Push pull converter

Check the capacitors and resistors in the push pull circuit.

Dickson.

There is no resistors or capcitors in the primary side.

Let me have ur cct. drawing.

Dickson

The circuit is very simple: The primary side of the transformer is centre-taped ( 2 equal coils) and a DC voltage is applied across one half of the primary for time "t" and the an equal but opposite voltage is applied to the other half of the primary for the same time period "t", then the process repeats it self.

What will be the peak of the triangle flux wave form in the core?

Does the flux will have only positive peaks or will have positive and negative peaks ?

Also check ur transformer primary windings polarity. If N1,N2 and N11,N22 are the two series coils of the primary winding, where N1,N11 and N2,N22 are the starts and ends of the two coils. Consider the polarities of these coil in relation to ur battery position.

would think its because the dc supply is not reference to ground...

i.e. 30Vdc can be -15V to +15V

The transformer, in the circuit drawing, may be phased. Check the drawing to see if the transformer has a dot over the primary winding with a corresponding dot over the secondary winding.

I am not sure why a "phased transformer would be called out....

apply high pass filter to output voltage

Your description is correct at start-up. But the transformer has winding resistance (effectively resistance in series with each of the coupled inductances), so the average flux over a cycle decays to zero.

In long-term operation the flux will vary between -F1/2 and +F1/2.

For amusement, consider what will happen if the transformer saturates below F1 (as is perfectly possible).

Fyz

P.S. Great teaching question...

PPS. the above is for an unloaded transformer - current in the load windings will accelerate the decay.

Have you some circuit in view or you are talking just as assumption.

Can you place a circuit diagram for discussion & then only the comments can be valid.

The transformer's flux will eventually average zero if the two voltage x time (Volt-Seconds) and the windings are perfectly balanced. If not, a DC current will build up and potentially saturate the transformer turning it into a short circuit (almost).

This is why, in most cases, a large AC capacitor is inserted in the common to integrate the transformer current and make sure that is is balanced.

Without this capacitor, the circuit might work but will not be reliable especially if the components have low losses to "balance" things out.

Good luck.

Thanks for help.

Where exactly that capacitor can be connected for push-pull converter ?

This capacitor is usually inserted between the transformer common and the supply. It must take all the load current with very little voltage increase. This translate into a large uF value. I = C dV/dT...

Some people have insured that the volt-second applied to the primary is balanced by placing a divider by 2 between the oscillator and the transistor drivers. They could do without this capacitor. Some current balancing with a current sensor could also be used but this becomes complicated.

The "simple" way to correct the balancing problem is to insert a dead time between transistors turn on to insure that the magnetic energy is returned to the supply through the free wheeling diodes. This reduces the max power transfer but is relatively simple and reliable.

At the end, nothing is free.

Marcot, for a push-pull core, if the primary DC voltage is 170V and the primary current is 10A (at full load) and the switching frequency is 100KHZ , what would be the capacitor value needed ?

Is it electrolytic capacitor or AC capacitor ?

Thanks for help

There is nowhere you can place a simple capacitor - unless the terminals are independently driven between the rails, in which case
a) you need two capacitors, typically one in series with each driver, and
b) the centre tap becomes both unnecessary and counter-productive.

Returning to the practical case where each balanced winding is shorted to rail in turn:
If the 50:50 mark-space is exact and the windings perfectly balanced, any flux offset will decay to zero under the influence of the winding resistance of the transformer together with the output resistance of the drivers. This is as previously stated (#9) - and the basis of the simplified text-book description. Even though the capacitor proposal is unsound, increasing the dead-time and using free-wheel diodes (as also mentioned by Marcot) works - but is no better than increasing the resistive losses.

You could place a capacitor in parallel with a resistor in series with each of the drivers. This is of course lossy, but (for a given loss) the reduction in flux offset can be doubled compared with using simple resistors.

Many balanced communications transformer drivers rely on this effect as well as quite a few "power inverters" - but be aware that such power inverters can need significant monitoring and control to keep them safe. Of course the problem is equivalent to that of switching a mains AC transformer mid-cycle - both rely absolutely on the net input signal having near-zero DC content.

So its worth being aware that the sensitivity to mark-space error increases together with the efficiency of your driver-transformer arrangement. Some balanced communications systems use current-limited drivers for just this reason.
Otherwise, you just(!) have to control the mark-space ratio.

I recommend getting hold of a copy of SPICE, trying some permutations and analysing the results. LTSpice is free and optimised for similar applications. But be sure to use realistic values for transformer winding inductance and leakage unbalance, or you could deceive yourself.

Astonished!

On which circuit all the comments are? No circuit diagram in view.

Zacky, U have one of the most subscribed question. I will sugest if the circuit ur trying is not coming out with a good result, U can try other circuits, there are so maney of them. Ur response indicates that U have a lot of insite in the proffession and project. Check the size of ur tansformer and see if saturation is the problem. Try other obtions as prescribed by other members or change ur design as recommended above. U can let us see ur drawing even though its a very simple one; lets have it.

Dickson.

The circuit on the paper, i do not know how to drew it on the computer and post it-what kind of software is suitable for that ?

Have ur cct scaned or photographed and place them in any of ur system folders, coppy this folder, right click message area of Forum Thread, select paste in the pup up menue.

Dickson.

Hello Dickson

Are you OK.

We hear about big electrical accident in Nigeria last week - are you ok

Guest, thanks for Ur concern. Actually 33 persons were sent to an untimely grave when a pole supporting 33kv line collapsed and drooped the lines on roof tops in a very high density area in the oil city of Port Harcourt Nigeria.

Yes, thanks for Ur cct diagram on our much deliberated topic. IF I may ask why is ur transformer secondary shifted? The cct drawing is ok, well theorised and depicts Zacky^,s cct explanation. The cct has its peculiar advantage of simplicity with reduced component but it is highly inefficient. Zacky^,s is that of saturation and transformer selection.

Dickson

We are different guests (not that it matters)...

The only reason that the secondary of the transformer appears shifted to the left is that this is the standard symbol in the schematics capture package. (It was probably originally prepared as a reduced version of a four-winding transformer)

I believe the circuit I have drawn is what Zacky was looking at. I would also note that this arrangement can only be really efficient if the DC-to-DC conversion ratio is known, and that the efficiency is very sensitive to the balance of currents in the primary (which needs active control if the primary is driven via switches). However, given an appropriate control circuit* and active switching on the output, the efficiency can be very high indeed. In the (increasingly rare) situations that they are suitable, circuits of this type can provide at least as high a power-efficiency as any other DC-to-DC conversion system.

Apart from the fixed (or at least only-discretely-switchable) conversion ratio, the only downside is that one of the three windings is inactive at any time - but it carries the upside that the supply and load currents have relatively low ripple (which means that this needs the least output filtering of any system), and the switch losses are intrinsically low.

*Power-MOS integrated circuits can adjust the M/S so that the average current in each branch is very closely matched)

P.S.1 From the wording, I doubt that Zacky was to expect this arrangement to saturate. Also, "simple", "derived by 555 timer chip" together with "half the primary" seemed to me to indicate the arrangement as drawn and analysed. However:
P.S.2 The same driver arrangement (combined with an appropriate controller) can be used in a balanced buck converter. This is shown (here) to give reduced conducted noise compared with unbalanced systems. There are of course disadvantages - including the increased Voltage rating required for the driver transistors. The boost mentioned in the paper would of course require reverse isolation to be provided for the driver transistors, but I scarcely see the point in this, given that a transformer is already used.

Do you have any views on this? (I'm not by trade a power-supply designer)

But what will happen if one of the two windings of the transformer primary are not equal ( one winding has 10% increase in the number of turns)

Is there any risk to burn the converter ?

Marcot, for a push-pull core, if the primary DC voltage is 170V and the primary current is 10A (at full load) and the switching frequency is 100KHZ , what would be the capacitor value needed ?

Is it electrolytic capacitor or AC capacitor ?

Thanks for help

Is the driver circuit essentially as shown - the centre tap of the primary connected to the supply and a single open-drain (or open-collector) device driving each end of the primary?
If your circuit is essentially different*, please provide details.
*(As presumably Marcot was assuming??)

If the situation is as above, any capacitor in series with the primary path will simply charge until the potential across it becomes equal to the supply Voltage, and there will be no long-term drive to the transformer.

Returning to the behaviour of this circuit (without series capacitors):
If the drivers are not very large relative to the load they are required to supply, they will behave as current sources and there will be an average flux that is considerably less than the peak. Similarly, the DC voltage across the transformer will decay over time to a small value (dependent on the mark-space of the drive and the mismatch of the current sources). The ultimate efficiency of this arrangement is limited by the saturation-Voltages of the transistors, though it can become much lower if the rectified load impedance is lower than the design value (and in some respects it can behave a bit like the large-driver case below if the rectified load impedance is larger than the design value - however, the peak flux will still be contained). [Note that current-source drivers are commonly used in balanced communications systems where the secondary current is not rectified and efficiency is secondary to signal integrity]
If the drivers are large and overdriven, they will behave as switches, and the average flux will grow over time to a value dependent on the mark-space of the drive and the values and mismatches of the windings and of the switch resistances. The design is quite critical. Worse, if one of the switches fails the current in the other could become very large; however, it can be safe if (for example) the switches have built-in over-current protection; unsurprisingly, power-MOS ICs are available that are suitable for this purpose. (Bipolar transistors used as current-multiplying mirrors with output-capability exceeding the required drive can also be used).

I've now done my best to keep your design sane, and will only respond if you either clarify how your system is different or ask specific questions.

Thanks for the detailed explanation.

Could you please explain further the following part:

"If the drivers are not very large relative to the load they are required to supply, they will behave as current sources and there will be an average flux that is considerably less than the peak."

What you mean for "drivers" , is it the DC supply voltage according to the drawing we have ?

I mean the transistors Q1 and Q2 - combined with how they are driven. (If we are using power MOSFETs, the supply shown could be anything up to half the continuous drain Voltage rating of these transistors.)

Hi Zacky,

if the circuit is as drawn by the guest above then you will get -F because the flux is a vector not a scalar quantity. This is a classic inverter circuit.

Assuming the two halfs of the primary are wound in the same direction (as drawn)then you can see that if the current in P1 produces a flux with North to the left of the diagram then current in P2 will produce a North at the right of the diagram. These are opposite polarities and and can be defined as +F and -F. Their effect on the secondary will be the positive and negative half cycles.

Chas