Let me summarize my understanding for you to see if I really get it.
The transistor start switching off(or start coming out of saturation) immediately when current flow in primary winding reaches the 'critical point' (that is to say that the current flow is no longer increasing exponentially at "I=V/L×t'' or in other words, no longer ramping up).
So the inductance of the primary winding decreases, and current flow in the primary winding will no longer be opposed,(as the primary winding now act as a normal conducting wires, current is limited only by "V/R").
So this in turn increases the current at the collector, which according to your first explanation will increase the collector-emitter voltage drop and force the transistor out of saturation and the transistor start switching off, then the primary winding will now induce a negative feedback voltage on the base winding which will reverse bias the CE junction and switch off the transistor.

At the beginning transistors fully ON (saturated) thus transistor collector voltage is then pulled down to near ground. And this means the primary winding will see full voltage across primary inductance (L1). Current will ramps upward, according to the ΔI = V/L*Δt
https://forum.allaboutcircuits.com/...lectromagnetic-induction.156639/#post-1354068
And the current will continues to rise in the primary winding until Ic < β*Ib. Now the transistor will start to leave the saturation region and collector voltage will start to rapidly rise. This reduces the voltage across the primary inductance. It also reduces voltage developed across Auxiliary winding and reducing the base current for the BJT's, which now has to lower its collector current even more.
So, if the base current was reduced the transistor will increase the impedance from collector to emitter, attempting to lower the collector current in response to the lower base current. Thus the primary winding voltage will reduce further and we will see a further reduction in the Auxiliary winding voltage...->dropping transistor base current even more. This quickly turns into a case where transistors turns-OFF entirely. Do you see the Positive feedback mechanism here?

 

Ok thank you.
Let me summarize my understanding for you to see if I really get it.
The transistor start switching off(or start coming out of saturation) immediately when current flow in primary winding reaches the 'critical point' (that is to say that the current flow is no longer increasing exponentially at "I=V/L×t'' or in other words, no longer ramping up).
So the inductance of the primary winding decreases, and current flow in the primary winding will no longer be opposed,(as the primary winding now act as a normal conducting wires, current is limited only by "V/R").
So this in turn increases the current at the collector, which according to your first explanation will increase the collector-emitter voltage drop and force the transistor out of saturation and the transistor start switching off, then the primary winding will now induce a negative feedback voltage on the base winding which will reverse bias the CE junction and switch off the transistor.

But according to my viewpoint, the transistor will not actually switch off,
not until the base winding receives a negative feedback voltage.

That last statement is correct, current will flow in the collector until the base current decreases to near zero.

However, i am not sure you understand the "forced out of saturation" yet so let me go a little farther with this.

First to know is that there is a fairly precise definition of saturation for a bipolar. That is when the base collector diode becomes forward biased. For active mode, the BC diode is reverse biased so that for an NPN the collector emitter voltage is higher than the base emitter voltage, and thus the base collector is reverse biased. When it goes into saturation, the base collector diode actually starts conducting in the forward mode and that means some of the base current makes it directly to the ouput and thus the Beta appears to decrease (because now it takes more current to make the same collector current) and that is why you often see the Beta drop from 100 to 10 when the transistor goes into saturation.
When in saturation, the Vce voltage is LOWER than the Vbe voltage which means the BC diode is forward biased. But as the collector current is increased, for any reason, that low Beta value can not hold the Vce voltage low because the collector current is more than iB*Beta (more or less) where iB*Beta is the max collector current that can be supported and still keep the transistor Vce low enough to be in saturation.
So say the Vbe is 0.7 which is typical, and say the Vce is 0.6v, that means the transistor is in sat. Now if Vce rises to 0.7v it starts to come out of sat, and at 0.71v it is otu of sat. Anyting higher like 0.75 and it stays out of sat.

In the circuit, it does not matter what the rate of change of the primary current is as long as the collector current reaches the limit where the transistor pulls out of sat and the voltage goes above 0.7 volts (as per previous example).
There are two possibilities:
1. The primary current increases linearly (a straight line ramp)
2. The primary starts out linearly, then shoots up quickly as the primary saturates.

Some circuits work by #1 but others work by #2 but either way the transistor begins to turn off and that causes the feedback which reverses the base drive to turn the transistor off completely.
To find out for sure which one it is you only have to look at the collector current. If it rises smoothly then it is #1 but if it ramps and then shoots up suddenly just before turning off then it is #2. The spike may be very short though and hard to see on a scope unless you look closely.

So to recap:
* Saturation is when the base collector diode becomes forward biased.
* High collector current causes the base collector to becomes reversed biased and thus pull out of sat although it then enters the active mode still with significant collector current until the base turns it off.
* The transistor is forced out of sat when the current reaches that critical value by either ramping up smoothly or ramping up and then suddenly shooting up just before the transistor turns off.
* After the transistor gets forced out of sat, usually the base is reverse driven so that the transistor turns off and does not overheat due to large collector current and high Vce in the active mode.

Example:
iCE=1 amp
iBE=0.1 amp
Beta=10
Vce=0.5v
Vbe=0.7v
Now we know:
Vce<Vbe so the transistor is in sat.

The iBE*Beta gives us a rough idea when it might pull out of sat, but Vce<Vbe is the defining criterion.

Now say the power supply voltage is 10.5 volts and the primary dynamic resistance is 10 Ohms, that means that we have exactly 1 amp collector current. Vbe-Vce=0.2 volts.
Now as time progresses, the dynamic resistance of the primary winding appears to decrease, let's say to 9.9 ohms. Now since the iBE*Beta is still 1 amp we can only support 1 amp in the primary, so the voltage Vce increases to 0.6 volts, which means now Vbe-Vce=0.1 volts. Now let's say the dynamic resistance decreases to 9.8 Ohms, now the Vce is 0.7v so now Vbe-Vce is 0v so the transistor is nearly out of saturation. Another decrease to 9.7 Ohms and now Vce is 0.8v and now Vbe-Vce= minus 0.1 volts so now the transistor is out of saturation because Vbe<Vce. As the primary winding tries to draw more current the Vce voltage rises more and then the feedback reverses the base drive and the transistor turns off and Vce shoots up to 10.5v and the collector current goes nearly to zero, thus completing that half of the switching cycle.

So from that example we can see that the collector current may not go up very high before the transistor pulls out of saturation, but when it does, the Vce voltage rises and that causes the feedback that ultimately turns the transistor off. In that example though 1 amp may have not been the normal operating current, it could have been just 0.8 amps or something. We started at 1 amp so it was quick to come out of sat. If it was 0.8 amps to start it would have taken more time to get up to 1 amp.

 

Am gonna go through everything, am kinda like getting a little bit confused, but I hope I'll understand it after I go through it over and over againgain.

 

The BJT's and the saturation region. You need to understand that BJT's will going into saturation isn't a property of the transistor itself, but instead a property of the circuit surrounding the transistor and the transistor, as part of it.



For these two circuits are you able to find the base current needed to saturate the BJT's?

The first one is 0.2mA, while the second one is 0.1mA.
That is, the current in the collector will be 10v/500ohms=0.02A.
Then since the beta or current gain is 100 the base will need 0.02A/100=0.2mA.
while in the second circuit, since the beta value is 200, the base will now only need 0.02A/200=0.1mA to saturate the transistor.
If am right, then that means my understanding is
really progressing.

 

The first one is 0.2mA, while the second one is 0.1mA.
That is, the current in the collector will be 10v/500ohms=0.02A.
Then since the beta or current gain is 100 the base will need 0.02A/100=0.2mA.
while in the second circuit, since the beta value is 200, the base will now only need 0.02A/200=0.1mA to saturate the transistor.

Very good. Also, notice that if we force Ib = 0.2mA = 200μA and the beat is 100 then if we change Rc resistance from 500Ω to 250Ω the BJT will come out of a saturation region (Vce = 5V).
Thus if the primary inductance current wants to increase the collector current beyond 20mA the BJT's cannot be in saturation region anymore for a large current in the collector.

 

Hello again,

Here is a transistor that is shown being forced out of saturation by the rising collector current.
The collector current level is shown along the x axis, the collector emitter voltage is the vertical axis for Vo1 and Vo2 shows the base collector voltage. The circuit marked "X" is just there to measure the BC voltage with gain of just 1 so whatever Vo2 shows that is the actual BC voltage.

As we can see, with a constant base drive of 1ma with lower collector currents the transistor collector emitter voltage is quite low, less than 1v easy. But as the collector current nears around 185ma in Fig 1 the collector emitter voltage starts to rise and the base collector voltage switches polarity meaning the BC diode becomes reversed biased which means it entered the active region.

In Figure 2 the transistor was assigned a somewhat lower intrinsic Beta, about two thirds of what it had in Figure 1. We can see it comes out of saturation sooner now with less collector current.

Comparing Fig 1 and Fig 2, we can see that the point where the transistor comes out of sat is closely related to the intrinsic Beta of the transistor, and so as temperature varies the point where it comes out of sat will change. This in turn will change the pulse width of the converter which means the frequency will vary somewhat. Typically the transistor will heat up and the time the primary is help 'on' will lengthen. Usually there is an RC timing circuit to hold the transistor 'off' which will only change a little so the frequency will decrease somewhat.

 

Very good. Also, notice that if we force Ib = 0.2mA = 200μA and the beat is 100 then if we change Rc resistance from 500Ω to 250Ω the BJT will come out of a saturation region (Vce = 5V).
Thus if the primary inductance current wants to increase the collector current beyond 20mA the BJT's cannot be in saturation region anymore for a large current in the collector.

Ok, so that means that Rc(collector resistor) will determine the collector current when the base is constant and it equivalent to the primary inductance in our smps circuit.
when Rc decreases, more collector current and more C-E voltage drop. Same thing as when the inductance in our smps circuit decreases.

 

Ok, so that means that Rc(collector resistor) will determine the collector current when the base is constant and it equivalent to the primary inductance in our smps circuit.
when Rc decreases, more collector current and more C-E voltage drop. Same thing as when the inductance in our smps circuit decreases.

Sounds good

 

Sounds good

Ok, please one more observation, we have 2 saturation states here. That of the transistor, and that of the primary winding. For the transistor to switch off, both itself and the primary must reach the point of saturation.
But the question is, which one saturate first?; is it the transistor that saturate before the primary winding or vice-versa.
I guess according to our explanation, it should be the transistor.
I just want to hear from you.

 

Very good. Also, notice that if we force Ib = 0.2mA = 200μA and the beat is 100 then if we change Rc resistance from 500Ω to 250Ω the BJT will come out of a saturation region (Vce = 5V).
Thus if the primary inductance current wants to increase the collector current beyond 20mA the BJT's cannot be in saturation region anymore for a large current in the collector.

Ok, please one more observation, we have 2 saturation states here. That of the transistor, and that of the primary winding. For the transistor to switch off, both itself and the primary must reach the point of saturation.
But the question is, which one saturate first?; is it the transistor that saturate before the primary winding or vice-versa.
I guess according to our explanation, it should be the transistor.
I just want to hear from you.

 

Ok, please one more observation, we have 2 saturation states here. That of the transistor, and that of the primary winding. For the transistor to switch off, both itself and the primary must reach the point of saturation.
But the question is, which one saturate first?; is it the transistor that saturate before the primary winding or vice-versa.
I guess according to our explanation, it should be the transistor.
I just want to hear from you.

Well maybe you are mixing something up because the transistor is in sat right after it is first turned on. It comes OUT of sat when the winding current increases to that critical point, and that point can be reached either by a linear rise of current or the primary saturating which causes a quick, sharp rise in current,, but either way the it is the primary current that pulls the transistor out of sat.

The sequence goes like this:
1. The transistor is turned on into saturation and that puts the +Vcc voltage across the primary.
2. The primary current starts to increase.
3. Now either one of these two:
3a. The primary current increases linearly to the point where the transistor starts to come out of sat.
3b. The primary current increases linearly to the point where the primary causes the core to start to saturate and so the primary current shoots up quickly and once it reaches the critical point the transistor comes out of sat fairly fast.
4. The base becomes reverse biased turning off the transistor.
5. Some RC time constant starts charging.
6. The RC time constant times out, turning the transistor back on and the whole cycle starts all over again.

 

Well maybe you are mixing something up because the transistor is in sat right after it is first turned on. It comes OUT of sat when the winding current increases to that critical point, and that point can be reached either by a linear rise of current or the primary saturating which causes a quick, sharp rise in current,, but either way the it is the primary current that pulls the transistor out of sat.

The sequence goes like this:
1. The transistor is turned on into saturation and that puts the +Vcc voltage across the primary.
2. The primary current starts to increase.
3. Now either one of these two:
3a. The primary current increases linearly to the point where the transistor starts to come out of sat.
3b. The primary current increases linearly to the point where the primary causes the core to start to saturate and so the primary current shoots up quickly and once it reaches the critical point the transistor comes out of sat fairly fast.
4. The base becomes reverse biased turning off the transistor.
5. Some RC time constant starts charging.
6. The RC time constant times out, turning the transistor back on and the whole cycle starts all over again.

Thank you so much. I really appreciate all your effort. I really learn a lot from you, hoping to learn more.
I'll quickly start building my first self oscillating flyback converter to see for my self.

 

Thank you so much. I really appreciate all your effort. I really learn a lot from you, hoping to learn more.
I'll quickly start building my first self oscillating flyback converter to see for my self.

Hello again,

You are welcome, and here is another circuit that works on the idea that the transistor gets forced out of saturation to make the circuit start to turn off. Q2 is the transistor that gets forced out of sat.
Almost cant believe we did this circuit 14 years ago now! Wow. That was probably about the time when white LEDs started to gain popularity. Brinkmann also used this circuit in one of their consumer flashlight with a single 5mm white LED, perhaps with some slightly different values and no diode D2. It used two AA batteries in series as the power source.

The diode was added later to help prevent Q1 from being reverse biased by too high of a reverse voltage and possibly damage the BE junction.

 

Hello again,

You are welcome, and here is another circuit that works on the idea that the transistor gets forced out of saturation to make the circuit start to turn off. Q2 is the transistor that gets forced out of sat.
Almost cant believe we did this circuit 14 years ago now! Wow. That was probably about the time when white LEDs started to gain popularity. Brinkmann also used this circuit in one of their consumer flashlight with a single 5mm white LED, perhaps with some slightly different values and no diode D2. It used two AA batteries in series as the power source.

The diode was added later to help prevent Q1 from being reverse biased by too high of a reverse voltage and possibly damage the BE junction.

But a PNP transistor needs a negative voltage to turn ON but here I see a positive voltage from the battery to the base of the PNP via a 10k resistor

 

But a PNP transistor needs a negative voltage to turn ON but here I see a positive voltage from the battery to the base of the PNP via a 10k resistor

The polarities are relative.

The emitter is more positive than the base which gets a more negative voltage through R1.