single phase ac-ac converters - philadelphia university · 2018-01-08 · the single phase ac...

Power Electronics Single Phase AC-AC Converters

1

Dr. Firas Obeidat

2

Table of contents

1 • Introduction

2 • The Single Phase AC Voltage Controller

3

• The Single Phase AC Voltage Controller - Resistive Load

4 • The Single Phase AC Voltage Controller - RL Load

Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University

Introduction

An ac voltage controller is a converter that controls the voltage, current, and average power delivered to an ac load from an ac source.

The phase-controlled ac voltage controller has several practical uses including light-dimmer circuits and speed control of induction motors.

In a switching scheme called phase control, switching takes place during every cycle of the source, in effect removing some of the source waveform before it reaches the load.

Integral-cycle control, the source is connected and disconnected for several cycles at a time.


The Single Phase AC Voltage Controller

Basic Operation

For the single phase AC voltage controller

shown, electronic switches are shown as parallel

thyristors (SCRs). This SCR arrangement

makes it possible to have current in either

direction in the load. This SCR connection is

called antiparallel or inverse parallel because

the SCRs carry current in opposite directions. A

triac is equivalent to the antiparallel SCRs.

Other controlled switching devices can be used

instead of SCRs.

Load current contains both positive and negative half-cycles. An analysis

identical to that done for the controlled half-wave rectifier can be done on a

half cycle for the voltage controller. Then, by symmetry, the result can be

extrapolated to describe the operation for the entire period.

S1 conducts if a gate signal is applied during the positive half-cycle of the

source. S1 conducts until the current in it reaches zero.

A gate signal is applied to S2 during the negative half-cycle of the source,

providing a path for negative load current.

vs

L

o

a

d

+ -vsw

S1

S2

+

-

+

-

io

vo


The Single Phase AC Voltage Controller

Basic Operation

Basic observations about this controller

The SCRs cannot conduct simultaneously.

The load voltage is the same as the source voltage when either SCR is on. The load voltage is zero when both SCRs are OFF.

The switch voltage vsw is zero when either SCR is ON and is equal to the source voltage when neither is ON.

The average current in the source and load is zero if the SCRs are on for equal time intervals. The average current in each SCR is not zero because of unidirectional SCR current.

The rms current in each SCR is 1/√2 times the rms load current if the SCRs are on for equal time intervals.


vs

+ -vsw

S1

S2

+

-

+

-

io

voR

The Single Phase AC Voltage Controller - Resistive Load

𝑣𝑠 𝜔𝑡 = 𝑉𝑚𝑠𝑖𝑛𝜔𝑡

𝑣𝑠 𝜔𝑡 = 𝑉𝑚𝑠𝑖𝑛𝜔𝑡 𝛼 < 𝜔𝑡 < 𝜋 𝑎𝑛𝑑 𝜋 + 𝛼 < 𝜔𝑡 < 2𝜋

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Let the voltage source be

Output voltage is

The rms load voltage is

𝑉𝑜,𝑟𝑚𝑠 =1

𝜋 (𝑉𝑚𝑠𝑖𝑛𝜔𝑡)2 𝑑𝜔𝑡

𝜋

α

=𝑉𝑚

2 1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋



The power factor of the load is

The rms current in the load and the source is

𝐼𝑜,𝑟𝑚𝑠 =𝑉𝑜,𝑟𝑚𝑠

𝑅=

𝑉𝑚

𝑅 2 1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋

𝑝𝑓 =𝑃

𝑆=

𝑉𝑜,𝑟𝑚𝑠2 𝑅

𝑉𝑠,𝑟𝑚𝑠𝐼𝑠,𝑟𝑚𝑠=

𝑉𝑜,𝑟𝑚𝑠2 𝑅

𝑉𝑠,𝑟𝑚𝑠(𝑉𝑜,𝑟𝑚𝑠 𝑅 )=

𝑉𝑜,𝑟𝑚𝑠

𝑉𝑠,𝑟𝑚𝑠=

𝑉𝑚2

1 −𝛼𝜋 +

𝑠𝑖𝑛2𝛼2𝜋

𝑉𝑚 2

𝑝𝑓 = 1 −𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋

The pf=1 for α=0, which is the same as for an uncontrolled resistive load, and

the power factor for α>0 is less than 1.

The average source current is zero because of half-wave symmetry.

𝐼𝑠,𝐴𝑣𝑔 = 𝐼𝑜,𝐴𝑣𝑔 = 0



The average SCR current is

𝐼𝑆𝐶𝑅,𝐴𝑣𝑔 =1

2𝜋

𝑉𝑚𝑠𝑖𝑛𝜔𝑡

𝑅𝑑𝜔𝑡

𝜋

𝛼

=𝑉𝑚2𝜋𝑅

(1 + 𝑐𝑜𝑠𝛼)

𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 =1

2𝜋 (

𝑉𝑚𝑠𝑖𝑛𝜔𝑡

𝑅)2 𝑑𝜔𝑡

𝜋

α

=𝑉𝑚2𝑅

1 −𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋

The rms SCR current is

𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 =𝑉𝑚

2 2𝑅 1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋=

𝐼𝑜,𝑟𝑚𝑠

2

πα 2π 2π+α 3πωt

iS1

vs/R


Example: The single-phase ac voltage controller has a 120-V rms 60-Hz source.

The load resistance is 15 Ω. Determine (a) the delay angle required to deliver

500 W to the load, (b) the rms source current, (c) the rms and average currents

in the SCRs, (d) the power factor.

𝑉𝑜,𝑟𝑚𝑠 =𝑉𝑚

2 1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋=

2 × 120

2 1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋

𝑃 =𝑉𝑜,𝑟𝑚𝑠

2

𝑅 → 𝑉𝑜,𝑟𝑚𝑠

2 = 𝑃𝑅 = 500 × 15 = 7500

𝑉𝑜,𝑟𝑚𝑠2 = 1202(1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋)

∴ 7500 = 14400(1 −𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋)

𝛼

𝜋−

𝑠𝑖𝑛2𝛼

2𝜋= 0.479

2𝛼 − 𝑠𝑖𝑛2𝛼 = 3.01

𝛼 = 1.54 rad = 88.1𝑜

(a)




(b)

𝐼𝑜,𝑟𝑚𝑠 =𝑉𝑜,𝑟𝑚𝑠

𝑅=

86.6

15= 5.77 A

𝑉𝑜,𝑟𝑚𝑠2 = 7500 → 𝑉𝑜,𝑟𝑚𝑠 = 86.6 From (a)

𝐼𝑆𝐶𝑅,𝐴𝑣𝑔 =𝑉𝑚2𝜋𝑅

1 + 𝑐𝑜𝑠𝛼 =2 × 120

2𝜋151 + 𝑐𝑜𝑠88.1 = 1.86 A

𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 =𝐼𝑜,𝑟𝑚𝑠

2=

5.77

2= 4.08 A

(c)

𝑝𝑓 =𝑃

𝑆=

𝑃

𝑉𝑠,𝑟𝑚𝑠𝐼𝑠,𝑟𝑚𝑠= 1 −

𝛼

𝜋+

𝑠𝑖𝑛2𝛼

2𝜋=

500

120 × 5.77= 0.722 (d)


The Single Phase AC Voltage Controller - RL Load

vs

+ -vsw

S1

S2+

-

+

-

io

vo

R

L

When a gate signal is applied to S1 at 𝜔t=α

in single phase AC voltage controller with

RL load, Kirchhoff’s voltage law for the

circuit is expressed as.

𝑉𝑚𝑠𝑖𝑛𝜔𝑡 = 𝑅𝑖𝑜 𝑡 + 𝐿𝑑𝑖𝑜 𝑡

𝑑𝑡

The solution for current in this equation

(as obtained in the controlled half wave

rectifier section) is

𝑖𝑜 𝜔𝑡 = 𝑉𝑚𝑍

sin 𝜔𝑡 − 𝜃 − sin (𝛼 − 𝜃)𝑒(𝛼−𝜔𝑡) 𝜔𝜏 𝛼 ≤ 𝜔𝑡 ≤ 𝛽

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

where

𝑍 = 𝑅2 + (𝜔𝐿)2 𝜃 = 𝑡𝑎𝑛−1(𝜔𝐿

𝑅)

The extinction angle 𝛽 is the angle at which the current returns to zero, when

𝜔𝑡= 𝛽,

𝑖𝑜 𝛽 =𝑉𝑚𝑍

sin 𝛽 − 𝜃 − sin (𝛼 − 𝜃)𝑒(𝛼−𝛽) 𝜔𝜏

(1)

(2)



A gate signal is applied to S2 at 𝜔t=π+α,

and the load current is negative but has a

form identical to that of the positive half-

cycle.

The above equation must be solved

numerically for 𝛽.

The angle (𝛽-α) is called the conduction

angle 𝛶.

𝛶 =𝛽-α

In the interval between π and 𝛽 when the

source voltage is negative and the load

current is still positive, S2 cannot be

turned on because it is not forward biased.

The gate signal to S2 must be delayed at

least until the current in S1 reaches zero,

at 𝜔t= 𝛽. The delay angle is therefore at

least 𝛽- π.

α≥𝛽- π



When α=θ, equation (2) becomes

sin 𝛽 − 𝜃 = 0

which has a solution

𝛽 − 𝜃 = π

Therefore

𝛶 = 𝜋 When α=θ

When 𝛶=𝜋, one SCR is always conducting, and the voltage across the load is

the same as the voltage of the source. The load voltage and current are

sinusoids for this case, and the circuit is analyzed using phasor analysis for AC

circuits. The power delivered to the load is continuously controllable between

the two extremes corresponding to full source voltage and zero.

The expression of rms load current is

𝐼𝑜,𝑟𝑚𝑠 =1

𝜋 𝑖𝑜

2(𝜔𝑡) 𝑑𝜔𝑡

𝛽

α

=1

𝜋

𝑉𝑚

𝑍sin 𝜔𝑡 − 𝜃 −

𝑉𝑚

𝑍sin (𝛼 − 𝜃)𝑒(𝛼−𝜔𝑡) 𝜔𝜏

2

𝑑𝜔𝑡

𝛽

α



The average output voltage can be found as

An other way to find the expression of rms load current

is


𝑅2 + (𝜔𝐿)2

𝑉𝑚

2

1

𝜋(𝛽 − 𝛼 −

1

2𝑠𝑖𝑛2𝛽 +

1

2𝑠𝑖𝑛2𝛼)

Power absorbed by the load is determined from

𝑃 = 𝐼𝑜,𝑟𝑚𝑠2𝑅

The power factor of the load is

𝑝𝑓 =𝑃

𝑆=

𝑉𝑜,𝑟𝑚𝑠𝐼𝑜,𝑟𝑚𝑠

𝑉𝑠,𝑟𝑚𝑠𝐼𝑠,𝑟𝑚𝑠=

𝑉𝑜,𝑟𝑚𝑠

𝑉𝑠,𝑟𝑚𝑠=

1

𝜋(𝛽 − 𝛼 −

1


1

2𝑠𝑖𝑛2𝛼)

𝑉𝑜,𝑟𝑚𝑠 =1

𝜋 (𝑉𝑚𝑠𝑖𝑛𝜔𝑡)2 𝑑𝜔𝑡

𝛽

α

=𝑉𝑚

2

1

𝜋(𝛽 − 𝛼 −

1


1

2𝑠𝑖𝑛2𝛼)



Each SCR carries one-half of the current waveform, making the average SCR

current

𝐼𝑆𝐶𝑅,𝐴𝑣𝑔 =1

2𝜋 𝑖𝑜(𝜔𝑡) 𝑑𝜔𝑡

𝛽

α

=1

2𝜋

𝑉𝑚𝑍

sin 𝜔𝑡 − 𝜃 − sin (𝛼 − 𝜃)𝑒(𝛼−𝜔𝑡) 𝜔𝜏 𝑑𝜔𝑡

𝛽

α

The rms current in each SCR is

𝐼𝑆𝐶𝑅,𝑟𝑚𝑠 =𝐼𝑜,𝑟𝑚𝑠

2

The average load current is zero,

𝐼𝑜,𝐴𝑣𝑔 = 0


Example: For the single-phase voltage controller with RL load, the source is

120Vrms at 60 Hz, and the load is a series RL combination with R=20Ω and

L=50mH. The delay angle α is 90. Determine (a) an expression for load current

for the first half-period, (b) the rms load current, (c) the rms SCR current, (d)

the average SCR current, (e) the power delivered to the load, and (f) the power

factor.

(a)


The extinction angle is determined from the numerical solution of i(β)=0 in the

above equation.



(b)

(c)

(d)

(e)

(f)


27.5× 120 ×

1

𝜋(3.83 − 1.57 −

1

2𝑠𝑖𝑛440 + 0) = 3.27 A


𝑅2 + (𝜔𝐿)2

𝑉𝑚

2

1

𝜋(𝛽 − 𝛼 −

1

2𝑠𝑖𝑛2𝛽 −

1

2𝑠𝑖𝑛2𝛼)

Or

single phase ac-ac converters - philadelphia university · 2018-01-08 · the single phase ac...

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