Switching power supplies

[Anditechnovire]

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.

 

[Diver300]

I don't think it is very complicated.

When the transistor is turned on, the top of the auxiliary winding (which is connected to the 220 Ohm resistor) is at a +ve voltage. Current will flow through the 0.004 uF capacitor, which will become charged, so the right of the capacitor will end up at a higher voltage than the left.

When the transformer saturates, the auxiliary winding will produce less voltage. Alternatively the 0.004 uF capacitor gets fully charged and the current to the base of the transistor reduces and the transistor starts to turn off, so there is less voltage on the input winding, so less on the auxiliary winding.

Either way, as soon as the voltage on the auxiliary winding falls, the voltage between the left and right of the capacitor can't change quickly, so when the voltage on the right of the capacitor, the output of the auxiliary winding, falls, so does the voltage on the left of the capacitor. A transistor needs about 0.6 V to turn on, so as soon as that voltage falls, it forces the transistor off very quickly.

The IN4148 prevents the transistor base going negative.

(In your circuit, I'm not sure what the zener and the 10uF capacitor do. I suspect that the capacitor is shown the wrong way round)

 

[Anditechnovire]

I don't think it is very complicated.

When the transistor is turned on, the top of the auxiliary winding (which is connected to the 220 Ohm resistor) is at a +ve voltage. Current will flow through the 0.004 uF capacitor, which will become charged, so the right of the capacitor will end up at a higher voltage than the left.

When the transformer saturates, the auxiliary winding will produce less voltage. Alternatively the 0.004 uF capacitor gets fully charged and the current to the base of the transistor reduces and the transistor starts to turn off, so there is less voltage on the input winding, so less on the auxiliary winding.

Either way, as soon as the voltage on the auxiliary winding falls, the voltage between the left and right of the capacitor can't change quickly, so when the voltage on the right of the capacitor, the output of the auxiliary winding, falls, so does the voltage on the left of the capacitor. A transistor needs about 0.6 V to turn on, so as soon as that voltage falls, it forces the transistor off very quickly.

The IN4148 prevents the transistor base going negative.

(In your circuit, I'm not sure what the zener and the 10uF capacitor do. I suspect that the capacitor is shown the wrong way round)

Ok thanks a lot, but does the transistor saturation play a part? And what about the idea that at a certain point, a negative voltage will be induced in the auxiliary winding which will aid in keeping the transistor off until the cycle repeat?

 

[Miracletech]

Let me make it simple.
1. The circuit before the transformer makes a different kind of AC signal, or let's say 'Pulsating DC' across the transformers primary winding.
2. The secondary winding steps down the AC voltage and then makes it safe to power simple devices.
3. The output diode rectifies that pulsating stepped down dc making it stable, and then filtered by the output capacitor.

The key about the SMPS is that the transistor is switching the transformer really fast ( 20-40khz) to produce an 'AC' signal.
The transistors saturation is when the transistor is fully 'ON', so that's how it works (simple version). ;-)

 

[Diver300]

Transistor saturation is where the transistor is turned on completely. In saturation the collector to emitter voltage is small and there is more base current than is needed for the collector current.

In a switching regulator the transistor is just about always either saturated or completely turned off. There will be a very small time where the transistor is in the process of turning on or turning off, where it's conducting but not saturated. During that time heat will be generated in the transistor so the design will try to make those time short.

When the transistor has just turned off, the auxiliary winding voltage will go negative, which will help turn the transistor off. However the 0.004 uF capacitor means that the transistor would be turned off whenever the voltage of the auxiliary winding falls, whether it goes negative or not.

I am now almost certain that the 10 uF capacitor connected via a diode to the auxiliary winding is shown the wrong way round. The phasing of the transformer isn't shown. There should be a dot at the top of the primary winding and the auxiliary winding, and a dot at the bottom of the secondary winding. If you get the transformer phasing wrong, the circuit won't work or the output voltage won't be regulated.

 

[Miracletech]

I suspected that, because the circuit looks strange to me.

 

[Anditechnovire]

Transistor saturation is where the transistor is turned on completely. In saturation the collector to emitter voltage is small and there is more base current than is needed for the collector current.

In a switching regulator the transistor is just about always either saturated or completely turned off. There will be a very small time where the transistor is in the process of turning on or turning off, where it's conducting but not saturated. During that time heat will be generated in the transistor so the design will try to make those time short.

When the transistor has just turned off, the auxiliary winding voltage will go negative, which will help turn the transistor off. However the 0.004 uF capacitor means that the transistor would be turned off whenever the voltage of the auxiliary winding falls, whether it goes negative or not.

I am now almost certain that the 10 uF capacitor connected via a diode to the auxiliary winding is shown the wrong way round. The phasing of the transformer isn't shown. There should be a dot at the top of the primary winding and the auxiliary winding, and a dot at the bottom of the secondary winding. If you get the transformer phasing wrong, the circuit won't work or the output voltage won't be regulated.

Ok thanks, I really appreciate your explanation. Pl....ease can you please put more highlight on how the 0.004uf capacitor aid in switching off the transistor, and also what causes this auxiliary winding current to drop.

 

[Anditechnovire]

Let me make it simple.
1. The circuit before the transformer makes a different kind of AC signal, or let's say 'Pulsating DC' across the transformers primary winding.
2. The secondary winding steps down the AC voltage and then makes it safe to power simple devices.
3. The output diode rectifies that pulsating stepped down dc making it stable, and then filtered by the output capacitor.

The key about the SMPS is that the transistor is switching the transformer really fast ( 20-40khz) to produce an 'AC' signal.
The transistors saturation is when the transistor is fully 'ON', so that's how it works (simple version). ;-)

I know the basic principles of a smps, but I really want to know deeply how this particular configuration works, because it looks simple, with fewer component.

 

[Diver300]

Ok thanks, I really appreciate your explanation. Pl....ease can you please put more highlight on how the 0.004uf capacitor aid in switching off the transistor, and also what causes this auxiliary winding current to drop.

The auxiliary winding never has much current in it. It is the voltage that is important.

I'll explain the circuit as I understand it. I'll have to guess some voltages because they depend on the turns ratio of the transformer.

When the circuit is energised, all the capacitors will be discharged. The voltage at 1 will be the same as V+, and all other voltages will be 0. Current will flow from V+ to 2 through the 1M resistor. That current will flow through C2, charging it up, and though the 220 Ohm resistor and the auxiliary winding. Neither the resistor nor the winding will have any significant voltage drop across them, so 3 and 4 will stay at 0 V.

When 2 gets to about 0.7 V, the transistor starts to turn on, so the voltage at 1 falls. As it does, the voltage at 4 and at 3 rise. C2 is still charged with 2 being about 0.7 V higher than 3, so this makes the transistor turn on more, so very soon, much less than a microsecond after the transistor started to turn on, the transistor is on fully.

Now 1 is at 0 V. I'll guess that 4 is at +10 V. 2 is still at about 0.7 V and 3 is still at about 0V, so there is 10 V across the 220 Ohm resistor and so around 40 mA flows, starting to charge C2 the other way, and keeping the transistor turned on.

With 40 mA flowing through C2, the voltage across it will be changing at about 10 V/us, so in about 1 us it will be charged to 10 V and there will be no current to keep the transistor turned on. If there is a lot of current in the primary winding, that current will be flowing through the 47 Ohm resistor, increasing the emitter voltage on the transistor, so 2 will have to be at a high voltage for the transistor to say turned on, so it will turn off sooner. Also the transformer may start to saturate, which will cause the current to increase faster, and the voltage on the auxiliary winding at 4 will reduce.

Anyhow, there is now a significant current flowing in the transformer primary winding, so there is significant magnetic energy in the core of the transformer, and the transistor is just starting to turn off. Because the voltage across the primary winding is reduced by the transistor turning off, the voltage at 4, which is produced by the auxiliary winding, is also reduced. Because C2 is now charged to about 10 V with 3 higher than 2, as soon as the voltage at 4 reduces, so does the voltage at 2, turning the transistor off very quickly.

Once the transistor is off, the voltage across the windings reverses, so 1 goes higher than V+, and 4 goes negative. The voltage at 6 will go positive, and current will flow through the IN5402 and start to charge the output capacitor. The output of the auxiliary winding also goes negative, so 2 goes negative, limited by the 1N4148, and C2 is charged the other way, so that 2 is at a more positive voltage than 3.

When the energy from the core of the transformer has all gone, the process starts again. Because 2 starts at a more positive voltage than 3, once the process has been run once, the current through the 1 M resistor isn't needed.

There are a couple of other points about this circuit. The auxiliary winding seems to have the same number of turns as the output, as they are labelled as having the same inductance. When the voltage at 6 is +ve, the voltage at 4 will be -ve, but the same magnitude. As the capacitor on the output is charged, the capacitor at 5 will be charged to the same voltage, but negative. (The capacitor is shown the wrong way round). When that gets to about -5 V, the zener diode will turn on and take current from the base of the transistor, preventing the transistor from turning on. That means that the output can't exceed +5 V, so the output is regulated. It may not be very good regulation.

The idea of the 1N4007 and the 10 nF, 1kV capacitor (labelled 103 1kV) is to absorb the switching spikes from the transformer. When the current in the transformer is stopped quickly, imperfections in the transformer, called "leakage inductance", means that some of the transformer's energy has to be absorbed from the primary. Most will be absorbed in the output capacitor, but some is on the primary. The capacitor absorbs that energy. The circuit is missing a resistor to discharge that capacitor. It will need maybe 100 kOhm in parallel, or it will get charged to a larger voltage each time the transistor switches off, so that would end up not absorbing the spikes and that transistor would be damaged.

I hope this helps.

 

[Anditechnovire]

The auxiliary winding never has much current in it. It is the voltage that is important.

View attachment 125399

I'll explain the circuit as I understand it. I'll have to guess some voltages because they depend on the turns ratio of the transformer.

When the circuit is energised, all the capacitors will be discharged. The voltage at 1 will be the same as V+, and all other voltages will be 0. Current will flow from V+ to 2 through the 1M resistor. That current will flow through C2, charging it up, and though the 220 Ohm resistor and the auxiliary winding. Neither the resistor nor the winding will have any significant voltage drop across them, so 3 and 4 will stay at 0 V.

When 2 gets to about 0.7 V, the transistor starts to turn on, so the voltage at 1 falls. As it does, the voltage at 4 and at 3 rise. C2 is still charged with 2 being about 0.7 V higher than 3, so this makes the transistor turn on more, so very soon, much less than a microsecond after the transistor started to turn on, the transistor is on fully.

Now 1 is at 0 V. I'll guess that 4 is at +10 V. 2 is still at about 0.7 V and 3 is still at about 0V, so there is 10 V across the 220 Ohm resistor and so around 40 mA flows, starting to charge C2 the other way, and keeping the transistor turned on.

With 40 mA flowing through C2, the voltage across it will be changing at about 10 V/us, so in about 1 us it will be charged to 10 V and there will be no current to keep the transistor turned on. If there is a lot of current in the primary winding, that current will be flowing through the 47 Ohm resistor, increasing the emitter voltage on the transistor, so 2 will have to be at a high voltage for the transistor to say turned on, so it will turn off sooner. Also the transformer may start to saturate, which will cause the current to increase faster, and the voltage on the auxiliary winding at 4 will reduce.

Anyhow, there is now a significant current flowing in the transformer primary winding, so there is significant magnetic energy in the core of the transformer, and the transistor is just starting to turn off. Because the voltage across the primary winding is reduced by the transistor turning off, the voltage at 4, which is produced by the auxiliary winding, is also reduced. Because C2 is now charged to about 10 V with 3 higher than 2, as soon as the voltage at 4 reduces, so does the voltage at 2, turning the transistor off very quickly.

Once the transistor is off, the voltage across the windings reverses, so 1 goes higher than V+, and 4 goes negative. The voltage at 6 will go positive, and current will flow through the IN5402 and start to charge the output capacitor. The output of the auxiliary winding also goes negative, so 2 goes negative, limited by the 1N4148, and C2 is charged the other way, so that 2 is at a more positive voltage than 3.

When the energy from the core of the transformer has all gone, the process starts again. Because 2 starts at a more positive voltage than 3, once the process has been run once, the current through the 1 M resistor isn't needed.

There are a couple of other points about this circuit. The auxiliary winding seems to have the same number of turns as the output, as they are labelled as having the same inductance. When the voltage at 6 is +ve, the voltage at 4 will be -ve, but the same magnitude. As the capacitor on the output is charged, the capacitor at 5 will be charged to the same voltage, but negative. (The capacitor is shown the wrong way round). When that gets to about -5 V, the zener diode will turn on and take current from the base of the transistor, preventing the transistor from turning on. That means that the output can't exceed +5 V, so the output is regulated. It may not be very good regulation.

The idea of the 1N4007 and the 10 nF, 1kV capacitor (labelled 103 1kV) is to absorb the switching spikes from the transformer. When the current in the transformer is stopped quickly, imperfections in the transformer, called "leakage inductance", means that some of the transformer's energy has to be absorbed from the primary. Most will be absorbed in the output capacitor, but some is on the primary. The capacitor absorbs that energy. The circuit is missing a resistor to discharge that capacitor. It will need maybe 100 kOhm in parallel, or it will get charged to a larger voltage each time the transistor switches off, so that would end up not absorbing the spikes and that transistor would be damaged.

I hope this helps.

Thank you so much, I really appreciate this, it really helped me a lot.

 

[audioguru]

The circuit looks strange because the transistor and transformer operate at double the mains frequency instead of at a high frequency like an ordinary SMPS.

 

[Diver300]

The circuit looks strange because the transistor and transformer operate at double the mains frequency instead of at a high frequency like an ordinary SMPS.

I think that the transistor operates much faster than twice the mains frequency. There is no capacitor to smooth the mains frequency, so the V+ rail will drop very low twice each cycle, but during each half-cycle, the transistor will switch on and off many times.

I don't think that it's a good circuit, and I've already mentioned some mistakes. I think that omitting the smoothing capacitor is another mistake. The would be quite a lot of time during each half-cycle when the oscillator stops, and the output smoothing capacitor would have to do all the work, and there would be no regulation. It would be better to have high-voltage smoothing so that the oscillator would keep going. A high-voltage smoothing capacitor would require a physically smaller capacitor to get the same ripple.

 

[Nigel Goodwin]

The circuit looks strange because the transistor and transformer operate at double the mains frequency instead of at a high frequency like an ordinary SMPS.

Not at all, it's still a high frequency switcher - just a cheap and low spec one - presumably used where requirements are very low.

 

[audioguru]

I see the simple transistor oscillator with a collector transformer winding and a base feedback winding. But the fullwave recifier from the mains is missing a filter capacitor so the oscillator produces only 100Hz bursts of its output. Then the low voltage output of the transformer is halfwave rectified and "filtered" by a very low value 10uF capacitor which filters away the high frequency but leaves the 100Hz pulses..

The circuit came from a website by Swagatam Innovations (I remember them from years ago) and the chat in the forum there has Swag saying that the circuit was copied from another site and a filter capacitor can be added after the fullwave rectifier. Swag could not answer a question about voltages in the circuit or details about the transformer.

 

[Anditechnovire]

I see the simple transistor oscillator with a collector transformer winding and a base feedback winding. But the fullwave recifier from the mains is missing a filter capacitor so the oscillator produces only 100Hz bursts of its output. Then the low voltage output of the transformer is halfwave rectified and "filtered" by a very low value 10uF capacitor which filters away the high frequency but leaves the 100Hz pulses..

The circuit came from a website by Swagatam Innovations (I remember them from years ago) and the chat in the forum there has Swag saying that the circuit was copied from another site and a filter capacitor can be added after the fullwave rectifier. Swag could not answer a question about voltages in the circuit or details about the transformer.

Yeah yeah you're all right. It was from Swagatam, and I ask him multiple questions concerning this related circuit, but he couldn't give me sure replies, that was why I brought it here.
My main concern is how to understand "A self oscillating flyback converter." Forget about this circuit I've posted, it was just an example circuit. Ok I'll just edit it.

 

[Anditechnovire]

I think that the transistor operates much faster than twice the mains frequency. There is no capacitor to smooth the mains frequency, so the V+ rail will drop very low twice each cycle, but during each half-cycle, the transistor will switch on and off many times.

I don't think that it's a good circuit, and I've already mentioned some mistakes. I think that omitting the smoothing capacitor is another mistake. The would be quite a lot of time during each half-cycle when the oscillator stops, and the output smoothing capacitor would have to do all the work, and there would be no regulation. It would be better to have high-voltage smoothing so that the oscillator would keep going. A high-voltage smoothing capacitor would require a physically smaller capacitor to get the same ripple.

I know it's not a good circuit, am not using it for any special purpose either, I wanted to understand it in order to lay the ground knowledge of how this kind of Oscillators work.
My main concern is how to understand "A self oscillating flyback converter." Forget about this circuit I've posted, it was just an example circuit. Ok I'll just edit it.

 

[KeepItSimpleStupid]

Part of the problem with understanding some circuits is getting "it" started. Sometimes part variability makes it start other times it has to be specifically "kicked".

 

[Anditechnovire]

Part of the problem with understanding some circuits is getting "it" started. Sometimes part variability makes it start other times it has to be specifically "kicked".

Please how do you mean.