Since my first Yr in 2018 enrolled in school of Engineering, this channel has been a great resource. Towards the finishing line and am still here. Awesome work guys
Fantastic explanation with a very easy to understand steps during the tutorial. Then with the math added to demonstrate the outcome. Great Job! I have read a couple other descriptions ahead of this video and this video completed my understanding. I graduated college in 1981 with a Biomedical Engineering Degree. However, have forgotten most of the rudimentary electronics over the years. This was really easy to understand and explained every element of the circuit well! Thank you!
VL = vin-vout. Not the way you showed, otherwise it would a negative voltage across the inductor. Good videos but this point got me confused so I just went back to KVL.
Thank you for all your help and for providing great content Julio. Question, do you ever respond to comments or make posts on your Community page? I'm sure there's a lot of viewers who would love to connect more and learn more about you. We definitely appreciate your work.
On discharge of the inductor, the voltage will *always* be sufficiently high to deliver current to the capacitor and/or load until it is fully discharged. This is a fundamental property of an inductor. The equation Vout = Vin • d applies *only* if the inductor current is continuous, that is, it is never drops to zero. Once the inductor current becomes discontinuous (spends some time a zero each cycle) the relationship of Vout to Vin has to be calculated based on the energy stored and delivered each cycle.
If the duty cycle is responsible for determining the output voltage, what is the role of inductor in this circuit to reduce the output voltage as you explained in the beginning like "if 5v dropped across the inductor, remaining 7v will be there across the RL" Appreciated if anyone help me to figure out this doubt.
Without the inductor, the RL load would get 12V 'spikes' controlled by the 555 timer instead of the 7V. Let's say RL is a 1.4V CPU, 12V spikes would kill the CPU instantly. You just 'turn off' the COIL before it could charge up to the full potential of the battery (12V).
The inductor stores energy in the magnetic field that develops when current flows through the coil. Buck converters are useful in solar power supplies.
So most step down bucks converters use LM2596 as it's regulator, with a max iput of 40V. If I wanted a down converter that can handle 6ov input, I could use a ? as the regulator. Very well explained brill.
Once you get into voltages and/or currents higher than the integrated switcher ICs can handle you would use a controller IC and a separate switch, typically a power MOSFET. Properly driving the "high side" switch for a buck converter is a little tricky. Even powering the controller IC requires some attention to detail.
It doesn't mention parasitics or ringing, or switching transients, or snubbers, but yeah, that would be overwhelming too early. There is a whole bunch of fun stuff to learn about switching converters. Good video.
I agree with everything you said EXCEPT; if S1 is static on (100% duty cycle), how do you figure you’ll get 7V out of that 12V supply at RL? With no PWM on S1, RL will see 12V after the lag time of your inductor and the R*C time constant of your cap and your load.
At the end of the video, you say that if the converter were 100% efficient, the voltage and current are both modified by a factor by 2. Is it always the case that the factor is the same?
Hi, this video can confuse us learners. Coz there was no parr of the ckt drawn that would provide the duty cycle effect. This cld be devastating to learning if the context is not proper. But I want to thank you for the effort.
If the output voltage is only dependent on the input voltage and the duty cycle as explained in the final part, does that mean a random inductor and capacitor can be chosen for the cycle irrespective of their respective inductance and capacitance?
Thanks for your explained... I need your help to undesrtand about solar panel control, so i can build for myself.. I need to fully understand everything about solar panel control.. Thank you sir..
@@alans2416 I meant it like this way. He has seen the coil as a source, so his voltage directions are the way he intended them to be, which means the voltage must be vin-vout
8 amp is a large value for output current....where it comes from? Show some computations, sir! From the Philippines... What are the values of the inductor, capacitor,diode and resistor?
Hi, I'm not an electrician, neither don't have any science base. I'd like to ask a question about "LM2596S Constant Current" buck converter. Let's say, we have 12V 1A adapter, power supply. Normally when we step down this to 3V, the current should increase in %100 efficiency to 4. I know this converter has 3A limit. So, this much decreasing the V can harm the converter? Additionally, when I regulate the converter to provide constant 0.5 or 1A, where the difference goes? Is this dangerous? I'm interested in modelling and going to use some LEDs. Therefore I'm trying to be cautious not to burn the house :) Thanks!
Very pleasant teaching style, thank you. What happens if your load doesn't use the extra current? does the voltage increase? Is a buck converter a current or a voltage source (or neither)?
In a a real circuit feedback to adjust the duty cycle to maintain the output voltage is essential. If the load doesn't require the energy stored in the inductor then the output voltage will rise and can become equal to the input voltage.
What about circuit with switch in series with coil, (buck type), and needing the 2nd flyback pulse to go forward also- as additive?The diode anode is facing the source plus voltage , and cathode is at end of other diode in series with coil L. In other words 3 diodes, diode ,coil, diode and diode across coil and diode - all cathodes are facing forward, and NO capacitor is present in my reference circuit found from 2008.
Well, and what will happen when the circuit hasn't D1 and switch S1 will open... what will be voltage on sides of the inductor :) Will reach voltage 2V and 7V or somthing different value and it will be related to quality factor of inductor and switching rise time?
If the circuit has not D1, the circuit could blow up. D1 here is called free wheeling diode, to make sure when S1 is open, there is still path for current to flow.
My daughter’s 12v electric Razor scooter is too fast for her. I like to drop the voltage down to 9v. Do you know a buck converter that can handle the scooter without overheating? It’s a 10 mph E90 Razor scooter. Send me a link if you would. Thanks.
Short answer: Since we switch on and of our input voltage periodically we've got an AC voltage which is rectangular in waveform Long answer: Yes and no: If you connect a constant voltage to an inductor in steady state (after an initial transient phase) not much happens. I think this is what you mean by "A DC voltage doesn't charge an inductor". But like mentioned above there is a transient phase: Let the voltage across the inductor be 0V. You then swith on your DC source and the inductor voltage jumps to V_in (input voltage) but the current that would start flowing through the wire is can't jump in an inductor. It starts increasing exponentionally which builds up a magnetic field. This is (roughly explained) the stored energy in the inductor. This transient phase is (depending on the inductance) very short. But since we switch our DC input voltage on and off quickly this happens often, so the stored energy in the inductor drives a current when the switch is off and the magnetic field in the inductor builds back up if the switch is on.
If an inductor is fully discharged (no energy is stored as a magnetic field) and you apply a voltage across it, the current through it will start at zero and rise linearly with respect to time. With an ideal inductor with no resistance there is no limit to the current that could be reached if you allowed sufficient time. With practical inductors that doesn't happen. The main limiting factor for inductors used in switch mode power supplies is the "core." The core can only handle a certain amount of magnetizing force before it "saturates." When the inductor is dicharging in the circuit, the current still flows in the same direction but the polarity of the voltage across it reverses. The discharge is also linear change (decline) of current with respect to time. Given enough time the current will drop to zero. di/dt = V/L the differential of current through the inductor with respect to time (or, the rate of change of current in the inductor) is equal to the voltage across the inductor divided by the inductance. i is in amperes, t is in seconds, V is in volts and L is in henries One very important thing that this equation says but maybe isn't really obvious is that you cannot instantaneously change the current through an inductor.
So when the switch is open the current will flow from negative towards positive of the inductor and so confusing how exactly a current flows through ground and back to the load?
So why do you need an inductor? If you just get a raw PWM signal and put it through a capacitor the capacitor just smooth out the voltage. I built a 555 circuit that is adjustable and the output of the connector to a capacitor and by adjusting the oscillation of 555 timer I can lower the output voltage.
Final Exams and Video Playlists: www.video-tutor.net/
Since my first Yr in 2018 enrolled in school of Engineering, this channel has been a great resource. Towards the finishing line and am still here. Awesome work guys
You are I both😁
Chemistry, Biology, Algebra, Calculus, Electronics, this guy does it all
its insane to me why lecturers dont explain like this. Soo good!
Great job!!! Nicely explained just how I learned it in school. It’s wonderful to get such explanations available on RUclips.
Fantastic explanation with a very easy to understand steps during the tutorial.
Then with the math added to demonstrate the outcome.
Great Job!
I have read a couple other descriptions ahead of this video and this video completed my understanding.
I graduated college in 1981 with a Biomedical Engineering Degree.
However, have forgotten most of the rudimentary electronics over the years.
This was really easy to understand and explained every element of the circuit well!
Thank you!
12 minutes JUST 12 MINUTES!!!! I thank you!
This is one of the best explanations of buck converters I have come across. Keep up the good work!
VL = vin-vout. Not the way you showed, otherwise it would a negative voltage across the inductor. Good videos but this point got me confused so I just went back to KVL.
Thanks for the correction ❤
Holy mother fuck man 😰
That's why I love youtube. You can learn the whole engineering concepts for free here that too from best teachers..
Can't believe an ochem tutor is helping me with my power electronics homework
Really great video, explained much better than my teachers managed.
The "switch" is actually more circuitry to get it to oscillate to produce the whole effect.
I can bet you're a good singer.
because he almost knows everything.
At 3:00 shouldn't it be VL = Vin - Vout considering the polarity of the voltage across the inductor?
Wow! great Explaination
through this video i understood the vrm section of motherboard
Thkkkku sir 😍
Thank you very much for your great explanation, it helped a lot! Greetings from Germany. ✌
101%
I really like your video. Writing and explanation are clear. It is easy to understand the concept s.
You always give the best explanations.
Thank you for all your help and for providing great content Julio. Question, do you ever respond to comments or make posts on your Community page? I'm sure there's a lot of viewers who would love to connect more and learn more about you. We definitely appreciate your work.
On discharge of the inductor, the voltage will *always* be sufficiently high to deliver current to the capacitor and/or load until it is fully discharged. This is a fundamental property of an inductor.
The equation
Vout = Vin • d
applies *only* if the inductor current is continuous, that is, it is never drops to zero.
Once the inductor current becomes discontinuous (spends some time a zero each cycle) the relationship of Vout to Vin has to be calculated based on the energy stored and delivered each cycle.
If the duty cycle is responsible for determining the output voltage, what is the role of inductor in this circuit to reduce the output voltage as you explained in the beginning like "if 5v dropped across the inductor, remaining 7v will be there across the RL"
Appreciated if anyone help me to figure out this doubt.
Without the inductor, the RL load would get 12V 'spikes' controlled by the 555 timer instead of the 7V.
Let's say RL is a 1.4V CPU, 12V spikes would kill the CPU instantly.
You just 'turn off' the COIL before it could charge up to the full potential of the battery (12V).
Thank you for the classes, they are very educational, I have a question:
Where did the 80% efficiency come from? Could you teach how to calculate?
you are the best one to understand keep going man and thanks to you i learn alot
So nice and educative like how you explained logic gates so nice
woah, that video really helped me, easy, understandable... Thanks!
Great video! Simple and complex all in one.
Best explanation I've seen, thank you!
The inductor stores energy in the magnetic field that develops when current flows through the coil.
Buck converters are useful in solar power supplies.
fantastic Video! thank you!
Thanks so much! Way better than my teacher!
Best explanation, m just curious with the period of the duty cycle
Excellent explanation
So most step down bucks converters use LM2596 as it's regulator, with a max iput of 40V. If I wanted a down converter that can handle 6ov input, I could use a ? as the regulator. Very well explained brill.
Once you get into voltages and/or currents higher than the integrated switcher ICs can handle you would use a controller IC and a separate switch, typically a power MOSFET. Properly driving the "high side" switch for a buck converter is a little tricky. Even powering the controller IC requires some attention to detail.
Nice job man, thank you for this explanation
It doesn't mention parasitics or ringing, or switching transients, or snubbers, but yeah, that would be overwhelming too early. There is a whole bunch of fun stuff to learn about switching converters. Good video.
5:27 why are you adding them? These two currents do not flow at the same moment of time, so you can't just take and sum them up. Please, clarify.
I think its supposed to be averaging
Hey thanks for a video, but are you sure that VL=Vout-Vinput? maybe that should be opposite like VL= Vinput-Vout?
Phenomenal explanation. Thank you.
Thanks for nice explanation sir.
Very nicely explained
Hi sir. Tnx for your good job. Please explain minus or negative voltage ⚡ and its applications. 🙏🙏🙏
simple great explanation
I agree with everything you said EXCEPT; if S1 is static on (100% duty cycle), how do you figure you’ll get 7V out of that 12V supply at RL? With no PWM on S1, RL will see 12V after the lag time of your inductor and the R*C time constant of your cap and your load.
At 3:00 is probably a small mistake just put the values in: 5V = 7V - 12V is not correct. It should be VL = Vin - Vout
Thanks for your information
Can you tell as witch program do you use ?
At the end of the video, you say that if the converter were 100% efficient, the voltage and current are both modified by a factor by 2. Is it always the case that the factor is the same?
Hi, this video can confuse us learners. Coz there was no parr of the ckt drawn that would provide the duty cycle effect. This cld be devastating to learning if the context is not proper. But I want to thank you for the effort.
Thanks for this video mate! Great explanation
If the output voltage is only dependent on the input voltage and the duty cycle as explained in the final part, does that mean a random inductor and capacitor can be chosen for the cycle irrespective of their respective inductance and capacitance?
Good lesson, answers alot of questions as well as creating questions.
This lesson is wonderful.
Great explanations! Thank you.
Awesome explaination 🔥
Thanks for your explained... I need your help to undesrtand about solar panel control, so i can build for myself.. I need to fully understand everything about solar panel control..
Thank you sir..
Explain very nicely 👌
great material
Thank you
3:01 On wikipedia, they say VL = Vin - Vout, which makes sense. Node voltage method. Are you sure you didn't make a typo? You showed VL = Vout - Vin
Yes it is Vin - Vout, when the inductor is seen as a source, which he did.
@@NIFUable No he didn't? I gave the timestamp too. When the switch is closed, VL = Vin - Vout
@@alans2416 I meant it like this way. He has seen the coil as a source, so his voltage directions are the way he intended them to be, which means the voltage must be vin-vout
8 amp is a large value for output current....where it comes from? Show some computations, sir! From the Philippines... What are the values of the inductor, capacitor,diode and resistor?
Hi, I'm not an electrician, neither don't have any science base. I'd like to ask a question about "LM2596S Constant Current" buck converter. Let's say, we have 12V 1A adapter, power supply. Normally when we step down this to 3V, the current should increase in %100 efficiency to 4. I know this converter has 3A limit. So, this much decreasing the V can harm the converter? Additionally, when I regulate the converter to provide constant 0.5 or 1A, where the difference goes? Is this dangerous? I'm interested in modelling and going to use some LEDs. Therefore I'm trying to be cautious not to burn the house :) Thanks!
Great explanation! very detailed.
are you saying the output voltage isn't affected by the value of the inductor and the switching frequency?
Very pleasant teaching style, thank you.
What happens if your load doesn't use the extra current? does the voltage increase?
Is a buck converter a current or a voltage source (or neither)?
In a a real circuit feedback to adjust the duty cycle to maintain the output voltage is essential.
If the load doesn't require the energy stored in the inductor then the output voltage will rise and can become equal to the input voltage.
Excellent! Thank you.
So in this example the Vout can never be more than 12v?
Best explanation! Keep it up!
What about circuit with switch in series with coil, (buck type), and needing the 2nd flyback pulse to go forward also- as additive?The diode anode is facing the source plus voltage , and cathode is at end of other diode in series with coil L. In other words 3 diodes, diode ,coil, diode and diode across coil and diode - all cathodes are facing forward, and NO capacitor is present in my reference circuit found from 2008.
So are Buck converter circuits very common? Because you don't want waste heat damaging the device?
5:27 ... Didn't understand it well. When you add A and add B, then the sum of A and B is larger than A? Clearly yes... what's the point here?
Please make a video of full circuit diagram. Thanks
Well, and what will happen when the circuit hasn't D1 and switch S1 will open... what will be voltage on sides of the inductor :) Will reach voltage 2V and 7V or somthing different value and it will be related to quality factor of inductor and switching rise time?
If the circuit has not D1, the circuit could blow up. D1 here is called free wheeling diode, to make sure when S1 is open, there is still path for current to flow.
shouldnt you have to calculate de RMS voltage in order to calculate the power on the load? which is Vin*sqrt(Duty cicle)
My daughter’s 12v electric Razor scooter is too fast for her. I like to drop the voltage down to 9v. Do you know a buck converter that can handle the scooter without overheating? It’s a 10 mph E90 Razor scooter. Send me a link if you would. Thanks.
Doesn't the source have to be ac for an inductor to charge? . but great video you are a natural academic
Short answer: Since we switch on and of our input voltage periodically we've got an AC voltage which is rectangular in waveform
Long answer:
Yes and no: If you connect a constant voltage to an inductor in steady state (after an initial transient phase) not much happens. I think this is what you mean by "A DC voltage doesn't charge an inductor". But like mentioned above there is a transient phase: Let the voltage across the inductor be 0V. You then swith on your DC source and the inductor voltage jumps to V_in (input voltage) but the current that would start flowing through the wire is can't jump in an inductor. It starts increasing exponentionally which builds up a magnetic field. This is (roughly explained) the stored energy in the inductor. This transient phase is (depending on the inductance) very short. But since we switch our DC input voltage on and off quickly this happens often, so the stored energy in the inductor drives a current when the switch is off and the magnetic field in the inductor builds back up if the switch is on.
@@KilRBass thanks im just a student nd i have a lot of iterest in this stuff .
If an inductor is fully discharged (no energy is stored as a magnetic field) and you apply a voltage across it, the current through it will start at zero and rise linearly with respect to time. With an ideal inductor with no resistance there is no limit to the current that could be reached if you allowed sufficient time. With practical inductors that doesn't happen. The main limiting factor for inductors used in switch mode power supplies is the "core." The core can only handle a certain amount of magnetizing force before it "saturates."
When the inductor is dicharging in the circuit, the current still flows in the same direction but the polarity of the voltage across it reverses. The discharge is also linear change (decline) of current with respect to time. Given enough time the current will drop to zero.
di/dt = V/L
the differential of current through the inductor with respect to time (or, the rate of change of current in the inductor) is equal to the voltage across the inductor divided by the inductance. i is in amperes, t is in seconds, V is in volts and L is in henries
One very important thing that this equation says but maybe isn't really obvious is that you cannot instantaneously change the current through an inductor.
@@KilRBass
No. The current in the inductor changes linearly, not exponentially with time when a constant voltage is maintained across the inductor.
wow!!!useful 12mins
0 voltage means negative voltage?
please make more videos about electronic 👍👍👍
Great video
What’s the formulae to find Vin ,when given Vout,Iout and Pout?
So when the switch is open the current will flow from negative towards positive of the inductor and so confusing how exactly a current flows through ground and back to the load?
Thank you very much
Does astable multi vibrator fit for this as a switch which is square wave ?
VERY GOOD
Thank you thank you thank you
Why is the voltage when turning off the switch double the amount of when turning on?
When he doesn’t say bye😭
How is there a voltage drop VL when the current is constant?
Question, if the input voltage is changing the output will also change or will it regulate?
Without a feedback circuit to adjust the duty cycle the output voltage would rise.
What if i remove diode and place capacitor across switch?
Holy s., you're good!
How do you calculate the inductor & cap values?
May I have digestive system and type of tooth video,I hope next will be these two
How the voltage across Inductor is 5V when switch is closed. Please clarify.
So why do you need an inductor? If you just get a raw PWM signal and put it through a capacitor the capacitor just smooth out the voltage. I built a 555 circuit that is adjustable and the output of the connector to a capacitor and by adjusting the oscillation of 555 timer I can lower the output voltage.
Whats a 555 circuit tho?
@@sirgalantoe6325 a circuit that uses a 555 timer. 555 Timers are very handy but I am not even Shure what they do.
Excellent!
Thanks alot sir
Why is it VL = Vout - Vin? Should it not be VL = Vin - Vout? Because Vin is equal to VL and Vout,.
Thant was great and clear ^^
You mean, duty=1ms/5ms, right?
No. The total period is 5 ms - 1 ms ON and 3 ms OFF.