Opamp controlled series regulator

[@manojboganadham9071]

Never enjoyed electronics like this before. Perfect explanation, sir!

[@akshatavdaggi2350]

Perfect explanation, Thank you for your all videos

[@memsuniverse]

Precious course thank you

[@awesomefish843]

Very clear explanation and very beautiful presentation. Thank you very much, Sir.

[@agstechnicalsupport]

Perfect explanation of opamp controlled series regulator !

[@bransoncamp192]

Best explanation on the internet thanks

[@CEO_Lifestyle]

Always fan of his lectures😀

[@abhisheksingh-hm1gk]

What a wonderful explaination sir !! U make it simple ... Tommorow is my University exam ... Thanks to clear my doubts ... Thank you sir

[@techgurukula]

Thank you Abhishek! Good Luck for your exam!

[@UMM-s6l]

Beautiful explanation thanks!!

[@MrRaghavaprasad]

Sir, my doubt is when Vz=V2 then output voltage of opamp will be zero, then will BJT turn off? I think Voltage regulation is a continues process, when voltage should be stabilised?

[@GrayeLara]

I have the exact same question, did you find the answer? A Op Amp as a comparator only have or not voltage/current at output depending on the voltages at its input terminals. Unless Op Amp output acts as a transistor, that in a rude manner, we could say acts as a current dependant resistor, in this case there is a continuous regulation and not a discrete one. Regards

[@diegoochoa572]

You can essentially already think of vz being equal to r2. Because both op amp inputs must be the same, the op amp will compare the inverting input to the zener reference, and adjust its output so that V2 *does* edual vz. Another way of looking at it is if you have the inverting input attached to the center Wiper arm of a pot that lies inbetween R1 and R2. If you adjust the pot so that the center is at the bottom and the inverting op amp input is in parallel with only R2 (we'll call this Vmax), then the op amp will regulate its output so that only V2 = Vz. If you adjust the pot so that the Wiper is at the top and the inverting input is in parallel with both the pot and r2 (we'll call this vmin), then the op amp will adjust its output so that only (vpot+v2) = vz. Say; vz = 3v, R1=1k, rpot=5k, and r2=2k. Therefore, in the vmax position ibranch = 3v/2k or 1.5mA. 1.5mA*6k = 9v 9v+3v=12v. Therefore Vout in that vmax position equals 12v In the vmin position: 3v/7k = 428.5 µA 428.5 µA * 1k = 428.5 mV 428.5mV + 3v = 3.428V Therefore Vout in the vmin position equals 3.4V. Sorry for the drawn out explanation, hope this helps

[@ancyjohn1121]

Thanks for this explanation. 👏

[@Userx29]

Very useful and educational! Merçi

[@drkazmawjiMScmedicine]

May be part 2 would show us the circuit and its application outcome on oscilloscope.

[@sandeeppinninti3087]

Thank you for super explanation

[@winstonegm]

Great work there

[@creativecorner1559]

Nice explanation sir

[@mrigankomandal710]

Perfect 👌!

[@sapankushwaha4069]

Thanky very much sir for uploading such amazing video

[@techgurukula]

It's my pleasure. Thanks a lot for the comment!! :) More videos to follow.

[@sapankushwaha4069]

Sir I am very very happy. When I saw your edc lacture they are amazing and I am waiting for upcoming vedio finally god fullfill my wishes and thanks again 😊

[@techgurukula]

@@sapankushwaha4069 Happy to know. Feeling blessed 🙏❤️.

[@132_pradyumnajoshi8]

Can this circuit be used for Battery Overcharge/Overdischarge protection as well?

[@40_usha80]

How is maximum load resistor selected here?

[@DarkDevil-dn3ot]

Hey,can anyone out there help me clear a doubt? At first,wr are using the formula of open loop config, Vo'=A*(V1-V2) But at last,we are assuming the close loop config and saying that V2=Vz

[@logeshwaran1637]

Thank you sir

[@hoytvolker3]

thanks bhai

[@mohammedelex4518]

Thank you

[@alexjunan]

Hi sir can i know how to find the power dissipated by BJT if load current is given? is that just use the Vo times load current?

[@user001sg9]

This has been explained please watch the video from 10.00 to 10:50. Neglecting the base current Ib (which is small compared to the collector current Ic), your load current becomes equal to the collector current. Multiply load current with BJT's Uce voltage, you have the power dissipated by the BJT.

[@nasrunhariyanto7393]

It's clear. Thans

[@shivendrayadav18]

Thanku sir

[@saint_and_holy_unicorn]

Super amazing

[@Yashu_mm]

Thank you sir 🙏

[@SAhellenLily]

Thank you