thank you sir... i didn't understand this ac resistance clearly in my engineering class . even i didn't find gud answer in book like JP Gupta, Sedra and soon .. thank u a lot.....you provide me a clear view of the topic . .... this tutorial completely match to my engineering syllabus , (Nepal)....
The ac ripple voltage is 1 V peak to peak. The dc (average) load voltage is 5 V. What is the Q-point current in the Zener diode? What is the maximum dynamic resistance allowed for the Zener diode if the output ripple is to be less than 10 mV peak to peak?
Sir why we does not take cutoff voltage in ac analysis of diode . I mean why we assume 0 cutin voltage for ac signal instead of 0.7 . It was my interview qus . Plz explain.
In case of ac source then the sine wave will be varying acc to time. So consider during positive cycle the p type is connected to the phase and during negative cycle neutral will be connected to p type of the diode .... so during the negative cycle diode won t conduct..... it will be reverse biased... so how can the resistance value can oscillate in this curve please explain.. i am getting confused.
what is the difference between AC resistance of diode and AC emitter Resistance? for ac resistance we have formula rd=26/IF where as ac emitter resistance it is r'e=25/IE. please help me. i want to request for derivation for AC emitter resistance.
Vt=26mV bcz it is called thermal voltage and its formulae is Vt=kT/q where,k=boltzmann constant T=temperature in kelvin generally 300 kelvin Q=electronic charge
if it is ac current or voltage , where is the negative cycle of ac ? .At negative cycle the diode should be reverse biased and the plotting should be at 3rd quadrant . I think the given graph is wrong as for AC
Thank you for providing such a good presentation. i have a doubt regarding ideality factor value. is it decided based on semiconductor or current through the diode?
thank you sir... i didn't understand this ac resistance clearly in my engineering class . even i didn't find gud answer in book like JP Gupta, Sedra and soon .. thank u a lot.....you provide me a clear view of the topic . .... this tutorial completely match to my engineering syllabus , (Nepal)....
sir ..soon provide video on op amp and timer 555, volage regulator lm317, LDR, and all other senser....
kun colz r faculty ho daju ko??
what a gargious explanation..........thank u sooo much sir
Thank you for providing these lectures on TH-cam
Great presentations
thank u so much
Thank you!! The final proof was very well explained💖👏
How could u get rd = 26mV/Id ? (4:43)
please explain more.....
thanks anyway.
vt=kt/e at room temp vt=26mv
Amazing Exlanation
sir explain about diode current equation
Very helpful lecture 🙂
Thank you so much sir
love from bangladesh
at last u said when i(d)is small then n=2 ,but please tell how much small in terms of value
VT changes when temp changes, VT is not constant. VT=kT/q
Great explanation and derivation, thanks!
smjh aa gya sir jiii ache se
at 1:39 why did you say that voltage is zero??
Not zero.
i think he is talking relatively to the point Q . He normalized with respect to point Q , i guess.
helpful, thank you very much !!!!!!!
very detailed and well explained .
The ac ripple voltage is 1 V peak to peak. The dc (average) load voltage is 5 V. What is the Q-point current in the Zener diode? What is the maximum dynamic resistance allowed for the Zener diode if the output ripple is to be less than 10 mV peak to peak?
Thank you so much for the amazing presentation Sir.
Why is rd=26mV/Id not applicable for the knee voltage region
if Id is low (at knee voltages) ,n=1 for germenium ,2 for silicon. but if I is large n=1 for both germenium and silicon
sir, where is that 26 mv coming from,?
Sir why we does not take cutoff voltage in ac analysis of diode . I mean why we assume 0 cutin voltage for ac signal instead of 0.7 . It was my interview qus . Plz explain.
Great work
What is Is?
In case of ac source then the sine wave will be varying acc to time. So consider during positive cycle the p type is connected to the phase and during negative cycle neutral will be connected to p type of the diode .... so during the negative cycle diode won t conduct..... it will be reverse biased... so how can the resistance value can oscillate in this curve please explain.. i am getting confused.
i am preparing for gate . is your lectures are enough for concept ?
So so good work sir
ThanK u Great Presentation
what is the difference between AC resistance of diode and AC emitter Resistance? for ac resistance we have formula rd=26/IF where as ac emitter resistance it is r'e=25/IE. please help me. i want to request for derivation for AC emitter resistance.
You r great👍🏻
Thanks ❤️
Thank you so much
Thanks a lot!
sir why r u saying that in graph that, vd = 0 and Id= 0 (initially)
It's just for AC he is saying
I think this equation can't be used to evaluate ac resistance in reverse bias???
great sir
Thanks a lot
3:56
Thank you sir
Thank you sir............!!!
sir which book would be best for above lecture
Why you told, Vd=26 mV!! ?? from what we know that.?
Vt=26mV bcz it is called thermal voltage and its formulae is Vt=kT/q where,k=boltzmann constant
T=temperature in kelvin generally 300 kelvin
Q=electronic charge
26mV means voltage in normal room temperature
Thermal voltage
How to calculate dynamic resistance in reverse direction ?
U don't have to. Because it blocks current in opposite directions.. And u can predict it's a large value.
I think I am right.... Am I right 🐱
dynamic resistance is also an ac resistance right?
Why you put vT =26mV?
This is thermal voltage ....and this was derived in earlier vdo
sir... for reverse bias condition......??
if it is ac current or voltage , where is the negative cycle of ac ? .At negative cycle the diode should be reverse biased and the plotting should be at 3rd quadrant . I think the given graph is wrong as for AC
Dear it's AC sitting on DC.. so when AC -ve half cycle comes the Q point shift down vertically... Which is shown in the graph..
you should change the title slightly u should put diode/transistor at the end
sir,how come you assume Vt=26mv and eta as 1
Vt =26mV only in room temp ....the formula to calculate vt=k*t/q ...where q is charge ,b is Boltzmann constant and t is temp in kelvin
how you assumed that Vt is equal to 26mV?
How small 'Id' should be to get replaced by n=2......???
+1
why Vd is equal to 1 ?
Thank you for providing such a good presentation. i have a doubt regarding ideality factor value.
is it decided based on semiconductor or current through the diode?
semiconductor. it is 1 for Ge and 2 for Si.
sir. if Id is not small then is eta=2
and is Resistance=52mv/Id?????
Vishal Choudhary eta=1 when Id is small and rd=26mV/Id as shown in the lecture
awesome
Why the value of neeta is taken 1 ?😓
لو احد يترجم هالمواضيع للعربي
Thank you sir