I want to make a DC-DC converter for 48 VDC to about 300 VDC at a = continuous=20 power of about 800W to drive a 1 to 2 HP three phase induction motor. I = have=20 some E55 cores and coilformers which should do the job, and also some=20 E47/20/15 which might work OK. The E55 is an Epcos type N27 and the E47 = is=20 probably the same. http://www.mouser.com/ds/2/136/e_47_20_16-75097.pdf http://www.mouser.com/ds/2/136/e_55_28_21-73714.pdf I found a website that shows a simple procedure to determine the turns=20 required for a transformer in a half-bridge topology using three = capacitors.=20 I plan to use the same circuit but not use the additional series = capacitor=20 and instead use two 20 uF 100 VAC PP capacitors in series across the DC = bus=20 and the center to one end of the primary. http://tahmidmc.blogspot.com/2013/02/ferrite-transformer-turns-calculatio= n_22.html The LTSpice for the basic circuit is on my server: http://enginuitysystems.com/pix/48V-320V_DCDC_HalfBridge_2Cap.asc I made a spreadsheet to automate the selection process and for the E55 = core=20 and 50 kHz it came up with 23 turns of #8 AWG for the primary at 33 amps = and=20 144 turns of #16 for the secondary at 5.3 amps. It appears to have 8 = watts=20 of ferrite losses and about 6 watts primary and secondary copper losses = for=20 total efficiency of about 97%. Here is the (OpenOffice) spreadsheet. Please have a look and see if it = is at=20 least close to being accurate: http://enginuitysystems.com/files/Ferrite_Transformer.ods Another question I have is if would be helpful to use heat shrink tape = to=20 compress the ferrite halves: http://www.mcmaster.com/#heat-shrink-tape/=3Dpjzuzt And, finally, I think it would be good to use Litz wire for this. I = found=20 some that seems to be a good deal: http://www.ebay.com/itm/370951676185 Thanks, Paul=20

# 800+ Watt DC-DC converter ferrite transformer design

Started by ●November 26, 2013

Reply by ●November 27, 20132013-11-27

P E Schoen wrote:> I want to make a DC-DC converter for 48 VDC to about 300 VDC at a > continuous power of about 800W to drive a 1 to 2 HP three phase > induction motor. I have some E55 cores and coilformers which should do > the job, and also some E47/20/15 which might work OK. The E55 is an > Epcos type N27 and the E47 is probably the same. > > http://www.mouser.com/ds/2/136/e_47_20_16-75097.pdf > http://www.mouser.com/ds/2/136/e_55_28_21-73714.pdf > > I found a website that shows a simple procedure to determine the turns > required for a transformer in a half-bridge topology using three > capacitors. I plan to use the same circuit but not use the additional > series capacitor and instead use two 20 uF 100 VAC PP capacitors in > series across the DC bus and the center to one end of the primary. > > http://tahmidmc.blogspot.com/2013/02/ferrite-transformer-turns-calculation_22.html > > > The LTSpice for the basic circuit is on my server: > http://enginuitysystems.com/pix/48V-320V_DCDC_HalfBridge_2Cap.asc > > I made a spreadsheet to automate the selection process and for the E55 > core and 50 kHz it came up with 23 turns of #8 AWG for the primary at 33 > amps and 144 turns of #16 for the secondary at 5.3 amps. It appears to > have 8 watts of ferrite losses and about 6 watts primary and secondary > copper losses for total efficiency of about 97%. > > Here is the (OpenOffice) spreadsheet. Please have a look and see if it > is at least close to being accurate: > http://enginuitysystems.com/files/Ferrite_Transformer.ods > > Another question I have is if would be helpful to use heat shrink tape > to compress the ferrite halves: > http://www.mcmaster.com/#heat-shrink-tape/=pjzuzt > > And, finally, I think it would be good to use Litz wire for this. I > found some that seems to be a good deal: > http://www.ebay.com/itm/370951676185 > > Thanks, > > PaulWord of warning regarding heatshrink: use the largest size that will do the job; use of the almost-smallest looks-good-on-paper size may result in cracked cores and split shrink.

Reply by ●November 27, 20132013-11-27

"Robert Baer" wrote in message news:8qhlu.262216$eE7.228412@fx21.iad...> Word of warning regarding heatshrink: use the largest size that > will do the job; use of the almost-smallest looks-good-on-paper > size may result in cracked cores and split shrink.Thanks for the advice. The heat shrink tape I found at McMaster is only=20 0.002" thick and is probably something like the shrink-wrap used for=20 shipping items on pallets, so I don't think it is very strong. Of course = it=20 would be a good idea to use several wraps. It seems like a good deal at = $13=20 for 300 ft of 3/4" wide tape. I plan to order some and try it. Paul

Reply by ●November 27, 20132013-11-27

On a sunny day (Tue, 26 Nov 2013 22:49:19 -0500) it happened "P E Schoen" <paul@peschoen.com> wrote in <l73q4c$jt6$1@dont-email.me>:>Another question I have is if would be helpful to use heat shrink tape = >to >compress the ferrite halves:Hi Paul, I have used cable ty ribbons in the past with success. You can put those in series too for bigger cores, and precisely adjust the strength, and or even connect the who assembly to the PCB using a few holes. Normal procefure would be a couple of long screws and a clamp plate.

Reply by ●November 27, 20132013-11-27

Waveform calculator: http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html If you don't know what you're doing as far as transformer specs, leave those fields alone (Lp, N2/N1, etc.). If you do know, you can enter them, hit Calculate, and it'll show what will happen with a (transformer/inductor) of those specs. Doesn't have push-pull (which is going to be best at a relatively low voltage), but you can play with a half or full bridge and adjust the primary volts, amps and turns respectively (PP has double the primary of H-bridge, which has twice the turns and half the amps of half bridge). Even has tables of core types; you should be able to enter parameters of your own core to see suitability as well. My website has a rather old article that I should rewrite for more clarity and stuff; http://webpages.charter.net/dawill/tmoranwms/Elec_Magnetics.html I think it still covers forward converter style transformer design, which is handy here (you probably wouldn't choose a flyback, let alone a buck, for these voltages, heheh). In short: core size depends on how many turns you can get around it, the peak flux density of the design, and the voltage and applied frequency: N = Vsq / (4 * F * Bmax * Ae) N = number of turns required for a given winding Vsq = square wave peak voltage applied to that winding F = frequency (50% duty cycle, no DC) Ae = cross sectional area of the core (i.e., area of the section the turns are wrapped around) Bmax = maximum design flux density For laminated iron at 50/60Hz, use Bmax = 1.2T or so. For ferrite under 100kHz or so, use 0.3T, or less at higher frequencies where losses bite more (depends on material). Or, rearrange the equation by algebra to solve for whichever parameter you need (e.g., you've set up a coil and want to know what Bmax it's actually achieving). Note this doesn't depend on gap; transformer action assumes gap is zero and permeability infinite. Other issues (like DC bias, startup transients, overload behavior, etc.) may affect the choice of gap (if any). This is one of the more important magnetic relations; if you don't know anything about magnetism, this should be helpful, and the rest is geometry and guessing how much copper can actually fit around the core. Tim -- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com "P E Schoen" <paul@peschoen.com> wrote in message news:l73q4c$jt6$1@dont-email.me... I want to make a DC-DC converter for 48 VDC to about 300 VDC at a continuous power of about 800W to drive a 1 to 2 HP three phase induction motor. I have some E55 cores and coilformers which should do the job, and also some E47/20/15 which might work OK. The E55 is an Epcos type N27 and the E47 is probably the same. http://www.mouser.com/ds/2/136/e_47_20_16-75097.pdf http://www.mouser.com/ds/2/136/e_55_28_21-73714.pdf I found a website that shows a simple procedure to determine the turns required for a transformer in a half-bridge topology using three capacitors. I plan to use the same circuit but not use the additional series capacitor and instead use two 20 uF 100 VAC PP capacitors in series across the DC bus and the center to one end of the primary. http://tahmidmc.blogspot.com/2013/02/ferrite-transformer-turns-calculation_22.html The LTSpice for the basic circuit is on my server: http://enginuitysystems.com/pix/48V-320V_DCDC_HalfBridge_2Cap.asc I made a spreadsheet to automate the selection process and for the E55 core and 50 kHz it came up with 23 turns of #8 AWG for the primary at 33 amps and 144 turns of #16 for the secondary at 5.3 amps. It appears to have 8 watts of ferrite losses and about 6 watts primary and secondary copper losses for total efficiency of about 97%. Here is the (OpenOffice) spreadsheet. Please have a look and see if it is at least close to being accurate: http://enginuitysystems.com/files/Ferrite_Transformer.ods Another question I have is if would be helpful to use heat shrink tape to compress the ferrite halves: http://www.mcmaster.com/#heat-shrink-tape/=pjzuzt And, finally, I think it would be good to use Litz wire for this. I found some that seems to be a good deal: http://www.ebay.com/itm/370951676185 Thanks, Paul

Reply by ●November 27, 20132013-11-27

"Tim Williams" wrote in message news:l74saa$guj$1@dont-email.me...> Waveform calculator: > http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html> If you don't know what you're doing as far as transformer specs, leave => those fields alone (Lp, N2/N1, etc.). If you do know, you can enter > them, hit Calculate, and it'll show what will happen with a=20 > (transformer/inductor) of those specs.> Doesn't have push-pull (which is going to be best at a relatively low=20 > voltage), but you can play with a half or full bridge and adjust the=20 > primary volts, amps and turns respectively (PP has double the primary > of H-bridge, which has twice the turns and half the amps of half =bridge).=20> Even has tables of core types; you should be able to enter parameters =of=20> your own core to see suitability as well.> My website has a rather old article that I should rewrite for more =clarity=20> and stuff; > http://webpages.charter.net/dawill/tmoranwms/Elec_Magnetics.html > I think it still covers forward converter style transformer design, =which=20> is handy here (you probably wouldn't choose a flyback, let alone a =buck,=20> for these voltages, heheh).> In short: core size depends on how many turns you can get around it, =the=20> peak flux density of the design, and the voltage and applied =frequency:> N =3D Vsq / (4 * F * Bmax * Ae)> N =3D number of turns required for a given winding > Vsq =3D square wave peak voltage applied to that winding > F =3D frequency (50% duty cycle, no DC) > Ae =3D cross sectional area of the core (i.e., area of the section the =turns=20> are wrapped around) > Bmax =3D maximum design flux density> For laminated iron at 50/60Hz, use Bmax =3D 1.2T or so. For ferrite =under=20> 100kHz or so, use 0.3T, or less at higher frequencies where losses =bite=20> more (depends on material).> Or, rearrange the equation by algebra to solve for whichever parameter > you need (e.g., you've set up a coil and want to know what Bmax it's > actually achieving).> Note this doesn't depend on gap; transformer action assumes gap is > zero and permeability infinite. Other issues (like DC bias, startup=20 > transients, overload behavior, etc.) may affect the choice of gap (if=20 > any).> This is one of the more important magnetic relations; if you don't =know=20> anything about magnetism, this should be helpful, and the rest is > geometry and guessing how much copper can actually fit around the =core. The calculator is very helpful. I found that the E55/28/21 is a bit too=20 small for 750W (300V 2.5A) but is very good for 625W (250V 2.5A). This = is=20 for 48-56V in and 50 kHz. However, it does not show the material of the=20 core, although maybe I can look up the Siemens part. The N27 I have = appears=20 to be better for low frequency (25 kHz) while the N87 is characterized = at=20 100 kHz. The calculator shows 8 turns primary and 88 turns secondary, while my=20 spreadsheet shows 23 and 144. I may not have properly calculated the = ratios=20 for the topology, and I used 1610 Gauss instead of 0.2 Tor. Making that=20 adjustment, I get 11T/57T, but the secondary should really be twice = that.=20 Then the ratio is 10.4 which compares to 11 for the calculator. I should = have used half the input voltage for the actual primary voltage with = this=20 topology. I suppose I should actually build one of these transformers and take=20 measurements. What do you think about the Litz wire? It is equivalent to #18 AWG so it = might be OK for the secondary, and the primary would need at least 8 or = 10=20 in parallel. #18 should be good for 4.7 amps at 342 CM/A or 5.8 A/mm^2. = The=20 calculator uses 3 A/mm^2, and if I use 600 CM/A or 3.3 A/mm^2, #18 is = good=20 for 2.7 A, and 8 in parallel will give 22 amps while I need 33. So = probably=20 12 in parallel would be needed, or else allow more temperature rise. I = will=20 probably need about the same total length of wire for primary and = secondary,=20 and I come up with about 20 ft. I think $35-$50 for 300 ft is a pretty = good=20 deal. Thanks, Paul

Reply by ●November 27, 20132013-11-27

"P E Schoen" <paul@peschoen.com> wrote in message news:l75jo3$44c$1@dont-email.me...> I suppose I should actually build one of these transformers and take > measurements.If nothing else, work it out on paper (or at worst, in the simulator). Square waves are easy. :)> What do you think about the Litz wire?Eh, maybe unnecessary. Over 500W it does start looking better, because the transformer becomes bulky enough to trap heat, especially a large one at low frequencies. Give or take construction, it might be a bit of a wash, economically speaking. What matters more is proximity effect and that. If you alternate layers between primary and secondary, any given layer is never more than one layer away from opposing currents, so the net field strength seen near a given layer is fairly weak. Weak fields means less crowding and more efficiency. (Downside being the added capacitance of interleaved layers. At 50 or even 25kHz, you can pretty well stomach the transients, or dampen them with a bit more snubbing than usual, and not care otherwise.) Stacking of currents by layer (and by conductor) is the reason Litz is useful, and also the reason you have to use much finer strands than the free space skin depth would suggest (i.e., 26AWG or so at these frequencies -- 34 or even 36AWG strands would be excellent in a large rope style conductor). If you plan on having single section windings (put down all the primary turns, layers of tape, then all the secondary), you'll definitely need Litz, and it may need to be finer even than what you've found (for best results, that is -- even bare stranded is better than solid, and Litz of the same dimension is even better). But you'd have shitty leakage doing that, probably too much to get the desired power output. A stack of P-S-P or P-S-PP-S-P (or swap P and S) is probably fine. This is just an efficiency thing, and you can always get away with crummy windings by using more of them. If you have a bunch of cores on hand, you could wind a few and save the cost of the Litz. Not like it's expensive or anything, especially compared to labor, if you're counting labor.> It is equivalent to #18 AWG so it > might be OK for the secondary, and the primary would need at least 8 or > 10 in parallel. #18 should be good for 4.7 amps at 342 CM/A or 5.8 > A/mm^2. The calculator uses 3 A/mm^2, and if I use 600 CM/A or 3.3 > A/mm^2, #18 is good for 2.7 A, and 8 in parallel will give 22 amps while > I need 33. So probably 12 in parallel would be needed, or else allow > more temperature rise. I will > probably need about the same total length of wire for primary and > secondary, > and I come up with about 20 ft. I think $35-$50 for 300 ft is a pretty > good > deal.If you go push-pull, remember each half is only used half the time, so you get a free sqrt(2) in the current capacity (but it's not 1/2, so the whole winding ends up taking more space; and note you want both ends of the CT primary themselves interleaved, so the windup gets messier too). Otherwise... yeah, that Litz would probably be a good fit. I normally crunch numbers at a higher A/mm^2, but usually on smaller things too, so that's not far off. Don't forget that the winding factor is smaller with Litz -- more air, so you need up to twice the winding area on the bobbin for a given windup. Tim -- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com

Reply by ●November 27, 20132013-11-27

On 11/27/2013 2:12 PM, P E Schoen wrote:> "Tim Williams" wrote in message news:l74saa$guj$1@dont-email.me... > >> Waveform calculator: >> http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html > >> If you don't know what you're doing as far as transformer specs, leave >> those fields alone (Lp, N2/N1, etc.). If you do know, you can enter >> them, hit Calculate, and it'll show what will happen with a >> (transformer/inductor) of those specs. > >> Doesn't have push-pull (which is going to be best at a relatively low >> voltage), but you can play with a half or full bridge and adjust the >> primary volts, amps and turns respectively (PP has double the primary >> of H-bridge, which has twice the turns and half the amps of half >> bridge). Even has tables of core types; you should be able to enter >> parameters of your own core to see suitability as well. > >> My website has a rather old article that I should rewrite for more >> clarity and stuff; >> http://webpages.charter.net/dawill/tmoranwms/Elec_Magnetics.html >> I think it still covers forward converter style transformer design, >> which is handy here (you probably wouldn't choose a flyback, let alone >> a buck, for these voltages, heheh). > >> In short: core size depends on how many turns you can get around it, >> the peak flux density of the design, and the voltage and applied >> frequency: > >> N = Vsq / (4 * F * Bmax * Ae) > >> N = number of turns required for a given winding >> Vsq = square wave peak voltage applied to that winding >> F = frequency (50% duty cycle, no DC) >> Ae = cross sectional area of the core (i.e., area of the section the >> turns are wrapped around) >> Bmax = maximum design flux density > >> For laminated iron at 50/60Hz, use Bmax = 1.2T or so. For ferrite >> under 100kHz or so, use 0.3T, or less at higher frequencies where >> losses bite more (depends on material). > >> Or, rearrange the equation by algebra to solve for whichever parameter >> you need (e.g., you've set up a coil and want to know what Bmax it's >> actually achieving). > >> Note this doesn't depend on gap; transformer action assumes gap is >> zero and permeability infinite. Other issues (like DC bias, startup >> transients, overload behavior, etc.) may affect the choice of gap (if >> any). > >> This is one of the more important magnetic relations; if you don't >> know anything about magnetism, this should be helpful, and the rest is >> geometry and guessing how much copper can actually fit around the core. > > The calculator is very helpful. I found that the E55/28/21 is a bit too > small for 750W (300V 2.5A) but is very good for 625W (250V 2.5A). This > is for 48-56V in and 50 kHz. However, it does not show the material of > the core, although maybe I can look up the Siemens part. The N27 I have > appears to be better for low frequency (25 kHz) while the N87 is > characterized at 100 kHz. > > The calculator shows 8 turns primary and 88 turns secondary, while my > spreadsheet shows 23 and 144. I may not have properly calculated the > ratios for the topology, and I used 1610 Gauss instead of 0.2 Tor. > Making that adjustment, I get 11T/57T, but the secondary should really > be twice that. Then the ratio is 10.4 which compares to 11 for the > calculator. I should have used half the input voltage for the actual > primary voltage with this topology. > > I suppose I should actually build one of these transformers and take > measurements. > > What do you think about the Litz wire? It is equivalent to #18 AWG so it > might be OK for the secondary, and the primary would need at least 8 or > 10 in parallel. #18 should be good for 4.7 amps at 342 CM/A or 5.8 > A/mm^2. The calculator uses 3 A/mm^2, and if I use 600 CM/A or 3.3 > A/mm^2, #18 is good for 2.7 A, and 8 in parallel will give 22 amps while > I need 33. So probably 12 in parallel would be needed, or else allow > more temperature rise. I will probably need about the same total length > of wire for primary and secondary, and I come up with about 20 ft. I > think $35-$50 for 300 ft is a pretty good deal. > > Thanks, > > Paul > >At 50Khz, use of litz wire made with #36 or # 38 would be the best to use. http://newenglandwire.com/products/litz-and-formed-cables/theory Mikek

Reply by ●November 27, 20132013-11-27

"amdx" wrote in message news:l75rb2$kag$1@dont-email.me...> At 50Khz, use of litz wire made with #36 or # 38 would be the > best to use. > http://newenglandwire.com/products/litz-and-formed-cables/theoryThat does appear to be the case, but it is difficult to determine the = actual=20 power and efficiency affected. I used a formula and some K factors from = the=20 following: http://www.electronicsteacher.com/alternating-current/reactance-and-imped= ance-inductive/more-on-the-skin-effect.php So for the same transformer, at 50 kHz, the AC resistance of the 8 turn = #10=20 AWG primary is 0.014 ohms with about 15 watts losses, and the 83 turn = #16=20 AWG secondary is 0.374 ohms with also 15 watts losses, and a total=20 efficiency of 95.4% At 100 kHz, the 4 turn primary has 0.010 ohms resistance and 11 watts, = and=20 the 42 turn secondary is 0.268 ohms and 11 watts losses, for total=20 efficiency of 96.4% So it appears that the skin effect losses are more than compensated for = by=20 the smaller number of turns at higher frequency. At 25 kHz the efficiency is 94.5%. I probably still have an error in the formulas, as the primary fill = factor=20 is 51% and the secondary is 103%. At 50 kHz it is 24% and 48%. They were = equal until I adjusted the formulas for the topology, which resulted in=20 twice the turns ratio for the same voltage ratio. I probably need to=20 increase the primary wire size. Bottom line, though, is that Litz wire may not be necessary for a = prototype,=20 although I think I may order some. I found some 100/38 (probably good = for=20 4.7A) for $20/50 ft. And 40 ft of 7x3x21/40 for $20, good for about 12 = amps. I calculated the cost per Ampere-Foot of various Litz wire, and it = ranged=20 from about $0.013 for the 300 ft of 13/30 at the minimum bid of $35, to=20 $0.142 for 30 ft of 200/38 at $40. I might low bid on the 13/30. The AWG = 30=20 is going to be a lot better than #22 AWG, which shows a 50% increase in = AC=20 resistance at 50 kHz. Paul=20

Reply by ●November 28, 20132013-11-28

"Tim Williams" wrote in message news:l74saa$guj$1@dont-email.me...> Waveform calculator: > http://schmidt-walter.eit.h-da.de/smps_e/smps_e.htmlI found what appears to be an error in the calculation of wire size for = the=20 primary. For my 48V 750W 50 kHz transformer it gives a primary of 7 = turns of=20 1.82 mm (2.6 mm^2 or about #13 AWG) but the primary current is about 33 = A=20 RMS which should be 11 mm^2 (about #6 AWG) at 3A/mm^2. I have updated my calculator for multiple primary strands in parallel, = with=20 appropriate skin effect calculation. I used 11 turns of 8 strands of #16 = AWG=20 (same as secondary), and the primary losses are now about 6 watts = instead of=20 15. Secondary is 138 turns and 15 watts losses. So the secondary would = also=20 benefit from multiple strands. http://enginuitysystems.com/files/Ferrite_Transformer.ods Paul