single input unbalanced output configuration || LICA by ushendra's engineering tutorials

[@familyforever8634]

it's useful., good explanation.

[@srijanisallyouneed]

9:54 the calculation of the 2nd loop should be taken in the same way as we did in *Dual input case* but...why it is taken as this way😨 ??? I didn't get it

[@nandhumadiri586]

Nice explanation sir 🥳

[@PrabhuKumar-xb4yh]

Good explanation

[@Bujjigadu-in-chennai]

What about Balanced condition

[@kudajohnthamous2457]

Nice sir

[@maaheg2346]

good

[@aryanpandita9529]

thanks sir for simplifying Ie2 also.

[@ushendrasengineeringtutori4259]

Tq🙏🙏. Keep supporting plz

[@saikrishnajakkula9]

@@ushendrasengineeringtutori4259 sure sir..

[@psnarayana1261]

Thank you sir

[@swaritmahalsekar5521]

Sir since 2nd input is not present so we don't calculate Ac (Common Mode gain) right?

[@ushendrasengineeringtutori4259]

Yes , u r absolutely correct.

[@swaritmahalsekar5521]

@@ushendrasengineeringtutori4259 Sir I have one more doubt like the other input is grounded so Rin = beta*Re, why factor 2 is present there?

[@swaritmahalsekar5521]

During Dual input Vin1 = Vs/2 and Vin2 = -Vin1 right? and during Single input, we directly apply Vin = Vs to any base of BJT right? So don't you think if Vin2 = zero so therefore Rin2 =0hms? Please guide me I'm trying to figure it with the help of H-pi model

[@srinueera4256]

nice explanation

[@sonttigayathri8474]

Tq💐

[@praveensetti5086]

Good explanation