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flyback converter driving

Discussion in 'Electronic Design' started by Jamie Morken, Oct 19, 2007.

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  1. Jamie Morken

    Jamie Morken Guest


    I have a flyback converter that takes an input of 100VDC to 400VDC and
    outputs ~15V at 100mA max. I am using ltspice to simulate it and
    it works fine, but I am not sure now how to drive the fet as the
    only available power supply is the 100V-400V input signal. I would be
    ok with using a fixed dutycycle driving the fet, but would prefer to use
    an IC like the MB3800 to drive the fet. Any ideas on how to get this
    flyback started so that it can power itself? :)

  2. Tom Bruhns

    Tom Bruhns Guest

    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.

  3. Jamie Morken

    Jamie Morken Guest

    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? :)

  4. Tom Bruhns

    Tom Bruhns Guest

    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

    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!


  5. Jamie Morken

    Jamie Morken Guest

    I found the LM5020 at

    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

  6. Jamie Morken

    Jamie Morken Guest

  7. Jamie Morken

    Jamie Morken Guest

    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..

  8. Jamie Morken

    Jamie Morken Guest

    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?

  9. Jamie Morken

    Jamie Morken Guest

    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.

  10. legg

    legg Guest

    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.

  11. Tom Bruhns

    Tom Bruhns Guest

    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.

  12. Tom Bruhns

    Tom Bruhns Guest

    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.

  13. Tom Bruhns

    Tom Bruhns Guest

    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.

  14. Jamie Morken

    Jamie Morken Guest

    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.

  15. Tom Bruhns

    Tom Bruhns Guest

    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.

  16. Jamie Morken

    Jamie Morken Guest

    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

  17. Jamie Morken

    Jamie Morken Guest

    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.

  18. Jamie Morken

    Jamie Morken Guest

  19. Tom Bruhns

    Tom Bruhns Guest

    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 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.

  20. Mach One

    Mach One Guest

    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

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