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flyback converter driving

J

Jamie Morken

Jan 1, 1970
0
Hi,

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? :)

cheers,
Jamie
 
T

Tom Bruhns

Jan 1, 1970
0
Hi,

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? :)

cheers,
Jamie

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
 
J

Jamie Morken

Jan 1, 1970
0
Tom said:
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.

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
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.

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:

http://www.nekrom.com/rocketresearch/new/flyback/flyback-stepdown.jpg

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
 
T

Tom Bruhns

Jan 1, 1970
0
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:

http://www.nekrom.com/rocketresearch/new/flyback/flyback-stepdown.jpg

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
 
J

Jamie Morken

Jan 1, 1970
0
Tom said:
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.

I found the LM5020 at national.com:
http://www.national.com/pf/LM/LM5020.html

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
 
J

Jamie Morken

Jan 1, 1970
0
Tom said:
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.

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
 
J

Jamie Morken

Jan 1, 1970
0
Tom said:
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.

Here is the current circuit:
http://www.nekrom.com/rocketresearch/new/flyback/flyback2.jpg

http://www.nekrom.com/rocketresearch/new/flyback/flyback2.asc
(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
 
J

Jamie Morken

Jan 1, 1970
0
Jamie said:
Here is the current circuit:
http://www.nekrom.com/rocketresearch/new/flyback/flyback2.jpg

http://www.nekrom.com/rocketresearch/new/flyback/flyback2.asc
(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?

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
 
L

legg

Jan 1, 1970
0
Jamie Morken wrote:

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.

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
 
T

Tom Bruhns

Jan 1, 1970
0
I found the LM5020 at national.com:http://www.national.com/pf/LM/LM5020.html

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
Actually, I was thinking more like the ST Electronics VIPer22A,
http://www.st.com/stonline/products/literature/ds/12050.pdf
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
 
T

Tom Bruhns

Jan 1, 1970
0
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
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
 
T

Tom Bruhns

Jan 1, 1970
0
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
 
J

Jamie Morken

Jan 1, 1970
0
Hi,
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.

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?

http://www.nekrom.com/rocketresearch/new/flyback/flyback3.jpg

http://www.nekrom.com/rocketresearch/new/flyback/flyback3.zip
(ltspice asc file and sub models for UC3842 and fet)
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.

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
 
T

Tom Bruhns

Jan 1, 1970
0
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?

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.
http://www.nekrom.com/rocketresearch/new/flyback/flyback3.jpg

http://www.nekrom.com/rocketresearch/new/flyback/flyback3.zip
(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.

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
 
J

Jamie Morken

Jan 1, 1970
0
Tom said:
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.

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. :)

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.

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
 
J

Jamie Morken

Jan 1, 1970
0
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. :)

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
 
T

Tom Bruhns

Jan 1, 1970
0
Hi,

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

http://www.nekrom.com/rocketresearch/new/flyback/flyback3-switch.jpg

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
 
M

Mach One

Jan 1, 1970
0
Hi Tom,


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 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.
 
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