Cosmic Feb 06, 2007 by Raja Reddy P

Transcription

1 osmic Feb 6, 7 by aja eddy P. ou() i() alculae ou(s)/(s). plo o(). calculae ime consan and pole frequency. ou ( e τ ) ou (s) ( s) Time consan (/) Pole frequency : ω p. i() n he above circui plo i I (), ()

2 i ( e τ ) τ i. ( e τ ) e τ 3. s p p s Boh circuis are equivalen. Express p and p erms of s and s. Fd ω range for which hese boh are equivalen. ssume high qualiy facor. p cp s where s jω s s p s On solvg his equaion we will ge p p p (Q ) s ( Q ) s where Q ω

3 4. sysem has an SN of 6 db. If an uncorrelaed noise of m is added a of signal o i, hen wha is he SN? 57 db 5. o() K 8K.6 s(ω).4s(ω) pf Fd o() he above circui If you observe he circui carefully, you can noice ha four resisors blue he figure below forms a wheasone bridge, and you can elimae he branch wih capaciance. o() K 8K.6 s(ω).4s(ω) pf The resulg circui is

4 o() K 8K.6 s(ω).4s(ω) Now you can solve i direcly. o.s(ω) 6. Tcq ns Tseup ps Thold 3 ps Tpd ps lk alculae he maximum clock frequency. T clk > Tcq Tpd Tseup ns ps ps.3 ns fmax /(.3ns) 7. Plo ou. K s(w) 5 5

5 This is a clipper circui Design a divide by 3 couner usg D flip flops. The duy cycle of he divided clock should be /3. S [] S [] S [] Presen sae Nex Sae Q Q D D o/p xx x From he able D D Q Q o/p Q Q D Q D Q O/P lk

6 9. Wha is he funcion of he followg circui? Ā Ē E Y Ā E Ē XO. Plo ou wih respec o he given pu waveforms. ou _ 5 pf 3 3 _ 5 ou

7 . Plo he oupu ou..s(w) ou The given circui is a full eave recifier. Observe he ga, i is equal o alculae he frequency of oscillaion. Wha is he mimum required Gm for oscillaion? G m G m G m G m In he given oscillaor here are four ga sages. One of hem operaes verg mode and he remag non verg mode. So he phase shif provided is 8 o and he remag 8 o phase shif should be provided by four he sages, o make he feedback posiive. The ransfer funcion of each sage is Gm H (jω) jω

8 So he phase shif provided by each sage is an (ω) 45 o. o an ( ω o) 45 ωo fo π and should be equal o fo he aenuaion provided by each sage is. So for loop ga o be greaer han or equal o one, each ga sage should a leas provide a ga of. G m G m 3. alculae he oupu impedance. MΩ 5KΩ 5KΩ OUT G m µ/ ds MΩ pply a small volage a he dra node and observe he curren flowg. Toal curren flow g o he dra (oupu) node is Δ Δ Δ/ ΔI ( Δ/)G OUT Δ ΔI ds ds m (/)G m 5KΩ MΩ Δ OUT OUT ds G MΩ MΩ MΩ MΩ 5KΩ m Δ/ 5KΩ ds

9 4. Boh ransisors are biased sauraion. alculae /. Neglec he body effec. D D M M M /g m Impedance seen lookg o his node is (/g m ) D /(g m r o ) /g m Now he simplified circui is a source follower wih a load resisance of /g m. So ga g m assumg g m g m g g m m

10 5. alculae he oupu if (i) ga v fiy (ii) ga v ou KΩ KΩ m ou v ( ) KΩ KΩ ou m ou Solvg he above wo equaions, we will ge ou v( ) v v ou v ou 5 ( 6 )

11 Inerview Quesions. In erview maly your knowledge abou circuis will be esed.. For me hey have asked abou my approach o solve he problems given he exam 3. bou my fal projec 4. Som e quesions abou oscillaors (maly We Bridge Oscillaor): a. how i works, and b. wha happens if you erchange he series and parallel branches We Bridge Oscillaor If you erchange series and parallel branches, here will be more posiive feedback a D and hence he oupu sauraes o osa. ***The erview will be very cool, hey will help you answerg he quesions!!!. nd he mos imporan hg is here is no separae H erview.