11/30/2010 Fundamentals of Microelectronics CH1 CH2 CH3 CH4 CH5 CH6 CH7 CH8 Why Microelectronics? Basic Physics of Semiconductors Diode Circuits Physics of Bipolar Transistors Bipolar Amplifiers Physics of MOS Transistors CMOS Amplifiers Operational Amplifier As A Black Box 1 Chapter 7 CMOS Amplifiers 7.1 General Considerations 7.2 Common-Source Stage 7.3 Common-Gate Stage 7.4 Source Follower 7.5 Summary and Additional Examples 2 1 11/30/2010 Chapter Outline CH7 CMOS Amplifiers 3 MOS Biasing RV VGS = −(V1 −VTH ) + V12 + 2V1 2 DD −VTH R1 + R2 1 V1 = W µnCox RS L Voltage at X is determined by VDD, R1, and R2. VGS can be found using the equation above, and ID can be found by using the NMOS current equation. CH7 CMOS Amplifiers 4 2 11/30/2010 Self-Biased MOS Stage I D RD + VGS + RS I D = VDD The circuit above is analyzed by noting M1 is in saturation and no potential drop appears across RG. CH7 CMOS Amplifiers 5 Current Sources When in saturation region, a MOSFET behaves as a current source. NMOS draws current from a point to ground (sinks current), whereas PMOS draws current from VDD to a point (sources current). CH7 CMOS Amplifiers 6 3 11/30/2010 Common-Source Stage λ=0 Av = −g m RD Av = − 2µnCox W I D RD L CH7 CMOS Amplifiers 7 Operation in Saturation RD I D < VDD − (VGS −VTH ) In order to maintain operation in saturation, Vout cannot fall below Vin by more than one threshold voltage. The condition above ensures operation in saturation. CH7 CMOS Amplifiers 8 4 11/30/2010 CS Stage with λ =0 Av = − g m RL Rin = ∞ Rout = RL 9 CS Stage with λ ≠ 0 Av = −gm ( RL || rO ) Rin = ∞ Rout = RL || rO However, Early effect and channel length modulation affect CE and CS stages in a similar manner. CH7 CMOS Amplifiers 10 5 11/30/2010 CS Gain Variation with Channel Length W L ∝ 2µnCoxWL ID λ ID 2µnCox Av = Since λ is inversely proportional to L, the voltage gain actually becomes proportional to the square root of L. CH7 CMOS Amplifiers 11 CS Stage with Current-Source Load Av = − g m1 (rO1 || rO 2 ) Rout = rO1 || rO 2 To alleviate the headroom problem, an active currentsource load is used. This is advantageous because a current-source has a high output resistance and can tolerate a small voltage drop across it. CH7 CMOS Amplifiers 12 6 11/30/2010 PMOS CS Stage with NMOS as Load Av = −gm2(rO1 || rO2 ) Similarly, with PMOS as input stage and NMOS as the load, the voltage gain is the same as before. CH7 CMOS Amplifiers 13 CS Stage with Diode-Connected Load Av = −gm1 ⋅ 1 (W / L)1 =− gm2 (W / L)2 1 Av = −gm1 || rO2 || rO1 gm2 Lower gain, but less dependent on process parameters. CH7 CMOS Amplifiers 14 7 11/30/2010 CS Stage with Diode-Connected PMOS Device 1 Av = − g m 2 || ro1 || ro 2 g m1 Note that PMOS circuit symbol is usually drawn with the source on top of the drain. 15 CS Stage with Degeneration Av = − RD 1 + RS gm λ =0 Similar to bipolar counterpart, when a CS stage is degenerated, its gain, I/O impedances, and linearity change. CH7 CMOS Amplifiers 16 8 11/30/2010 Example of CS Stage with Degeneration Av = − RD 1 1 + gm1 gm2 A diode-connected device degenerates a CS stage. CH7 CMOS Amplifiers 17 CS Stage with Gate Resistance VR = 0 G Since at low frequencies, the gate conducts no current, gate resistance does not affect the gain or I/O impedances. CH7 CMOS Amplifiers 18 9 11/30/2010 Output Impedance of CS Stage with Degeneration rout ≈ gmrO RS + rO Similar to the bipolar counterpart, degeneration boosts output impedance. CH7 CMOS Amplifiers 19 Output Impedance Example (I) 1 1 Rout = rO11+ gm1 + gm2 gm2 When 1/gm is parallel with rO2, we often just consider 1/gm. CH7 CMOS Amplifiers 20 10 11/30/2010 Output Impedance Example (II) Rout ≈ g m1rO1rO 2 + rO1 In this example, the impedance that degenerates the CS stage is rO, instead of 1/gm in the previous example. CH7 CMOS Amplifiers 21 CS Core with Biasing Av = R1 || R2 − RD R1 || R2 ⋅ , Av = − gm RD RG + R1 || R2 1 + R RG + R1 || R2 S gm Degeneration is used to stabilize bias point, and a bypass capacitor can be used to obtain a larger small-signal voltage gain at the frequency of interest. CH7 CMOS Amplifiers 22 11 11/30/2010 Common-Gate Stage Av = gmRD Common-gate stage is similar to common-base stage: a rise in input causes a rise in output. So the gain is positive. CH7 CMOS Amplifiers 23 Signal Levels in CG Stage In order to maintain M1 in saturation, the signal swing at Vout cannot fall below Vb-VTH. CH7 CMOS Amplifiers 24 12 11/30/2010 I/O Impedances of CG Stage Rin = 1 gm λ =0 Rout = RD The input and output impedances of CG stage are similar to those of CB stage. CH7 CMOS Amplifiers 25 CG Stage with Source Resistance Av = RD 1 + RS gm When a source resistance is present, the voltage gain is equal to that of a CS stage with degeneration, only positive. CH7 CMOS Amplifiers 26 13 11/30/2010 Generalized CG Behavior Rout = (1 + g m rO )RS + rO When a gate resistance is present it does not affect the gain and I/O impedances since there is no potential drop across it ( at low frequencies). The output impedance of a CG stage with source resistance is identical to that of CS stage with degeneration. CH7 CMOS Amplifiers 27 Example of CG Stage vout gm1RD = vin 1 + ( gm1 + gm2 )RS 1 Rout ≈ gm1rO1 || RS + rO1 || RD gm 2 Diode-connected M2 acts as a resistor to provide the bias current. CH7 CMOS Amplifiers 28 14 11/30/2010 CG Stage with Biasing vout R || (1/ gm ) = 3 ⋅ gmRD vin R3 || (1/ gm ) + RS R1 and R2 provide gate bias voltage, and R3 provides a path for DC bias current of M1 to flow to ground. CH7 CMOS Amplifiers 29 Source Follower Stage Av < 1 CH7 CMOS Amplifiers 30 15 11/30/2010 Source Follower Core vout r || R = O L vin 1 +r || R O L gm Similar to the emitter follower, the source follower can be analyzed as a resistor divider. CH7 CMOS Amplifiers 31 Source Follower Example Av = rO1 || rO 2 1 + rO1 || rO 2 g m1 In this example, M2 acts as a current source. CH7 CMOS Amplifiers 32 16 11/30/2010 Output Resistance of Source Follower Rout = 1 1 || rO || RL ≈ || RL gm gm The output impedance of a source follower is relatively low, whereas the input impedance is infinite ( at low frequencies); thus, a good candidate as a buffer. CH7 CMOS Amplifiers 33 Source Follower with Biasing 1 W 2 ID = µnCox (VDD − I DRS −VTH ) 2 L RG sets the gate voltage to VDD, whereas RS sets the drain current. The quadratic equation above can be solved for ID. CH7 CMOS Amplifiers 34 17 11/30/2010 Supply-Independent Biasing If Rs is replaced by a current source, drain current ID becomes independent of supply voltage. CH7 CMOS Amplifiers 35 Example of a CS Stage (I) 1 Av = − g m1 || rO1 || rO 2 || rO 3 g m3 1 Rout = || rO1 || rO 2 || rO3 g m3 M1 acts as the input device and M2, M3 as the load. CH7 CMOS Amplifiers 36 18 11/30/2010 Example of a CS Stage (II) Av = − rO2 1 1 + || rO3 gm1 gm3 M1 acts as the input device, M3 as the source resistance, and M2 as the load. CH7 CMOS Amplifiers 37 Examples of CS and CG Stages Av _CS = −gm2 [(1+ gm1rO1 )RS + rO1 ] || rO1 Av _ CG = rO 2 1 + RS gm With the input connected to different locations, the two circuits, although identical in other aspects, behave differently. CH7 CMOS Amplifiers 38 19 11/30/2010 Example of a Composite Stage (I) Av = RD 1 1 + gm1 gm2 By replacing the left side with a Thevenin equivalent, and recognizing the right side is actually a CG stage, the voltage gain can be easily obtained. CH7 CMOS Amplifiers 39 Example of a Composite Stage (II) vout2 vin 1 || rO3 || rO4 g m3 =− 1 1 || rO 2 + gm2 gm1 This example shows that by probing different places in a circuit, different types of output can be obtained. Vout1 is a result of M1 acting as a source follower whereas Vout2 is a result of M1 acting as a CS stage with degeneration. CH7 CMOS Amplifiers 40 20