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 :(

This was a key piece for me as well. We're just charging a battery then sticking it in series with the source.

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).

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.

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.

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

@@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?

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.

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.

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.

It was me I was 13 seconds early or it might be a glitch or something

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

Get used to conventional current. It is universally used and isn't going to go away any time soon.

@d614gakadoug9 it just seems like something that should be fixed. I see both used about 50/50, it makes it confusing as hell to read schematics in one when you use the other, and conventional is literally just incorrect.

@@MrCleanAteMyWife Yes, I can see it being confusing if someone is actively explaining something and hasn't said at the outset whether they are talking about electron current or conventional current. Usually it is pretty obvious which as long as there is a voltage or current source with polarity marked in the circuit. In the best explanations I've seen in videos the presenter will explain both. I've seen a video where animations were used. Many people commented on how great they thought the animations were. I thought they were absolutely terrible because they created a false impression of what really happened in the circuit. Electrons were shown entering one end of a solenoid-wound coil (like you'd get if you wound wire in a single layer around a straight round rod). Slowly the electrons went around one turn and a magnetic field was shown around that turn, then the electrons continued to move from one end of the coil to the other, making magnetic fields around each turn as they went. That isn't what happens at all. If one electron enters one end of the coil other electrons throughout the entire circuit move at the same time time. If they didn't the first electron couldn't go anywhere (well it could, but in an extremely local sense). Over time the the number of electrons per unit of time increases as the magnetic field strength increases, but that is quite hard to animate in a way that's easy to follow. The way that was used was electron current, but it was completely wrong. I haven't watched many electronics videos. I've just been doing it to see how well done they are. They are highly variable. Some are horrible for teaching what they are intended to teach and the presenter gets a lot of things wrong. Some are very good. For most, most of the commenters will say how great the video is, even when it isn't. THIS video has a number of errors and/or poor explanations and only two or three people have noticed them. Of course those commenters are people who already know the material, not those trying to learn it for the first time.

@@MrCleanAteMyWife One thing I should have mentioned: The arrow in schematic symbols for semiconductors usually points in the direction of conventional current flow. There are exceptions, like zener diodes, that are operated in "reverse breakdown" where conventional current flows the opposite way - "into" the cathode and "out of" the anode.

boomer if I saw one