Designing variable choke coil

1. DESIGNING VARIABLE
CHOKE COIL
1

2. STEP 1
 Bg
2Ai=
µ𝑉𝑝ℎ 𝐼𝑝ℎ
2𝜋𝑓𝐿𝑔
= k
 Where Lg=2lg’=2(length of each gap) for 1-Φ variable choke coil
 =1.5 lg’ for 3-Φ variable choke coil
2

3. STEP 2
 Take various values of Bg from 0.2 to 0.8T and calculate corresponding values
of 𝐴𝑖
 𝐴𝑖 =
𝐾
𝐵𝑔
 plot a curve Bg vs 𝐴𝑖
3

4. 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.
4

5. STEP 4
 Gross core area ,
 Agi=
𝐴𝑖
𝐾𝑠
 assume stacking factor Ks=0.9
 Because core is laminated and therefore gross area is larger then net iron
area Ai
5

6. STEP 5
 If we assume square cross–section of the iron core then width of limb ,
A= 𝐴𝑔𝑖
a
a 6

7. STEP 6
 Maximum mmf required for the air gap
𝐵𝑔𝐿𝑔
𝜇
=
𝐵𝑔×2𝑙𝑔′
𝜇
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
7

8. STEP 7
 Number of turn per coil, N=
𝐴𝑇
1
 There are three such coil having n turn for each in 3- Φ variable choke coil.
 There are 2 such coils having
𝑁
2
turn for each in case of 1- Φ variable choke
coil.
8

9. STEP 8
 Synthetic enameled round copper conductors are used for the winding.
 Area of bare conductor A=
1
𝛿
𝑚𝑚2
 Assume 𝛿=2.3 to 2.5 A/𝑚𝑚2
 1= current per phase for 3- Φ variable choke coil
 Diameter of bare conductor , D=
4𝑎
𝜋
mm
9

10. 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’=
𝜋
4
𝑑12
10

11. STEP 10
 Calculate the area of window
 Average value of space factor , Sf=0.8((
𝑑
𝑑1
)2is usuallu used for round
conductor
 Sf=
𝑎𝑐𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎
𝑔𝑟𝑜𝑠𝑒 𝑎𝑟𝑒𝑎
=
𝑎𝑐𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑖𝑛 𝑤𝑖𝑛𝑑𝑜𝑤
𝑤𝑖𝑛𝑑𝑜𝑤 𝑎𝑟𝑒𝑎
 =
2×
𝑁
2
×𝑎′
𝐴′𝑤
for in 1- Φ variable choke coil.
 =
2×𝑁×𝑎′
𝐴′ 𝑤
𝑓𝑜𝑟 in 3− Φ choke coil.
 (one window compare 2 coil section having N turns)
 :. Window area =𝐴′𝑤 =
2×
𝑁
2
×𝑎′
𝑆𝑓
for in 1- Φ variable choke coil.
 =
𝑁×𝑎′
𝑆𝑓
𝑓𝑜𝑟 in 3− Φ choke coil.
 :. Gross area of window Aw=1.2A’w
11

12. STEP 11
 Calculate the depth and height of the coil with zero air gap position
 assume =
𝐻𝑤
𝑊𝑤
=
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑛𝑑𝑜𝑤
𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑤𝑖𝑛𝑑𝑜𝑤
=2
 Aw=Hw*Ww=2𝑊𝑤2
Ww=
𝐴𝑤
2
∶. 𝐻𝑤 = 2𝑊𝑤
 Available height for winding h’f=Hw-20 mm
 Turn per layer, Nh=
ℎ𝑓′
𝑑1
 Turn per layer, Nd=
𝑁
𝑁ℎ
:. Depth of coil df’=Nd*d1
 Actual depth of coil df=df’+5mm
 Actual depth of coil hf=hf’+2*5 =hf’+10mm
12

13. STEP 12
Distance between the two coils in
window
Dc=Ww-2df
13

14. STEP 13 STEP 14
Total width of core =D+A
=Ww+2A
D=Ww+
𝐴
2
+
𝐴
2
=Ww+A
14

15. STEP 14 STEP 16
Impedance of the coil with max air gap
Z=
𝑣 𝑝ℎ𝑎𝑠𝑒
𝐼 𝑝ℎ𝑎𝑠𝑒
Ω
Total width of core =D+A
=Ww+2A
STEP 15
Overall height of core with max
airgap
=Hw+2A+lg’
15

16. 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=
𝜌𝑙
𝑎
ohm where a=area of c/s of bare conductor
𝜌=resistivity of copper=1.72*10^-8ohm m
 Reactance of coil with max airgap Xl= 𝑍2 + 𝑅2 ohm
 Inductance of the coil with max airgap L=
𝑋𝑙
2𝜋𝑓
𝐻
 where f= supply frequency
16

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