Timeline for How to safely generate 200V from low voltage supply
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Dec 12, 2018 at 21:29 | answer | added | Andy aka | timeline score: 2 | |
| Dec 12, 2018 at 20:46 | comment | added | Andy aka | Aim for no more than 0.3 Tesla for Bmax is my advice. What peak current did you calculate for peak energy transfer. What switching frequency also? What is the mean circumference of the toroid and what permeability has it got. That tells you everything plus what Neil said. | |
| Dec 12, 2018 at 20:36 | comment | added | Neil_UK | @deltamac For a given power throughput W at a given frequency f, you need x=W/f joules stored in the magnetic field. Your core will have a given max flux before saturation Bmax. Calculate what weight of core you need to store x joules with a gap, and without a gap, and compare (hint,energy goes as H*B, Bmax is fixed, you can increase H by gapping). Only then start thinking about turns and current. | |
| Dec 12, 2018 at 19:59 | comment | added | deltamac | @Andyaka, the turns ratio is 20:1, the primary has 2 turns and the secondary has 40. Sorry about the confusion. | |
| Dec 12, 2018 at 19:58 | comment | added | deltamac | @Andyaka, I think you just clarified one major thing I was missing. I think I have been using the formulae for flux linkage, not flux. Flux linkage (N * flux) is the same in both cases, but flux is reduced with increased turns and same inductance. | |
| Dec 12, 2018 at 19:51 | comment | added | Andy aka | Higher up you said your transformer was 20 turns. But you will need a gap. | |
| Dec 12, 2018 at 19:49 | comment | added | deltamac | @Andyaka, the secondary in 40 turns on a 2.88uH/T^2 core. I expected 4.6mH on the high voltage winding. LIkewise, the primary is 2 turns and gives, 11.5uH which is what I measure. Is this wrong for some reason I am not seeing? | |
| Dec 12, 2018 at 19:37 | comment | added | Andy aka | If you have wound 20 turns on a core that is a 3000 nH per turn squared, the inductance you have wound is 1.2 mH and not what you think. | |
| Dec 12, 2018 at 19:30 | comment | added | Andy aka | @deltamac your flux argument is total pants. If gapping reduces the permeability to (say) one quarter then to retrieve the inductance you double the turns. The net effect is that with the gapped core and double the turns, the flux density is halved. Reason: permeability has quartered, turns have doubled, flux is therefore half. Inductance is proportional to turns squared in case you didn’t know. | |
| Dec 12, 2018 at 18:52 | comment | added | deltamac | @Neil_UK , you can probably poke holes in my arguments. In my mind, if I gap, then wind more turns, that can only lead to more leakage inductance and more capacitance. | |
| Dec 12, 2018 at 18:49 | comment | added | deltamac | @TimWescott, I ran into this exact issue. Energy stored is .5*L*I^2. If I have the same L and the same I... stored energy should be the same. | |
| Dec 12, 2018 at 18:42 | comment | added | TimWescott | I can't remember the materials-science detail of why you need the gap. But I do remember that I once ran the numbers, and found that the majority of the energy stored by the coil right before you cut the current is in the gap. The basic design of the flyback circuit is that the source charges up the inductor, then the inductor discharges into the load -- so unlike other topologies, the inductor must be able to store a chunk of energy. | |
| Dec 12, 2018 at 18:41 | comment | added | deltamac | @TimWescott My transformer is 20 turns evenly spaced around circumference in a single layer. Then I wrapped that layer in some teflon tape and did my low voltage windings over top of those. I used this core link | |
| Dec 12, 2018 at 18:35 | history | edited | deltamac | CC BY-SA 4.0 |
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| Dec 12, 2018 at 18:34 | comment | added | deltamac | To continue, if ungapped I need only a few turns to get the necessary inductance. If I do gap it, I need more turns to get same inductance. Either way, I need the same inductance and the same peak current, so the same peak flux, so the same flux density regardless of if there is a gap or not. | |
| Dec 12, 2018 at 18:33 | comment | added | deltamac | I can believe that I have overlooked something crucial about the core. My logic on the flyback transformer was this... the design equations for flybacks ends up requiring a particular primary inductance, and a peak primary current. Together those define a peak flux. Therefore, I need a core with enough area not to saturate at maximum flux density. That's how I picked my core. | |
| Dec 12, 2018 at 18:00 | history | tweeted | twitter.com/StackElectronix/status/1072914189012885504 | ||
| Dec 12, 2018 at 17:50 | comment | added | TimWescott | @Neil_UK: He's a student. He's not ready to design anything. That's what this project is for -- to provide him with blocks to stumble over, so that he can learn. | |
| Dec 12, 2018 at 17:49 | comment | added | TimWescott | I'm suspecting inter-winding capacitance. Post a picture of your output transformer? Maybe do some measurements with a signal generator? | |
| Dec 12, 2018 at 17:49 | comment | added | Neil_UK | No, a gapped core is essential for flybacks, at least for powers at the watts level, otherwise you end up with far too oversized core, and so parasitics far too high, which seems to be one of your problems. If you don't know why you need a gapped core, you're not ready to design a flyback. | |
| Dec 12, 2018 at 17:49 | comment | added | mike65535 | Have you tried the various output windings, separately? | |
| S Dec 12, 2018 at 17:45 | history | suggested | mike65535 | CC BY-SA 4.0 |
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| Dec 12, 2018 at 17:27 | review | Suggested edits | |||
| S Dec 12, 2018 at 17:45 | |||||
| Dec 12, 2018 at 17:23 | history | edited | deltamac | CC BY-SA 4.0 |
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| Dec 12, 2018 at 17:17 | history | asked | deltamac | CC BY-SA 4.0 |