As the current rises in the Base Windings opening the collector and permitting more current in the Collector Windings.
Thus, the flux rises in the Collector Windings and the Base Windings rising the current in the base of the Transistor and opening the collector a little more and allowing more current in the Collector Windings up to saturation of the transistor.
Then the flux collapses...
Is that the oscillation?
The base winding generates a voltage, not a current. This voltage adds or subtracts from the battery voltage and switches the transistor on and off.
Wikipedia has a good description on how the oscillation works:
http://en.wikipedia.org/wiki/Joule_thief#Description_of_operationThe input voltage is one volt?
In your simulation, yes.
My question was about there is no capacitor to calculate a thank oscillation.
Let's say the resistor is 200 Ohms, the inductance 20uH: a time constant 1*10^-7 sec
5 timesconstants: 5 *10^-7 sec ==> 2 000 000 Hz
Can we start from that?
It is more complicated:
First start with the base current:
The turns ratio is about 3:1, so the base winding generates about 0.3V. This adds to the input voltage -> 1.3V
The base resistor is 280 ohms, Vbe is around 0.8V, that gives around 1.7mA base current. The ZTX690B has a hfe of about 900. This gives a peak current of 1.5A. Your simulation shows about 0.75. It is at least not completely off.
The current rise in the 280uH inductor at 1V to 1.5A requires 0.42ms. Your simulation shows about 0.26ms.
The off time depends on the load voltage. A white led has a Vf of about 3.1V. 3.1V - 1V supply voltage = 2.1V at the inductor. The current falls from 1.5A to 0A in about 0.2ms. This gives a total period of 0.62ms. Compared with the 0.35ms from you simulation thats pretty good. As I said, it is only a rough estimation using ideal components.