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DESIGNING VARIABLE CHOKE

COIL

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STEP 1

Bg2Ai= = k

Where Lg=2lg’=2(length of each gap) for 1-Φ variable choke coil =1.5 lg’ for 3-Φ variable choke coil

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STEP 2

Take various values of Bg from 0.2 to 0.8T and calculate corresponding values of

plot a curve Bg vs

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STEP 4

Select critical value of Bg from the curve and calculate corresponding value of Ai

• Higher value of Bg means larger number of turns and more copper is reqired.

• Lower value of Bg means high cross-section Aiand hence large length of turn and resulting more copper.

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STEP 4

Gross core area , Agi= assume stacking factor Ks=0.9 Because core is laminated and therefore gross area is larger then net

iron area Ai

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STEP 5

If we assume square cross–section of the iron core then width of limb , A=

a

a

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STEP 6

Maximum mmf required for the air gap AT (for 1-Φ variable choke coil)

Totale AT required= AT required for the air gap +AT required for the iron part

Assume Ati=10% to 20% of Atg :. Total AT per coil = 1.1 to 1.2 ATg

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STEP 7

Number of turn per coil, N= There are three such coil having n turn for each in 3- Φ variable choke

coil. There are 2 such coils having turn for each in case of 1- Φ variable

choke coil.

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STEP 8

Synthetic enameled round copper conductors are used for the winding.

Area of bare conductor A= Assume =2.3 to 2.5 A/ 1= current per phase for 3- Φ variable choke coil Diameter of bare conductor , D=mm

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STEP 9 A standard size of conductor is selected from the table Nearest standard conductor has nominal diameter, d and diameter

of conductor with insulation d1 are selected from table :. Area of conductor with insulation a’=

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STEP 10 Calculate the area of window Average value of space factor , Sf=0.8(is usuallu used for round

conductor Sf= = for in 1- Φ variable choke coil. (one window compare 2 coil section having N turns) :. Window area = = for in 1- Φ variable choke coil. = :. Gross area of window Aw=1.2A’w

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STEP 11 Calculate the depth and height of the coil with zero air gap

position assume ==2 Aw=Hw*Ww=2 Ww= Available height for winding h’f=Hw-20 mm Turn per layer, Nh= Turn per layer, Nd= :. Depth of coil df’=Nd*d1 Actual depth of coil df=df’+5mm Actual depth of coil hf=hf’+2*5 =hf’+10mm

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STEP 12

Distance between the two coils in window Dc=Ww-2df

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STEP 13 STEP 14

Total width of core =D+A =Ww+2A

D=Ww++=Ww+A

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STEP 14 STEP 16Impedance of the coil with max air gapZ=

Total width of core =D+A =Ww+2A

STEP 15 Overall height of core with max airgap=Hw+2A+lg’

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STEP 17 Inner dimentions of the square coil=(A+10mm)*(A+10mm) Outer dimensions of the square coil=(A+2df)*(A+2df) Length of mean turn, Lmt=4(A+df) Total length of wire, l=N*Lmt per coil R= ohm where a=area of c/s of bare conductor =resistivity of copper=1.72*10^-8ohm m Reactance of coil with max airgap Xl= ohm Inductance of the coil with max airgap L= where f= supply frequency

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