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Instrumentation amplifiers are ideally suited as the name suggests for Instrumentaion applications as they have many specialties like hIgher CMRR etc. but using it in circuits that require just one opamp doesn't make sense.
i can suggest an easy method to find the kind of feedback employed in any circuit.Finding out the mixing is very easy:if the signal from the feedback path meets the input signal at a node then its current mixing,otherwise its voltage.finding out the type of sampling requires a little more...
ya sridhar i agree what i meant was becoz of width reducing it reduces the w/l of the finger transistors further in layout if the difference in size of W & L is not too high it gives rise to symmetric and nice layout with a good aspect ratio.
nmos finger multiplier
fingers are made when a mos havinga alarge w/l is split up into smaller Moses of lesser w/l's in parallel.the equivalent w/l and hence gm of the combination remains the same but using fingers dramatically reduces the parsitics aasociated withe the junctions as well as...
take the buffer as an example. the gain with feedback is =Ao/(1+AoB).since B=1 we have Af=Ao/(1+Ao), now if Ao -->inf Af-->1 ,hence we require high gain in opamp to reduce the dependence of closed loop gain on the open loop gain.The gain of an ideal opamp must depend only on its feedback...
sry i misunderstood the question,as the other ppl have pointed out ,the gain increases when we use a current source load but a problem tat arises is that the o/p CM level is not well defined .you need to use a CMFB circuit to define the O/P CM level
Re: bandgap reference
u can use the .TF comand instead and specify the nodes at which u want to check the resistance ,i n the chi file u can then get the transfer function along with the output resistance.
i think u mean to say the current source ISS is replaced by an ideal one. in this case,the gain increase slightly and the o/p CM level becomes completely independent of the i/p CM level
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