Instead of a single diode at the output let's imagine there's a full-bridge rectifier(The rest is just the same.). What happens then? I've actually tried this and surpsingly nothing changed and I got te same output voltage. How can you explain such a thing?
First I'll say that diagnosing issues specific to someone's implementation is challenging, even in person. However, I can say from a circuit theory perspective I would expect a change in operation. Assuming CCM, one pair of diodes in the full-bridge would function as the original diode would. The second pair of diodes provide a path for the input voltage to directly connect to the output voltage when the main switch is on. Obviously this current can only flow when Vg reflected to the secondary side is greater than Vout, when D > 0.5. This means the output voltage has a lower limit equal to the reflected input voltage. Beyond D = 0.5, the converter should operate as a conventional flyback because the second pair of diodes would never turn on. Below D = 0.5, the inductor current would likely fall into DCM and could go negative as the output voltage (charged directly to the reflected Vg) would discharge it more than it is charged by the input voltage (D < D" in this case). While operating in this mode, the output voltage would actually rise beyond Vg as you continue to transfer energy to the output through the magnetizing inductance. Depending on the load current, the output voltage would vary. If you experienced something otherwise, I'd like to hear it. I'm just throwing out thoughts I haven't simulated.
First of all, thank you for your response and your channel which is full of quality contents. In the circuit I mentioned D is always lower than 0.5. I couldn't understand the D below 0.5 part of your explanation exactly. I explain this circuit in this way: when the switch is on the voltage is reflected to the secondary by n times the input voltage and charge the output capacitor with that voltage. By the way D is 0.45 during the rising time of the output voltage. But I don't think D matters for this configuration(as long as it stays below 0.5). And when the switch is off sudden change in current through the primary creates high flux density through the secondary and that effect may contribute the voltage if it's high enough to forward bias the diode. Last when the voltage is at the desired level the feedback mechanism takes place and makes D=0.01 I think this is for compensating the output drop which is caused by the feedback. Does my explanation make sense to you?
@@resattr7425 I would say trying to derive the operation of a new circuit can be made more challenging if you close a control loop around the plant. Your explanation is not quite precise and misses several key aspects of the operation, notably the discontinuous current of the magnetizing inductance. Consider operating with a fixed duty ratio and varying the load current.
Good question. The converter is named after the transformer. After digging around the flyback transformer gets its name from one of it's most well known application: producing and controlling the electron beam in a CRT television. The beam scans across the screen, then flies back to start again. I have no way of proving this, but it sounds good! www.watelectrical.com/flyback-transformer/
The time you waste "writing" needs to be eliminated. Write it and say simultaneously, otherwise it becomes annoying, watching you spell words. Or have the power point slide pre-printed. This video is not an effective use of my time.
This presentation is a complete nonsense! The flyback transformer does NOT have magnetizing inductance!? The very definition of magnetizing inductance in a real transformer is that it DOES NOT store energy. In fact, the beauty of the 60Hz isolation transformer and High Frequency switching converters is that is that its magnetizing current (core without any air-gap!) a is huge, thus making the magnetizing current extremely small! Yet this small circulating AC current enables a huge 99% of input power to be transferred immediately to output WITHOUT ANY STORAGE! Flyback "transformer" is in fact, two winding inductances which stores all input power during on-time and releases it during off time. The consequence is that only small current and power can be used with large size magnetics and very low efficiency! You must have learned it from Professor Emeritus Ben Yaakov! However, he has 40,000 followers on his UA-cam channel so his damage to Power Electronics community is exponentially much worse.
What are you talking about? I watched the whole thing and didn't hear him say even once that a flyback transformer doesn't have magnetizing inductance. I also searched the the transcript for "magnetizing inductance" and he accurately describes several times how the magnetizing inductance is used to store energy during the on-time and then release that stored energy to the secondary side during the off-time. You can't even write out a comment without contradicting yourself and making errors that would probably cause Professor Sam Ben Yaakov to give you an F and tell you to pick a new major if you were one of his students. You stated... "In fact, the beauty of the 60Hz isolation transformer and High Frequency switching converters is that is that its magnetizing current (core without any air-gap!) a is huge, thus making the magnetizing current extremely small!" So, is the magnetizing current "huge" or "extremely small"? I know you probably meant to write that the magnetizing INDUCTANCE is huge, thus magnetizing current is extremely small. Even this is still an overly broad statement though. You also stated, "The very definition of magnetizing inductance in a real transformer is that it DOES NOT store energy." This is false. Try running a simple single ended forward converter at a duty cycle greater than 50% without any sort of reset winding, active clamp, resonant reset capacitor, etc... and see what happens. REAL transformers don't have infinite inductance, IDEAL transformers do. That's why reset circuits exist. Again you can't even type out a comment without making an error. You also also stated "Yet this small circulating AC current enables a huge 99% of input power to be transferred immediately to output WITHOUT ANY STORAGE!". This completely neglects the fact that these converters need an output inductor. That's hardly "WITHOUT ANY STORAGE" as you chose to put in all caps for some reason. I'm just some guy who is interested in power electronics and doesn't have any sort of degree/formal training and even I was able to spot all these issues with your comment. I'm not even going to get into your absurd assertions that forward mode converters actually deliver 99% efficiency in the real world (only at certain narrow line/load conditions) or that flybacks are always bulky, inefficient, and low current. There are obviously reasons why different applications with varying line/load conditions, budgets, size/weight/performance requirements, etc... would cause a designer to go with one topology vs another. If this is how poorly you communicate about power electronics in writing I can't even imagine how much you get wrong when speaking in your videos. That's probably why hardly anybody watches them.
Instead of a single diode at the output let's imagine there's a full-bridge rectifier(The rest is just the same.). What happens then?
I've actually tried this and surpsingly nothing changed and I got te same output voltage. How can you explain such a thing?
First I'll say that diagnosing issues specific to someone's implementation is challenging, even in person. However, I can say from a circuit theory perspective I would expect a change in operation. Assuming CCM, one pair of diodes in the full-bridge would function as the original diode would. The second pair of diodes provide a path for the input voltage to directly connect to the output voltage when the main switch is on. Obviously this current can only flow when Vg reflected to the secondary side is greater than Vout, when D > 0.5. This means the output voltage has a lower limit equal to the reflected input voltage. Beyond D = 0.5, the converter should operate as a conventional flyback because the second pair of diodes would never turn on. Below D = 0.5, the inductor current would likely fall into DCM and could go negative as the output voltage (charged directly to the reflected Vg) would discharge it more than it is charged by the input voltage (D < D" in this case). While operating in this mode, the output voltage would actually rise beyond Vg as you continue to transfer energy to the output through the magnetizing inductance. Depending on the load current, the output voltage would vary.
If you experienced something otherwise, I'd like to hear it. I'm just throwing out thoughts I haven't simulated.
First of all, thank you for your response and your channel which is full of quality contents.
In the circuit I mentioned D is always lower than 0.5.
I couldn't understand the D below 0.5 part of your explanation exactly.
I explain this circuit in this way: when the switch is on the voltage is reflected to the secondary by n times the input voltage and charge the output capacitor with that voltage. By the way D is 0.45 during the rising time of the output voltage. But I don't think D matters for this configuration(as long as it stays below 0.5).
And when the switch is off sudden change in current through the primary creates high flux density through the secondary and that effect may contribute the voltage if it's high enough to forward bias the diode.
Last when the voltage is at the desired level the feedback mechanism takes place and makes D=0.01 I think this is for compensating the output drop which is caused by the feedback.
Does my explanation make sense to you?
@@resattr7425 I would say trying to derive the operation of a new circuit can be made more challenging if you close a control loop around the plant.
Your explanation is not quite precise and misses several key aspects of the operation, notably the discontinuous current of the magnetizing inductance. Consider operating with a fixed duty ratio and varying the load current.
Why it is named as flyback converter
Good question. The converter is named after the transformer. After digging around the flyback transformer gets its name from one of it's most well known application: producing and controlling the electron beam in a CRT television. The beam scans across the screen, then flies back to start again. I have no way of proving this, but it sounds good!
www.watelectrical.com/flyback-transformer/
The time you waste "writing" needs to be eliminated. Write it and say simultaneously, otherwise it becomes annoying, watching you spell words. Or have the power point slide pre-printed. This video is not an effective use of my time.
This presentation is a complete nonsense! The flyback transformer does NOT have magnetizing inductance!? The very definition of magnetizing inductance in a real transformer is that it DOES NOT store energy. In fact, the beauty of the 60Hz isolation transformer and High Frequency switching converters is that is that its magnetizing current (core without any air-gap!) a is huge, thus making the magnetizing current extremely small! Yet this small circulating AC current enables a huge 99% of input power to be transferred immediately to output WITHOUT ANY STORAGE! Flyback "transformer" is in fact, two winding inductances which stores all input power during on-time and releases it during off time. The consequence is that only small current and power can be used with large size magnetics and very low efficiency!
You must have learned it from Professor Emeritus Ben Yaakov! However, he has 40,000 followers on his UA-cam channel so his damage to Power Electronics community is exponentially much worse.
What are you talking about? I watched the whole thing and didn't hear him say even once that a flyback transformer doesn't have magnetizing inductance. I also searched the the transcript for "magnetizing inductance" and he accurately describes several times how the magnetizing inductance is used to store energy during the on-time and then release that stored energy to the secondary side during the off-time.
You can't even write out a comment without contradicting yourself and making errors that would probably cause Professor Sam Ben Yaakov to give you an F and tell you to pick a new major if you were one of his students.
You stated... "In fact, the beauty of the 60Hz isolation transformer and High Frequency switching converters is that is that its magnetizing current (core without any air-gap!) a is huge, thus making the magnetizing current extremely small!" So, is the magnetizing current "huge" or "extremely small"? I know you probably meant to write that the magnetizing INDUCTANCE is huge, thus magnetizing current is extremely small. Even this is still an overly broad statement though.
You also stated, "The very definition of magnetizing inductance in a real transformer is that it DOES NOT store energy." This is false. Try running a simple single ended forward converter at a duty cycle greater than 50% without any sort of reset winding, active clamp, resonant reset capacitor, etc... and see what happens. REAL transformers don't have infinite inductance, IDEAL transformers do. That's why reset circuits exist. Again you can't even type out a comment without making an error.
You also also stated "Yet this small circulating AC current enables a huge 99% of input power to be transferred immediately to output WITHOUT ANY STORAGE!". This completely neglects the fact that these converters need an output inductor. That's hardly "WITHOUT ANY STORAGE" as you chose to put in all caps for some reason.
I'm just some guy who is interested in power electronics and doesn't have any sort of degree/formal training and even I was able to spot all these issues with your comment. I'm not even going to get into your absurd assertions that forward mode converters actually deliver 99% efficiency in the real world (only at certain narrow line/load conditions) or that flybacks are always bulky, inefficient, and low current. There are obviously reasons why different applications with varying line/load conditions, budgets, size/weight/performance requirements, etc... would cause a designer to go with one topology vs another. If this is how poorly you communicate about power electronics in writing I can't even imagine how much you get wrong when speaking in your videos. That's probably why hardly anybody watches them.