sin(𝛼 − 𝜙)𝑒

−𝑅𝜔𝐿(𝜔𝑡−𝛼)])

sin(A-B)=sin A cos B - cos A sin B, thus,

0 =𝑉𝑚

𝑍[cos(𝜔𝑡 − 𝜙) +

sin(𝛼−𝜙)𝐿

𝑅

𝑒−𝑅

𝜔𝐿(𝜔𝑡−𝛼)] , at ωt=λ

𝐿

𝑅cos(𝜆 − 𝜙) = −sin(𝛼 − 𝜙) 𝑒

−𝑅𝜔𝐿(𝜆−𝛼)

Hence, by substituting ωt=λ, to io, the maximum current via L can be obtained.

𝐼𝑚𝑎𝑥 =𝑉𝑚𝑍[sin(𝜆 − 𝜙) − sin(𝛼 − 𝜙)𝑒

−𝑅𝜔𝐿(𝜆−𝛼)]

TO DERIVE AN EXPRESSION FOR AVERAGE (DC) LOAD VOLTAGE

Average DC Load Current

The RMS load voltage is (if ωt=θ)

The RMS load current:

𝐼𝑜(𝑟𝑚𝑠) = √1

2𝜋∫ (

𝑉𝑚𝑍[sin(𝜔𝑡 − 𝜙) − sin(𝛼 − 𝜙)𝑒

−𝑅𝜔𝐿(𝜔𝑡−𝛼)])

2

𝑑𝜔𝑡𝛽

𝛼

Iteration needed

for solving λ



10

Input power factor pf can be obtained as:

C. RL-Load with Freewheeling Diode (FWD)

With a RL load it was observed that the average

output voltage reduces. This disadvantage can

be overcome by connecting a diode across the

load as shown in figure.

The modes operation of the rectifier is shown

below:

H.W: What are the reasons results in SCR and FWD turned

off during the periods ωt = 0 → α and ωt = β→ 2π

At ωt=π, vS falls to zero and as vS becomes

negative, FWD is forward biased and SCR is

turned OFF.

Stored energy in L maintains the load

current flow through R, L, and the FWD.

During the period π to ωt=β, the load

current flows through FWD (freewheeling load current) and decreases

exponentially towards zero at ωt=β.

The load is shorted by the conducting FWD and the load voltage is almost zero,

if the FWD forward voltage drop is neglected.

The mean load voltage (hence mean output current) for all conduction cases, with

a passive L-R load, is

State SCR FWD vo vT1 iT1 IFWD

For ωt =

0 → α OFF OFF 0 V1 0 0

For ωt =

α →π ON OFF vS 0 iO(t) 0

For ωt =

π → β OFF ON 0 V2 0 iO(t)

For ωt =

β→ 2π OFF OFF 0 Vm 0 0

V2



11

RMS output voltage for both continuous and discontinuous load current is:

NOTE: The following points are to be noted.

If L value is not very large, the energy stored in its able to maintain the

load current only up to ωt=β, where π <β< 2π, where before the next gate

pulse, the load current tends to become discontinuous.

During the conduction period (δ) α to π, io is carried by the SCR and during

the freewheeling period π to β, io is carried by the freewheeling diode.

The value of β depends on R and L and the forward resistance of the FWD.

If the value of L is very large, io does not decrease to zero during the

freewheeling time interval and the ripple in io waveform decreases.

To find load current, the same equation used in Part B can be used. However, current via

SCR is different where this current occurs between α<ωt≤π



12

D. RLE-Load

The load circuit consists of a dc source ‘E’ in addition

to resistance and inductance.

SCR will be forward biased for anode supply voltage

greater than the load dc voltage (i.e. vS>E).

If vS<E, SCR will be revered biased.

The value of ωt at which the supply voltage increases and becomes equal to

the load circuit dc voltage can be calculated as:

o For trigger angle α < γ, the SCR conducts only from ωt= γ to β.

o For trigger angle α > γ, the thyristor conducts from ωt= α to β.

The load current can be found as:



13

The average or dc load voltage

Note that: Vo(dc)= E+IoR

RMS Output Voltage can be calculated by using the expression

For L=0H, β= π- γ, the average load current can be given as:

𝐼𝑜 =1

2𝜋𝑅𝑉𝑚(𝑐𝑜𝑠𝛼 + cos𝛾) − 𝐸(𝜋 − 𝛾 − 𝛼)

The RMS load current for L=0H can be given as:

𝐼𝑜𝑟 = √1

2𝜋∫ (

𝑉𝑚𝑠𝑖𝑛𝜔𝑡 − 𝐸

𝑅)2

𝑑𝜔𝑡𝛽

𝛼

𝐼𝑜𝑟 =1

√2𝜋𝑅(𝑉𝑠

2 + 𝐸2 )(𝛽 − 𝛼) −𝑉𝑠2

2(𝑠𝑖𝑛2𝛽 − 2𝑠𝑖𝑛2𝛼)− 2𝑉𝑚𝐸(𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛽)

1/2

Power deliver to the load is equal to:

𝑃𝑑𝑐 = 𝐼𝑜𝑟2 𝑅 + 𝐼𝑜𝐸

And the input power factor is:



14

𝑝𝑓 =𝐼𝑜𝑟2 𝑅 + 𝐼𝑜𝐸

𝑉𝑠𝐼𝑜𝑟

To find Ior, Io for half-wave diode rectifier, α = γ

The PIV= Vm+E

II. SINGLE PHASE FULL WAVE CONTROLLED RECTIFIERS

Single-phase full wave rectifier can be divided into the following configurations:

Centre tapped rectifier.

Semi-Converter rectifier.

Full-bridge rectifier.

Full-wave uncontrolled converter

Bridge Rectifier split rail dc supplies Bridge Rectifier voltage doubler

Full-wave half-uncontrolled converter

Full-wave mid-point controlled converter

Full-wave Full -controlled converter

① ②

③ ④

⑤ ⑥



15

Single phase half wave controlled rectifiers are

rarely used in practice as they give low dc output and low dc output power. They

are only of theoretical interest. Therefore, various full-wave rectifier

configurations are studied here.

NOTE: All previous configurations give a full-wave two-pulse output

voltage.

A) Configurations ① and ③

Both circuits appear identical as far as the load and supply are concerned.

Two fewer diodes can be employed in Config. ①, but this circuit requires

a centre-tapped secondary transformer where each secondary has only a

50% copper utilisation factor.

Each of the secondary windings in ① must have the same RMS voltage

rating as the single secondary winding of the transformer in ③.

Rectifying diodes in ① experience twice the reverse voltage, (2 √2Vs), as

that experienced by each of the four diodes in the circuit of ③, (√ 2 Vs).

The operation modes of two configurations above are shown below for

continuous load current operation:

Full-wave half-controlled

semi-converters

⑦ ⑧

⑨



16

The average and RMS output voltage can be derived as shown below:

𝑉𝑜 =1

𝜋∫ 𝑉𝑚

𝜋

0

𝑠𝑖𝑛𝜔𝑡𝑑(𝜔𝑡) =2𝑉𝑚𝜋

𝑉𝑜𝑟 = √1

𝜋∫ 𝑉𝑚

2𝜋

0

𝑠𝑖𝑛𝜔𝑡2𝑑(𝜔𝑡) =𝑉𝑚

√2= 𝑉𝑠

Note: Vm is the maximum of secondary winding voltage

With an inductive passive load, (no back EMF) continuous load current

flows, which is given by:

Appropriate integration of the load current squared, gives the RMS current:

Vm for full-

bridge, 2Vm for

centre-tap

Vs



17

This current is the same one for all full-wave uncontrolled rectifier

configurations.

H.W.: Derive an expression to find the average and RMS value of diode

currents.

Hint: the current via each diode whether centre-tap or full-bridge

configuration is for the following period 0≤ ωt≤π or π ≤ ωt≤2π.

B) Configurations ② and ④

The main difference between the configurations ① and ③ and the rectifiers

shownin②and④is the possibility of control the output voltage using SCR.

The operation modes of ②and④aresimilarto① and ③ except of when

thevoltageappearacrosstheloadisdependingonfiringangleα:

There are two types of operations possible.

ƒ Discontinuous load current operation, which occurs for a purely resistive

load or an RL load with low inductance value.

ƒ Continuous load current operation which occurs for an RL type of load

with large load inductance.

Discontinuous Load Current Operation (for low value of load inductance)

The main waveforms with equivalent state are shown below



18

The average and RMS output voltage can be derived as shown below:

The output voltage waveforms and DC output

voltage with FWD are shown below:

H.W: Derive an expression for RMS output

voltage for full-bridge controlled rectifier with low value of load inductance.

① ② ③ ④ ⑤

②

①

③

④

⑤

Same output current equation

with RL load can be obtained as

shown for half-wave rectifier

except her the equation should be

multiplied by 2.

ـــ

ـــ ـــ

ـــ



19

Continuous Load Current Operation (for high value of load inductance)

The main waveforms are shown below

Generally the load current is continuous

for large load inductance and for low trigger

angles.

The load current is discontinuous for low

values of load inductance and for large

values of trigger angles.

The DC output voltage can be obtained as:

The above equation can be plotted to

obtain the control characteristic of a single

phase full wave controlled rectifier with RL

load assuming continuous load current

operation.

For trigger angle α in the range of 0 to ≤900

Vdc is positive and the circuit operates as a

controlled rectifier.

For α> 900, Vdc becomes negative but io

flows in the same positive direction. Hence

the output power becomes negative. This means that the power flows from the

load circuit to the input ac source. This is referred to as line commutated inverter

operation. During the inverter mode load energy can be fed back from the load

circuit to the input ac source.

The RMS thyristor current can be calculated as

The average thyristor current can be calculated as



20

C) Configurations ⑦,⑧, and⑨

These rectifier topologies are semi-converters since they have diode devices in

addition to the SCRs. The key waveforms of these configurations are shown

below; the circuits ⑦ to ⑨ don’t require additional FWD since they

inherently can remove the effect of load inductance:

The DC output voltage for circuits ⑦,⑧ can be obtained as shown in circuits

②,④withFWDwhichisequalto:

The DC output voltage for circuit ⑨isdifferentandcanbeobtainedasshown

below:

For α=00, this circuit operate like full-bridge uncontrolled rectifier.

III. Single-phase full-wave bridge rectifier circuit with a C-filter and R-load

The capacitor reduces the ripple voltage, so large voltage-polarised capacitance

is used to produce an almost constant dc output voltage.



21

The average voltage

The capacitor charging current

period θc around the ac supply

extremes is short, giving a high

peak to rms ratio of diode and

supply current.

α: The start of diode conduction,

β: The diode current extinction angle

Θc :diode conduction period= β-α

Vs rises

Supply provides load and simultaneously

charges the capacitor

charging current period

capacitor voltage > Vs

At ωt= β, D1-D2/D3-D4 OFF

capacitor supplies the load current

voltage decreases with an R-C

time constant until ωt = π +α



22

The average voltage

Diodes conducting Diodes non-conducting

α ≤ ωt ≤ β β ≤ ωt ≤ π+α

vo(ωt) vo(ωt)

Vm sinωt

See Next Page



23

The output voltage for (β ≤ ωt ≤ π+α) can be found as:

𝑉𝑠 = 𝐼𝑅 +1

∫ 𝑖𝑑(𝜔𝑡)

+ 𝑉𝑐(𝜔𝑡 = 𝛽)During all diodes are turned off C supply the required current.

𝑉𝑠 = 0

0 = 𝐼𝑅 +1

∫𝑖𝑑(𝜔𝑡)

+ 𝑉𝛽 𝑉𝛽 = 𝑉𝑚𝑠𝑖𝑛𝛽

β ≤ ωt ≤ π+α

𝑉𝛽

Taking LT:

0 = 𝐼(𝑠)𝑅 +𝐼(𝑠)

𝑠 +𝑉𝛽𝑠

By simplifying the above equation:

=1

𝑅 Taking LTI:

𝑖 𝑡 =𝑉𝛽𝑅𝑒−(𝜔𝑡−𝛽)𝜔𝑅 𝑉𝑜 𝜔𝑡 = 𝑉𝛽𝑒

−(𝜔𝑡−𝛽)𝜔𝑅

𝑖D(ω𝑡)= 𝑖c(ω𝑡)+ 𝑖R(ω𝑡) at ωt= β, 𝑖o=0

𝑖𝑅(ω𝑡) =𝑉𝑜(ω𝑡)𝑅

=𝑉𝛽

𝑅=𝑉𝑚𝑠𝑖𝑛𝛽

𝑅 𝑖 (ω𝑡) = ω 𝑑𝑉𝑜(ω𝑡)

𝑑ω𝑡=𝑉𝑚𝑐𝑜𝑠𝛽

0= 𝑉𝑚𝑐𝑜𝑠𝛽

+ 𝑉𝑚𝑠 𝛽

𝑅𝑉𝑜 ω𝑡 = 𝑉𝛽𝑒

−(𝜔𝑡−𝛽)𝑡𝑎 𝛽 for (β ≤ ωt ≤ π+α)

The average voltage

𝑉𝛽 = 𝑉𝑚𝑠𝑖𝑛𝛽 𝑒 𝑒

𝑡 𝑛𝛽 = 𝜔𝑅

𝑉𝑜 =1

2𝜋∫ sinωt 𝑑ωt𝛽

𝛼

+∫ 𝑉𝛽𝑒−(𝜔𝑡−𝛽)𝑡𝑎 𝛽 𝑑ωt

𝜋 𝛼

𝛽

𝑉𝑜 = 𝑉𝑚1 − 𝑐𝑜𝑠 𝑐−𝑐𝑜𝑠𝛽

𝐼𝑜 =𝑉𝑜𝑅

𝑐= 𝛽- α



24

The maximum output voltage occurs at

Vm

The minimum output voltage occurs at 𝑉𝑚𝑠𝑖𝑛𝛼

output peak-to-peak ripple voltage

Vm ــــ 𝑉𝑚𝑠𝑖𝑛𝛼Δvo= = 𝑉𝑚(1-𝑠𝑖𝑛𝛼)

The Output voltage Ripple

The Output voltage Ripple

At ωt = π+α

vo(ωt)=vo(π+α)=Vm sin(π+α)

𝑉𝑜 ω𝑡 = 𝑉𝛽𝑒−(𝜔𝑡−𝛽)𝑡𝑎 𝛽 𝑉𝑜 π+α = 𝑉𝛽𝑒

−(π+α−𝛽)𝑡𝑎 𝛽

Sine voltage start and exp voltage end both at the same value.

Vm sin(π+α) = 𝑉𝛽𝑒−(π+α− )

The two equation sides gives maximum value when α and =0.5π

𝑉𝑜 π+α =Vm 𝑉𝑜 π+α = 𝑉𝑚𝑒−(π)𝑡𝑎 𝛽𝑉𝛽=Vmsin 𝛽= Vmsin 0.5π=Vm

Δvo=Vm-𝑉𝑚𝑒−(π) 𝑒

−(π)𝑡𝑎 𝛽 1 −

𝜋

𝜔𝑅 Δvo=

𝑉𝑚 𝑅



25

The average capacitor current is zero

𝑖D(ω𝑡) = 𝑖c(ω𝑡)+ 𝑖R(ω𝑡), 𝐼o= 𝐼D, the average diode current is the same as the average load current because

Since the diode is on for a short time in each cycle, the peak diode current is generally much larger than the average diode current

𝑖D𝑃𝑒 = 𝑖c𝑃𝑒 + 𝑖R𝑃𝑒 ,

𝑖c𝑃𝑒

𝑖R𝑃𝑒

Peak current occurs when the diode turns on

𝑖 ω𝑡 = ω 𝑑𝑉𝑜 ω𝑡

𝑑ω𝑡 𝑖c𝑃𝑒 =

𝑉𝑚𝑐𝑜𝑠𝛼

𝑖𝑅(ω𝑡) =𝑉𝑚𝑠𝑖𝑛ω𝑡

𝑅 𝑖 𝑃𝑒 =

𝑉𝑚𝑠𝑖𝑛𝛼𝑅

𝑖D𝑃𝑒 = 𝑉𝑚𝑐𝑜𝑠𝛼

+𝑠 𝛼

𝑅

The source current equal to diode current

Example

A single-phase, full-wave, diode rectifier is supplied from a 230V ac, 50Hz voltage sourceand uses a capacitor output filter, 1000µF, with a resistor 100Ω load. Ignoring diode voltagedrops and assume the diode start conducting at 66.5 degree, determine:

Expressions for the output voltage

output voltage ripple ∆vo and the % error

Expressions for the capacitor current

Diode peak current

The average load voltage and current

Assuming the output ripple voltage is triangular, estimate:

The average output voltage and rms output ripple voltage

Capacitance C for ∆ vo = 2% of the maximum output voltage



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IV. Rectifier performance with input stray inductance

It’s shown that for single-phase bridge controlled rectifier when incoming SCRs

Tl and T2 are fired outgoing SCRs T3 and T4 get turned off due to the application

of reverse voltage and the current shifts to SCRs Tl and T2 instantaneously. This

is possible only if the voltage source has no internal impedance.

If the source impedance is Rs→ IoRs voltage drop occur → Vo(dc) reduced by IoRs

This drop voltage must be considered. However, the source impedance could be

consist of internal resistance Rs and stray inductance Ls. Hence, source inductance

the outgoing and incoming SCRs to conduct together results in zero voltage

across the load. This period of commutation called overlap period. the effect of

source inductance is :lower the mean output voltage, and distort the output

voltage and current waveforms.

Capacitance C for ∆ vo = 2% of the maximum output voltage

Δvo=𝑉𝑚 𝑅



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The effect of source inductance on the operation of single-phase full-bridge

controlled rectifier will be explained below. Same analysis can be done for

uncontrolled rectifier simply by applying α=00.

When T1 , T2 are triggered at a firing

angle α , the commutation of already

conducting T3, T4 begins. Because of

the presence of source inductance

Ls" the current via outgoing devices

T3, T4 decreases gradually to zero

from its initial value of io ; whereas

in incoming SCRs T1, T2 ; the current

builds up gradually from zero to full

value of io,

Hence, the full-bridge equivalent

circuit during this period is look like the

following circuit:

During overlap period μ:

The current flow via Ls, T1, Load, and T2

The current flow via T3, load, T4, and Ls

At beginning of μ , Ls stored energy by i2,

while then realise energy by i1.

By applying KVL for abcda loop:



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It can be re-expressed Dc output voltage to make a relation between Vo and load

current and value of source inductance:

Vo without overlap is equal to:

The maximum DC voltage occur at α=00. Thus,

Hence, output voltage with overlap period is can be found as:

From load current equation it can be obtained that:

from this equation μ can be obtained

Substitute the above equation in Vo, then



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It can be seen from equation above that

the voltage drop due to Ls, is proportional

to Io and Ls.

Note: for μ<π, Vo is reduces due to Ls,

for μ=π, Vo=0V since all SCRs are

conducting.

Note: if α=00, then Vo can be controlled

between μ≤α≤π

Note: The maximum value of firing

angle can be (1800- μ).



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Exercise:

A fully controlled bridge converter is connected to a 400V, 50Hz supply having

a source reactance of 0.3Ω. The converter is operating at a firing angle of quarter

of the one cycle. Determine the average load voltage and the overlap angle when

the converter is supplying a steady-state current of 60A.

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Lecture Note 7 - · PDF file · 2016-12-23Lecture Note 7 Uncontrolled and Controlled Rectifiers ... The main difference between ... The type of commutation used in controlled rectifier