flux balancing capacitor in SMPS ?

[Adam S.]

I'm building a 100W half bridge power supply, and is my my first time of
off line switching supply. To get some ideas on practical design, I've
being looking at PC supplies because they use push pull topology. My
question is why do these supplies have a 1 to 2 uF capacitor in series with
the primary winding. I figured its for flux balancing of the transformer,
however shouldn't the two main input electrolytics correct for transformer
flux imbalance since the center voltage of the two main capacitors can
shift up or down, thus nulling DC primary winding current ?

The rectified mains voltage is typically applied across 2 x 200V capacitors
in series, so that's 265V*1.4142 = 360V, or 360/2 = 180V per capacitor.
That the center voltage is allowed to deviate up to +-20V.

 

[Genome]

I'm building a 100W half bridge power supply, and is my my first time of
off line switching supply. To get some ideas on practical design, I've
being looking at PC supplies because they use push pull topology. My
question is why do these supplies have a 1 to 2 uF capacitor in series with
the primary winding. I figured its for flux balancing of the transformer,
however shouldn't the two main input electrolytics correct for transformer
flux imbalance since the center voltage of the two main capacitors can
shift up or down, thus nulling DC primary winding current ?

The rectified mains voltage is typically applied across 2 x 200V capacitors
in series, so that's 265V*1.4142 = 360V, or 360/2 = 180V per capacitor.
That the center voltage is allowed to deviate up to +-20V.

It's down to how fast you can restore the volt-second imbalance applied to
your transformer's magnetising inductance. It is the current due to this
volt second imbalance which charges the C. The smaller C requires less
energy to charge to the required voltage.

There's probably some energy balance equation knocking around in there that
would allow you to determine an 'optimum' value for the small C. Something
like work out the magnetising inductance. Work out what current it would
take to saturate it. Work out what the equivalent energy it is. Work out
what your worst case volt second imbalance will be. Work out what DC offset
is required to counter it. Work out what value of C has that energy stored
in it for that value of voltage. Pick the next smallest one.

Otherwise you suck it and C.

DNA

Adam Seychelle?

 

[Kevin Aylward]

It's down to how fast you can restore the volt-second imbalance
applied to your transformer's magnetising inductance. It is the
current due to this volt second imbalance which charges the C. The
smaller C requires less energy to charge to the required voltage.

Its the ac ps ripple voltage applied to the transformer that is usually
the main problem

There's probably some energy balance equation knocking around in
there that would allow you to determine an 'optimum' value for the
small C. Something like work out the magnetising inductance. Work out
what current it would take to saturate it.

Not really. Assuming this is a conventional SMPS, the current is
irrelevant. A transformer is *voltage* driven/operated. The equation
ET=nB.A is the one to use. What ever current flows is all in the wash.

Work out what the

equivalent energy it is.

Not nessesary.

Work out what your worst case volt second
imbalance will be.

Calculate ET due to the ripple voltage.

Work out what DC offset is required to counter it.
Work out what value of C has that energy stored in it for that value
of voltage. Pick the next smallest one.

It has nothing directly to do with energy. The capacitor and the
reflected load resistance form a simply H.P. filter. This filter reduces
the amplitude of ac ripple voltage at the transformer input. The
capacitor can then be sized based on what lf voltage is a reasonable
small proportion of ET=nB.A. It is true that off load the l.p filter is
less effective, but in this case the ripple voltage tends to zero.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Kevin Aylward]

I'm building a 100W half bridge power supply, and is my my first time
of off line switching supply. To get some ideas on practical design,
I've being looking at PC supplies because they use push pull
topology. My question is why do these supplies have a 1 to 2 uF
capacitor in series with the primary winding. I figured its for flux
balancing of the transformer, however shouldn't the two main input
electrolytics correct for transformer flux imbalance since the center
voltage of the two main capacitors can shift up or down, thus nulling
DC primary winding current ?

The rectified mains voltage is typically applied across 2 x 200V
capacitors in series, so that's 265V*1.4142 = 360V, or 360/2 = 180V
per capacitor. That the center voltage is allowed to deviate up to
+-20V.

There is a low frequency ripple voltage (~sawtooth/triangular) at the
centre tap. This low frequency ripple is what saturates the transformer.
The cap is sized large enough to pass the h.f. switching frequency, but
small enough to block the l.f. p.s. ripple.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[R.Legg]

Its the ac ps ripple voltage applied to the transformer that is usually
the main problem

It doesn't matter where the problem originates - vt imbalance is just
as likely to be generated the control and drive circuit, as the
topology is usually voltage-mode (I believe there is a school prepared
to demonstrate that current mode is physically impossible).

The practical design rule is to use as small a C value as can be
tolerated - 5 or 10% ripple at full load is a common rule of thumb -
keeping in mind that the part must be rated for the current ( C x
dV/dT rating and thermal rise due to ESR x Irms^2), that it costs
money and that it takes up space. You also want to use parts that are
easily obtained.

The LF capacitors cannot be counted on to counter flux imbalance,
particularly if they are used as part of input rectifier voltage
doubler. In the later case their voltage is independent of converter
operation during significant parts of the input voltage's phase.

The HF coupling capacitor can be returned to either rail, not just the
one center-tap point - if correctly biased to avoid start-up and
transient misbehavior. Power circuit EMI and capacitor current stress
can be reduced if the part is split into two between the rails, and
their 'center-tap' used as the transformer return point. The voltage
rating for the parts in the independent configuration is always equal
to the supply - LF centertap configuration requires only half this
voltage rating, so costs have to be weighed for the specific choices
made.

In some topologies, this position functions in a series-resonant
circuit, in which case other considerations must obviously addressed.

RL

 

[CBarn24050]

Hi, you need about 0.47uf for a 100kHz design at 100w.

 

[Genome]

Its the ac ps ripple voltage applied to the transformer that is usually
the main problem


Not really. Assuming this is a conventional SMPS, the current is
irrelevant. A transformer is *voltage* driven/operated. The equation
ET=nB.A is the one to use. What ever current flows is all in the wash.

Work out what the

Not nessesary.


Calculate ET due to the ripple voltage.


It has nothing directly to do with energy. The capacitor and the
reflected load resistance form a simply H.P. filter. This filter reduces
the amplitude of ac ripple voltage at the transformer input. The
capacitor can then be sized based on what lf voltage is a reasonable
small proportion of ET=nB.A. It is true that off load the l.p filter is
less effective, but in this case the ripple voltage tends to zero.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

I quote all you say. It means you are an ignorofuck.

DNA

 

[Kevin Aylward]

I quote all you say. It means you are an ignorofuck.

I see you can make well reasoned arguments.


Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Kevin Aylward]

It doesn't matter where the problem originates - vt imbalance is just
as likely to be generated the control and drive circuit,

Yes. But my point here was that even in the ideal case of perfect
balance, the ps ripple *will* be still there, usually at several volts.
I consider this a "first" order calculation, with asymmetry being a
"second" order calculation. Of course, all factors need to be accounted
for.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.