+

Single Phase, Full wave, R load

_ VS RVL

IL

I1

I2

T1

T2

ωt

vs

iG1

iG2

iG1 iG1 iG1

iG2 iG2

ωt

VL

α

α

VLrms= Vm √ 1/ π ( α∫πsin2 (ωt) d(ωt ))

= Vm √ 1/(2π) (α∫π(1-cos(2 ωt) d(ωt )) = Vm √ 1/(2π) × (π – α +½ sin(2α))

vS= Vm sin ( ωt )π

Isolation of the Gate

D1

D2

D3

D4

T

T1 T2

D1D2

Single Phase, Full wave, R-L load

+

_ VS

R

VL

IL

I1

I2 T2

iG1

iG2

L

T1

ωtiG1 iG1 iG1

iG2iG2

VS

i1

ωt

ωt

i2i1(ωt) = Vm / │Z │ sin(ωt – φ) + A exp-[(ωt-α)/ωζ]for α ≤ ωt ≤ α+γ

i1(α) = 0

A = - Vm / │Z │ sin(α – φ)

i1(ωt) = Vm / │Z │ sin(ωt – φ) - sin(α – φ) exp-[(ωt-α)/ωζ] for α ≤ ωt ≤ α+γi1(ωt) = 0 for α+γ ≤ ωt ≤ 2π+α

vS(ωt) = Vm sin(ωt)

Where │Z │= √(R2+ω2L2) φ= tan-1(ωL/R) ζ = L/R

ωt

iL

ωt

vL

αφ

γ

VLrms= √ Vm2 / π α∫(α+γ) sin2(ωt) d(ωt)

= Vm √ 1/(2π) α∫(α+γ) [1-cos(2 ωt)] d(ωt)

= Vm √ 1/(2π) [ωt - ½ sin(2ωt)]αα+γ

= Vm √ 1/(2π) [ γ - ½ sin(2α+2γ) + ½ sin(2α)]

Note that we need that α > φ so that γ < π

ie T1stops conducting before firing T2

When α = π

VLrms= Vm / √2 ( no control)

Calculation of γ

i1(α+ γ) = 0

sin(α+γ– φ) - sin(α – φ) exp-[(γ)/ωζ] = 0

Transcendental equation

The relation between γ and α

1- Pure R load

Φ= 0

i1 = Vm / R × sin(ωt)

γ = π - α

γ

α

γ = π - α

π

π

2- pure L load

Φ = 90o

i1 = Vm / (ωL) × ( - cos(ωt) + cos α )

I1 = 0 @ ωt = α + γ

cos(α + γ) = cos α

α + γ = 2π – α

γ = 2π – 2απ/2

2π

pure Rload

pure Lload

R-L load

3-phase ac Voltage Controller

ωt

vanvbn vcnT1

T4

T3

T6

T5

T2

R

R

R

nn

van

vbn

vcn

α1=0 α3=0 α5=0

α4=0α2=0 α6=0

Control ranges

0 ≤ α < 60o

60o ≤ α < 90o

90o ≤ α < 150o

vanL

ωt

van vbn vcn

½Vab

½Vac

α

For 0o ≤ ωt ≤ α

Assume that T5 and T6 are conductingSince neither T1 or T4 is conducting, then vanLis zero.

(α= 30o)

vanL

For α ≤ ωt ≤ 60o

T1 conducts in addition to T5 and T6. Note that van and vcn are +ve, while vbn is –ve.This means that the load is connected to the 3 supply lines. vanL= van in this interval.

60o

For 60o ≤ ωt ≤ 60o+ α

@ 60o T5 ceases to conduct since the load is pure resistive. In this interval only T6 and T1 conduct. This means that vanL= ½vab in this interval.

For 60o+α ≤ ωt ≤ 120o

@ 60o+α T2 conducts. In this interval T6, T1 in addition to T2 are conducting. van is +ve, while vdn and vcn are –ve.This means that the load is connected to the 3 supply lines. vanL = van in this interval.

For 120o ≤ ωt ≤ 120o+ α

@ 120o T6 ceases to conduct since the load is pure resistive. In this interval only T1 and T2 conduct. This means that vanL= ½vac in this interval.

For 120o+α ≤ ωt ≤ 180o

@ 120o+α T3 conducts. In this interval T1, T2 in addition to T3 are conducting. Van and vbn are +ve, while vcn is –ve.This means that the load is connected to the 3 supply lines. vanL = van in this interval.

ωt

van vbn vcn

½Vab

½Vac

α

For 180o ≤ ωt ≤ 180o+α

T1 ceases to conduct while T2 and T3 are conducting. Since neither T1 or T4 is conducting, then vanLis zero.

(α= 30o)

vanL

For 180o+α ≤ ωt ≤ 240o

T4 conducts in addition to T2and T3. Note that vbn is +ve, while van and vcn are –ve.This means that the load is connected to the 3 supply lines. vanL= van in this interval.

60o

For 240o ≤ ωt ≤ 240o+ α

@ 240o T2 ceases to conduct since the load is pure resistive. In this interval only T3 and T4 conduct. Since vbn is +ve while van and vcn are –ve, vanL= ½vab which is –ve in this interval. For 240o+α ≤ ωt ≤ 300o

@ 240o+α T5 conducts. In this interval T3, T4 in addition to T5 are conducting. van is -ve, while vbn and vcn are +ve.This means that the load is connected to the 3 supply lines. vanL = van in this interval.

For 300o ≤ ωt ≤ 300o+ α

@ 120o T3 ceases to conduct since the load is pure resistive. In this interval only T4 and T5 conduct. Van is -ve, while vcn is +ve. This means that vanL= ½vac in this interval, and is -ve.For 300o+α ≤ ωt ≤ 360o

@ 300o+α T6 conducts. In this interval T4, T5 in addition to T6 are conducting. van and vbn are -ve, while vcn is +ve.This means that the load is connected to the 3 supply lines. vanL = van in this interval. At ωt=360o, T4 ceases to conduct, this means that T5 and T6 remain conducting. This coincides with oue assumption during the interval 0o ≤ ωt < α

For 0 ≤ α ≤ 60o

VLrms = Vm√{1/(2π){ π/3 - ½ sin120o – α + ½ sin(2α) + π + 6α – 3 sin(180o+2α) – π + 3 sin(180o) + 2π/3 - ½ sin(240o) – π/3 – α + ½ sin(120o+2α) + π - ½ sin(360o) - 2π/3 – α + ½ sin(240o+2α) } }

= Vm√{1/(2π) { - π - √3 /4 + √3 /4 + 3α + ½ sin(2α) + ½ sin(2α+120o) + ½ sin(2α-120o) } }

= Vm√{1/(2π) { - π + 3α + ½ sin(2α) + ½ [sin(2α) cos(120o) + cos(2α) sin(120o) + sin(2α) cos(120o) - cos(2α) sin(240o)] } }

= Vm√{1/(2π) { - π + 3α + ½ sin(2α) + sin(2α) cos(120o) } }

= Vm√{1/(2π) { - π + 3α + ½ sin(2α) - ½ sin(2α) }

Transistor Switches•Very small reverse voltage capability•Fully controlled

BJT (Bipolar Junction Transistor)

Power MOSFETMetal Oxide Semiconductor

Field Effect Transistor

IGBTInsulated Gate Bipolar Transistor

C

E

B NPNN channel MOSFET

G

S

D C

E

G

current controlled voltage controlled

DC to DC Converters ( Choppers)

Chopperconstantvoltagedc source

variablevoltagedc load

PWM control