Actually, it's a beautiful piece of work. The Joules per unit time (power) is independent of the inductance of the transformer, so you don't mess things up if you wind it too many times or too few. All that affects is the frequency of operation. A simple resistor sets the power. Just a very few components, too. It's sweet.
The point is that saturation isn't required and will actually waste power. So I think that saturation would be a problem, actually.
Assume the collector winding saturates the core. This happens as Ic is rising on a V/L ramp and, for purposes you are putting it to, occurs by definition well __before__ Ic/Ib gets anywhere near the beta limit of the BJT. So let's say this happens when Ib=Ic, just to keep it easy. So the BJT is in deep saturation, still. And the core suddenly decides "that's it, I'm tapped out!" At that point, it isn't the BJT, but the core that decides the voltage across it must cease. So the collector winding voltage goes to zero, suddenly, which means Vce on the BJT suddenly rises to the fully battery voltage -- meaning VERY BAD dissipation. You are right. The base winding also goes to zero volts and the base current takes a hit. But it goes to about 1/2 the earlier value. Since Ic doesn't change, but only Vce did, the beta is now 2. Which means the BJT can still support the Ic required (which hasn't yet changed -- the collector winding is zero volts, not negative, just the Vce has jumped up.)
At this point, we see the BJT with beta 2 (which is still deep saturation) and a base current that is 1/2 of what it just was but is still well more than enough. So the BJT's Vce times Ic gets dissipated by the BJT, which is now heating up big-time. (Beta capability actually rises with temperature increases, memory serving.)
If inductors were perfect (superconducting?), I suspect things would just sit there with beta=2, the BJT heating up until it reaches some stable point, Ic remaining fixed, Ib remaining fixed, and that would be that. But the resistance in the inductor is instead gradually (slowly) eating up the field's energy, causing a voltage reversal... probably on the order of timing less than a second but not nearly at the design frequency rate. This opposes the battery voltage to the base, of course, and the whole thing does wind down. But I suspect that saturation is NOT a good thing here because of all this. In other words, it is not "a friend" to the process. It's to be avoided.
But I've only two or three weeks ago started studying magnetics design for the first time. And I may have something wrong. But that's the way it looks to me.
I think in a profoundly different mode. But yes.
I don't think so. Read what I wrote and see if you can find fault with it. I'm curious about learning this stuff better.
Well, I didn't take your starting point so conclusions from it still don't flow from it, for me.
Actually, I think it is an incredible arrangement.
Look. You can design this entirely with the idea of just knowing how much current on the other end you require. Knowing the current, you can compute the peak current you want in the collector winding. It is just:
Ipeak = Iout*[2*(Vout+Vd-Vceon)/(Vbattery-Vceon)]
(These assume that diode I suggested to the OP, so Vd is the forward voltage of it during BJT-off times. Vceon and Vceoff will be rough numbers used and aren't all that critical. I use Vceon=0.1 [as an average value between 0.0V and 0.2V during on-time] and use Vceoff of between 0.4V and 0.7V depending on just how big Ic happens to be.)
Notice that there is no inductance here, no base resistor value, etc. There is a pure number ratio (bracketed) times the desired Iout. Very simple and independent of a lot of stuff you don't want to have to worry about. This makes design easy.
Then, you compute the inductance of the collector winding. At this point, you need to know some idea about the desired frequency. The beauty of this is that you can now worry about volt-seconds and that saturation problem. To avoid volt-second problems, choose a faster frequency. The frequency will have NO effect on the power/Iout that is delivered. You get to choose it independently and not worry about its effects on Iout. Wonderful, for designing. So to avoid huge volt-second figures, pick a frequency that is high. But to avoid dealing with BJT capacitance and charge mobility issues, choose a frequency that is low. Between these is a nice region in the tens of kHz... so I like 50kHz as a good place to be. At this point, compute L: (Vbattery-Vceon)*(Vout+Vd-Vbattery) L = ------------------------------------- (Vout+Vbattery-Vceon)*Ipeak*frequency
At this point, Rbase can also be calculated: Rbase = beta*[N*(Vbattery-Vceoff)+Vbattery-Vbeon]/Ipeak
Here, N is the winding ratio, with higher values occuring when the base winding has more turns. Normally, N=1. Also, beta is selected from the datasheet at the value for Ic=Ipeak, roughly. Also, Vbeon is picked up from the datasheet (or estimated.)
Your peak base current isn't simply the Ipeak divided by the chosen beta, here. It's more like:
Ibaseavg = [N*(Vbattery-Vceon)+Vbattery-Vbeon]/Rbase
Rbase is determined using a higher Vceoff calculation because the value of beta won't apply well, if applied using the lower Vceon.
But with the freewheeling diode and a capacitor, there is no need. And that arrangement provides a nice smooth current. The issue with pulsing is that the LED voltages rise with high currents flowing and this wastes energy uselessly. What the diodes really want is a steady DC at the right level, not huge pulses of current. Best efficiency is not necessarily at the nominal current for the LEDs, but it usually isn't too far from it. Pulsing with duty cycles not unlike 1:10 means
10-fold current increases and that is way past their efficiency curve peak. It wastes power. I think the diode/cap almost suggests itself.On the surface, it seems easier and better to just add a diode and cap than to wind another winding in a tiny bead.
Jon