ELE 704EE8502 Analog CMOS Integrated Circuits Assignment1

							   ELE 704 / EE8502 Analog CMOS Integrated Circuits
                                       Assignment 1

                                Instructor : Dr. Fei Yuan


Q1. The common-source amplifier is shown in Fig.1. Assume the transistor is in saturation.
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Derive the expressions of the input-referred noise voltage generator vn and input-referred noise
current generator i2 of the amplifier at low frequencies. Derive the noise figure expression of
                   n

the amplifier at the input at low frequencies.


                                                    RD

                                  Input source                vo

                                              Rs

                                  vs




                             Figure 1: Common-source amplifier.

Q2. The schematic of the common-source amplifier with a diode load is shown in Fig.2. Derive
the expression of the voltage gain of the amplifier at low frequencies.


                                                   M2
                                                         vo


                                       v in        M1




                     Figure 2: Common-source amplifier with diode load.

Q3. The schematic of a widely used biasing circuit is shown in Fig.3. The aspect ratio of
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the current mirror M1-M2 is k. Neglect the effect of channel length modulation. Derive the
expression of the biasing current Ib .



                                           M4

                                           M3                 Ib

                                               M1          M2



                                  Figure 3: Biasing current source.

Q4. The schematic of the common-source amplifier with a current source load is shown in
Fig.4. Note that all transistors are biased in the saturation. Show that the maximum output
voltage of the amplifier is given by vo,max = Vb + |Vtp |, where Vtp is the threshold voltage of
pMOS transistors. The key advantage of common-source amplifiers with current-source loads
is that if we set Vb = VDD − |Vtp |, then vo,max = VDD . Since vo,min = Vsat , where Vsat is the
pinch-off voltage of nMOS transistors, the swing of the output voltage can therefore be nearly
rail-to-rail, i.e. from Vsat to VDD .



                                           Vb       M2
                                                             vo

                                        v in        M1



                  Figure 4: Common-source amplifier with current source load.




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