ECE3050 Assignment 121. The figure sho ws a differential amplifier. Each MOSFET has the parameters gm=1/400 S,r0=40kΩ,andχ =0.5.ItisgiventhatRD=10kΩ and RQ=20kΩ. The answ ers givenassume the r0approximations.(a) Show that the resistance seen looking out of each source is Rts= 263.16 Ω, the resistanceseenlookingintoeachdrainisrid=79.74 kΩ, and the output resistance seen lookinginto the two outputs with vi1= vi2=0is rout1= rout2=8.886 kΩ.(b) For th e differential inputs vi1= vid/2 and vi2= −vid/2, show that vo1/vid= −vo2/vid=−11.11.(c) For the common-mode inputs vi1= vicmand vi2= vicm, show that vo1/vicm= vo2/vicm=−0.147.2. For a MOSFET diff amp, show that the body effect cancels out with differential inputs ifthere is no external resistor in series with the source leads, i.e. RS=0.Hint:Fordifferentialinputs, the currents can be calculated with the sources connected to signal ground. This isthe case in the preceding problem.3. The figure shows a MOSFET differential amplifier. For the MOSFETs, it is given thatk0=0.008 A/ V2, W/L =1, VTO=1.5V, λ =0,andχ =0.4. Forthecircuit,itisgiventhatV+=20V, V−= −20 V, IQ=2.5mA, RG=5kΩ, RS= 100 Ω,andRD=12kΩ. Assumethe dc components of each input are zero, i.e. VI1= VI2=0.(a) With vI1= vI2=0, calculate VGSfor each transistor. Answer: VGS=1.956 V.(b) With vI1= vI2=0, calculate VDfor each transistor. Answer: VD=5V.(c) UsethevalueofVGSfound in part (a) and the value of VDfound in part (b) to solv e forVDSfor each transistor. Verify that VDS>VGS− VTO.Answers:VDS=6.956 V andVGS− VTO=0.456 V.(d) Calculate gm, rs, r0s, r0, and the resistance Rtsseen looking out of eac h source. Answers:gm=5.477 mS, rs= 182.6 Ω,andr0s= 130.4 Ω, r0= ∞,andRts= 330.4 Ω.1(e) With vi2=0, show thatvo1vi1= −vo2vi1= −18.6 rout=12kΩ(f) With vi1=0, use the Norton drain circuit to show thatvo2vi2= −vo1vi2= −18.6 rout=12kΩ(g) For the differential input signals vi1= vid/2 and vi2= −vid/2, show thatvo1vid= −vo2vid= −18.6(h) For the common mode inputs vi1= vi2= vicm, why is the common-mode gain equal tozero? Answer: Because the tail supply is a perfect current
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