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### Homework Assignment 7 Solution

Course: ELECTRICAL ee2021, Spring 2011
School: National University of...
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UNIVERSITY NATIONAL of SINGAPORE Department of Electrical and Computer Engineering EE2005 Electronics Homework Assignment #7 Solution (No need to hand in homework assignment) 1. Consider the current mirror shown in Fig. 7-1. The transistors have Kp=100A/V2, VTHP=-0.5V, p=0.01 and no body effect. (a) Select R1 such that the output current Iout is 0.2mA. (b) Find maximum output DC voltage (VD) for all transistors...

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UNIVERSITY NATIONAL of SINGAPORE Department of Electrical and Computer Engineering EE2005 Electronics Homework Assignment #7 Solution (No need to hand in homework assignment) 1. Consider the current mirror shown in Fig. 7-1. The transistors have Kp=100A/V2, VTHP=-0.5V, p=0.01 and no body effect. (a) Select R1 such that the output current Iout is 0.2mA. (b) Find maximum output DC voltage (VD) for all transistors to still operate in saturation region and the corresponding AC small signal parameters of the transistors. (c) Find the small-signal AC output resistances (Rout). (d) Write down the netlist for the schematic in Fig. 7-1. (e) What is the simulated AC output resistance (Rout)? (Hints: Treat VD as voltage source that has DC bias of 0V and AC amplitude of 1V. You can then obtain output resistance by doing ac analysis and find out the small signal vd/i(vd).) You may assume the following model for the PMOSFET : .model Psimple_mos PMOS level=1 w=289u l=30u vto=-0.5 tox=100n lambda=0.01 +cgdo=1.73n 10V 10V M1A M1 M2A M2 M3A VR1 M3 Rout Iout R1 VD Fig. 7-1 (a) I D , M 1 A = I D , M 2 A = I D , M 3 A = I D ,M 1 = I D , M 2 = I D , M 3 = I out = 0.2mA ( I D , M 1 A = K p VGS , M 1 A VTHP ) 2 = 0.2mA VGS ,M 1 A = 1.914 VGS , M 1 A = VGS ,M 2 A = VGS , M 3 A = 1.914 I D,M 1A = V R1 10 VGS , M 1 A VGS , M 2 A VGS , M 3 A = = 0.2mA R1 = 21.3k R1 R1 (b) 1 In order for all transistors to operate at saturation V D < 10 VGS , M 1 A VGS , M 2 A VGS , M 3 A + VGS , M 3 VDSSAT ,M 3 V DSSAT , M 3 = VGS , M 3 VTHP = 1.414 V D < 4.76 g m , M 1 A = g m , M 2 A = g m , M 3 A = g m , M 1 = g m , M 2 = g m , M 3 = 4 K p I out = 283A / V 2 ro , M 1 A = ro , M 2 A = ro , M 3 A = ro ,M 1 = ro ,M 2 = ro , M 3 = (c) 1 = 500k p I out Rx1 Rx1 M1A M2A M3A VR1 Rx2 Rx3 M1 Rx3 M3 R1 M3 Rout Iout Rout Iout M2 M3 Rx2 M2 M1 M2 M1 VD VD Ry2 Ry1 Rout VD Table 2 Configuration B R y 2 = ro , M 1 = 500k Table 2 Configuration G R y1 = ro , M 2 (1 + g m , M 2 R y 2 ) = 71.25M Table 2 Configuration G Rout = ro , M 3 (1 + g m , M 3 R y1 ) = 10G (d) *Cascode Current Source vdd vdd 0 10 M1A A A vdd vdd Psimple_mos M2A B B A A Psimple_mos M3A C C B B Psimple_mos M1 D A vdd vdd Psimple_mos M2 E B D D Psimple_mos M3 vd C E E Psimple_mos R1 C 0 21.3k Vd vd 0 0 ac 1 .model Psimple_mos PMOS level=1 w=289u l=30u vto=-0.5 tox=100n lambda=0.01 cgdo=1.73n .ac dec 10 0.1 100 .probe .end (e) Simulated Rout=11.562G. 2 2. Consider the CS amplifier shown in Fig. 7-2. and no body effect. The transistors have K=100A/V2, VTH=0.5V Design R1 and RB such that M1 has drain current (ID,M1) of 500A and the amplifier has an input resistance (Rin) greater than 100k. (b) Identify the source, load the and two-ports network. Estimate the two-ports equivalent parameters, Rin, Rout, Gm. (c) Estimate the overall gain (AV=vo/vs). (d) Verify your gain through PSPICE simulation. You may assume the following model for the NMOSFET : .model Nsimple_mos NPMOS level=1 w=289u l=30u vto=0.5 tox=100n lambda=0.01 +cgdo=1.73n 10V RD R1 10k vo RB M1 M1A (a) RS vs 1k Cin 1 Fig. 7-2 (a) 3 I D , M 1 A = I D , M 1 = 500 = K (VGS , M 1 A VTH ) VGS , M 1 A = 2.74 2 I D,M 1A = V DD VGS , M 1 A R1 = 500 R1 = 14.5k g m , M 1 A = 4 KI D , M 1 A = 447 A / V 2 1 Rin = RB + // R1 = R B + 1.9k > 100k g m,M 1A Choose RB = 100k (b) 10k RB RL Rin RS 1k vo M1 M1A Two-Ports Network vs Load RD R1 Rout Source g m , M 1 = 4 KI D , M 1 = 447 ro , M 1 = 1 I D , M 1 = 200k Amplifier Configuration is CS G m = g m , M 1 = 447 1 Rin = R B + // R1 = 101.9k g m,M 1A Rout = R D // ro , M 1 = 9.5k (c) Source Rin vi = v s vs Rin + RS vo = Gm vi Rout = g m , M 1 Rout v s vi RS vs 1k vo = g m , M 1 Rout vs = 4.25 (d) 4 vo Rin Gmvi Rout 102k 447vi 9.5k *CS Amplifier vdd vdd 0 10 M1A A A 0 0 Nsimple_mos R1 vdd A 14.5k RB A B 100k M1 out B 0 0 Nsimple_mos RD vdd out 10k vs vs 0 ac 1 RS vs C 1k Cin C B 1u .model Nsimple_mos NMOS level=1 w=289u l=30u vto=0.5 tox=100n lambda=0.01 cgdo=1.73n .ac dec 10 100 1g .probe .end (e) Simulated AV=-4.37 3. Analyze the following circuit. Find out all the driving point resistance. effect. 5 Neglect body R6 M2 M3 I R5 M1 VB2 Rin Q4 I R4 R3 VB1+vs R1 RL I R7 VB1 Q1 Q2 vout Q3 R8 R9 R2 Rout R10 2I Fig. 7-3 R6 M3 M2 I M1 Rin Q4 RL R4 R3 R7 Q1 Q2 R1 R8 R9 R2 R10 = r ,Q 4 (1 + g m ,Q 4 RL ) [Table R8 = r ,Q 3 (1 + g m ,Q 3 R10 ) [Table [ vout Q3 R5 Rout R10 1 Configuration E ] 1 Configuration E ] ] = r ,Q 3 1 + g m ,Q 3 r ,Q 4 (1 + g m ,Q 4 RL ) [Table R7 = ro ,M 3 R9 = 1 g m ,Q 3 Rout = R1 = + 1 g m ,Q 4 1 g m ,Q1 [ 2 Configuration B ] ro , M 3 R7 1 = + + 1 g m ,Q 3 + 1 + [Table 1 Configuration F ] ro ,M 3 R9 1 11 + = + + 1 g m ,Q 4 + 1 g m ,Q 3 + 1 [Table [Table 1 Configuration F ] 1 Configuration C ] ] [Table R3 = ro ,Q 2 1 + g m ,Q 2 (r ,Q 2 // R1 ) 1 Configuration G ] 1 2ro ,Q 2 = ro ,Q 2 1 + g m ,Q 2 r ,Q 2 // g m ,Q1 6 [Q g m ,Q1 = g m ,Q 2 ] R5 = ro , M 1 (1 + g m , M 1 R3 ) = ro , M 1 (1 + 2 g m , M 1 ro ,Q 2 ) [Table 2 Configuration G ] R6 = R4 = R2 = 1 g m,M 2 1 g m,M 1 // R5 ro , M 1 + 1 g m,M 2 [Table 2 Configuration D ] 1 g m,M 2 ro , M 1 1 ro ,Q1 + R4 g m ,Q 2 ro ,Q1 1 g m,M 1 1 g m ,Q 2 [Table 2 Configuration H ] [Table 1 Configuration H] Rin = r ,Q1 (1 + g m ,Q1 R2 ) 2r ,Q1 [Table 1 Configuration E ] 7
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