Forums

transformer design for smps

Started by panfilero March 8, 2012
I have a couple questions, one is about identifying smps topology and
the 2nd is about the 3rd winding in an example I found. I need to
design a smps that can take

Vin: 70V to 105V and output 5V at 1A

I've chosen to use the mic9130 http://www.micrel.com/_PDF/mic9130.pdf

On the front page of the datasheet there is a "Typical Application"
schematic, which is not to far off from what I need to do, it's
Vin: 36V to 72V
Vout: 3.3V @ 4A

So, I think I'll copy that design, but now I have to design a
transformer... from the reading I've done about this, looks like one
of the first things I need to know is the topology of my smps.  That's
my question, can anyone tell me the topology of this smps example?
Looks like an isolated synchronous buck converter... or maybe a
forward converter?  Also I don't see how the third winding is
supplying the stead voltage to Vcc, according to the datasheet Vcc
operates between 9V -18V (pg 3) confusingly on pg 8 it says Vcc is
typically 8.5V....

much thanks!
On Thu, 08 Mar 2012 11:01:42 -0800, panfilero wrote:

> I have a couple questions, one is about identifying smps topology and > the 2nd is about the 3rd winding in an example I found. I need to design > a smps that can take > > Vin: 70V to 105V and output 5V at 1A > > I've chosen to use the mic9130 http://www.micrel.com/_PDF/mic9130.pdf > > On the front page of the datasheet there is a "Typical Application" > schematic, which is not to far off from what I need to do, it's Vin: 36V > to 72V > Vout: 3.3V @ 4A > > So, I think I'll copy that design, but now I have to design a > transformer... from the reading I've done about this, looks like one of > the first things I need to know is the topology of my smps. That's my > question, can anyone tell me the topology of this smps example? Looks > like an isolated synchronous buck converter... or maybe a forward > converter? Also I don't see how the third winding is supplying the > stead voltage to Vcc, according to the datasheet Vcc operates between 9V > -18V (pg 3) confusingly on pg 8 it says Vcc is typically 8.5V.... > > much thanks!
An isolated synchronous buck converter is a kind of forward converter, and that's what this is. That auxiliary winding on the transformer supplies a DC voltage that's roughly a diode drop and some resistance loss below 1/4 of the input voltage. Every time the main transistor turns on, the dotted end of the auxiliary winding jumps up to (turns ratio) * (input voltage), more or less. The 200 ohm resistor insures no huge current spikes, the diode rectifies, and the result is an unregulated power supply to the chip. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
On Mar 8, 1:26=A0pm, Tim Wescott <t...@seemywebsite.please> wrote:
> On Thu, 08 Mar 2012 11:01:42 -0800, panfilero wrote: > > I have a couple questions, one is about identifying smps topology and > > the 2nd is about the 3rd winding in an example I found. I need to desig=
n
> > a smps that can take > > > Vin: 70V to 105V and output 5V at 1A > > > I've chosen to use the mic9130http://www.micrel.com/_PDF/mic9130.pdf > > > On the front page of the datasheet there is a "Typical Application" > > schematic, which is not to far off from what I need to do, it's Vin: 36=
V
> > to 72V > > Vout: 3.3V @ 4A > > > So, I think I'll copy that design, but now I have to design a > > transformer... from the reading I've done about this, looks like one of > > the first things I need to know is the topology of my smps. =A0That's m=
y
> > question, can anyone tell me the topology of this smps example? Looks > > like an isolated synchronous buck converter... or maybe a forward > > converter? =A0Also I don't see how the third winding is supplying the > > stead voltage to Vcc, according to the datasheet Vcc operates between 9=
V
> > -18V (pg 3) confusingly on pg 8 it says Vcc is typically 8.5V.... > > > much thanks! > > An isolated synchronous buck converter is a kind of forward converter, > and that's what this is. > > That auxiliary winding on the transformer supplies a DC voltage that's > roughly a diode drop and some resistance loss below 1/4 of the input > voltage. =A0Every time the main transistor turns on, the dotted end of th=
e
> auxiliary winding jumps up to (turns ratio) * (input voltage), more or > less. =A0The 200 ohm resistor insures no huge current spikes, the diode > rectifies, and the result is an unregulated power supply to the chip. > > -- > Tim Wescott > Control system and signal processing consultingwww.wescottdesign.com
Now, a forward converter usually has some kind of reset winding right? But, would I be right to think that this circuit doesn't need the reset winding because there are bidirectional switches that allow for some current to keep flowing into the secondary, to keep that flux going and not violate anything? I think that's right... so I think I'll approach my transformer design then assuming I'm designing for a forward converter... Vo =3D nDVin much thanks, appreciate the help
wait, I just looked at this circuit again, and i thought that in order
to keep the flux flowing through the transformer core that current had
to be able to flow into one of the dots at all times, but when i
analyze this circuit for the switching transistor open, I believe this
would force Si4800DY to turn on and Si4884DY to turn off, which would
break the circuit for the bottom end of our second winding... so I
don't see current able to flow into any dot of this transformer during
the off cycle
On Fri, 09 Mar 2012 08:33:41 -0800, panfilero wrote:

> wait, I just looked at this circuit again, and i thought that in order > to keep the flux flowing through the transformer core that current had > to be able to flow into one of the dots at all times, but when i analyze > this circuit for the switching transistor open, I believe this would > force Si4800DY to turn on and Si4884DY to turn off, which would break > the circuit for the bottom end of our second winding... so I don't see > current able to flow into any dot of this transformer during the off > cycle
I'm not sure what you're thinking. The flux in the core is equal to the sum of the turns * current for all the windings -- so as far as the core is concerned, the flux is the same whether there's 1A in a 1-turn primary, or 0.1A in a 10-turn secondary. In a forward converter current comes out of the secondary (or secondaries) proportional to the current going into the primary, for the most part, except for magnetizing current. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
On Mar 9, 9:57=A0pm, Tim Wescott <t...@seemywebsite.please> wrote:
> On Fri, 09 Mar 2012 08:33:41 -0800, panfilero wrote: > > wait, I just looked at this circuit again, and i thought that in order > > to keep the flux flowing through the transformer core that current had > > to be able to flow into one of the dots at all times, but when i analyz=
e
> > this circuit for the switching transistor open, I believe this would > > force Si4800DY to turn on and Si4884DY to turn off, which would break > > the circuit for the bottom end of our second winding... so I don't see > > current able to flow into any dot of this transformer during the off > > cycle > > I'm not sure what you're thinking. =A0The flux in the core is equal to th=
e
> sum of the turns * current for all the windings -- so as far as the core > is concerned, the flux is the same whether there's 1A in a 1-turn > primary, or 0.1A in a 10-turn secondary. > > In a forward converter current comes out of the secondary (or > secondaries) proportional to the current going into the primary, for the > most part, except for magnetizing current. > > -- > Tim Wescott > Control system and signal processing consultingwww.wescottdesign.com
I was just pointing out that there needs to be a path for current to flow into the transformer in order to support the flux in the core at all times. When current flows into the dot of a winding it's able to support the flux. This schematic has an error, during the off portion of the PWM there is no way for current to enter a dot, no way for current to support the flux.
panfilero <panfilero@gmail.com> wrote:

>On Mar 9, 9:57=A0pm, Tim Wescott <t...@seemywebsite.please> wrote: >> On Fri, 09 Mar 2012 08:33:41 -0800, panfilero wrote: >> > wait, I just looked at this circuit again, and i thought that in order >> > to keep the flux flowing through the transformer core that current had >> > to be able to flow into one of the dots at all times, but when i analyz= >e >> > this circuit for the switching transistor open, I believe this would >> > force Si4800DY to turn on and Si4884DY to turn off, which would break >> > the circuit for the bottom end of our second winding... so I don't see >> > current able to flow into any dot of this transformer during the off >> > cycle >> >> I'm not sure what you're thinking. =A0The flux in the core is equal to th= >e >> sum of the turns * current for all the windings -- so as far as the core >> is concerned, the flux is the same whether there's 1A in a 1-turn >> primary, or 0.1A in a 10-turn secondary. >> >> In a forward converter current comes out of the secondary (or >> secondaries) proportional to the current going into the primary, for the >> most part, except for magnetizing current. >> >> -- >> Tim Wescott >> Control system and signal processing consultingwww.wescottdesign.com > >I was just pointing out that there needs to be a path for current to >flow into the transformer in order to support the flux in the core at >all times. When current flows into the dot of a winding it's able to >support the flux. This schematic has an error, during the off portion >of the PWM there is no way for current to enter a dot, no way for >current to support the flux.
Because it is a flyback. Don't model the 'transformer' like a transformer but like an inductor! The transformer has an air gap in which the energy builds up. When the primary FET opens, the inductor wants to keep the current flowing. The only way is through the secondary coil. -- Failure does not prove something is impossible, failure simply indicates you are not using the right tools... nico@nctdevpuntnl (punt=.) --------------------------------------------------------------
On Sat, 10 Mar 2012 11:34:37 GMT, the renowned nico@puntnl.niks (Nico
Coesel) wrote:

> >Because it is a flyback. Don't model the 'transformer' like a >transformer but like an inductor! The transformer has an air gap in >which the energy builds up. When the primary FET opens, the inductor >wants to keep the current flowing. The only way is through the >secondary coil.
It's not a flyback- the relative 'polarity' of the windings is wrong for that. It's a resonant-reset forward converter- the 'missing' current flows are through parasitic capacitances. Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote:

>On Sat, 10 Mar 2012 11:34:37 GMT, the renowned nico@puntnl.niks (Nico >Coesel) wrote: > >> >>Because it is a flyback. Don't model the 'transformer' like a >>transformer but like an inductor! The transformer has an air gap in >>which the energy builds up. When the primary FET opens, the inductor >>wants to keep the current flowing. The only way is through the >>secondary coil. > >It's not a flyback- the relative 'polarity' of the windings is wrong >for that. > >It's a resonant-reset forward converter- the 'missing' current flows >are through parasitic capacitances.
I think you're right. The extra output inductor should have rung a bell :-) IMHO its not the easiest design to start with. The OP needs 5W output power. A flyback is the easiest topology to achieve his goal. -- Failure does not prove something is impossible, failure simply indicates you are not using the right tools... nico@nctdevpuntnl (punt=.) --------------------------------------------------------------
On Sat, 10 Mar 2012 17:21:33 GMT, the renowned nico@puntnl.niks (Nico
Coesel) wrote:

>IMHO its not the easiest design to start with. The OP needs >5W output power. A flyback is the easiest topology to achieve his >goal.
If he wants to give it a try, here's a relevant article:- http://powerelectronics.com/mag/510PET22b.pdf Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com