Optimizing the design of a magnetic circuit actuator

Replies

• 

  lal

  Definitely the flux will not be divided equally between the limbs, due to non uniform reluctance in the paths.
  
  But, I assume by carefully designing the core, a satisfactory result could be achieved. By increasing the cross sectional area of top limb, RcoreBCDA could be reduced, thus allowing more magnetic flux to flow through that limb. Further, a thinner bottom limb would increase its reluctance.
  
  I am not quite sure how well that would turn out to be. Yet, let's try to work a way out.
• 

  Agar

  lal

  Definitely the flux will not be divided equally between the limbs, due to non uniform reluctance in the paths.
  
  But, I assume by carefully designing the core, a satisfactory result could be achieved. By increasing the cross sectional area of top limb, RcoreBCDA could be reduced, thus allowing more magnetic flux to flow through that limb. Further, a thinner bottom limb would increase its reluctance.
  
  I am not quite sure how well that would turn out to be. Yet, let's try to work a way out.

  Thanks for replying, ok let me explain a bit more in detail. The lamination's are of constant thickness of 0.5 mm and the maximum thickness of the core is 16 mm
  
  These are the constraints:
  12V dc input supply
  Current not to exceed 1A in the coil
  Maximum depth of core is 16 mm
  Lamination type, hot rolled silicon steel
  
  Parameters under my control:
  Number of turns on coil
  Area of the core
  Top air gap width
  
  The objective is to get the highest force/weight ratio
  
  Here's the dimensions:
  
  
  All dimensions are in mm. The 0.5mm is the thickness per lamination.
  
  I will show some of my work here what iv'e done so far:
  
  The first thing is to calculate the current in the coil:
  
  Using, R = p*length of wire/Area of wire , p = resistivity of copper wire
  
  Length wire = Number of turns*Perimeter of bobbin , Perimeter of bobbin = 74mm
  
  Area of wire = (pi*d^2)/4, where d is diameter of wire = 0.32 mm
  
  I am thinking I should be aiming for maximum current in the coil to get maximum mmf ?
  So I chose as a start:
  Number of turns = 775
  Using V = IR, where V = 12V (fixed)
  I = 0.97 A
  
  So, Total mmf = NI
  = 775*0.97
  = 751.75 At
  
  For the air gap, I am thinking I should make it small so as to get a higher H value ?
  So I chose as a start:
  Top air gap = 0.5 mm
  Using similar triangles:
  lgt/40 = lgc/20 where lgt = top air gap (0.5mm) and lgc = centre air gap
  lgc = 0.25mm
  
  I can now calculate the air gap reluctances, as follows:
  Rgt = Lgt/u*Agt
  Rgc = Lgc/u*Agc
  
  where Rgt and Rgc are the top and centre air gap reluctances
  
  For the area, i'm not too sure. If I add too many laminations the mass will increase
  So I chose as a start:
  16 laminations*0.5mm thickness = 8mm depth
  
  So, Agt = 8*8 = 64 mm^2
  Agc = 16*8 = 128 mm^2
  
  Substituting I get:
  Rgc = 1.55*10^6 At/Wb
  Rgt = 6.22*10^6 At/Wb
  
  The next part I need to estimate the mmf drops in the air gap, so I estimated 90% drop in the combined Rgt and Rgc. This part I am not sure off but I know that mmf will be greatest in the air gap
  
  So, 0.9*751.75 (Total mmf) = 676.575 At for Rgt and Rgc
  
  I then divided that mmf individually between Rgt and Rgc using the ratio:
  20:80 , ie 20% for Rgc and 80% for Rgt. I am also not too sure about this part
  
  So, 0.2*676.575 = 135.315 = mmf of Rgc (Fgc)
  and 0.8*676.575 = 541.26 = mmf of Rgt (Fgt)
  
  Now, I found B using the formula
  Bgc = Fgc*u/lgc
  Bgt = Fgt*u/lgt
  
  Substituting I get Bgc = 0.68 T = magnetic field of Rcore because area and flux is same
  
  and, Bgt = 1.36 T = magnetic field of RcoreBCDA because area and flux is same
  
  Using the BH curve for hot rolled silicon steel, I get
  Hgt = 660 At/m
  Hgc = 40 At/m
  
  Now using the formula F = HL, I can find the remaining mmf core drops
  
  I hope that wasn't too long, can you please see if i did it correctly, thanks
• 

  Agar

  Can I use the KVL technique to calculate the mmf drops and fluxes for each loop ?
• 

  lal

  I tried a quick calculation. I got mmf drop in Rgt to be around 89 and that in Rgc to be 660.
  
  By the way, I tried to calculate flux in each limb after I calculated the resistivities. Relative permeability of lamination was taken as 4000 (from wiki).
  
  Edit: Yes, I followed the same method as in KVL, analogous to electrical circuits.
• 

  Agar

  lal

  I tried a quick calculation. I got mmf drop in Rgt to be around 89 and that in Rgc to be 660.
  
  By the way, I tried to calculate flux in each limb after I calculated the resistivities. Relative permeability of lamination was taken as 4000 (from wiki).
  
  Edit: Yes, I followed the same method as in KVL, analogous to electrical circuits.

  Okay, but wouldn't the mmf drop in Rgt be higher than Rgc because Rgt has a larger gap width ?
  
  This is my KVL equation:
  Top loop:
  
  Ftotal = (phi1*RcoreAB) + (phi1*Rgc) + (phi2*Rgt) + (phi2*RcoreBCDA)
  
  Bottom loop:
  
  Ftotal = (phi1*RcoreAB) + (phi1*Rgc) + (phi3*RcoreBEFA)
  
  If I simplify by equating them and saying that RcoreBEFA = RcoreBCDA, I get this expression
  
  phi3/phi2 = (Rgt/RcoreBEFA) + 1
  
  I'm not sure if that's right ?, also will RcoreBCDA = RcoreBEFA ?, thanks
• 

  lal

  A larger gap would mean higher reluctance. But that would also mean less flux passing through that branch (path ABCD) as there is a very much low reluctance path (ABEF) through the bottom limb.
  
  The above expressions you derived are correct. RcoreBEFA and RcoreBCDA are equal.
  
  If you substitute values for Rgt and RcoreBEFA, you'll get
  Ø3 = 28.678 x Ø2
  [Ø2 being the flux through BCDA and Ø3 through BEFA]
  
  Now,
  Fgc = Rgc x (Ø2+Ø3) ----(1)
  Fgt = Rgt x Ø2 -----(2)
  Eqn (1)/(2) and substituting values for Rgc and Rgt,
  Fgc/Fgt = 7.18
  That is, mmf drop in Rgc is 7.18 times greater than the drop in Rgt.
  
  The bottom limb not having an air gap is actually forcing too much less flux to flow through top limb. May be a small gap there could actually be aiding the situation.
• 

  Agar

  lal

  A larger gap would mean higher reluctance. But that would also mean less flux passing through that branch (path ABCD) as there is a very much low reluctance path (ABEF) through the bottom limb.
  
  The above expressions you derived are correct. RcoreBEFA and RcoreBCDA are equal.
  
  If you substitute values for Rgt and RcoreBEFA, you'll get
  Ø3 = 28.678 x Ø2
  [Ø2 being the flux through BCDA and Ø3 through BEFA]
  
  Now,
  Fgc = Rgc x (Ø2+Ø3) ----(1)
  Fgt = Rgt x Ø2 -----(2)
  Eqn (1)/(2) and substituting values for Rgc and Rgt,
  Fgc/Fgt = 7.18
  That is, mmf drop in Rgc is 7.18 times greater than the drop in Rgt.
  
  The bottom limb not having an air gap is actually forcing too much less flux to flow through top limb. May be a small gap there could actually be aiding the situation.

  Yeah, I guess so. But I can't change the design, although I can change the top airgap. I'm guessing I should make it as small as possible so that the reluctance would be smaller so that a larger flux can flow through the top ?
  
  Also, R = l/u*a, so if I increase the area by adding more lamination's, the reluctance would decrease but this would increase the mass. What do you think is better ?
  
  And if I increase the number of turns in the coil, this will increase the resistance and for a fixed voltage of 12V the current will decrease. If I increase the number of turns too much then the current would decrease, so should I be aiming for lower or higher number of turns ?
  
  I have a question about the permeability, The one listed on wikipedia is for "Silicon Steel" and my laminations are "Hot rolled silicon steel", are they the same ?, because I have a BH curve and they do not actually have the same properties
• 

  lal

  Agar

  But I can't change the design, although I can change the top airgap. I'm guessing I should make it as small as possible so that the reluctance would be smaller so that a larger flux can flow through the top ?

  It wouldn't be sensible to reply without knowing the application. You could try to figure an alternate mechanical coupling to whatever you are trying to actuate.
  

  Agar

  And if I increase the number of turns in the coil, this will increase the resistance and for a fixed voltage of 12V the current will decrease. If I increase the number of turns too much then the current would decrease, so should I be aiming for lower or higher number of turns ?

  You could use thicker copper wire and more number of turns, although that would increase the weight too.
  

  Agar

  I have a question about the permeability, The one listed on wikipedia is for "Silicon Steel" and my laminations are "Hot rolled silicon steel", are they the same ?, because I have a BH curve and they do not actually have the same properties

  The value most probably is different. I used permeability of Silicon steel for calculations only because I couldn't find the permeability of hot rolled silicon steel 😁
• 

  Agar

  Then I would have to predict the mmf drop in the top gap ?
  
  Also if B = flux/Area, then is BcoreBCDA = Bgt because they have the same flux and area ?
  
  The method i'm using is to estimate the mmf in Rgt then calculate Bgt = BcoreBCDA and then use the BH curve to get H. Then I use F = HL
• 

  lal

  Agar

  Also if B = flux/Area, then is BcoreBCDA = Bgt because they have the same flux and area ?

  I'm not quite sure about that. But I suspect flux density in air will be quite less than that in the laminations as from what I know, flux lines will have a tendency to be spread apart in air. We cannot clearly define an area at the gap there. It might be enough to consider area to be same as that of metallic core for the sake of calculations, but probably with a correction factor. Not really sure though.
• 

  Agar

  lal

  I'm not quite sure about that. But I suspect flux density in air will be quite less than that in the laminations as from what I know, flux lines will have a tendency to be spread apart in air. We cannot clearly define an area at the gap there. It might be enough to consider area to be same as that of metallic core for the sake of calculations, but probably with a correction factor. Not really sure though.

  I heard fringing is minimal for small air gaps. I changed my top gap to 1.2 mm and middle gap is now 0.6 mm because the middle part was saturating before. Now Bgc is 1.43 T and saturation starts at 1.5 T according to the curve. I'm worried that fringing might increase it and cause it to saturate, will it ?
• 

  lal

  I cannot comment on that now. Will have to find my text books and start reading again 😁
• 

  Agar

  Okay, thanks for all your help 😀