These driver networks are kind of non-intuitive at first glance, but T3sl4co1l has a good explanation from this thread:
https://www.eevblog.com/forum/projects/half-bridge-bipolar-transistor-psu-base-drive-transformer-design-data/For half bridge circuits, base drive transformers give you a floating high-side supply in the absence of a "bootstrapped" high-side supply that are common in new controller ICs. Also, this chip was intended to drive BJTs, so the requirements are different. I suppose BJTs were the better choice 10-20 years ago, in terms of performance/price.
There are different designs used with BJTs, so an exact answer depends upon the circuit in question.
This should serve as a useful example:
The FT37-43 (actually a Fair-Rite 5943000201) ferrite bead is a small NiZn part, somewhere around, uh, Ae = 10mm^2 and Bmax = 0.3T I think. (Ed: actually more like 7 mm^2, and 0.2T due to remenance and unipolar drive.)
That means, for a winding of 60 turns, we can get a maximum flux of (10 mm^2/t) * (0.3 uVs/mm^2) * (60t) = 180 uVs on the primary. From a 12V supply, that's a square pulse 15 us long (although in practice, both voltage and duration will be shorter due to the series 51 ohm supply resistor).
When the primary switch turns on, the output transistor is forced on. With a 4:1 ratio, and about 1V Vbe(on), the primary actually only drops about 4V (not 12V, hence the 51 ohm resistor serves as current limiting, to about 150mA).
When the output transistor turns on and load current is drawn, this current also flows through a tertiary winding on the drive transformer. The 1:5 ratio between load current and base winding gives a forced hFE = 5 condition. This is positive feedback, so the transistor remains switched on, even if the primary side transistor turns off.
In a typical ATX computer supply (half bridge configuration), the base transformer is driven in shorting mode commutation with a pair of transistors, so that the base voltage can be shunted to zero, forcing the output transistors off.
In this example, turn-off is provided by the drive transformer saturating. The transformer has some inductance still (it's not a super high permeability core), which acts to shunt more and more drive current (increasing the forced hFE) as time goes on. Eventually, two things can happen: 1., the transformer saturates, and its current draw goes way up, shunting base voltage off; or 2. forced hFE rises above the transistor's hFE(sat), the transistor comes out of saturation (while still carrying full load current -- this method costs more turn-off switching loss), load current commutates to the clamp diode (UF5404), and with no current in the tertiary winding, the transistor turns off. Because the inductance is still reasonably high and the transistor has no shortage of hFE under this load, the first case occurs here. (The second case is very common in blocking oscillators, like the "joule thief" circuit, where a non-saturating inductor is used.)
Transformer core calculations are all about ratios, so it's pretty easy stuff. Follow the units (dimensional analysis).
As for transformer winding design, where you place the wires matters -- the further apart any two wires / windings are, the higher the leakage inductance. Probably not a big issue for a small drive transformer, but this can be critical for large power transformers, especially at low impedances.
Tim
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