Electronics > Projects, Designs, and Technical Stuff

ATX Flyback Transformer

<< < (10/14) > >>

T3sl4co1l:
Simulations can be misleading like that -- the data comes in thousands of times slower than real time, so it can seem to be working, but if you've not checked something (like SS, Ith or Vcc) in that time, it can keep on doing things that you maybe weren't expecting.

The power supply stopping and restarting isn't unintended behavior, much of the time; it's advertised as "hiccup" overload response.  What does seem to be unintended, is you've got nominal supply and load, and it should be running fine.  What's up with that?

Datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/3873fb.pdf

The controller isn't magic, it needs power to run.  It's got that 150k pullup from, whatever it is, 160V bus? (Can't seem to see the AC input section anymore, that attachment got mangled. :palm: )  That gives about 1mA.  Which will charge the 10uF bypass to startup voltage (p.3, UVLO rising: 8.4V) in about 84ms.

Assuming you've set it to zero; you may want to speed up simulations by setting initial conditions on the DC link and VCC caps, and remove AC ripple for now by using a DC source, not AC + rectifier.  Fewer things to mix up on the sim, eh?

And the controller needs 45-400uA to run, depending on condition; in and of itself.  That should be plenty, right, you've got ~1mA?

Well hold on there, that's with no load attached.  Useless right?  Well, it wouldn't be very useful if they characterized it with one particular load, so they don't do any.  Page 9 is your first hint on this I think, "since RSTART cannot supply enough current to operate the external MOSFET..."

How much current does it need?  Depends on the gate.  The STP18NM60N claims 35nC total gate charge, at 10V.  Supply max is 9.5V, I guess that's close enough.  (The min/max will differ quite a bit anyway, and it's rather irritating they don't give limits.  It's probably more like 50nC max.)  To fully charge and discharge a capacitance, to some charge Q, at frequency F (~200kHz), requires current I = Q F.  Or 10mA.

Which is 10 times the startup current, so we'd expect it starts up, runs for about 4ms, then stops, waits 40ms, and so on.  (Your simulation seems to be doing much better than this, which is interesting!)

Referring to page 1, they have a flyback configuration, with a diode D1 from output to VCC.  Since the output in that example is between 5-9V, they can use a single diode and that's that.

If your output voltage is higher, you might consider using a voltage divider, sized to sink >>10mA, and giving a Thevenin voltage of say 8-9V, and then connect a diode from the divider to VCC.  (You can't use just a divider, because that would load the startup resistor and nothing happens.)  The voltage divider is poor efficiency, but expedient.  (Better efficiency can be obtained say with an LDO, one that has low shutdown current and tolerates backwards voltages.  Works the same way, sourcing current from the output to VCC once the output starts up, while not drawing the excess current a divider does.)

On page 9, they show an auxiliary winding providing VCC; this is out in the present case (your transformer doesn't have one), but quite useful when available.


As for Ith: that's the error or compensation node.  When output voltage (sensed by Vfb) differs from the setpoint, a current proportional to that error is driven in/out the pin.  This charges the capacitor, which integrates the current into a voltage; for small and brief disturbances, not much happens, but for large disturbances over long times, V(Ith) changes dramatically, and consequently the setpoint current (this is a peak current mode controller, so this is the peak switch current).  Thus responding to load.  Simple enough, right.  Often a resistor is placed in series (e.g. page 1) to afford better compensation.  In general, an R+C is much better, and you can simply adjust values up and down until you zero in on a good step response.


You may want a resistor in series with SW, to introduce slope compensation and filtering.  This is discussed on page 10.

Vfb divider is also shown on page 10.


The L2-D8 strategy is an equivalent winding-flux-reset network, obviously only an option if you have the extra winding of course.  Since L2 and L9 are 1:1, it behaves the same as a 2-switch converter does.  The RCD clamp shown earlier was drawn correctly, also.


Hehheheh... also... page 18, can you spot the error?  :-DD Ehhh...yikes!

Tim

engrguy42:
Tim,
Thanks much. I really appreciate your getting into this.

In response to your wonderfully detailed response...HUH?  :D  Mostly over my head.

Though I did extract something that might be pertinent. I plotted the Vcc voltage across the 10u cap, and superimposed my Vout voltage on the load, and it seems that the shutdown is occurring when/because Vcc is dropping to 4V, and starting/re-starting occurs when it gets up near 9V.

So I guess I'll disconnect the AC/DC bridge and insert a DC source of 160V and see if that fixes it. Which begs the question:

Why so much wigglies in the DC supply?

engrguy42:
BTW, here's a zoom-in of the input circuit, including a trace of the Vin feeding the controller circuit.

T3sl4co1l:
I have covered everything.  Take your time, read it by paragraph if you have to.  I kept the paragraphs short too, which should help.

Cheers!

Tim

engrguy42:
Anyone else notice the image attachments are all screwed up? Sometimes it shows the correct thumbnail, but when you expand it's a different image.  :wtf:  I think the server was having issues yesterday with uploading too.

Anyway, I replaced the whole bridge input section with a rock solid 160V DC supply to power this, and the shutdown cycles are still there.

I don't get it.

Navigation

[0] Message Index

[#] Next page

[*] Previous page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod