Question

4) Consider the MOSFET differential amplifier shown below, with Io-2 mA, and RL- 10 kS2, Rss-100 k2, VDD- +8V and Vss--8V. The NMOS transistors in the circuit are nominally identical, with kn 2 mA/V2, VTn 1.0 V and ro 100 k2. The PMoS transistors in the circuit are nominally identical, with kp 2 mA/V2, [VTpl 1.0 V and ro 100 kΩ M3 M4 0 M1 M2 a) First consider the DC bias point. Assuming that the current mirror requires at least 1V (across Io source) in order to operate, what are the maximum and minimum common-mode DC input voltages (Vi-V2) over which the differential amplifier can operate with both transistors in saturation mode? b) Consider ac small signal operation for differential input (vi+Av/2 and v2--Av/2) and the indicated output (vo across RL). Find the overall voltage gain (vo/Av), including sign. c) Find the input resistance and output resistance corresponding to this mode (output d) Next, consider ac small signal operation for common-mode input (vi 2Av) and e) Find the common-mode rejection ratio (CMRR) resistance is looking to left from capacitor) the indicated vo. Find the overall voltage gain ((Vo /Av) corresponding to this mode.

Answer 1

a) As seen in the circuit M3 and M4 form a current mirror circuit. Draw the modified circuit VDD M4 Vo D2 C M2 M1 Vs S RL -V sS For part (a), Rss MB

According to the question, V1 V X NOTE: The voltage applied after the current source -V should be negative in the given circuit That means, V should have the value 8 V. Which is given wrong in the question -V SS SS -8 SS 8 V V SS

From the figure, -V-V S SS -VV V-8 V 1V S SS S V-7 V Write down the drain current equation for the saturation region 1, W (VGs-V S ) I D 2 Tn L 1 W k k, (1) Tn D n 2 L From the Figure, I IpIp2 + D1 As, Ip Ip2 I 21p D Ip 2 2 mA ID 2 I 1 mA

Substitute the values in equation (1) and calculate Ves I k (VesV) 1x103 2x103 (VGSI -1) Tn GS1 n 1 GS1 2 1 VGS1 1 2 VGS1 1.707 VGS1 VGS2 1.707 Find the source to gate voltage of PMOS(M3 and M4) IpsVs3-) 2x 103 (VsG3-1) SG3 1x10 (VSG3 -1) 2 1+ 2 SG3 V31.707 SG3 V SG4 1.707 SG3

Apply KVL in the circuit (MODFET M3 to M1) VDDVSG3VSVs 4ISA VDSVsGVDVs SG3 VDS1 +1.707 V 8-(-7) DS1 13.293 V DS1 VD213.293 V

Apply KVL from input to -Vs: -Vcs -Vs-sVs CM,min GS CS SS Here, is the voltage across current source Find the minimum common mode DC input voltages VCM,min = VGS1 +Vcs-Vss 1.707 V 1V-8 -5.293 V Now, find the drain voltage, VD Vps V-Vs V VpSiVs 13.293 V+(-7 V) DS1 -6.293 V Find the maximum common mode DC input voltages. Vсм нх VV Tn ax - 6.293 V 1V - 7.293 V Hence, the range will be -5.293 V < V 7.293 V cm

(b) From equation (1) k(Ves-V) G: Differentiate with respect to VGS

al n) 2k, (VGsV GS ате gm - 2k, (V6s-V) gm Tn GS = 2x 2 x 103(1.707 -1) gm gm 2.828 mA/V The transconductance is, G gm2 Draw the small signal model.

r04 gm Vgst Vg3= Vg4 1||ro3 gm3 g3,g4 gi Vid/2 r02 Via 2 ro1 id gm2 Vid gm 2 Figure 2 Output resistance is, 04 o2 02 0 01 AV AV =-gml2 02 2 2 =8m,2 (02 || 74)AV 4

Hence, the open circuit differential voltage gain is, AV Vid Ad Vid G R 8m1,2 (02 || 04) 3 =-2.828 x 10(100k || 100k since, 2o4 = = 100k2 -141.4 (c) Find the input resistance 1 Rid gm 1 2.828x 10 353.6 Q Find the output resistance. R3 0204 100k ||100k = 50 kQ

(d) Draw the small signal model for common mode ro4 A i ro3 gm3 Vo ro2 ro1 gm2 Viem gml Vicm Figure 3 From the circuit. Calculate the common mode gain

VO Agm V cm 1 104 2Rss 1+ gm304 SS 1 (gm3'04>>1 and ro3 =o4 = o) 2RSS gm30 1 (gms m) 2gmRsS SS u 1 2(2.828 x 103x100 x103 --0.001768 (e) Calculate CMRR AD Acm CMRR -141.4 -0.001768 - 79,977

AD CMRRI 20log10 Aem "ст - 20log10 79,977 -98.059 dB