# Push pull core saturation

I want to build a circuit similar to this one:

I read that using push pull without per-pulse current limit or series with trafo cap can cause core to saturate because two pulses aren't perfectly symmetric.

But there is hundreds of circuits without any limit or cap and they work ok. When does this saturation process starts being noticeable? Should I care about it when making 35Watts dc-dc or it's noticeable only when power is big?

## 3 Answers

If one half of the primary is slightly more "on" over time than the other side then there is a risk of saturation but far, far less than say a fly-back converter.

With any transformer coupled DC converter you have to design the basic magnetics to suit the operating frequency and the input voltage limits. More turns on the primary means much more primary inductance (L is proportional to turns squared) and much more primary inductance means much less magnetization current and it is this current (not load current) that causes a core to saturate.

Start by calculating the H field. It is generated by the amps and the turns and the effective length of the magnetic path in the core (H = A.t/metre): -

The above picture shows the effective length ($l_e$) but any ferrite part supplied by a reputable manufacturer will have this detailed in the data sheet. The value of current used has nothing to do with load current - it's the magnetization current alone that causes saturation so, to calculate this you have to know what your operating frequency is and what the inductance of your winding is (one half winding in the split primary example) and your highest incoming supply voltage.

Again, the supplier comes to the rescue because they always state the value $A_L$ and this tells you the inductance based on one turn. For tens turns the inductance is 100 times more i.e. there is a square law relationship between turns and inductance.

The basic relationship between current, inductance, voltage and time is: -

$V=L\dfrac{di}{dt}$ and this converts to I = $\dfrac{V}{2fL}$

I is the ramping current through the inductor due to the voltage applied for one half period of frequency f. If inductance is 10 uH, f is 100 kHz and V is 12 V, current I will linearly rise to 6A over the period dt. Given that you already know the number of turns and the mean length of the core, H (magnetic field strength) is just another bit of number crunching.

So you have your H field and you need to look at the ferrite data sheet to see if this H field is going to saturate the core. The graph looks something like this: -

If it looks like saturating then increase the number of turns. Doubling the turns quadruples the inductance and this quarters the magnetization current so the new value of ampere turns is halved and so is H.

So, if the basic half cycle current hasn't created sufficient ampere-turns to significantly cause saturation then it's very unlikely on such a low powered set-up that a slight mismatch in the position of the centre connection or the mismatch in duty applied consecutively to both halves is going to have any noticeable effect.

• Ah a downvote, any explanations? – Andy aka Oct 19 '15 at 20:05

Push-Pull converter is one of the oldest topologies around, so it's hard to even think about how many are out there working. It's true though, a side to side imbalance can cause the core to walk into saturation. How do people keep this from happening?

Well, here are some things to do (and not to do) to keep the core balanced:

• Absolutely do not use a core that has a square BH loop. These have extremely high $\mu$ and will quickly end up saturated.

• Use a ferrite core. Something like and EE, or EP, or Pot core.

• Gap the core. Sometimes just putting the two core halves together can be enough of a gap, but not usually. Expect to have need of some kind of added gap. The gap linearizes and stretches out the H of the BH curve allowing it to take more coercive force (H). That's another way of saying the gap allows the core to take some DC bias. Also it will minimize the variation of $\mu$ found in ferrite, which can easily be +/-20%. Now, a gapped core will have higher magnetizing current, so loss will be a little higher. But, by tayloring the gap to your needs, a highly efficient power supply is still possible. Also, if FETs are used as switches, as an imbalance of magnetizing current occurs, the FET exposed to the higher current will warm up, raising $R_\text{ds-on}$ of that FET and causing currents to rebalance.

Here's an article that talks about voltage mode control and Push-Pull converters "Voltage Mode Converters Deserve a Second Look".

You could also use current mode control, which uses the feedback loop to keep currents in balance.

Current-mode control is probably the most bulletproof way to fix any possible flux imbalance problem. Other measures include:

• Gapped core: In the event of some small amount of flux imbalance, a gapped core is much less likely to saturate.
• Low-value resistor in series with primary windings: This will act as negative feedback to help counteract any possible increase in volt-second product as a result of higher current peaks. You may have to do some empirical testing to determine an optimal value, but generally you're looking at something less than 0.5Ω.
• Use MOSFET devices to drive transformer: It is much easier to drive two MOSFETs in a symmetric fashion than it is two BJTs since there is no storage time effect. Furthermore, a MOSFET will provide a crude form of negative feedback due the positive temp-co of the Rdson, providing the same effect as adding a low-value resistor in series with the winding.
• Thanks for advices. I considered using sg3525 pwm controller with two irfz44 mosfets. I'll also try to implement simple current limiting using 0.1Ω resistor between mosfet sources and gnd, when voltage on it will exceed 0.6v transistor will open and pull down sg3525's shutdown pin. Will this work? – user1940679 Oct 19 '15 at 19:59