The best explanation of how these converters work that I have seen. I'm trying to build a 40v to 400v DC-DC converter for charging an EV from vehicle mounted solar. Thank you, you have a new subscriber
Excellent teacher and educator. When he mentions that there is no free lunch, he wanted to educate the conservation of energy, very rarely others may have touched upon. 🙏🙏
One correction to note: The output current is reduced because of P = IV where P must be equal or less than the input power, not because some is lost to switching. The loss to switching is what causes a reduction in efficiency of the boost converter. Some of the best modern boost circuits run at about 95% efficiency.
The maker of this video didn't have a really firm grasp on the fundamental concepts. I've seen much worse, but he still makes some misleading statements. I simply can't make sense of what he is trying to say at a couple of points. I got around 95% efficiency or better across the input voltage range for a universal input active power factor correction stage that was delivering around 1.2 kW, and that was in a design I did nearly two decades ago. FETs have improved quite a bit since then. Although other topologies are possible, active PFC is usually implemented with a boost converter with the interesting requirement that the instantaneous input current be directly proportional to the instantaneous input voltage (as you want for good power factor) but the average input current must inversely proportional to the average input voltage (because of the negative input resistance characteristic of a switchmode converter).
Intuition which led to invention of boost converter :- When current passing through inductor is suddenly stopped, huge voltage develops across inductor.
Nycc,i think of coming up with a website where my clients will be able to see these tutorials,my views will be of benefit to you,thats what i can attribute to your hard work.thanks again 👍👍👌
A boost converter without a load can produce an arbitrarily high voltage regardless of the switching frequency or the inductance. In practical circuits there is a limit because of of losses, primarily those due to switching transitions that are finite. Of course eventually something would go _Bang!_ due to excessive voltage stress. At the instant the switch opens, the current through the inductor is *exactly* the same as it was the instant before the switch was opened. This is the most fundamental aspect of the nature of an inductor in this sort of circuit (where you are considering energy storage and delivery as opposed to considering it in terms of impedance in an AC circuit). The voltage across the inductor will rise to whatever voltage is required to allow that current to flow. At opening on the very first cycle that voltage will be equal to the diode forward voltage if there is no load on the circuit because the capacitor voltage will be equal to the battery voltage, having charged before the switch was closed (provided there was sufficient time after the input power was connected before the first cycle began). With each additional cycle the voltage on the capacitor will increase, hence the voltage across the inductor will increase. Again, unless there is a load connected and ignoring losses, there is no limit to the voltage to which the capacitor will charge regardless of the inductance, duty cycle or switching frequency. Vdt = Ldi or the alternative arrangements says that in equilibrium with a load the rise and fall of current in the inductor is linear ramps. If the inductor is not allowed to fully discharge with each cycle, these ramps "ride" on a DC level.
Great video. I use those sealed marine grade 12v to 36v (5a) step up boosters to convert 12v battery power to power Bluetooth digital chip amplifiers in my speaker builds. They work flawlessly. Here's my question, why do my amplifiers (running at 36v) sound better (more power, fidelity) vs using their stock (included) AC/DC adapters that come with these little digital chip amplifiers? I don't have an audio dyno or distortion tester...but, I can "hear" and "see" the difference when the woofers on my speakers are pounding harder. Thanks.
My guess is that the larger power supplies are better able to handle the high and relatively long transient current required for loud parts. The voltage of the small supplies probably "droops" under heavy load because they are overloaded. When the supply voltage drops the audio amplifier tires to correct but it may not be able to, resulting in "clipped" waveforms. It MIGHT be possible to make them sound better by adding capacitance across the output of the supply for better transient handling, but that does nothing for prolonged excessive current demand. It would be quite easy to confirm or reject my hypothesis with an oscilloscope.
I don't know what you mean by "a hi tolerant diode." The voltage rating needs to be no higher than the output voltage of the circuit. The current rating needs to be no higher than the input current of the circuit (it will see instantaneous current higher than the average input current, but that rarely is much of a concern). Fast reverse recovery time (the time it takes the diode to switch off when the reverse bias is applied) is important in practical boost converter circuits. Schottky diodes are often used if their voltage rating is adequate.Their forward voltage is lower than that of a similarly-rated PN junction diode and they are extremely fast (very low reverse recovery time).
Thank you very much, easy to understand . Btw , is it still needed feedback to switch ? What's happen when there is no feedback ? Thankyou because i want make hardware from boost converter
Would there still be an increased voltage output without the transistor? Perhaps using the most powerful inductor and highest useful frequency. Thanks Alan
How the circuit will work/functional? and How the voltage and current waveforms at the output?, If we keep the load as a inductor or capacitor instead of resistor?. Please could you answer my question?.
Sir, i have used MT3608 step up booster and connected with 5v Lithium Battery with rated of 3000mah and set constant output as 12v, however when i connect to any dc device like wifi router, settop box or dc mother 555, automatically the output from 12v its drops to 4v, Not able to understand. Could you please give some insight in what sceniaro could lead into such issue.
This video has a bunch of problems. The output voltage may have nothing to do with the inductance or di/dt as long as the inductor current is continuous. Your point is also correct, the current should not suddenly change from large to small.
@@henryfordson4787 the current should change to small of course because the energy should be the same, this is a logical physic phenomenon. but the wrong thing I saw in this video that the increase of voltage is depend on the frequency and this is not right. i think it's dependent on the rise time of changing of the state from high to low and low to high.
With no load connected the current through the inductor would decay to zero once the output capacitor was charged, before any switching started. However, if there were a load connected the current, again before switching started, would be equal to the (input supply voltage minus the diode forward voltage) divided by (the inductor winding resistance plus the load resistance). You cannot protect a boost converter against short circuiting of the output without additional circuitry.
So, the switch has to be closed then opened again? Do we have to close-open-close-open continuously, or do we just do it once? *I don't have electronics knowledge 😶
@@computifyguy Thank you. So, the result is not a flat voltage, but more like a wave? Or is it the current that is going up and down? How can I get that PWM?
sir iam a student in srilanka , sir thank you for the video , sir i want to know that when out put voltage increase how does it reduce the current , when we take ohms law voltage is equal to current doesnt it happen in here sir
In general terms, switchmode converters are power converters - the power from the output is equal to the power at the input, ignoring the losses (things like resistance of the inductor winding, losses in the core of the inductor, loss at each transition of the switch, diode forward voltage, etc.). Because of this, a switchmode power supply actually has a "negative resistance" input characteristic. That doesn't mean it actually behaves like some true resistance of negative magnitude, but that the slope of current versus voltage at the input is opposite what it would be for a resistor if you were to make a graph. It a normal resistor if you increase the voltage across the resistor the current through it will increase in direct proportion. With a negative resistance, if you increase in voltage the current decreases in proportion. For example with a switcher, if the input had 10 volts applied and the current were 4 amperes (so 40 watts) and you increase the input voltage to 15 volts, the input power remain the same and the input current would become 40 watts / 15 volts = 2.67 amperes. In practical circuits the relationship won't be perfect because the losses in the circuit change in non-proportional ways with varying voltage and current. Maximum efficiency might come at some particular input voltage with lower efficiency for both lower and higher voltage.
My understanding is that to get to the required voltage you need more than one cycle to add up voltages with the inductor, and to add up the voltage it needs to be stored somewhere between the cycles which is why we use a capacitor.
@@louistiticaramel6848yes you are right from what I understand. Also if there wasn't a capacitor to smoothen the signal, the output would just oscillate.
I think the current will be increased due to the charge accumulated In the output capacitor and the inductor current, not as you have explained because no way to add the current of the inductor and the input source when the switch is off ( circuit ON)
Nice explanation on boost converters, thanks for making this video. I was confused about one part, though--at 7:38, you say that increasing the frequency applied to the mosfet gate will increase the dI/dt and hence V across L. I was thinking about how all of the (various few) buck/boost converter designs I've seen use PWM control, and not FM control. Wouldn't dI/dt depend on the rise time of the MOSFET rather than the switching frequency? Seems like a square wave would give you the minimum rise time and hence maximum dI/dt.
I didn't really follow what he was trying to get at. It seemed rather muddled.11 Increasing the frequency, all other things being equal, actually decreases di/dt in the inductor. The rate of change of current in an inductor is proportional to the voltage across it. If the input voltage, output voltage and duty cycle all remain constant, increasing the frequency results in dt being smaller in each cycle thus di is proportionally smaller. If you want to increase di under those conditions you must make the inductance smaller. EDIT TO CLARIFY: di/dt as such does not imply any fixed time, and in fact the opposite - a way of expressing what happens with infinitesimal change in time. It really isn't correct to use di/dt for some fixed amount of time - that would be expressed as Di/Dt where the capital D really should be a capital Greek delta (a triangle Δ; the d in di/dt really should be a small Greek delta δ thus Δi/Δt rather than δi/δt - characters may not render properly). If delta t is shortened, delta i will be reduced in proportion for the same voltage applied across the inductor. You can do a boost converter with FM though it isn't common in general purpose circuits. It can increase efficiency a bit at light loads because you have fewer switching cycles per unit of time with more or less fixed loss each cycle. Variable frequency operation is moderately common with active power factor correction circuits.Most active PFC circuits are boost converters with the interesting property that average input current must be inversely proportional to average input voltage but instantaneous input current must be directly proportional to instantaneous input current.
@@fiddlyphuk6414 A decrease in frequency with no change in duty cycle will allow greater ripple voltage on the capacitor but the average voltage will remain the same unless the capacitor is allowed to fully discharge each cycle. No one with the slightest clue about how to design a switching regulator would do that.
@@d614gakadoug9 The greater the ripple voltage the lower the average DC voltage across the load will be. It's the same way in linear power supplies. The original question regarded how frequency alone can affect the output voltage as stated by robanada. An increase in frequency alone will increase the voltage but, in reality, only up to a point depending on the size of the inductor. For that reason PWM makes for far better control of the output voltage.
@@fiddlyphuk6414 No. Greater ripple voltage does not imply lower average voltage. It simply means that there is a greater difference between the minimum, average and peak voltages. A triangle wave swinging between 1 volts and 9 volts has exactly the same average voltage as one swinging between 4.5 volts and 5.5 volts but 8 volts peak-to-peak ripple versus 1 volt peak to peak ripple. It is not at all the same as in linear supplies because a linear supply uses capacitors to "filter" what is a not-very-good VOLTAGE source while in a switcher the capacitors filter a CURRENT source. An ideal voltage source has zero impedance. An ideal current source has infinite impedance. The voltage in a buck or boost converter is dependent ONlY on the duty cycle if the input voltage is fixed and the inductor current is continuous. for a buck converter Vout = Vin x duty cycle for a boost converter Vout = Vin / (1 - duty cycle) Again, frequency simply does not appear in the expressions. ONLY ONLY ONLY if a change in frequency results in the inductor current changing from discontinuous to continuous does a change in frequency change the output voltage.
This video has a bunch of problems. The output voltage may have nothing to do with the inductance or di/dt as long as the inductor current is continuous.
No. There are ways to do limited voltage boosting (typically nominally doubling) with just capacitors in a "charge pump" circuit, but the switching is more elaborate and such circuits are typically only useful at currents of no more than a few tens of milliamperes. With actual AC input you can do voltage multiplication with diodes and capacitors.
Very good explanation...and here is the actual result ...ua-cam.com/video/XHfp97ZlrnQ/v-deo.html .I was able to get from 12 to 50 volts DC with an efficiency between 70 to 94 %. Thanks for the video.
@@yasyasmarangoz3577 At 19:38...in the video there is a table that shows the output current at 4.55 Amps for 103 watts from an input of 119 watts. Cheers!!!
Crazy idea for you. Do some videos on communication. Half for us old guys who can’t understand why students today can’t figure out how to use things like email and telephone. Half for students explaining how and why email and telephone are essential for an actual career. Don’t get me wrong, I appreciate your work. I am frustrated seeing my ex-students fail on their first job. It is simply insane to ask your employer to use UA-cam and Instagram. I’m serious. Graduates who depended on social media their whole life are actually failing.
Final Exams and Video Playlists: www.video-tutor.net/
How come Organic Chemistry tutor explains so well the electronic circuits? Kudos 🙏🙏🙏
The best explanation of how these converters work that I have seen. I'm trying to build a 40v to 400v DC-DC converter for charging an EV from vehicle mounted solar. Thank you, you have a new subscriber
Damn thats a pretty intense thing to try to build on your own if your not very familiar with electronics! how did it turn out?
Did you end up making the circuit? if yes, how did it turn out?
He died :(
Excellent teacher and educator. When he mentions that there is no free lunch, he wanted to educate the conservation of energy, very rarely others may have touched upon. 🙏🙏
smooth and clear explanation of the heart of how it works, without stuffing equation, exactly what one needs to understand , all the best
thank you so much, just by saying that the interruptor is there for shorting the circuit i finally understood this
This was a key piece for me as well. We're just charging a battery then sticking it in series with the source.
Best explanation of this type of circuit I have seen.
The way of explaination is the best in youtube
I generally didn't get english accent
But you explain really well
best explanation on UA-cam, thank you!!
I wish u were my teacher.. none of my proffesors explained material in such a simple, clear and gentle way!
DC to DC? More like "thank you for helping me!" These videos are amazing; keep 'em coming!
You know how to make things soooo easy for others to understand...👍☺
Such an amazing explanation - I needed a refresher and this was perfect! Thank you
Thank you, I don't know if I learned more about eletronics how to teach, or how to communicate
Best explanation on the internet. Thank you so much.
Your channel has helped me so much with studying, that I felt like I should owe you money :)
thanks a lot, I was looking for this video for more than a week.
Great video, looking forward for an math example of the boost converter calculation. Thanks!
One correction to note: The output current is reduced because of P = IV where P must be equal or less than the input power, not because some is lost to switching. The loss to switching is what causes a reduction in efficiency of the boost converter. Some of the best modern boost circuits run at about 95% efficiency.
You didn’t say anything new. 😕
So where is the current?
@@yasyasmarangoz3577 It gets exchanged for the higher voltage per P=IV
The maker of this video didn't have a really firm grasp on the fundamental concepts. I've seen much worse, but he still makes some misleading statements. I simply can't make sense of what he is trying to say at a couple of points.
I got around 95% efficiency or better across the input voltage range for a universal input active power factor correction stage that was delivering around 1.2 kW, and that was in a design I did nearly two decades ago. FETs have improved quite a bit since then.
Although other topologies are possible, active PFC is usually implemented with a boost converter with the interesting requirement that the instantaneous input current be directly proportional to the instantaneous input voltage (as you want for good power factor) but the average input current must inversely proportional to the average input voltage (because of the negative input resistance characteristic of a switchmode converter).
good video!Explain clearly than anyone!thx
Great VDO, Easy to understand.
Intuition which led to invention of boost converter :- When current passing through inductor is suddenly stopped, huge voltage develops across inductor.
Nycc,i think of coming up with a website where my clients will be able to see these tutorials,my views will be of benefit to you,thats what i can attribute to your hard work.thanks again 👍👍👌
A boost converter without a load can produce an arbitrarily high voltage regardless of the switching frequency or the inductance. In practical circuits there is a limit because of of losses, primarily those due to switching transitions that are finite. Of course eventually something would go _Bang!_ due to excessive voltage stress.
At the instant the switch opens, the current through the inductor is *exactly* the same as it was the instant before the switch was opened. This is the most fundamental aspect of the nature of an inductor in this sort of circuit (where you are considering energy storage and delivery as opposed to considering it in terms of impedance in an AC circuit). The voltage across the inductor will rise to whatever voltage is required to allow that current to flow. At opening on the very first cycle that voltage will be equal to the diode forward voltage if there is no load on the circuit because the capacitor voltage will be equal to the battery voltage, having charged before the switch was closed (provided there was sufficient time after the input power was connected before the first cycle began). With each additional cycle the voltage on the capacitor will increase, hence the voltage across the inductor will increase. Again, unless there is a load connected and ignoring losses, there is no limit to the voltage to which the capacitor will charge regardless of the inductance, duty cycle or switching frequency.
Vdt = Ldi or the alternative arrangements says that in equilibrium with a load the rise and fall of current in the inductor is linear ramps. If the inductor is not allowed to fully discharge with each cycle, these ramps "ride" on a DC level.
You are the best!!!!! 😭😭😭😭
I love your videos, so simple, short and easy to understand. Please can you make a video on DC to AC converters?
You are amazing. Thank you!
Very cool video though I wish he included the MOSFET transistor discussion with this video!
Very well explained,thanks a lot
So nice and great
This guy is responsible for a 1/2 point increase in my gpa.
Well explained !
THANK YOU VERRY MUCH !!!!
so good explanation
Excellent video.
oh man, how helpful that was
Thanks , It's very helpful
thanks sir u are a legend
Great video. I use those sealed marine grade 12v to 36v (5a) step up boosters to convert 12v battery power to power Bluetooth digital chip amplifiers in my speaker builds. They work flawlessly. Here's my question, why do my amplifiers (running at 36v) sound better (more power, fidelity) vs using their stock (included) AC/DC adapters that come with these little digital chip amplifiers? I don't have an audio dyno or distortion tester...but, I can "hear" and "see" the difference when the woofers on my speakers are pounding harder. Thanks.
My guess is that the larger power supplies are better able to handle the high and relatively long transient current required for loud parts. The voltage of the small supplies probably "droops" under heavy load because they are overloaded. When the supply voltage drops the audio amplifier tires to correct but it may not be able to, resulting in "clipped" waveforms. It MIGHT be possible to make them sound better by adding capacitance across the output of the supply for better transient handling, but that does nothing for prolonged excessive current demand.
It would be quite easy to confirm or reject my hypothesis with an oscilloscope.
Excepcional vídeo... thank you for sharing knowledge 👍
great job
Thank you
Thanks
Very very helpful
Your other videos are far better
How is it determined how much energy flows to the capacitor & voltmeter ?
Thx
Great videos
simle but effective. you need a hi tolerent diode though
I don't know what you mean by "a hi tolerant diode." The voltage rating needs to be no higher than the output voltage of the circuit. The current rating needs to be no higher than the input current of the circuit (it will see instantaneous current higher than the average input current, but that rarely is much of a concern). Fast reverse recovery time (the time it takes the diode to switch off when the reverse bias is applied) is important in practical boost converter circuits. Schottky diodes are often used if their voltage rating is adequate.Their forward voltage is lower than that of a similarly-rated PN junction diode and they are extremely fast (very low reverse recovery time).
you should always have wiggle room or a slight spike blows the thing. @@d614gakadoug9
Thank you very much, easy to understand . Btw , is it still needed feedback to switch ? What's happen when there is no feedback ? Thankyou because i want make hardware from boost converter
amazing MAN!
Would there still be an increased voltage output without the transistor? Perhaps using the most powerful inductor and highest useful frequency. Thanks Alan
How the circuit will work/functional? and How the voltage and current waveforms at the output?, If we keep the load as a inductor or capacitor instead of resistor?. Please could you answer my question?.
New sub added😊 great explanation
So nice
Sir, i have used MT3608 step up booster and connected with 5v Lithium Battery with rated of 3000mah and set constant output as 12v, however when i connect to any dc device like wifi router, settop box or dc mother 555, automatically the output from 12v its drops to 4v, Not able to understand. Could you please give some insight in what sceniaro could lead into such issue.
well done👍👍
Is it possible to increase the voltage from 12V to 350V in this way?
Very good
Great video!
Great 👏👏
1:52 why it is small current? even switch off whole current will flow through inductor L?
please reply
This video has a bunch of problems. The output voltage may have nothing to do with the inductance or di/dt as long as the inductor current is continuous. Your point is also correct, the current should not suddenly change from large to small.
@@henryfordson4787 the current should change to small of course because the energy should be the same, this is a logical physic phenomenon. but the wrong thing I saw in this video that the increase of voltage is depend on the frequency and this is not right. i think it's dependent on the rise time of changing of the state from high to low and low to high.
With no load connected the current through the inductor would decay to zero once the output capacitor was charged, before any switching started. However, if there were a load connected the current, again before switching started, would be equal to the (input supply voltage minus the diode forward voltage) divided by (the inductor winding resistance plus the load resistance). You cannot protect a boost converter against short circuiting of the output without additional circuitry.
it is possible to create a self current generating circuit using this?
amazing
So, the switch has to be closed then opened again?
Do we have to close-open-close-open continuously, or do we just do it once?
*I don't have electronics knowledge 😶
@@computifyguy Thank you.
So, the result is not a flat voltage, but more like a wave? Or is it the current that is going up and down?
How can I get that PWM?
Hello teacher can you tell me what the programs that you use to draw your thumnails.
I want more explanation about converters could you please help me
Sir I need just 0.3 v more out of 2 volts input and 2.3 v output how to do it
Can we increase the amps?
sir iam a student in srilanka , sir thank you for the video , sir i want to know that when out put voltage increase how does it reduce the current , when we take ohms law voltage is equal to current doesnt it happen in here sir
In general terms, switchmode converters are power converters - the power from the output is equal to the power at the input, ignoring the losses (things like resistance of the inductor winding, losses in the core of the inductor, loss at each transition of the switch, diode forward voltage, etc.).
Because of this, a switchmode power supply actually has a "negative resistance" input characteristic. That doesn't mean it actually behaves like some true resistance of negative magnitude, but that the slope of current versus voltage at the input is opposite what it would be for a resistor if you were to make a graph. It a normal resistor if you increase the voltage across the resistor the current through it will increase in direct proportion. With a negative resistance, if you increase in voltage the current decreases in proportion. For example with a switcher, if the input had 10 volts applied and the current were 4 amperes (so 40 watts) and you increase the input voltage to 15 volts, the input power remain the same and the input current would become 40 watts / 15 volts = 2.67 amperes. In practical circuits the relationship won't be perfect because the losses in the circuit change in non-proportional ways with varying voltage and current. Maximum efficiency might come at some particular input voltage with lower efficiency for both lower and higher voltage.
Why is the capacitor necessary? Surely the voltage can be used up by the load?
My understanding is that to get to the required voltage you need more than one cycle to add up voltages with the inductor, and to add up the voltage it needs to be stored somewhere between the cycles which is why we use a capacitor.
@@louistiticaramel6848yes you are right from what I understand. Also if there wasn't a capacitor to smoothen the signal, the output would just oscillate.
I think the current will be increased due to the charge accumulated In the output capacitor and the inductor current, not as you have explained because no way to add the current of the inductor and the input source when the switch is off ( circuit ON)
instead of a motor I am going to try using a computer fan so that I can simultaneously cool the circuit :P
I watched it.
Nice explanation on boost converters, thanks for making this video. I was confused about one part, though--at 7:38, you say that increasing the frequency applied to the mosfet gate will increase the dI/dt and hence V across L. I was thinking about how all of the (various few) buck/boost converter designs I've seen use PWM control, and not FM control. Wouldn't dI/dt depend on the rise time of the MOSFET rather than the switching frequency? Seems like a square wave would give you the minimum rise time and hence maximum dI/dt.
I didn't really follow what he was trying to get at. It seemed rather muddled.11
Increasing the frequency, all other things being equal, actually decreases di/dt in the inductor. The rate of change of current in an inductor is proportional to the voltage across it. If the input voltage, output voltage and duty cycle all remain constant, increasing the frequency results in dt being smaller in each cycle thus di is proportionally smaller. If you want to increase di under those conditions you must make the inductance smaller.
EDIT TO CLARIFY: di/dt as such does not imply any fixed time, and in fact the opposite - a way of expressing what happens with infinitesimal change in time. It really isn't correct to use di/dt for some fixed amount of time - that would be expressed as Di/Dt where the capital D really should be a capital Greek delta (a triangle Δ; the d in di/dt really should be a small Greek delta δ thus Δi/Δt rather than δi/δt - characters may not render properly). If delta t is shortened, delta i will be reduced in proportion for the same voltage applied across the inductor.
You can do a boost converter with FM though it isn't common in general purpose circuits. It can increase efficiency a bit at light loads because you have fewer switching cycles per unit of time with more or less fixed loss each cycle. Variable frequency operation is moderately common with active power factor correction circuits.Most active PFC circuits are boost converters with the interesting property that average input current must be inversely proportional to average input voltage but instantaneous input current must be directly proportional to instantaneous input current.
A decrease in frequency will allow more time for the capacitor to discharge through the load thus bringing its average voltage down.
@@fiddlyphuk6414
A decrease in frequency with no change in duty cycle will allow greater ripple voltage on the capacitor but the average voltage will remain the same unless the capacitor is allowed to fully discharge each cycle. No one with the slightest clue about how to design a switching regulator would do that.
@@d614gakadoug9 The greater the ripple voltage the lower the average DC voltage across the load will be. It's the same way in linear power supplies. The original question regarded how frequency alone can affect the output voltage as stated by robanada. An increase in frequency alone will increase the voltage but, in reality, only up to a point depending on the size of the inductor. For that reason PWM makes for far better control of the output voltage.
@@fiddlyphuk6414
No. Greater ripple voltage does not imply lower average voltage. It simply means that there is a greater difference between the minimum, average and peak voltages. A triangle wave swinging between 1 volts and 9 volts has exactly the same average voltage as one swinging between 4.5 volts and 5.5 volts
but 8 volts peak-to-peak ripple versus 1 volt peak to peak ripple.
It is not at all the same as in linear supplies because a linear supply uses capacitors to "filter" what is a not-very-good VOLTAGE source while in a switcher the capacitors filter a CURRENT source. An ideal voltage source has zero impedance. An ideal current source has infinite impedance.
The voltage in a buck or boost converter is dependent ONlY on the duty cycle if the input voltage is fixed and the inductor current is continuous.
for a buck converter
Vout = Vin x duty cycle
for a boost converter
Vout = Vin / (1 - duty cycle)
Again, frequency simply does not appear in the expressions.
ONLY ONLY ONLY if a change in frequency results in the inductor current changing from discontinuous to continuous does a change in frequency change the output voltage.
What happen when input power is higher than the output power?
This video has a bunch of problems. The output voltage may have nothing to do with the inductance or di/dt as long as the inductor current is continuous.
Hello video
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can you replace the inductor with a capacitor?
No.
There are ways to do limited voltage boosting (typically nominally doubling) with just capacitors in a "charge pump" circuit, but the switching is more elaborate and such circuits are typically only useful at currents of no more than a few tens of milliamperes.
With actual AC input you can do voltage multiplication with diodes and capacitors.
The frequency does NOT change the output voltage. It is a function of the duty cycle and NOT the frequency
Saturation point DOES IT!
HOW DO YOU KNOW EVERTHING???!!!
Very good explanation...and here is the actual result ...ua-cam.com/video/XHfp97ZlrnQ/v-deo.html .I was able to get from 12 to 50 volts DC with an efficiency between 70 to 94 %. Thanks for the video.
How is the current drop?
@@yasyasmarangoz3577 At 19:38...in the video there is a table that shows the output current at 4.55 Amps for 103 watts from an input of 119 watts. Cheers!!!
@@learningpower9437 Thanks alot
Crazy idea for you. Do some videos on communication. Half for us old guys who can’t understand why students today can’t figure out how to use things like email and telephone. Half for students explaining how and why email and telephone are essential for an actual career.
Don’t get me wrong, I appreciate your work. I am frustrated seeing my ex-students fail on their first job. It is simply insane to ask your employer to use UA-cam and Instagram. I’m serious. Graduates who depended on social media their whole life are actually failing.
boomer if I saw one
Try speaking faster
Amazing