lot better than my professor. thanks!

omg...u really explained it well than my intructor!! its like he doesn't know what he is talking about!! and thank you for calculating it too in a simple way(step by step) p.s. i calculated the stuff and isn't it suppose to be 11.4?

While the problems are accurate (except ex. 2 due to the simple error of the voltage being 11.4 instead of 11.3) they are somewhat misleading because 0.7 is the knee voltage for a silicon diode meaning it requires that amount to pass any current. After enough voltage is applied to achieve current flow, the current is the circuit grows exponentially as the graph demonstrates until it eventually overheats. This is why he can assume the voltage of the diode as a given. BUT THIS IS ONLY ACCURATE FOR AN IDEAL DIODE. In an actual circuit, "body" resistance and "contact" resistance of the diode itself will affect current. Lastly because this is only for ideal circuits, saturation current was not taken into account.

Great video for understanding the basics. Way better explanation than my professor

You explained diodes in 10 minutes better than my uni prof in an hour and 20, gg

Most useful video in youtube for me.

thank you so much for ur video!!! it took me a day to figure out this question i had for hw but learned it from u in 5 mins

You explained it in a simple way . Thank you for your efforts

Yes the answer is 11.4v not 11.3v otherwise your battery is 11.9v not 12v like you showed

11.4V in the 2nd example, instead of 11.3V :-)

do you have anything besides this one? like Half Wave Rectifier, Op Amps, Transistors, Combination Logic Modules, J-K FlipFlop and much more? Your explanation of this is so much better than any of my professors! I hope you make more videos. Thanks!

Better than my prof, thanks a lot!

Thank you so much for explaining some basic solving tools! Now I can understand my homework! :D

Thanks alot alot ,, u really saved me :) please if you have more stuff about electronics and their problems please do more videos since you are better than my prof

Very helpful! Thanks very much!

Really appreciate your work

i love the way u teach sir

at 5:28, shouldn't that be 11.4 instead of 11.3 since the diode is 0.6v

Thanks man!!! Your video is a HUGE help!!!:)

Thank you so much. You have answered my question.

you just saved my uni life.

Grande tío, me salvas!!!

Thanks man , this helped me a lot 👍👍

This was not what I was looking for but it was interesting never the less.

Very well explained

Very helpful!

Thank you!

thank you so much sir 💚

In the last segment, why is the current flowing through the parallel resistor at all given that forward bias diode gives a short circuit ?

Wow... Co.mplex. I am just a hobbyist, but I am learning.

thank you so much, u explained it well

excellent!!!!!! thanks.

THANK YOU!!!!!

Hey just a quick question, in series voltage is the same and parallel voltage is different? thanks in advance

Thank you so much 🙏🏻

Thank you ..

excellent

Nice sir

thank you very much

This helped me

if there is no current in reverse biased conditions then how does potential drop is measured accross pn diode as there is no current in circuit

Thanks sir

at 4:15, why are we using 11.3 V instead of 12 V to find the current in the resistor? Shouldn't all components in series have equal current?

so resistors take on vf values if its in parallel to something??? wouldnt it depend on whatever the load is?? or do resistors even have that value??? im learnin all this on my own so im sure im missin a lot of very critical elementary things.. im tryin tho n im not stoppin till i figure it out

what happens if same circuit, but reversed diod?

Thanks

what if you have 2 voltage source ? and 2 diodes?

a lot of people seem to be getting this confused, the range of a diode is 0.6< Vd < 0.8 (this does NOT include 0.6) so for the lesson in the video it is all correct, it is 11.3 because 12-0.7 = 11.3 not 11.4, he even draws on the graph 0.7 as the point of Vd.

thanks

isn't the diode supposed to not let any current through for the second example? since its voltage is less than the 0.7 V required to turn it on.

Why does no current flows through a diode in Reverse bias mode?

11.4V

4:25 legit math right there

For when you are finding the resistance (where you got R = 1883 ohms) shouldn't the math be 11.4 / 6e-4 since it's .6 ma, not just 6 ma? I got R to be 18K, 19K ohms

if silicon is .7v and Germanium .3 what volts has in Carbon? pls answer

I prefer learning you tube than class room.. May be these kind of video is liable.. :p Thanks... Cool explanation..

it is me or every instructor on youtube did a better than job than my professor. god and i pay 15k a year for these people who gave absolutely zero fuck

What if there is 2 diodes

:*

I was looking for a current with 2 diodes in the circuit but thanks anyways 😔

There is a lot wrong here as some have stated. V drop should be 11.4Volts .6mA is 6*10^-4 answer should be 19K ohms

You needed a calculator to work out 93 - 7 ? ....really?

hELLO HOW DIODE WORK DIODE PRICE IS 1 HOW TO EARN 1MILLION

Thank you!