flyback converter driving

A typical way is to use a circuit with a low-voltage lockout; in lockout, it draws only very low current. A capacitance on its power supply pin is charged from the DC input (100--400V) through a large resistance so power dissipation in that resistance is negligible. When the converter power input reaches a high enough voltage, it turns on, and the cap holds the voltage up long enough to get things started. Once started, the converter, through another winding on the core, provides power to run the circuit. Under some circumstances, that winding can also be used to sense the output voltage and no other feedback from the output side is necessary to maintain acceptable regulation. Tight control of the output may require other techniques.

Cheers, Tom

I guess this will work as long as the IC isn't drawing power before it reaches it startup threshold voltage?

Once started, the converter, through another winding on the

Are these types of small ferrite transformers usually off the shelf parts? I only need a tiny one for this circuit as my output power is only about 1.5watts maximum.

I posted a pic of the ciruit so far:

I don't know if this is really called a flyback as it is stepping down the voltage.

I am looking for a fet driver with voltage feedback dutycycle control still for this circuit, most of the ones I have seen have the fet integrated into the IC which is no good as I am dealing with up to

800Volts on the fet drain with a max of 400V input, so it is better to have an external fet I think. Any linear/ltspice parts that would work for this? :)

cheers, Jamie

As drawn, it is, I believe, a forward converter, not a flyback. You'd need to reverse the polarity of one of the windings for it to be flyback. The 'acid test' is to note when the output capacitor is charging. If it is charging when the FET on the primary side is conducting, then it's a forward converter; if it's charging when the FET turns off, it's a flyback. The fact that it steps the voltage down doesn't matter.

Since you only need low power, I'd suggest that you look into a switching regulator part that's designed to do as much of the whole job as you reasonably can. A lot of small "wall wart" power supplies these days will supply a few watts happily, and do it from AC inputs from under 100V to 250V, which translates to perhaps 130VDC to about

350VDC after rectification. I know there are ICs that implement practically everything, including directly driving the transformer primary. The applications notes will show you just how to use them. You can probably take a reference design and make very minor modifications to get the output voltage you want. The 400V input may be a little high for them, but you may well find one that can handle that. Seems worth a look. Maybe Linear Technology, but I'd look at NXP, ST and some of the others that are more oriented toward consumer electronics.

You can find some off-the-shelf transformers for that sort of application, but I suppose you'll have trouble finding one that gives you the output voltage you want. The good news is that the number of turns isn't huge since the switching frequency is high. Much better than winding your own 50Hz/60Hz transformers!

Cheers, Tom

Cheers, Tom

I found the LM5020 at national.com:

Is this the type of IC that would be best to use? I will need to power it with the high resistance series resistor and storage cap from the

400VDC rail as you suggested I think. Now I just need to find its spice model..

cheers, Jamie

Hi,

The circuit works with the LT1619 PWM controller:

(ltspice simulation)

Now I just need to add the startup circuit so the LT1619 can be powered from the high voltage directly instead of the 20V in the simulator. Any suggestions on the best way to do this?

cheers, Jamie

I added the big series resistor and capacitor for this startup function, but I would have to use a 10K resistor or so to get it working. I think the problem is that the IC Vin pin is drawing over 200uA, even during shutdown mode, when the datasheet says its should be drawing only 15uA in shutdown mode.. so I think it could just be a simulator error but it is hard to know. I tied the shutdown pin to ground and the IC is still drawing 200uA into the Vin pin, I checked with a 1 ohm series resistor on it. Any ideas on this, if I should just ignore the 200uA and assume it is really only drawing 15uA? Dangerous assumption maybe..

cheers, Jamie

Here is the current circuit:

(ltspice file)

That circuit currently works and outputs 12VDC with R7 being 10k, but when I switch it to 100k (to reduce the power dissipation) the shutdown pin starts oscillating so that the device is being enabled/disabled, and this doesn't allow the output voltage to ramp up. Any ideas on why this shutdown pin is oscillating?

cheers, Jamie

I think I understand what is wrong, the circuit starts up when shutdown pin goes over the startup threshold, and then the chip starts drawing a lot more current, which pulls the supply voltage down far enough that the shutdown pin gets turned off before the circuit can run long enough to charge up the supply.. so the problem is to find a way to keep the shutdown pin above its threshold, the current zener/resistor doesn't work. Thanks for any help.

cheers, Jamie

The LT1619, with a 1V8 minimum start-up voltage, is not intended for off-line applications. You are correct in observing that all you are doing is modulating operation through the SS pin. In your schematic\\simulation, adjusting the D5 zener voltage or resistors in this string regulate the chip supply, regardless of the presence of an output voltage.

The Lin Tech version of the commodity UC3842 current mode controller is the LT1246. You'll have better luck sticking this in your simulations, though the compensation pin voltage is not accurate in the currently available model.

Such a low power flyback may not justify a discrete controller or switch. You could probably do it with a Power Integrations 8pin dip or smaller part, by itself.

I'm thinking that simulation at this power level may be waste of time, if not actually misleading.

RL

Actually, I was thinking more like the ST Electronics VIPer22A,

It has a 400V max rating; at the low power you'll be running and because of the simplicity of the design using such a part, it could be worth adding a simple circuit to limit the maximum input rail to, say,

350 volts to give a little headroom. There well may be other such parts from other vendors, or even others from ST; that just happens to be one I know of. Do a search for things like "off line flyback ..." where ... might be "regulator" or "driver" or "controller".

On the other hand, since limiting the max input to a lower voltage would require some parts, it may be just as efficient to use a flyback controller IC driving a high voltage mosfet. LOTS of manufacturers make those, including Linear Technology (which means LTSpice will have models already built in). And if you want to use a National part, you can just use their "WebBench" simulation tool.

An alternative way of doing all this that may ultimately be simpler: International Rectifier makes a self-oscillating half-bridge driver that I'm pretty sure supports voltages to over 400. You could use it in a dog-simple circuit to drive a transformer to get to a much lower voltage, say 0.12 times the input DC voltage (i.e., 12 to 48 volts out) and use something like a National SimpleSwitcher buck regulator to get from that down to your desired 10V output. It's kind of a wide input voltage range, but seems like it should be possible.

I'm gathering that simplicity is what you're after, but I don't think you really ever did say much about your goals. If you're uncomfortable working with stuff like this, I'd highly recommend getting a local guru that can help you a bit in person. I have the feeling that you're on a steep learning curve, and at times like that, you're likely going to miss some fairly important point. 400V is nothing to take lightly.

Cheers, Tom

Bigger capacitance at C6? Another way to do it would be to tie a resistor from /SHDWN to the output voltage so that as soon as the output voltage starts to rise, there is a component of input to /SHDWN that causes it to rise.

Cheers, Tom

By adding a 1 meg from the output to the /SHDN, I was able to raise R7 to 470k and get rid of C6; just the 10uF C5 there now.

But as I've posted in a different followup, you can probably find an IC that integrates more, and especially has the output FET included. RL suggested another source of the same sort of IC. Realize that the "wall wart" power supplies that operate on US and Euro line voltages are almost all switchers, built to a very low price point. They do practically what you want, and typically put out 5-10 watts at full rated load.

Cheers, Tom

Hi,

Thanks that works now! The limit of this circuit is the draw of the IC during startup, which requires a long startup time if a large resistor is used so I would rather use the UC3842 as RL suggested to use instead as it doesn't draw current till a voltage threshold is reached.

I found an example circuit for it that is pretty well what I would like, even has isolation. The only problem is the feedback resistors draw to much power so the circuit won't start up if I increase the value of R8. Is there a way to wire this up to prevent the feedback circuit from drawing power during startup?

(ltspice asc file and sub models for UC3842 and fet)

It is nice to have the fet outside the IC as it lets you have more flexibility for high voltages, I will need at least an 800V rated fet for this circuit eventually, 2x the input voltage because of the transformer primary voltage doubling on the fets drain.

cheers, Jamie

I suppose you can connect the feedback to the output of the aux winding and rectifier directly, but put a diode between C6 and U1's Vcc pin. Be sure to add a bypass to U1's Vcc pin, too. R8 goes to the Vcc pin. That way, that pin can pull up before anything gets applied to the feedback divider.

OK, there's a little point you need to grok about flyback supplies...the primary voltage does not always fly back to twice the supply voltage. The DC output voltage plus the output rectifier drop equals the peak voltage the secondary flys back to. The primary flys back to that voltage times the turns ratio. So for example, with 70uH in the secondary and 1mH in the primary, the turns ratio is 1:0.265 (the square root of the inductance ratio). If the output is 12VDC, and the diode drops 0.5 volts when conducting, the primary will be at about 47 volts. I would expect you would want a bit higher turns ratio than that, but you get the picture...the primary flyback is the same, independent of the primary supply voltage. You DO need to allow for some uncoupled inductance; the transformer won't be perfect, and you're liable to get some spikes possibly considerably above the nominal flyback voltage calculated as above. It's not unusual to add a bit of damping across the primary to mitigate that sort of thing. But you certainly shouldn't need to accommodate 800 volts at the FET drain, even with a 400V supply.

Cheers, Tom

Thanks that works now, I noticed that the Vcc start up current draw of the UC3842 is about 500uA max which is more than the LT1619, which shows about 266uA in the simulator. But the UC3842 pulls less current until it gets closer to its turn on threshold while the LT1619 pulls the same current always (probably a spice error I Am guessing)

Because of the reduced start up current I think the LT1619 is a better IC to use for this as the startup resistor will be rated at a lower wattage as it needs to supply less current during startup and/or it will give a faster startup time. Startup time is already over 500ms, as both these chips draw too much current to start up quickly without using a high wattage startup resistor. I guess I Could put a switch in series with the startup resistor to and then it would only be on for 50ms or so then the current would be cut by the switch before it smokes. :)

Thanks for the info! I will probably protect the fet drain to source with a TVS just in case there is a transient spike that the fet can't handle.

cheers, Jamie

In the LT1619 datasheet the shutdown supply current is a max of 40uA, up to a Vin of 18V, so the ltspice model of the LT1619 is incomplete I guess. I think the LT1619 is probably fine to use with a bigger startup resistor than the simulator allows. The only feature missing from the LT1619 that the UC3842 has is variable frequency operation by setting an external R and C which is nice to have.

cheers, Jamie

Hi,

I made an attempt to add a switch to turn off the current flow through R8 once the flyback starts up.

How can I turn M1 off once the flyback has started up, or should I use another type of switch for this?

cheers, Jamie

Geez, this is getting awfully complicated for a circuit that only has to deliver 1 watt or so. ;-)

Is M1 a depletion-mode mosfet? If not, you won't ever get much current through it...and it looks like it's turned upside down: a P channel device would normally have its source and substrate tied to a more positive terminal than the drain...and there's more than that wrong there, I'm afraid.

Actually, what I see in the data sheet for the LT1615 is that the Vcc current is 1uA max with the /shutdown at 0; the /shutdown pin takes it out of shutdown by by the time it reaches 1 volt, and the current into /shutdown at that point appears to be less than 10uA. Sounds like the simulation is wrong. But at some point you need to get some parts and play with it on the bench; that's how you find out how it really works.

Back to the switch to shut off the bootstrap bias current: if the LT1615 really is so low current, I'd say it's not worth worrying about. If you need even 20uA, at 100V that's 4.7 megohms. At 400 volts, it becomes 80uA, and about 32 milliwatts dissipation. If you do need to switch it off, I'd be inclined to use an NPN transistor rated for 500V or so. (Seems like that starts to get big...I know of some SOT23 parts rated to 350 or 400V, but I'm not sure you'll get 500 in a small package). Then I'd tie the base to a zener to ground, and a high-value resistor to the +supply input. The emitter supplies current to wherever the resistor did. The collector connects to the

+supply input, either directly or through a resistor that will limit the maximum current--i.e. the same value or a bit lower than you'd have used without the switch. If the zener is a lower voltage than the aux supply (C5 in that latest jpg schematic), the transistor emitter will be pulled up by that supply and turn the transistor off. You could do the same with a mosfet, but the gate threshold voltage is not as sharply defined as the base-emitter voltage of a bipolar transistor, so you give up margins elsewhere if you use a mosfet. But in that suggestion, beware of the zener voltage at very low currents...it may be quite a bit less than the rated voltage of the part. 10V zeners have a very sharp knee, but 5.6 volt zeners have a very "soft" knee.

Are you going to be building a lot of these, or what? I'm trying to get a sense for why you're going through all this; I'm sure it's a good learning experience for you, but I'm not convinced it's the most practical way to make a little low-power supply.

Cheers, Tom

Hi Tom,

I think the polarity shows what industry normally calls a flyback. When M2 is on, the dot of L1 is positivr. The corrresponding dot on L2 should be positive. Thus, D1 anode should be negative, holding D1 off.

A forward converter would usually have an inductor and freewheel rectifier in secondary.