Simple rcc flyback switching power supplies

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Please am studying about switching power supplies, starting from the simple ones. I've just concluded on joule thief and blocking oscillator, now am progressing to Self oscillating flyback converter(or RCC) and am finding a bit difficulty on how the transistor switches OFF.
It seems the transistor switch off concept is quite complex than the switch ON process. Some say the operating principle is very much the same to that of blocking oscillator, but i think the are some contrast.

https://homemade-circuits.com/wp-content/uploads/2012/05/cheapestsmpscircuit.png
Here is an example of the RCC smps diagram.
 
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Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Please am studying about switching power supplies, starting from the simple ones. I've just concluded on joule thief and blocking oscillator, now am progressing to Self oscillating flyback converter(or RCC) and am finding a bit difficulty on how the transistor switches OFF.
It seems the transistor switch off concept is quite complex than the switch ON process. Some say the operating principle is very much the same to that of blocking oscillator, but i think the are some contrast.

https://homemade-circuits.com/wp-content/uploads/2012/05/cheapestsmpscircuit.png
Here is an example of the RCC smps diagram.
Here I only understand that a small current flows through the 1M resistor to the base of the transistor, to initiate the turn on process. As the transistor is partially ON(active region), collector current will flow and energize the primary winding, as a result, a same polarity voltage will be induce in the feedback winding which will create a positive regenerative feedback and conduct more current to the base of the transistor, and the transistor turn on fully(saturation).
That's all I know. I don't know how the transistor will switch OFF. I might make some assumptions based on my knowledge of the joule thief circuit, but am not sure.
I don't also know how the RC circuit affect the switching frequency.
 

OBW0549

Joined Mar 2, 2015
3,457
I'm not too familiar with this kind of circuit, but my guess would be that the transistor is switched off by the reversal of drive polarity when the core of T1 saturates; or it may be that the transistor itself comes out of saturation. With the collector current no longer increasing, the drive then disappears.

But that's just a wild guess.
 

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
I'm not too familiar with this kind of circuit, but my guess would be that the transistor is switched off by the reversal of drive polarity when the core of T1 saturates; or it may be that the transistor itself comes out of saturation. With the collector current no longer increasing, the drive then disappears.

But that's just a wild guess.
Yeah your pretty close to the ones I sometimes read about.
I really appreciate your effort in responding.If at all you have a collegue that is familiar with it, you might also want to link him, if that will be ok with you.
Thanks.
 

MrAl

Joined Jun 17, 2014
7,672
Here I only understand that a small current flows through the 1M resistor to the base of the transistor, to initiate the turn on process. As the transistor is partially ON(active region), collector current will flow and energize the primary winding, as a result, a same polarity voltage will be induce in the feedback winding which will create a positive regenerative feedback and conduct more current to the base of the transistor, and the transistor turn on fully(saturation).
That's all I know. I don't know how the transistor will switch OFF. I might make some assumptions based on my knowledge of the joule thief circuit, but am not sure.
I don't also know how the RC circuit affect the switching frequency.
Hi, i just saw this thread,

In a circuit like that there are two ways that the transistor could turn off.
The first is the inductance of the core and the capacitance MAY form a resonant circuit and when the polarity changes so does the base drive. For this scenario the transistor drive comes from the resonant cycling alone.
The second is the primary current increase as a result of the applied voltage and the inductance of the windings and the time it stays on, and as the current increases the collector current increases, and as the collector current increases the transistor eventually can not sustain the collector current due to limited base drive and limited Beta, so the transistor is forced out of saturation. As it comes out of sat, the increasing voltage changes the base drive polarity (eventually or quickly) and so the transistor is now turned off completely by the base drive change in polarity. There is also a chance that the primary saturates and that pushes the collector current up quickly and so the transistor comes out of sat quickly and with the change in collector voltage (rising) the base drive polarity changes and turns the transistor off completely.

So in one case the transistor turns off by virtue of the cycling due to the resonant frequency, and in the other case the transistor turns off because it gets forced out of saturation by the increasing collector current. The latter is the case with many different converters like these, the resonate case can only happen if the inductance and added capacitance make it resonate.

I just noticed the RC coupling to the base. That brings up a third possibility which is the RC time constant times out and reduces base drive.

To find out which one you'd have to do some simulations and look carefully at what is happening at a few nodes and some current levels in a few places.
It may be hard to see if it is the case where the transistor is forced out of sat because at that point everything changes.
 
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Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Hi, i just saw this thread,

In a circuit like that there are two ways that the transistor could turn off.
The first is the inductance of the core and the capacitance MAY form a resonant circuit and when the polarity changes so does the base drive. For this scenario the transistor drive comes from the resonant cycling alone.
The second is the primary current increase as a result of the applied voltage and the inductance of the windings and the time it stays on, and as the current increases the collector current increases, and as the collector current increases the transistor eventually can not sustain the collector current due to limited base drive and limited Beta, so the transistor is forced out of saturation. As it comes out of sat, the increasing voltage changes the base drive polarity (eventually or quickly) and so the transistor is now turned off completely by the base drive change in polarity. There is also a chance that the primary saturates and that pushes the collector current up quickly and so the transistor comes out of sat quickly and with the change in collector voltage (rising) the base drive polarity changes and turns the transistor off completely.

So in one case the transistor turns off by virtue of the cycling due to the resonant frequency, and in the other case the transistor turns off because it gets forced out of saturation by the increasing collector current. The latter is the case with many different converters like these, the resonate case can only happen if the inductance and added capacitance make it resonate.

I just noticed the RC coupling to the base. That brings up a third possibility which is the RC time constant times out and reduces base drive.

To find out which one you'd have to do some simulations and look carefully at what is happening at a few nodes and some current levels in a few places.
It may be hard to see if it is the case where the transistor is forced out of sat because at that point everything changes.
Ok I can get the sense of your explanation. You know, one reason I had difficulty understanding was because I thought every kind of RCC converters operate with the same principle.
Now I can know that any circuit designer can choose any of the available ways, provided it accomplish the job.
And also there is this one particular point from your explanation I find common to other ones I read.
It about the transistor 'coming out of saturation'. How does this happen? Does it mean that the transistor moves back to it active region, or is kind of oversaturated.
 

MrAl

Joined Jun 17, 2014
7,672
Ok I can get the sense of your explanation. You know, one reason I had difficulty understanding was because I thought every kind of RCC converters operate with the same principle.
Now I can know that any circuit designer can choose any of the available ways, provided it accomplish the job.
And also there is this one particular point from your explanation I find common to other ones I read.
It about the transistor 'coming out of saturation'. How does this happen? Does it mean that the transistor moves back to it active region, or is kind of oversaturated.
Hello again,

I think the resonate ones are the most efficient but it has been some 40 years since i worked with one like that.

The transistor comes out of saturation in two ways but first how does the transistor operate.
Using the current controlled model in the saturation region the collector current is the base current multiplied by the Beta. So if we have a Beta of 10 and 100ma through the collector we need AT LEAST 10ma into the base (base emitter). IF we have less than 10ma, we may no longer get 100ma in the collector we might only get 90ma. Now if 100ma kept the collector emitter voltage low like 0.5 volts, 90ma might cause that voltage to rise to 0.7 volts depending on the impedance of the collector circuit.
But we can look at this the other way around too. What if we drive the base AND collector with a curent source. With the 10ma and 100ma example we see everything as before with 0.5v CE voltage. But as we increase the collector current now 10ma is not enough and the CE voltage rises to 0.7 volts. So maybe 120ma causes the CE voltage to go up to 0.7v. But then if we increase the collector current more like maybe 140ma, the base current is not enough to keep the transistor in saturation so the CE voltage rises even more, maybe to 1 volt. Further increase causes the CE voltage to rise even farther and thus the transistor is now completely out of saturation and is now in the active region.

So the main idea here is the collector current is deciding whether or not the transistor stays in saturation not the base current because the base current is constant and so can not cause a change in the operating mode.
The change in operating mode causes the collector voltage to rise as the transistor comes out of saturation, and so the rest of the circuit begins to react to that change. That reaction then causes the base drive to flip polarity and then the base emitter becomes reverse biased and that turns the transistor off and thus it enters the second half of the switching cycle where the transistor is completely off. Later, other timing components cause the transistor to turn back on and that starts the beginning of the first half of the switching cycle again.

In order to see this effect though you might have to first test the transistor to see what current level will force it out of saturation with the expected constant (or slowly changing) base drive. That way you can tell if it is the collector current pulling it out of sat or the base current, and it should be the collector current that does it.

As mentioned before, the base is capacitively coupled so the turn off mechanism may be that the base RC times out and thus the base current actually reduces and that causes the end of the saturation mode. To tell which it is, you have to do some measurements or do some careful circuit analysis.
 
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Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Hello again,

I think the resonate ones are the most efficient but it has been some 40 years since i worked with one like that.

The transistor comes out of saturation in two ways but first how does the transistor operate.
Using the current controlled model in the saturation region the collector current is the base current multiplied by the Beta. So if we have a Beta of 10 and 100ma through the collector we need AT LEAST 10ma into the base (base emitter). IF we have less than 10ma, we may no longer get 100ma in the collector we might only get 90ma. Now if 100ma kept the collector emitter voltage low like 0.5 volts, 90ma might cause that voltage to rise to 0.7 volts depending on the impedance of the collector circuit.
But we can look at this the other way around too. What if we drive the base AND collector with a curent source. With the 10ma and 100ma example we see everything as before with 0.5v CE voltage. But as we increase the collector current now 10ma is not enough and the CE voltage rises to 0.7 volts. So maybe 120ma causes the CE voltage to go up to 0.7v. But then if we increase the collector current more like maybe 140ma, the base current is not enough to keep the transistor in saturation so the CE voltage rises even more, maybe to 1 volt. Further increase causes the CE voltage to rise even farther and thus the transistor is now completely out of saturation and is now in the active region.

So the main idea here is the collector current is deciding whether or not the transistor stays in saturation not the base current because the base current is constant and so can not cause a change in the operating mode.
The change in operating mode causes the collector voltage to rise as the transistor comes out of saturation, and so the rest of the circuit begins to react to that change. That reaction then causes the base drive to flip polarity and then the base emitter becomes reverse biased and that turns the transistor off and thus it enters the second half of the switching cycle where the transistor is completely off. Later, other timing components cause the transistor to turn back on and that starts the beginning of the first half of the switching cycle again.

In order to see this effect though you might have to first test the transistor to see what current level will force it out of saturation with the expected constant (or slowly changing) base drive. That way you can tell if it is the collector current pulling it out of sat or the base current, and it should be the collector current that does it.

As mentioned before, the base is capacitively coupled so the turn off mechanism may be that the base RC times out and thus the base current actually reduces and that causes the end of the saturation mode. To tell which it is, you have to do some measurements or do some careful circuit analysis.
Thank you very much, I'll quickly set up the circuit, using the information you give me and also do some testing evaluations.
You have always been a source of help to me .
I really appreciate that.
 

MrAl

Joined Jun 17, 2014
7,672
Thank you very much, I'll quickly set up the circuit, using the information you give me and also do some testing evaluations.
You have always been a source of help to me .
I really appreciate that.
Hello again,

Oh sure you are welcome. When you study and work in a field for as long as i have you kind of like to share the knowledge you've learned over the years. Sometimes you dont remember everything though :)
 

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Hello again,

Oh sure you are welcome. When you study and work in a field for as long as i have you kind of like to share the knowledge you've learned over the years. Sometimes you dont remember everything though :)
It will be my pleasure to invite you again, anytime I have a problem
 

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Feel free to mention whatever it is that you dont get yet.
Hi, check out this statement from the site.

"As Ip=Vp/Lp * t, the current in PRI winding (5-6) path begins increasing linearly, and Ip will reach the critical point finally – Ip=Ib*B (beta), beyond this point, according to transistor characteristic curves,
A little Ic (equals to Ip here) increasing will cause a bigger Vce increasing, this means the Q1 will leave saturation region, and Vce will increase dramatically.
Therefore, the voltage on PRI winding will begin to decrease, the voltage on AUX winding will decrease too, this will make transistor base current decreasing, it is positive feedback process again, and will lead to transistor turn-off quickly."

So please when is the transformer said to reach a "critical point"?
Is it when it is saturated?, and how does it also affect the collector current?

How does the transistor characteristics curve also relate to the behaviour of the collector-emitter voltage (Vce)?
I'll really get it now if i understand this things.
 

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
And these "things" are? What bothers you?
Hi, check out this statement from the site.

"As Ip=Vp/Lp * t, the current in PRI winding (5-6) path begins increasing linearly, and Ip will reach the critical point finally – Ip=Ib*B (beta), beyond this point, according to transistor characteristic curves,
A little Ic (equals to Ip here) increasing will cause a bigger Vce increasing, this means the Q1 will leave saturation region, and Vce will increase dramatically.
Therefore, the voltage on PRI winding will begin to decrease, the voltage on AUX winding will decrease too, this will make transistor base current decreasing, it is positive feedback process again, and will lead to transistor turn-off quickly."

So please when is the transformer said to reach a "critical point"?
Is it when it is saturated?, and how does it also affect the collector current?

How does the transistor characteristics curve also relate to the behaviour of the collector-emitter voltage (Vce)?
I'll really get it now if i understand this things.
 

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
Hello again,

I think the resonate ones are the most efficient but it has been some 40 years since i worked with one like that.

The transistor comes out of saturation in two ways but first how does the transistor operate.
Using the current controlled model in the saturation region the collector current is the base current multiplied by the Beta. So if we have a Beta of 10 and 100ma through the collector we need AT LEAST 10ma into the base (base emitter). IF we have less than 10ma, we may no longer get 100ma in the collector we might only get 90ma. Now if 100ma kept the collector emitter voltage low like 0.5 volts, 90ma might cause that voltage to rise to 0.7 volts depending on the impedance of the collector circuit.
But we can look at this the other way around too. What if we drive the base AND collector with a curent source. With the 10ma and 100ma example we see everything as before with 0.5v CE voltage. But as we increase the collector current now 10ma is not enough and the CE voltage rises to 0.7 volts. So maybe 120ma causes the CE voltage to go up to 0.7v. But then if we increase the collector current more like maybe 140ma, the base current is not enough to keep the transistor in saturation so the CE voltage rises even more, maybe to 1 volt. Further increase causes the CE voltage to rise even farther and thus the transistor is now completely out of saturation and is now in the active region.

So the main idea here is the collector current is deciding whether or not the transistor stays in saturation not the base current because the base current is constant and so can not cause a change in the operating mode.
The change in operating mode causes the collector voltage to rise as the transistor comes out of saturation, and so the rest of the circuit begins to react to that change. That reaction then causes the base drive to flip polarity and then the base emitter becomes reverse biased and that turns the transistor off and thus it enters the second half of the switching cycle where the transistor is completely off. Later, other timing components cause the transistor to turn back on and that starts the beginning of the first half of the switching cycle again.

In order to see this effect though you might have to first test the transistor to see what current level will force it out of saturation with the expected constant (or slowly changing) base drive. That way you can tell if it is the collector current pulling it out of sat or the base current, and it should be the collector current that does it.

As mentioned before, the base is capacitively coupled so the turn off mechanism may be that the base RC times out and thus the base current actually reduces and that causes the end of the saturation mode. To tell which it is, you have to do some measurements or do some careful circuit analysis.
At the first glimes, I didn't really get this answer, but as I went through this answer over again, it is now I fully understand this explanation. I really appreciate this, it even make me understand more about transistor operating principle.
Thanks a lot.
Sometimes, in science you just have to be patient, and read it over and over again.
Am just a teenage beginner in electronics, and this experience has really enlighten me.
 

MrAl

Joined Jun 17, 2014
7,672
At the first glimes, I didn't really get this answer, but as I went through this answer over again, it is now I fully understand this explanation. I really appreciate this, it even make me understand more about transistor operating principle.
Thanks a lot.
Sometimes, in science you just have to be patient, and read it over and over again.
Am just a teenage beginner in electronics, and this experience has really enlighten me.
You're welcome and here is a little more info...

Yes the transistor is a three terminal device so it is quite a bit more complicated than a two terminal device like a resistor.
From the standpoint of voltage alone the two terminal resistor can be driven 2 different ways:
+-
-+

the transistor because of just one more terminal can be driven 8 different ways::
---
--+
-+-
-++
+--
+-+
++-
+++

The circuit we had been talking about is driven with the base and with the collector, and using the collector in this way is very unusual because most circuits are either:
common emitter: base driven
common collector: base driven
common base: emitter driven

so this kind of circuit is a more unusual case where we have to consider what happens when we keep the base and emitter constant but change the collector current.

BTW sometimes the primary current increases linearly for a while but hen shoots up quickly as the core of the transformer or inductor starts to enter saturation and that becomes the point where the transistor comes out of sat.

When we look at the saturation characteristic we have to consider four things:
1. Base current iB
2. Collector current iC
3. Beta (sometimes written as 'B' or upper case Greek beta)
4. Collector emitter voltage (Vce).
To discover how this works in more detail we could set up a simulation with a transistor with a current source for the base current and a second current source for the collector current, and as we increase collector current we watch the collector emitter voltage and see how it starts to climb as it is forced out of saturation.
 

Thread Starter

anditechnovire

Joined Dec 24, 2019
93
You're welcome and here is a little more info...

Yes the transistor is a three terminal device so it is quite a bit more complicated than a two terminal device like a resistor.
From the standpoint of voltage alone the two terminal resistor can be driven 2 different ways:
+-
-+

the transistor because of just one more terminal can be driven 8 different ways::
---
--+
-+-
-++
+--
+-+
++-
+++

The circuit we had been talking about is driven with the base and with the collector, and using the collector in this way is very unusual because most circuits are either:
common emitter: base driven
common collector: base driven
common base: emitter driven

so this kind of circuit is a more unusual case where we have to consider what happens when we keep the base and emitter constant but change the collector current.

BTW sometimes the primary current increases linearly for a while but hen shoots up quickly as the core of the transformer or inductor starts to enter saturation and that becomes the point where the transistor comes out of sat.

When we look at the saturation characteristic we have to consider four things:
1. Base current iB
2. Collector current iC
3. Beta (sometimes written as 'B' or upper case Greek beta)
4. Collector emitter voltage (Vce).
To discover how this works in more detail we could set up a simulation with a transistor with a current source for the base current and a second current source for the collector current, and as we increase collector current we watch the collector emitter voltage and see how it starts to climb as it is forced out of saturation.
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.
 

Jony130

Joined Feb 17, 2009
5,176
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.

I think a reasonably simple yet useful description of the saturation process goes something like this:

Think of a transistor as a device that tries to create a situation in which the collector current, Ic, is β*Ib. The mechanism by which it does this is by controlling the "resistance" of the collector-emitter path through the device. If the present resistance is such that the collector current is less than β*Ib, then the transistor reduces the resistance. This generally results in a lower collector-emitter voltage which means that the voltage dropped across the external components increases which generally allows more collector current to flow. The reverse happens if the present collector current is more than β*Ib in order to reduce it.

But this process can only be carried so far. The transistor can only lower the collector-emitter voltage so far and, once that point is reached, the transistor no longer has any tools in its bag to increase the collector current up to the point it would like it to be, so the collector-emitter voltage stops dropping and the collector current is less β*Ib.

Ideally this saturation voltage would be 0V and the collector-emitter junction would simply look like a short circuit allowing any current to flow as long as it doesn't go above β*Ib (since, at that point, the transistor will come out of saturation and start increasing Vce in order to hold the current to β*Ib).

In practice, real transistors have a saturation voltage that is greater than 0V, but it is generally only a few hundred millivolts.

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

 
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