Please note: the figures refered to in the text are those printed in the original article (December 2002 issue of Practical Wireless).
The circuit above shows a (simplified) equivalent
circuit of a real transformer.
Rc is the series resistance of the primary.
Ll is the leakage inductance of the primary.
Lp is the inductance of the primary
(the ideal transformer has infinite inductance).
Rp represents the losses in the core.
Cd is the capacitance of the primary winding.
The secondary will also have series resistance,
inductance, leakage inductance and capacitance,
but they're not a problem in our case.
There's no secondary equivalent of Rp (core losses);
that's only associated with the primary.
Transformers are by no means perfect. In order to predict their behaviour in real circuits, they can be modeled electrically as a perfect transformer within a network of (also perfect) resistors, capacitors and inductors.
The series resistance of the windings, the losses in the core material, and the primary inductance are mentioned elsewhere. In the main, these govern the efficiency of the transformer.
The perfect transformer that exists inside a real transformer has perfect coupling between primary and secondary. All the magnetic flux set up by current flowing in the primary winding, cuts - or intersects - the secondary winding. Any load placed across the secondary winding is reflected back into the primary winding, modified only by the ratio of the transformer.
In a real transformer, this is not strictly true. The flux set up by the primary winding does not completely cut the secondary winding. There will always be a small measure of flux that misses the secondary winding, and therefore, is not affected by any load placed across the secondary. Not surprisingly, this is called the leakage flux.
The leakage flux gives rise to a property of a real transformer called leakage inductance. It is modeled electrically as a small inductor placed in series with the primary winding (Ll in the circuit above). If a short circuit is placed across the secondary of a real transformer, the primary appears (to a first approximation) as a low value resistor - the resistance of the windings - in series with this small inductor.
Small mains transformers used by electronics enthusiasts often have a high leakage inductance, particularly those with split bobbins. That's the kind where the primary and secondary are wound side by side, separated by a barrier, rather than wound one on top of the other. Leakage inductance is usually of no consequence in normal operation at mains frequencies.
Leakage inductance does become important when transformers are handling very high power. In small transformers, at 50Hz or 60Hz, the leakage reactance (2*pi*f*L) is small, just a fraction of the resistance of the primary winding. Not so large power transformers. They have very low resistance windings carrying hundreds, even thousands of amps. Under such conditions, any leakage inductance is of concern.
High-fidelity audio output transformers don't handle megawatts (usually!), but do handle frequencies up to 20kHz and beyond (whether intentionally or unintentionally). At these higher frequencies, the leakage inductance becomes significant; if not for the magnitude of the reactance, then for the phase shift it causes.
In our inverter design you may think leakage inductance won't bother us. After all, we're only driving the transformer at 160Hz. This is true, but the MOSFETs are switching very quickly, and that can mean trouble.
As the MOSFET switches on, 12V suddenly appears across one half of our primary. Fortunately, at this point, nothing nasty happens. Current flows in our primary winding due to the supply 'seeing' the reflected load on our secondary. All the leakage inductance may do is slow the rate of rise of current in the primary; you can't change the current through an inductor instantaneously!
Half a cycle later, at the end of the 'on' period, the MOSFET switches off in less than a microsecond. Now, current is flowing in the transformer primary, predominantly in the reflected - and resistive - load that's due to the real load across our secondary. But it's also 'flowing' through the leakage inductance. Here we go again, you can't change the current....
This situation is very similar to what occurs in a switched-mode regulator. Indeed, the leakage inductor has similarities with L1 in Fig. 3. Recall that when the transistor switch inside IC1 opens, the current flowing in L1 can't just stop, so it diverts through D4, the only place it can go. As the current can continue flowing, no nasty high voltages are developed. This is why you always put a back e.m.f. diode across a relay coil. If you don't, and you de-energise the relay, a high voltage will be developed which can destroy a transistor or burn the contacts of a mechanical switch.
Back to our inverter; as the MOSFET switches off, there's nowhere for the current flowing in the leakage inductance to go. A fast, high voltage ramp (stray capacitance will slow it a little) is generated which appears across the MOSFET, drain to source (drain will be positive). As it reaches the MOSFET's maximum drain to source voltage rating, the MOSFET will break down and the voltage spike will be clamped. Current will flow until the energy stored in the leakage inductance is dissipated (we're talking microseconds).
In the main, MOSFETs can survive these spikes, but it's not good engineering practice. Of more concern to us, the spikes also generate interference (which we definitely don't want - this is a radio, remember). Fortunately, there are two ways of dealing with the problem.
The best way, and the one shown in Fig. 8, is to add what are called 'snubbers' across each half of our primary. A snubber is simply a resistor and a capacitor in series. The resistor, capacitor and the leakage inductance form a lossy tuned circuit where energy flows - maybe several times - between the capacitor and the leakage inductance. During this time, the energy that was stored in the leakage inductance is dissipated in the resistor.
If you know, in detail, all the parameters of the transformer, have a good idea of the stray capacitance in the circuit (and have a mathematics degree), you can calculate the values of the resistor and capacitor. In the real world, you make a guess and use trial and error.
To arrive at the correct values you need an oscilloscope, and a good one at that. With the circuit working normally, monitor the waveform at the drain of one of the MOSFETs. The spike will be very narrow and be about the same voltage as the breakdown voltage of the MOSFET (50V for the BUZ10).
Using crocodile clips, connect a 22nF capacitor in series with a 1kohm variable resistor and wire them across one half of our primary (the side you've got the 'scope on!). Adjust the resistor until the spike reduces in amplitude and broadens. You will see a little damped oscillation if the transformer's leakage inductance is high. Repeat the whole thing again with a 47nF capacitor and again with a 10nF capacitor.
By now you should have an idea which capacitor plus resistor-setting combination produces the 'best' looking waveform. I can't be more descriptive; you'll know it when you see it! If none of the capacitor/resistor-settings seem to work you can try other values. Putting big (470nF or so) capacitors straight across our primary will kill the spike, but they'll also cause nasty currents to flow and efficiency will drop. Don't do it.
When you've chosen a capacitor and measured the value of the variable resistor (when adjusted to give the best waveform), substitute the nearest preferred value for the resistor and duplicate the pair on the other side of our primary. If no combination of resistor and capacitor seems to work - and I didn't find any suitable combination when using a toroidal transformer - then there is a 'brute force' way to tackle the problem.
Forget the snubber approach and wire a 1W, 40V zener diode between drain and source of each MOSFET. Anode to source, cathode to drain. You may have problems getting 40V zeners, so connect two 20V zeners in series to make each 40V zener. The zeners will clamp the drain at 40V, thus protecting the MOSFETs but doing nothing to stop the interference the spikes generate. Can't win them all....