Volt second imbalance in push pull converter.

[mitrarka]

In a push pull converter we often worry about volt second imbalance. But it seemed to me that when both the FETs are off the diodes in the transformer secondary demagnetizes the core. In that case ther is no flash x build up in the core. But there must be some catch with this conclusion as in all the resources explicit ways to mitigate this problem is discussed. Can anyone please explain this phenomenon?

 

[KlausST]

Hi,

I´d say the diodes work the other way round.

Demagnetizing means reducing the magnetic field.
But the filed is proportional to the current. (unless remanence)
So as long as there is current there also will be magnetic field.

The diodes (in detail it depends on the circuit) limit the voltage thus keep the current.
They prevent the current to collapse. Thus the prevent the field to collapse.
(also known as "slow decay")

If you want a fast collapse of the filed you need to allow high voltage.
So a zener is more useful.
Theoretically a "OPEN" circuit leads to fastest collapse, but also to very high voltage.
This high voltage may kill semiconductors (MOSFETs, BJTs, IGBTs..) .. thus you need to limit the voltage ... at least it does not hurt the semiconductors.

For more detailed discussion we urgently need to see your schematic.

Klaus

 

[mitrarka]

Hi. Thanks for the reply. I am attaching the schematic for more clarity.

Here when Q1 is on the dot end of Np1 is positive and current flows from drain to source in Q1. Now during switch off of Q1 the current through L1 forward biases diode D1 and D2. But magnetising current through transformer can also flow through D2 ( all the dot ends during this time becomes negative). Now if the magnetising current decays during this off period then volt second balance is automatically ensured and no question of flux Imbalance arises. If that is not the case, then this current must decay through the switch Q2 in the next switching instant before magnetising current can increase in the reverse direction. In that case Q2 must support reverse current initially. But in case of BJT that's not possible. And push pulls are widely implemented with BJT
So which of the above assumption is true ?

 

[KlausST]

Hi,

I recommend to add junction dots on your schematic. Else one has to guess which line is connected and which one not.

Good that you posted the schematic. I had a different idea (in pist#2).

Your post: (blue)
* Here when Q1 is ON: the dot end of Np1 is positive and current flows from drain to source in Q1.
(D1 is forward bypassed)
* Now during switch OFF of Q1: the current through L1 forward biases diode D1 and D2.
* Now if the magnetising current decays during this off period then volt second balance is automatically ensured and no question of flux Imbalance arises. (The current may decay, but most probably not to zero. In detail it depends on deat_time. What´s your dead time setup?)
* If that is not the case, then this current must decay through the switch Q2 in the next switching instant before magnetising current can increase in the reverse direction. (this is what I expect)
* In that case Q2 must support reverse current initially. But in case of BJT that's not possible. And push pulls are widely implemented with BJT (The voltage is limited because of the current through D1. But you should expect voltage peaks due to stray inductance. To prevent the BJT form beeing destroyed you may connect a diode-zenerdiode combination across each BJT´s EC. Mind: Zener voltage must be higher than 2 x DC bus voltage.).

Many push pulls are equipped with MOSFETs, too. Here the internal body diode prevents from overvoltage (of the other MOSFET). But still the leakage inductance problem exists.

More detail information is provided by semiconductor manufacturers in Application notes. They are for free and they are for you.

Klaus

 

[c_mitra]

In a push pull converter we often worry about volt second imbalance. But it seemed to me that when both the FETs are off the diodes in the transformer secondary demagnetizes the core. In that case ther is no flash x build up in the core. But there must be some catch with this conclusion as in all the resources explicit ways to mitigate this problem is discussed. Can anyone please explain this phenomenon?

It is more complicated than that!

The magnetization produced by the coils is perfectly determined by the current and the number of turns (NI). It has nothing to do with the volt-section you mention (but hey, wait a moment and see below).

And that will be correct for an air core transformer but not for a common magnetic core. Common magnetic cores have hysteresis. The figure is from https://electronics.stackexchange.c...-with-hysteresis-doesnt-result-in-a-highly-no

You see that when the current becomes zero, there is still some magnetism left. Sooner or later, the diode will get the current close to zero but there will be considerable magnetism still left. To remove that, you need a current in the reverse direction. If you cycle the current in the positive quadrant only, the curve will be different. The residual magnetism will tend to accumulate over time.

About the volt-second: To pass current in an inductor, you need a voltage source. But an inductor develops a back emf and resists external voltage. When you apply a voltage pulse to an inductor, the current builds up linearly. But the core saturates after some time and the back emf falls and the pulse must be stopped before the core saturates and the current must be limited. During this linear period, the volt-sec is equal to the max current in the coil and this is value you need to use for all practical purposes.

The important lesson here is that you need to exercise the current in both directions in equal measure so that the magnetic field follows the curve. If you don't, you may reach in the saturation region of the curve.

I have omitted some details but hope this is clear for this purpose.

 

[mitrarka]

"Sooner or later, the diode will get the current close to zero but there will be considerable magnetism still left. To remove that, you need a current in the reverse direction. If you cycle the current in the positive quadrant only, the curve will be different."
But we do exactly that in a forward converter. There the core only operates in the first quadrant. How does the demagnetization take place there?

 

[KlausST]

Hi,

When you activate Q1, then Np1 is active. PLUS is a the dot. Core is forward magnetized.
When you activate Q2, then Np2 is active. PLUS is a the non_dot. Core is reverse magnetized.

And the operation scheme should be:
Q1 | dead_time | Q2 | dead_time | Q1 | dead_time | Q2 | dead_time | ... and so on.

So you continoulsy do forward then backward magnetize the core.

Klaus

 

[RCinFLA]

A permanent magnet with a coil wrapped around it will have no voltage or current in coil. This is what you will have if you allow the residual magnetism to build up in transformer core by running imbalanced switching currents-seconds. This reduces the AC flux induced delta before core saturation, effectively giving you a smaller transformer core.

Ferrites have an almost vertical rectangular B-H curve so just a little residual magnetism will saturate the core material.

A power transformer does not (should not) store energy in core (in contrast to a flyback transformer) so a vertical rectangular B-H curve is actually desireable for a power transformer but you must be careful of residual magnetism.

 

[c_mitra]

A permanent magnet with a coil wrapped around it will have no voltage or current in coil. This is what you will have if you allow the residual magnetism to build up in transformer core by running imbalanced switching currents-seconds. This reduces the AC flux induced delta before core saturation, effectively giving you a smaller transformer core.

Lots of misconceptions here.

A coil with an iron core (that is magnetized; a permanent magnet) shall still have considerable inductance (more than the air-core coil) but certainly less than the soft-iron core. In other words, when you apply a voltage pulse, the current will increase (approx) but the magnitude will be much higher.

Ferrites have an almost vertical rectangular B-H curve so just a little residual magnetism will saturate the core material.

Ferrites are optimized for a small area for the hysteresis loop because the energy dissipated per cycle is proportional the area of the hysteresis loop. There are both soft and hard ferrites and they are used for different purposes.

A power transformer does not (should not) store energy in core (in contrast to a flyback transformer) so a vertical rectangular B-H curve is actually desireable for a power transformer but you must be careful of residual magnetism.

A power transformer is equivalent to a pair of coupled inductors. All inductors store energy in the core (even if that is a vacuum) and that is true for any power transformer. Optimum working range is the linear region (B-H curve) and a lower area of the hysteresis loop is the important desirable parameter (also the effective permeability) for the core material.

In physics, it is called the field; the capacitor stores energy in the dielectric and an inductor stores energy in the core. Energy is associated with the electric and magnetic field (creation and destruction) respectively.

 

[RCinFLA]

A coil with an iron core (that is magnetized; a permanent magnet) shall still have considerable inductance (more than the air-core coil) but certainly less than the soft-iron core. In other words, when you apply a voltage pulse, the current will increase (approx) but the magnitude will be much higher.

Ferrites are optimized for a small area for the hysteresis loop because the energy dissipated per cycle is proportional the area of the hysteresis loop. There are both soft and hard ferrites and they are used for different purposes.

A power transformer is equivalent to a pair of coupled inductors. All inductors store energy in the core (even if that is a vacuum) and that is true for any power transformer. Optimum working range is the linear region (B-H curve) and a lower area of the hysteresis loop is the important desirable parameter (also the effective permeability) for the core material.

In physics, it is called the field; the capacitor stores energy in the dielectric and an inductor stores energy in the core. Energy is associated with the electric and magnetic field (creation and destruction) respectively.

For its primary function, I would refer to a flyback transformer as coupled inductors but not a power transformer. Power transformer is coupled inductors but inductance of the inductors is counter to functional objective of a power transformer.