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!

 

I believe you will find that this topic violates AAC policy.

 

You missed a guess. That circuit on page 1 looks like a flyback converter, which basically operates like a boost converter but has extra windings on the inductor core to reduce the output voltage and provide isolation. So the core has to be able to handle the peak primary current without saturating.

Pg 8 (10?) is taking about the startup voltage from the input power, not the operating voltage from the auxiliary winding.

According to the notation on the schematic on page one, the nominal operating voltage from the auxiliary winding is 12V.

 

I would have to disagree, a flyback converter which is based on a buck-boost topology, would imply Vo = Vin*D/(1-D).... which means that our inductor is switching between Vin and Vout (which means that the inductor is parallel with the input voltage during part of the cycle). This inductor is in series with Vin and Vout, which means Vo = DVin, which is a buck converter (no boosting)

sorry, yes page 10, sorry. On pg 10 it says Vcc is typically 8.5V, and the point of the auxillary winding is to feed Vcc... but, Vcc should be between 9V and 18V

 

I would have to disagree, a flyback converter which is based on a buck-boost topology, would imply Vo = Vin*D/(1-D).... which means that our inductor is switching between Vin and Vout (which means that the inductor is parallel with the input voltage during part of the cycle). This inductor is in series with Vin and Vout, which means Vo = DVin, which is a buck converter (no boosting)

sorry, yes page 10, sorry. On pg 10 it says Vcc is typically 8.5V, and the point of the auxillary winding is to feed Vcc... but, Vcc should be between 9V and 18V

Disagree if you like. I said it's similar to a boost configuration since that has the inductor connected to the input power and its output periodically connected to ground by the switching FET, the same as the flyback. A buck converter has the FET between input power and the inductor, which is not the case here.

I my post I stated that the voltage they were referring to is the startup voltage from the input startup circuit, not the voltage from the auxiliary winding, which is shown on the schematic to be 12V.

 

not a boost, not a flyback

you're right, a boost has one end of the inductor connected to positive voltage in and the other to ground through a switch. this circuit does not have an inductor periodically shorted to ground through a switch. And a flyback has no inductor, it uses the secondary coil of the transformer (a flyback is derived from a buck-boost not a boost). this circuit steps voltage down, not up, no boosting going on in here

for the Vcc, your right it's referring to taking voltage straight from the line, and not using the auxillary winding... I find this part confusing, I'll have to go over it some more

thanks

 

you're right, a boost has one end of the inductor connected to positive voltage in and the other to ground through a switch. this circuit does not have an inductor periodically shorted to ground through a switch. And a flyback has no inductor, it uses the secondary coil of the transformer (a flyback is derived from a buck-boost not a boost). this circuit steps voltage down, not up, no boosting going on in here

for the Vcc, your right it's referring to taking voltage straight from the line, and not using the auxillary winding... I find this part confusing, I'll have to go over it some more

You are confused about a flyback's mode of operation. A flyback does indeed have an inductor and cannot operate without one. It uses the primary inductance of the flyback transformer to store energy when the transistor is conducting and the current builds up. When the transistor suddenly turns off this inductive energy tries to keep the current flowing but, since the transistor blocks the current in the primary, it is shifted to the secondary by transformer action.

The flyback may not be performing a boost function, but the basic inductive energy storage and transfer mechanism is identical. The energy is stored during the ON time of the transistor and transferred to the output during the OFF time. The main difference is that the boost converter inductor has a single winding, whereas the flyback inductor has multiple windings, making it a transformer.

During startup you need a bootstrap circuit to get things going. This is provided by a small current directly from the input. After startup the chip current is then provided by the auxiliary winding.

 

This site explains the various smps modes pretty well: http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html

 

no, a flyback has a transformer not an inductor see this link

http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html

unless you're trying to say that the secondary is an inductor, but it's not, it's the secondary winding of a transformer, being used as a transformer winding, thr circuit we're discussing has a transformer and an inductor.

if you look at the circuit in the datasheet it looks like to me that when the switching transistor is closed the voltage across the inductor is

(Vin - Vo)*D*Ts (where Vin is the voltage across the secondary)

when the switching transformer is open the voltage across the inductor is

-Vo*(1-D)*Ts

because the average voltage across an inductor across one time period Ts (on and off switching) is equal to zero, then

0 = (Vin - Vo)*D*Ts + (-Vo)*(1-D)*Ts

solve this for Vo

Vo = D*Vin

insert your turns ratio and

Vo = (N2/N1)*D*Vin

because D is less than or equal to 1 this converter can only buck, it is not a flyback or any derivative of a buck-boost or boost (it can't boost)

maybe I'm wrong somewhere? But to me, this is obviously not a flyback or boost topology

 

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

 

This is a flyback circuit which uses the inductance of the transformer to store the energy during the transistor ON time which is transferred to the secondary during the transistor OFF time. All transformers have inductance. The only difference between a transformer and an inductor is the number of windings. Read this for more info.

Any circuit that applies voltage only in one direction to a transformer, such as this, must be operating as a flyback. To use a transformer in its normal mode, you must apply a equal value plus and minus voltage to the transformer terminals.

The transistors on the secondary act as rectifiers and the output inductor is a filter.

 

This is a flyback circuit which uses the inductance of the transformer to store the energy during the transistor ON time which is transferred to the secondary during the transistor OFF time. All transformers have inductance. The only difference between a transformer and an inductor is the number of windings. Read this for more info.

Any circuit that applies voltage only in one direction to a transformer, such as this, must be operating as a flyback. To use a transformer in its normal mode, you must apply a equal value plus and minus voltage to the transformer terminals.

The transistors on the secondary act as rectifiers and the output inductor is a filter.

"The only difference between a transformer and an inductor is the number of windings."

no, big difference between them is that there is more than one coil on a transformer, and your inducing a voltage across another coil.... how could you say the only difference is the number of windings?

as far as this being a flyback.... i guess you didn't go through my previous post, I explained why this is definitely not a flyback. Yes the link you sent gives good info on flybacks, this circuit can not produce a Vo greater than Vin, it can not operate as a flyback, plus you are completely ignoring the inductor on the output, and it is storing the energy and delivering it to the output when the switch is open.

Just looking at my previous post and the link you sent shows that this is not a flyback.

And this circuit has no way of letting current flow into the transformer to support the flux in the core when the switch is off

 

You don't believe it's a flyback, fine. Call it what you like. But if it looks like a flyback and operates like a flyback, then it's a flyback.

You say that "big difference between them is that there is more than one coil on a transformer, and your inducing a voltage across another coil..." But that is simply due to the number of windings. An inductor induces voltage in its single winding. A transformer induces voltage in its multiple windings. So the only difference is the number of windings, which was my original statement.

Of course it can produce an output greater (or smaller) then Vin. It just depends upon the turns ratio of the transformer.

I didn't ignore the inductor on the output. I previously stated it was a filter for the rectified pulses generated by the flyback output.

Good look in doing your design. You're going to need it.

 

do the analysis and see for yourself, it can't produce a Vo greater than Vin, just follow the steps in my post

 

There is no purpose in following an analysis when the outcome predicts an erroneous result. Since you now think you understand the circuit, you certainly don't need any further input from myself.

 

Why do you think your analysis would produce erroneous results?

If you can't analyze the circuit how can you be so confident you've identified the right topology?

Just take the voltage across the output inductor during both on and off portions of the switching cycle, then set that equal to zero and solve for Vo. Then you will understand the topology, if you can't do that then your just attempting to make an educated guess.

I've been discussing this question with others on another forum and it seems obvious to everyone in there that this is a forward converter, one person mistakenly referred to it as a flyback but was quickly corrected and the person recognized his mistake. Stubbornness is fine as long as you can back it up, maybe we're all looking at the circuit wrong and only you are right.

Here's the link
http://groups.google.com/group/sci.electronics.design/browse_thread/thread/58bfc1a644c4ef42#

Anyway debating the topology with you has made me take a close look at the circuit and helped me become more confident in its topology and I thank you for that, debating a circuit is helpful

cheers

 

Mea culpa. OK, I forgot to look at the transformer polarity marks. If they are correct as marked, then it cannot be a flyback. So I apologize for my previous remarks.

They are apparently using the transformer to reduce the voltage to the buck inductor to give a more reasonable duty-cycle.

But one point. You can get a voltage out greater than the input with that circuit. It just depends upon the transformer turns ratio.