flyback converter driving

I was using the LT1619. Datasheet shows 40uA max current draw in shutdown mode. Simulator shows about 265uA :)

I think the current draw will need to be greater in order to charge the capacitor up within ~100ms max too.

Thanks! I put in that new circuit, I think it is working fine, but the simulation gets stuck at 147ms for some reason doing "transient analysis" so I don't know for sure. I think I might have to increase the inductance of the L2 coil so that Vcc gets to at least the zener voltage (20V) to turn off Q2, there isn't much draw off of the L2 coil, is there a way to estimate what inductance to use, since the simulator won't run so I can't trial and error it like usual!

Here are the current schematic and ltspice files:

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I guess a simpler solution could involve using a PTC thermistor, but a restart would have to wait for the resistor to cool down.

Also if there was an optocoupler that is normally on, then turns off when the voltage on the input increases that may work too. I don't know if they have optocouplers that work this way? Maybe one built with a depletion mode n channel fet would work?

I want to put the circuit into my SMPS, it is to generate a 12V rail to run the uC etc.

cheers, Jamie

Reply to
Jamie Morken
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Hi,

By changing C5 to 20uF the simulation runs, it prevents Vcc from going below a threshold which was causing the simulator to stall - thanks Helmut from ltspice yahoo group :)

Q2 is still not shutting off though, the zener voltage on Q2's base is staying higher than the emitter voltage so that is keeping it turned on.

cheers, Jamie

Reply to
Jamie Morken

On Oct 21, 2:40 am, Jamie Morken wrote: ...

OK about the part number. I'm having a hard time keeping up. ;-) I've trimmed a whole lot the last post off to keep from using quite so much net bandwidth... ...

Why are you using a 20 volt zener?? I think you'll find that a 10 volt zener will have a sharp enough knee that it will work well with just a few microamps. It's been years since I looked carefully at one, though, so I could be wrong. Surely you don't need 20 volts into the controller chip to get it going.

Re-read what I wrote about the flyback voltages. If you have turns ratios pri:sec1:sec2, the voltages across the windings (assuming perfect coupling) will be always in that ratio. During flyback, the output side goes to the regulated output voltage (or gets there after charging the cap, anyway), plus a diode drop. That tells you the flyback voltage on the other windings. And inductance ratios are the square of turns ratios: with perfect coupling of the turns, the inductance goes as the square of the number of turns.

Cheers, Tom

Reply to
Tom Bruhns

Yes, you are absolutely right. Thanks for noticing that. Either Jamie changed the polarity between the time I looked at it and when you looked at it, or I had my head in a dark, smelly place when I looked at it. Most likely the latter.

Cheers, Tom

Reply to
Tom Bruhns

I was worried that the high voltage rated fet would need to have a large initial gate drive voltage to avoid it overheating, but looking at the datasheet of the 900V fet, 2SK3565 it seems to be ok with 7 volt gate drive.

Also the start-up threshold voltage for the UC3842/LT1246 is 16Volts, so the zener has to be higher than that. The UC3843/LT1247 has a 8.4Volt start-up threshold, but its minimum operating voltage is only 7.6Volts which doesn't give much room for the circuit to start unless the startup resistor is small enough to keep the Vcc rail above 7.6Volts during startup, still should work though either way I guess.

Thanks that works now!

I used this formula, and substituted voltages for turns since they are equivalent:

coil_turns(inductance) = (1 / (fullcoil_turns / coil_turns(turns))^2) * fullcoil_inductance

coil_turns(inductance) = (1 / (12v / 22v)^2) * 70uH =235uH

Also changed the feedback voltage divider, using 2.5V reference in the IC the formulas are:

Vo=2.5v(1+R1/R2) R1=R2(Vo/2.5v-1)

R1=4.7k(22v/2.5v - 1)

R1 = 36.6k R2 = 4.7k

That gives 11.3v regulated output and ~21.3v to power the IC and shut off the npn transistor now. Thanks for all your help! :)

cheers, Jamie

Reply to
Jamie Morken

On Oct 21, 1:41 pm, Jamie Morken wrote: ...

I'm not sure at this point which IC that is, but if it's the LT1519, that's more than the data sheet rating of the part, I believe. It's generally a good idea to run ICs well below their maximum rated voltage... Maybe it's for that other part; I don't know the ratings for that one. Also, a lot of mosfets have a maximum gate voltage of

20V, so you'd want to be careful about running the driver on a voltage that could be higher than the gate voltage rating of some part you may pick in the future. Since the mosfet almost certainly won't turn on much "harder" with 20V gate drive than with 10V, the extra voltage mainly means that you dissipate more power than you need to in the gate drive.

Cheers, Tom

Reply to
Tom Bruhns

Hi,

I tested the circuit from 20V input to 450V input, the circuit starts to operate at about 30V (below that it just doesn't start up) and the startup time is about 550ms at 30V input. The startup time at 450V input is about 18ms, measured by when the input switch shuts off. So using an input switch series resistor of 20k, this resistor is dissipating about 10watts for 18ms, and then 0watts once the circuit is started. Is that safe to use a 1206 package or even smaller resistor for this?

Also the fet dissipates about 1.5watts over the whole input voltage range, and at 450V input, there are very short 700watt power spikes going through the fet. Is this ok for this fet to handle?

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It says 5amp continuous current, but 700watt spikes seem high :)

I think I could increase the switching frequency and/or increase the inductance to reduce these spikes but the low voltage operation may be effected by this. Right now at 30Volts input, the circuit is only outputting about 8.5volts, I would like to increase this to at least 10V if possible to do without increasing the heat dissipation too much at 450V.

Are these 3 coil ferrite transformers an off the shelf part? I would rather modify the circuit to work with an off the shelf transformer than to get one custom wound to my trial and error "specifications" :)

cheers, Jamie

Reply to
Jamie Morken

Tom Bruhns snipped-for-privacy@msn.com posted to sci.electronics.design:

It happens to everyone who is honest with themselves. I am pretty darn close to making it to membership.

Reply to
JosephKK

Ouch, 1.5 watts dissipation in a part in a supply that's only supplying a watt?? Something needs to change there, I think. I don't have time to run the simulation and analyze it, but you should probably look more at the max current in the MOSFET. If it stays within the rating, you should be OK.

You're asking for a huge input voltage range. Why so large? Most commercial switching modules operate over a 2:1 range or a little more, and they are designed by people with quite a bit of experience. It's not that it can't be done, but you need to make tradeoffs that will almost certainly cause you to pick parts that will lower the efficiency: you have to handle higher currents at low input voltage, and still handle high voltages. Back when you were suggesting 100 to

400 volts, that seemed at least feasible, but now you're talking about going to 30 volts on the low end?

You MAY be able to find an off-the-shelf transformer, but especially since you want such a high ratio of max:min input voltages, the design of the transformer is going to be a lot touchier than if you are doing a narrow range. I've pretty much always just wound my own for protos, and/or contracted with a transformer manufacturer. I notice CoilCraft has a few stock switcher transformers, but I have no idea if any will be even close to what you need.

I'm going to need to bow out of this for a while. I hope you're able to get to where you want with this, but I suspect you'll need to make some compromises to get something workable. I still think for such low power that one of the chips designed for use in compact off-line switchers, with built-in mosfet switch, will be a good solution. I think they've thought through many of the things you've been struggling with. EVEN IF you have to put a circuit in front of it to boost low input voltages a bit, it will likely be a simpler and more efficient solution.

Cheers, Tom

Reply to
Tom Bruhns

;-) Yes, and if you can't laugh at yourself for having done it, the membership is not worth a lot.

Reply to
Tom Bruhns

That was with a 3ohm load on the 12V output, with a 100ohm load the fet dissipation is only 33mW at 30V input, and 486mW at 450Vin.

Thanks for all the help, just need to source the transformer now!

cheers, Jamie

Reply to
Jamie Morken

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