In other words: take your time to explain things... :-)

@@user-ge9yh6qc3hHe was clear and concise. If he is explaining it to fast maybe try playing the video at a slower speed ?

Can I help with an explanation or anything? :)

It's called Every Circuit, it's great for visualising current flow :)

It's called Every circuit!

The resistor simulates a load on the transformer

Like @taroc9752 says, the resistor adds a load to the transformer creating a complete circuit loop and making it easier to monitor!

You need all but the bottom segment!

I mean A line , B line,CDEFG please tell me how to show @@electronicshandbook8788

You are completely right, my concept is slightly dramatic. But this circuit offers a way to have a sort of simulated negative resistance which can be very useful in niche oscillators and amplifiers.

It is a program called every circuit, its great for simple circuit simulation.

Yay I have it

Nice observation!

114, there’s 14 sets of repeats Edit: plus the blank so 113

Depends,if a psycho put 1 in the left

Good spot

Wow yeah

@@electronicshandbook8788 Where are H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z?

Thank you 🙂

Thanks!

Very helpful!

A: ABCEFG B: CDEFG

Glad you found it helpful!

Hi, the site used is called Every Circuit, there is a link in the description.

Glad it was helpful!

Hi, Its called every circuit, there is a link in the description to check out the circuits in the video :)

Hi!

Thanks!

oh interesting, I'd love to know more about it?

@@electronicshandbook8788 kzfaq.info/get/bejne/fremqKyim8y8das.html they also have a version without a split segment but this one is my favorite

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Adam I/R'a voltaj demiş amk biz de izliyoz bi şey anlatcak diye

Thanks!

Thank you!

I see PWM as a way to turn pulsed DC into a varying or alternating signal (AC). It would be clearer if I could use smoothing capacitors and show off a sinewave, but the simulator software I use cant run fast enough for it to be clear to a viewer so I went for the motor demonstration. Hope that helps

They really are so useful, even if just for sheer practicality like you say, not needing huge current wires everywhere.

Hi, That's the current flow, in order to calculate that you need to add the bottom two resistors in parallel, Then using that new resistance and first 10K resistor youd be able to calculate the current flowing through each. I have a video on voltage dividers here if thats helpful: kzfaq.info/get/bejne/fdKihZdoz7e5f6c.html And a video here of adding resistances in series and parallel: kzfaq.info/get/bejne/fJqPkpycspmuoH0.html Hope that helps.

Hi, Every diode data sheet will list something like IF ( Forward Current ), which could be a large range of values depending on the diode you're using, for example 1 amp. To be clear that is just the maximum current that can pass through the diode, you can of course pass less current if you wish. Hope that helps, Ben

Im glad you're learning something, any questions feel free to ask!

Hi, Just for clarity, a 2V drop is very common but it is absolutely not always the case, there are definitely LEDs out there that I have used before that have a different drop, that's why its always worth checking the component data sheet if you're not sure.

Hi, Im glad you're picking up a new hobby, hopefully you'll start to love electronics as much as I do. The reason we need use a resistor as a load is because you generally don't want to make a circuit without some kind of load simply because the wire itself has a very small resistance maybe 0.1 Ohms, so we don't have control over the current, that might be a bit confusing so i think the best way to understand would be to try the following LED circuit yourself: everycircuit.com/circuit/6020702572118016 In this circuit we know the LED only want about 20mA, which we control using ohms law by having a 12V source and taking off 2V for the LED voltage drop, and a 500 ohm resistor, V/R=I or 10/500=20mA Try seeing if you can make the circuit work without a load resistor, you'll notice the LED will keep blowing up, the only way you'll be able to make it work is if you set the supply to 2V which is the voltage drop of the LED. Essentially we like to try and always use a load resistor because without it you don't really have control over the current. Onto your next main question, the way a diode works is below 0.7 Ohms its resistance is extremely high, but above that value its resistance lowers very quickly, this is why we call them semi-conductors as they conduct electricity some of the time. Forward current is a measure of the maximum current you can flow through a diode when above that 0.7V, it will be different for every diode. The graph has voltage on the X axis and current on the y axis so its a graph of both current and voltage, the reason it has no values is because the graph will differ for every diode you find, some will have lower breakdown voltages for example, some will have high forward current. There are even diodes that have different voltage drops, I will be doing a video on them soon. I hope that I have been helpful, let me know if you have any questions in response, happy to help :)

@@electronicshandbook8788 This is excellent! Thank you for responding. A circuit with no load on it is a short circuit, right? And that is a bad thing. I get it. I have practiced doing circuit analysis. I find math fascinating. Yesterday I soldered up my first circuit. It was a push button and an LED, running on 9 volts with a terminal block to add wires to get the 9V into the board. It worked perfectly. I have bought an Arduino Super Starter Kit and I have been loving it. Now I'm looking for simple but useful circuits to build. I think I'm hooked. :-D

Hi, I may do a video on DC regulating or DAC/ADC at some point :)

Hi, Coupling Coefficient in simple terms is basically a measure of transformer efficiency. If you have a coupling coefficient of 1 that means that 100% of the magnetic flux or 'power' on the primary is converted to the secondary. A coupling coefficient of 1 is impossible in real life though because there are loads of small losses and inefficiencies, however I simulated a coupling coefficient of 1 in the video as its easier to understand as a viewer who is learning. A more realistic coupling coefficient would be 0.99 or 0.98 but you will notice the output dip slightly. Thanks for the great feedback and I hope that helps!

That's a great point!

Thanks!

Thanks! 😄

Hi, Thanks for the great feedback, i'm glad to hear you enjoyed the content. The 10K ohm resistor is a pull-down resistor in the N-Channel circuit, or a pull-up resistor in the P-Channel Circuit. The purpose of a pull-down resistor is to help keep the transistor base pin low when the transistor should be off, and allows the small amount of current flowing from the supply to go somewhere. The pull-up resistor is the same but inverted, it helps ensure the transistor base pin is high when it should be. Pull-down/up resistors are used all over electronics and whilst not always strictly necessary, they are often recommended. I hope that helps

@@electronicshandbook8788 thanks bro

Hi Mario, Great questions, the gate pin of a MOSFET in an ideal world draws no current at all, in practice it draws a few nanoamps. However because there is no current flowing through the 1kOhm resistor there is also no voltage drop across the resistor. This is because Ohms law teaches us that voltage is equal to current times Resistance and if there is no current and the resistance is fixed then there cannot be a voltage drop. I hope that makes sense!