Common Source Amplifiers - Current Source Load

www.eneradi.com

[@asafabou]

thanx for the video. if i understood correctly, you need to increase L toghether with L, but if you do so so R will remain the same, isn't it?

[@TahaMehrabiMusic]

very clear and interesting.thank you, are you gonna be putting up common gate too?

[@pyroBlastM]

Good explanation Eneradi. It was a nice explanation on trade-off. I am trying to design this amplifier on Cadence but without success. I want to get a gain of Av = 100 but with my experiments I am not getting out more that 12. Can you give me some good references or maybe you could give me a tip where to start? Regards.

[@MohamedElsheikh22]

Thanks very much excellent explanation ,you made life simpler , I hope some day I sit in a class in your university :)

[@narendraprasad244]

thank u very much sir for giving such a nice lecture. i want to request u to upload a lecture on exercise problems on cmos amplifier.

[@DwiNoviantoUntidar]

Can u explain PMOS Current Source Load as a current controled? I am still confused to calculate it. thanks before.

[@akhimakhi93]

+Dwi Novianto should be mathematically similar to diode connected load if I'm not mistaken.

[@ramkrishna3256]

Dear sir I have doubt in this, Please solve, We made a bias voltage there to make M2 in triod region, right.? Then there exist Triode resistance instead of r0. But you have not taken Triode resistance here.why?

[@ΓιωργοςΛαμψιδης]

he made the bias voltage to keep M2 in saturation ( 3:48 ) (equivalent to common source amps with diode n or p Mos). With both M1,M2 in saturation, their resistances (ro1,ro2) are comperable so the gain equation cant be simplified as with the Rd

[@Souraneel]

although i dont like how u drawing the mosfets, great video man. ur explanations easy to understand

Thank you for very clear explanation..

[@43maruko]

what the relation between transconductant (gm) and DC voltage? please explain.. thankyou

[@joaquincury8378]

Hi Eneradi, Really good the video ! I have two questions !. The first is related to the Rd in the common source amplifier. I found in many books that if you want to improve your gain Av, you should make higher the resistance Rd, but the disadvantage doing this is that the transistor will be more closer to the triode region. Am I correct ? So in this configuration with a current source as a load, why should I use this configuration if it provides the highest resistance in CMOS process ? The other question is related to a good video you posted where you say the reason why the transistor should be in a saturation mode. That is for the amplification but maybe I am wrong but if I make a configuration where my transistor works at a lineal region, it will amplify. So why do you say that in the lineal region the transistor will act as a switch ? I would be very grateful if you can help me ! Best regards ! Joaquin

[@sivananduri2933]

yes Joaquin, if the Rd(increases) => Mosfet will get triode region. In this current source load configuration, it is ideal current source, Resistance is infinite, so we are using channel resistance here, so channel resistance is high but that is also finite

[@sivananduri2933]

coming to your 2nd question: transistor acts in 3 regions: 1 cut off (open switch) 2, linear mode (closed switch) 3. saturation (closed switch and amplifier), means constant current there is no open switch in saturation.

[@mahima7638]

Great video! Thanks.

[@ajinkya16]

thanks

[@obtron]

nice vids but, right ear feels rejected :(