With the reference to the book by McLyman .
Skin depth is : 6.62/ Sqrt(frequency)
So in our case.
6.62/Sqrt(100Khz) = 0.021 cm
So, Wire diameter is 2 X skin depth = 0.021 x 2 = 0.041cm
So, calculate bare wire area. Aw.
(Pi x sqr(wire diameter )) /4 = 3.14 x 0.00175 / 4 = 0.00137 sqcm.
next finding wire of AWG.
Using chart given here Wire Gauges
and with conversion 100Sqmm = 1Sqcm.
We select #26 AWG. which is 0.13Sqmm = 0.0013Sqcm.
So we use minimum wire size for this design #26AWG.
Tony, please correct me if I am wrong. (I am just trying to save your efforts , If I can .)
If this is ok we move to next.
Skin depth is : 6.62/ Sqrt(frequency)
So in our case.
6.62/Sqrt(100Khz) = 0.021 cm
So, Wire diameter is 2 X skin depth = 0.021 x 2 = 0.041cm
So, calculate bare wire area. Aw.
(Pi x sqr(wire diameter )) /4 = 3.14 x 0.00175 / 4 = 0.00137 sqcm.
next finding wire of AWG.
Using chart given here Wire Gauges
and with conversion 100Sqmm = 1Sqcm.
We select #26 AWG. which is 0.13Sqmm = 0.0013Sqcm.
So we use minimum wire size for this design #26AWG.
Tony, please correct me if I am wrong. (I am just trying to save your efforts , If I can .)
If this is ok we move to next.
Magnetic Path Length is the average path length the flux lines take through the core to complete the circuit. If we were dealing with a toroid (as an example) the MPL is approximated by PI()*D. This is an approximation but in the case of someone not providing the data then that's what I use till I have better data. This is sometimes termed Le in the
Mean Length Per Turn. If you take a Cylinder and wrap 1 turn around the outside of the cylinder, then the Mean length per turn would be PI()*CylDia+WireDiam. If you wrapped 3 layers of windings around that cylinder then the MLT is calculated from Layer 2. If you add a bobbin in between the windings and the cylindrical core then you would add the thickness of the insulator. This number will be used later to calculate the copper wire losses in the windings. This value is usually on the data sheet for the bobbin.
Ae is basically the PI(Radius)^2 area of the cross section of the cylinder.
Transformer size and winding area is tough. Basically you can use any core to push any amount of power as long as you can get enough turns on it. That is pretty much what the design equations will tell you. You problem is that you have to get those turns into the winding area allowed by the core. McLyman covers this fairly well and we will cover it in more detail as we design the transformer. For little cores that need a lot of windings then you have to use thin wire. as long as the equations work out and you can fit the windings on the bobbin then you are OK (assuming copper losses are small enough to hit your efficiency requirements).
McLyman uses an Area Product approach which is OK but doesn't fully utilize the core. It's a good approximation and probably better than most other ways I have seen.
The final method for the approximation of the appropriate core/copper selection is an iterative approach which is really the method I use. When we get done with our calculations we will have calculated the core/copper/gap losses. These power losses can be plugged into a standard black body calculation to estimate the DeltaT rise of the core. Basically this is the air/core interface area. If you know the ambient operating temperature of the enclosure you are in you can then estimate how hot the core will be when we are at full power. Design simulations using ICEPAK or Flowtherm will get you a more accurate number (finite element analysis) but I can usually get within 10 degrees or so using the black body calculation.
This leads into one of your other questions on core loss and operating points. Rule one is that transformers are designed to run HOT!. I can design a transformer than runs cool by over sizing the wire and the Core Ae and it may even be more efficient. However, what I did was over-engineer the device. This is OK if you are a DIY and not OK if you are a power supply engineer.
Take for example the 3700 gauss number. You will find a lot of design examples where they use 2200 gauss as their operating point. They will also ignore the effects of fringing flux. This is OK if you are only using half the core (saturation flux density at 4000 and they are using 2200). However, that isn't the right way to do. One of these days they will screw things up and get a transformer that is saturating on them.
What I try and do is design the transformer one time. I run all the calculations so that when I get done, I know exactly how that transformer will operate in the end application.
Now for your question on operation at 90C. From the core loss curves you see that the material was engineered to be its most efficient at 90C. This is the point where most power supply engineers will try and operate the core. When we do our measurements on transformer put the transformer at max load and its lowest input voltage. The down side to operating at this temperature is that the copper losses increase by about 40% because of the increased temperature. What we have to watch for is the wire temperature right over the gap. The Flux lines cutting through the area and the gap will heat the wire up even higher at that point. Normally the max wire temp needs to be kept below 105C which is the enamel limit on the wire. Usually we will put a thermal couple in the center of the core and make sure that we are within the safety limits for the design.
Tony
If I miss some questions or you need more clarification please post it again. I will try to not miss any of your questions.
1) Ae covered above.
2) Typically you will wind the center leg. Lg is the total gap separation between the core halves. Typically this is done by grinding the center leg down. Since making an exact gap distance is difficult usually you will make a small gap then assemble the core haves onto a wound bobbin and test the inductance with an LCR. The alternate way is the way that was mentioned above. You take a piece of shim material and you use it to gap the outside legs. The only issue For an EE core you will have 3 gaps if you gap the two outside legs. I will warn you that this method is very prone to radiated EMI and is not the preferred method. Wonce you have the core assembled and gapped appropriately (within about 10% of you target inductance) glue the cores in place.
4) The window width is the leg that you wind the winding around. In the case of the EE core it is the center leg.
Tony
I will want to add some additional information to this a little later (current desnity and copper losses).
Tony
I got this but was trying to understand, which method did you use to reach value of MLT=53
That was a very important input. I should be learning this well while we actual implement this in our design.
"I screwed up on the Minimum Vac. If we use 90Vdc on the output then I will need a turns ration less than 1 (we will talk about this later). This would mean a 1kV Mosfet so we will use Vac Min as 190Vac. That gives us a Min Bulk voltage of about 184V."
Why we need 1kv mosfet ? is it because of back emf of the primary coil of transformer? or will be discussing this later ?
Ok... I did the calculations several years ago for all the standard cores. We'll see if I remember how I did it.
A) RM10 outer winding diameter of the core is about 21.2
B) The post diameter is about 10.9
C) bobbin wall is about 0.8mm
D) You need about 0.15mm clearance around the ferrite
This gives (((A/2)-D)-((B/2)+C+D))=4.55 for the average height of the winding stack.
MLT=PI*(4.55+B+2*C+2*D)=54 If you go to the Philips website you can read on page that they calculate it at 52.3. So... pretty close.
http://www.ferroxcube.com/prod/assets/rm10i.pdf
Tony
A) RM10 outer winding diameter of the core is about 21.2
B) The post diameter is about 10.9
C) bobbin wall is about 0.8mm
D) You need about 0.15mm clearance around the ferrite
This gives (((A/2)-D)-((B/2)+C+D))=4.55 for the average height of the winding stack.
MLT=PI*(4.55+B+2*C+2*D)=54 If you go to the Philips website you can read on page that they calculate it at 52.3. So... pretty close.
http://www.ferroxcube.com/prod/assets/rm10i.pdf
Tony
I may have to eat that off the cuff statement. Let me run through the rest of the math and see what I get. It's actually burried in my mathcad file somewhere.
Tony
That's a good Idea, Calculate or get from some other Core mfg the values. and use them for life time. I too would create a chart to make calculations handy.
Thanks Tony,
OK.. for 100KHz operation at 61Vdc on the bulk cap that gives us 42V for 90W and 100KHz. Ipk=8.3A and Lp=30uH.
Let me Check and see what QR operation would be... For QR operation down to 35KHz, Lp=87uH Still too low.
Cor CCM operation usign 100KHz, and 25% deltaI then Ipk drops to 4.2A and Lp needs to be 231uH. Still not good.
Lets drop he output power by 5W to 85W pk. Iout=0.944 and Vo=90V, Vbulkmin=56V, and we will use a 10% tolerance bulk cap instead of 20%.
Now we get Lp=455uH for CCM at 100KHz, 700uH for operation at 65KHz (this to mee is where you need to be), QR for Fswmin=35KHz Ipk=6A and Lp=135uH, DCM 100KHz=48uH.
From above there are some non-obvious design tradeoffs. First the small inductance for DCM/QR operation means that at means that your switching losses as a percentage of on time will be terrible as will the edge rates so your EMI will be poor. At 230Vac and QR mode you will be at a VERY high frequency.
I post the above so you see some of the non-obvious trade-offs.
Tony
Let me Check and see what QR operation would be... For QR operation down to 35KHz, Lp=87uH Still too low.
Cor CCM operation usign 100KHz, and 25% deltaI then Ipk drops to 4.2A and Lp needs to be 231uH. Still not good.
Lets drop he output power by 5W to 85W pk. Iout=0.944 and Vo=90V, Vbulkmin=56V, and we will use a 10% tolerance bulk cap instead of 20%.
Now we get Lp=455uH for CCM at 100KHz, 700uH for operation at 65KHz (this to mee is where you need to be), QR for Fswmin=35KHz Ipk=6A and Lp=135uH, DCM 100KHz=48uH.
From above there are some non-obvious design tradeoffs. First the small inductance for DCM/QR operation means that at means that your switching losses as a percentage of on time will be terrible as will the edge rates so your EMI will be poor. At 230Vac and QR mode you will be at a VERY high frequency.
I post the above so you see some of the non-obvious trade-offs.
Tony
I could guess you must be busy. But sometimes even at this age I get impatient. Strange, I know
Tony tell me are you into service industry? Cause people working in service industry has wide knowledge ( I have been reading you in other topics too).
cause their domain knowledge needs to be updated with projects.
where as people from mfg unit, working on development of a single product spend most on one product making limiting their knowledge.
let me understand whats Quasi resonant first.
I definitely have lot of questions about typologies. Which I have decide to ask you after the example design is done.
Initially 90V was selected, based on example given by you. of course I am going to make a prototype after design calculations are complete.
But I think This moment we stick to what we started with. Flyback with DCM. ( I consider flyback with DCM to be easiest choice for learning).
For me everything is going to be a first time.
We would be avoiding changes and maintain the learning flow. This is what I feel. What would advice ?
let me understand whats Quasi resonant first.
I definitely have lot of questions about typologies. Which I have decide to ask you after the example design is done.
Initially 90V was selected, based on example given by you. of course I am going to make a prototype after design calculations are complete.
But I think This moment we stick to what we started with. Flyback with DCM. ( I consider flyback with DCM to be easiest choice for learning).
For me everything is going to be a first time.
We would be avoiding changes and maintain the learning flow. This is what I feel. What would advice ?
90W at 230Vac for a DCM is OK. The inductance is in a range we can deal with. 90W at 90Vac would not be OK. You would need to be CCM for 90-264V operaton (full range).
Hello Tony,
I had read other varients of flyback converter. Probably CCM would work more effeciently. But I would wish to stick to a design that teaches me DCM flyback.
If we need to change voltage and values for the purpose, I am open to it.
but preferably not the other topology we have targeted. ( although I would surely refer them when ever they need to be, like you pointed in the previous thread. that gives me new direction to think.)
Well do you any design (which you did previously.) a complete power supply design. that uses flyback with DCM and that can operate at 230vac ( we have here 230vac mains supply for testing). so that by using that existing design as a refrence we would not have to think of altering the values and rethink over typologies. with the help of per-executed example we would be going smooth.
Thanks..
I had read other varients of flyback converter. Probably CCM would work more effeciently. But I would wish to stick to a design that teaches me DCM flyback.
If we need to change voltage and values for the purpose, I am open to it.
but preferably not the other topology we have targeted. ( although I would surely refer them when ever they need to be, like you pointed in the previous thread. that gives me new direction to think.)
Well do you any design (which you did previously.) a complete power supply design. that uses flyback with DCM and that can operate at 230vac ( we have here 230vac mains supply for testing). so that by using that existing design as a refrence we would not have to think of altering the values and rethink over typologies. with the help of per-executed example we would be going smooth.
Thanks..
OK... Back to you Skin depth calculation...
6.62/ Sqrt(frequency) This needs to have a Kcu term added to it.
This constant is the SQRT(Rho.e/Rho) where Rho.e is the adjusted electrical resistivity in Ohm-cm and Rho is the term at room temp.
Rho.e=Rho(1+Alpha(WireTemp-Wireampient) - This will be used later in the wire calculations)
Rho=1.7241*10^-6 ohm-cm Per Nema 1000
Alpha=0.0039 temperature compensation for Cu (use a different value if you are using Aluminum)
The Kcu=Sqrt(Rho.e/Rho)
Delta(which is skin depth)=6.62*Kcu/Sqrt(Fsw)=.24mm
MaxWireDia=2*Delta or 0.48mm Round up to .5mm or approximately 24AWG. Using JIS3032 2UEW 0.5mm magnet wire gives us a OD of 0.54mm (added insulation thickness).
We would next look at the length of the winding window in the bobbin. This is approximately 10mm (See bobbin data sheet). So our maximum number of turns is on a single layer using 0.5mm wire is 10/0.54 or 18 turns. However, we have a creapage / clearance issue to take care of. We need 5.1mm of margin (2.55mm on either side - Most will round the 5.1mm to 6 to provide more margin) inside the transformer for creapage distance. so that really only leaves 5mm for the winding area (using standard margin tape). If you are using tripple insulated wire on the secondar (or primary) you do not need margin tape.
At this point we have a decision to make. The lowest leakage transformers will give you the best efficiency and the best EMC performance. The lowest leakage is achieved by sandwiching layers of primary and secondary. For every sandwich layer we cut the leakage in half. Your target lakage should always be <1% of the primary indictace. There are some reasons for this but I won't go into that here.
So... what to use and how to construct it? I think the best construction is to use 3 layers of 10 turns each for a series number of turn=30. We will then sandwich 2 layers of secondary. To make the 10 turns happen we will need to reduce the wire diameter to 0.45mm so that the OD is now 0.49mm. So... build up your margin tape on either side and add a teflon sleeve to the start winding, wrap the lead around the bobbin pin and bring it across the margin boundary and wrap 10 turns, cross the other boundary and hand the extra wire outside the bobbin. Add 2 layers of transformer tape. Add the secondary layer (be sure to teflon sleeve the secondary side where it crosses the margin tape) and add 2 layers of tape. Bring the primary winding back across the margin tape and add 2 more layers of tape. Add another layer of secondary with tubing, tape, and then add the final layer of primary to give us the required 30 turns. The primary wire is now on the opposite side of the transformer. Bring it back across the transformer and add your final tape.
What we have now constructed is a Primary side referenced transformer. Meaning that you will need the same 5.1mm of clearance to all conductive surfaces on the secondary side that you had inside the transformer. You could have made it secondary side referenced in which case the tubing would have been on the primary wire and the clearance to the conductive surfaces would be for primary side parts.
We still may have to change our strategy once we calculate the copper resistance and power dissipation on the winding.
The Area of the wire is Awb=PI()(r)^2
The Lp resistance is RcuPri=(Rho.e/Awb)*MLT*Np (told you we would eventually use it).
We will approximate the RMS current (there are precise calculations to do that and I will let you find them) by using the PawbPri=(Po/(n(90Vac*sqrt(2))))^2*(RcuPri) where n is the efficiency. If this number is acceptable (about 115mW for the primary winding).
Next is the secondary side...
Ns1=(Vo+Vfd)*(1-DutyCycle)*Np/(DutyCycle*Vbulkmin*Kcoupling)
Vo is the output voltage, Vfd is the forward voltage drop of the output rectifier Npis the primary turns and Vbolkmin we calculated previously.
Kcoupling=Sqrt(1-Lk/Lp) If we used 10% leakage then we would have had a 5% error term that would have had to have been factored in.
Ns1=10
I'll leave you to do the wire selection for the secondary. Be sure you look at the overall bobbin height to make sure you fit in the bobbin withe all the other layers and tape.
Tony
6.62/ Sqrt(frequency) This needs to have a Kcu term added to it.
This constant is the SQRT(Rho.e/Rho) where Rho.e is the adjusted electrical resistivity in Ohm-cm and Rho is the term at room temp.
Rho.e=Rho(1+Alpha(WireTemp-Wireampient) - This will be used later in the wire calculations)
Rho=1.7241*10^-6 ohm-cm Per Nema 1000
Alpha=0.0039 temperature compensation for Cu (use a different value if you are using Aluminum)
The Kcu=Sqrt(Rho.e/Rho)
Delta(which is skin depth)=6.62*Kcu/Sqrt(Fsw)=.24mm
MaxWireDia=2*Delta or 0.48mm Round up to .5mm or approximately 24AWG. Using JIS3032 2UEW 0.5mm magnet wire gives us a OD of 0.54mm (added insulation thickness).
We would next look at the length of the winding window in the bobbin. This is approximately 10mm (See bobbin data sheet). So our maximum number of turns is on a single layer using 0.5mm wire is 10/0.54 or 18 turns. However, we have a creapage / clearance issue to take care of. We need 5.1mm of margin (2.55mm on either side - Most will round the 5.1mm to 6 to provide more margin) inside the transformer for creapage distance. so that really only leaves 5mm for the winding area (using standard margin tape). If you are using tripple insulated wire on the secondar (or primary) you do not need margin tape.
At this point we have a decision to make. The lowest leakage transformers will give you the best efficiency and the best EMC performance. The lowest leakage is achieved by sandwiching layers of primary and secondary. For every sandwich layer we cut the leakage in half. Your target lakage should always be <1% of the primary indictace. There are some reasons for this but I won't go into that here.
So... what to use and how to construct it? I think the best construction is to use 3 layers of 10 turns each for a series number of turn=30. We will then sandwich 2 layers of secondary. To make the 10 turns happen we will need to reduce the wire diameter to 0.45mm so that the OD is now 0.49mm. So... build up your margin tape on either side and add a teflon sleeve to the start winding, wrap the lead around the bobbin pin and bring it across the margin boundary and wrap 10 turns, cross the other boundary and hand the extra wire outside the bobbin. Add 2 layers of transformer tape. Add the secondary layer (be sure to teflon sleeve the secondary side where it crosses the margin tape) and add 2 layers of tape. Bring the primary winding back across the margin tape and add 2 more layers of tape. Add another layer of secondary with tubing, tape, and then add the final layer of primary to give us the required 30 turns. The primary wire is now on the opposite side of the transformer. Bring it back across the transformer and add your final tape.
What we have now constructed is a Primary side referenced transformer. Meaning that you will need the same 5.1mm of clearance to all conductive surfaces on the secondary side that you had inside the transformer. You could have made it secondary side referenced in which case the tubing would have been on the primary wire and the clearance to the conductive surfaces would be for primary side parts.
We still may have to change our strategy once we calculate the copper resistance and power dissipation on the winding.
The Area of the wire is Awb=PI()(r)^2
The Lp resistance is RcuPri=(Rho.e/Awb)*MLT*Np (told you we would eventually use it).
We will approximate the RMS current (there are precise calculations to do that and I will let you find them) by using the PawbPri=(Po/(n(90Vac*sqrt(2))))^2*(RcuPri) where n is the efficiency. If this number is acceptable (about 115mW for the primary winding).
Next is the secondary side...
Ns1=(Vo+Vfd)*(1-DutyCycle)*Np/(DutyCycle*Vbulkmin*Kcoupling)
Vo is the output voltage, Vfd is the forward voltage drop of the output rectifier Npis the primary turns and Vbolkmin we calculated previously.
Kcoupling=Sqrt(1-Lk/Lp) If we used 10% leakage then we would have had a 5% error term that would have had to have been factored in.
Ns1=10
I'll leave you to do the wire selection for the secondary. Be sure you look at the overall bobbin height to make sure you fit in the bobbin withe all the other layers and tape.
Tony
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