hw12_sol - ECE ECE342 HW#12 Solutions 442 HW#13 Solutions...

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Unformatted text preview: ECE ECE342 HW#12 Solutions 442 HW#13 Solutions 1. gm1 = gm1 VOV 1m × 0.2 2ID ⇒ ID = = = 100µA VOV 2 2 RO = (gm2 ro2 )ro1 Since the transistors are identical with the same current flowing through both, ro1 = ro2 = ro and gm1 = gm2 = gm ; as specified in the problem: 2 RO = gm ro = 400k Ω ⇒ ro = 20k Ω Since ro = VA L ID , we have: L= gm = 2µn Cox W × L 100µ × 20k ID ro = = 0.4µm 1 VA 5µ ID ⇒ 2 W gm (1000µ)2 = = 12.5 = L 2µn Cox ID 2(400µ)(100µ) For the maximum negative swing at the output, we want the MOSFETs to be biased so that each transistor can reach VDS = VOV = 0.2V . ⇒ VG2 = Vtn + VOV + VOV = 0.5 + 0.2 + 0.2 = 0.9V Minimum output voltage will then be Vout = 2VOV = 0.4V . 2. From the small-signal schematic, we can write the following KCL equations, noting that the transistor M1 reduces to just a resistance ro1 , since the control voltage vgs1 = 0, and the control voltage vgs2 = 0−vy = −vy : ix = −(gm2 + gmb2 )vy + ⇒ vx − vy vy = ro2 ro1 ro1 1 vy ro1 ≈ ≈ = vx ro2 (1 + (gm2 + gmb2 )ro1 ) ro1 ro2 (gm2 + gmb2 ) gm2 ro2 Therefore, vy is smaller than vx by a factor of gm2 · ro2 . 3. a) For the same I in cases (a) and (b): I= gm ( W )a V2 1W2 L kn VOV ⇒ OV b = W 2 2 L VOV a ( L )b 2 For the same I, if W/L is divided by 4, then VOV is multiplied by 4, or equivalently VOV is doubled, W = µn Cox L VOV . Thus, gm for the circuit (b) is half of the one for circuit (a). AO = gm ro = 2ID VA 2V L × =A VOV ID VOV Thus, if L is multiplied by 4, and VOV is halved, then AO is doubled for circuit (b). In summary, for circuit (b), VOV is doubled, gm is halved, AO is doubled. 1 P6. ...
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This note was uploaded on 11/06/2011 for the course ECE 342 taught by Professor Nareshshanbhag during the Spring '11 term at University of Illinois, Urbana Champaign.

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hw12_sol - ECE ECE342 HW#12 Solutions 442 HW#13 Solutions...

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