Master Micro’s Post

The answer is d Doubling the bias current at constant gm/id will double the transconductance gm but it will also double the device capacitances because the width is doubled to keep the same gm/id and double the current . So speed(GBW)=gm/Ctot remains unchanged as both are doubled And rms noise =KT/Ctot. Decreases

First time you read the question, you may think about speed boosting; however, the key word is "limited by MOSFET capacitance". To keep gm/ID bias point constant, gm should be doubled too. To double gm, you must double the width where the doubled current is not enough to double gm. Here comes the trade-off, MOSFET parasitics would also be doubled as they are proportional to the width of the channel even if some are independent on the length. Assuming first order system, the speed is defined as follow: GBW = gm/Cout. In our case, Cout is doubled, and gm is also doubled, so GBW is almost constant: no speed boosting. Noise? Since it's a baseband analog circuit, the noise mostly depends on the capacitance: KT/C noise. Then, the rms Integrated noise will decrease with the increase in capacitance, and this leave us with the correct answer, which is D ✓.

The answer is D By doubling the bias current at a constant bias point (gm/ID, V*) that is led to double the width of the device (doubling the device caps)and doubling the gm of the device too so, rmsThermal Noise = ( 4𝑘𝑇𝛾/𝑔𝑚)*BWn BWn=( π/2)*fp=1/4RC GBW = gm/Ctotal -by doubling the width the caps will double also so the GBW will remain the same. -the rmsThermal Noise will decrease. I hope to be correct

If you go by the basic equation dt = Cdv/I, we see here the Current is Doubled keeping the Q point same which means W/L has been doubled which leads to Double the capacitances so dt is constant for a specific dv, so speed will remain the same, now noise is nothing but KTB, Where B is BW is inversely proportion to the Time constant which has RC in it…So As cap doubles, BW decreases thus, noise decreases!

Speed is directly proportional to BW, if gm/Id is same, that means gm is also made twice. And bandwidth is gm/C(self), both will increase as gm is increased by W/L. But same effect will be seen of self caps as there is no external capacitance connected. So BW remaining same. Now thermal noise is inversely proportional to capacitance loading so it will reduce. Hence D

D id doubled but gm/id unchanged, so Width is doubled. Cap is doubled. Speed is defined by I/C so speed unchanged. Since gm is doubled, input referred noise is defined by 4ktr/gm, so noise is decreased

I guess D should be the answer