Recall Lecture 17 MOSFET DC Analysis Using GS (SG) loop to calculate VGS Remember that there is NO gate current! Assume in saturation Calculate ID using saturation equation Find VDS (for NMOS) or VSD (for PMOS) Using DS (SD) loop Calculate VDS sat or VSD sat Confirm that VDS > VDS sat or VSD > VSD sat Confirm your assumption!

LOAD LINE, ID versus VDS Common source configuration i.e source is grounded. It is the linear equation of ID versus VDS Use KVL VDS = VDD – IDRD ID = -VDS + VDD RD RD

ID (mA) VDS (V) VGS VDS ID Q-POINTS y-intercept x-intercept

DC Analysis where source is NOT GROUNDED For the NMOS transistor in the circuit below, the parameters are VTN = 1V and Kn = 0.5 mA/V2.

Calculate the value of VGS KVL at GS loop: R D = 2 k S G 24 k + 5 V 1 V I 5 V Assume the transistor is biased in the saturation region, the drain current: Calculate the value of VGS KVL at GS loop: 0 + VGS+ 1(ID) -5 +1 = 0 VGS = 4 - ID 1 k ID = 6.646 mA VGS= -2. 646 V Replace in VGS equation in step 1 VGS = 4 - ID ID = 1.354 mA VGS = 2.646 V Why choose VGS = 2.646 V ? Because it is bigger than VTN

Calculate VDSsat = VGS – VTN = 2.646 – 1 = 1.646 V R D = 2 k S G 24 k + 5 V 1 V I 5 V Use KVL at DS loop Calculate VDSsat = VGS – VTN = 2.646 – 1 = 1.646 V Confirm your assumption: VDS > VDS sat , our assumption is correct IDRD + VDS + IDRS – 5 – 5 = 0 1.354 (2) + VDS + 1.354 – 10 = 0 VDS = 10 – 1.354 – 2.708 = 5.938 V 1 k

Answers: R1 = 422 kΩ R2 = 248 kΩ RD = 2.84 kΩ EXERCISE 2 Assume that the transistor parameters are VTN = 0.8 V and Kn = 0.80 mA/V2. Given VDD = 5V and R1+ R2 = 670 kΩ, design the circuit such that ID = 0.88 mA and VDS = 2.5 V. Confirm any assumptions you make during your analysis. Answers: R1 = 422 kΩ R2 = 248 kΩ RD = 2.84 kΩ

CHAPTER 7 Basic FET Amplifiers

For linear amplifier function, FET is normally biased in the saturation region.

Need to do DC Analysis first to find ID AC PARAMETERS where   Need to do DC Analysis first to find ID vgs gmvgs

The MOSFET Amplifier - COMMON SOURCE The output is measured at the drain terminal The gain is negative value Three types of common source source grounded with source resistor, RS with bypass capacitor, CS

Common Source - Source Grounded A Basic Common-Source Configuration: Assume that the transistor is biased in the saturation region by resistors R1 and R2, and the signal frequency is sufficiently large for the coupling capacitor to act essentially as a short circuit. + vgs - gmvgs vs

EXAMPLE The transistor parameters are: VDD = 5V RD = 10 k 520 k 320 k 0.5 k The transistor parameters are: VTN = 0.8V, Kn = 0.2mA/V2 and  = 0. vs Voltage Divider biasing: Change to Thevenin Equivalent RTH = 198 k VTH = 1.905 V

gm = 0.442 mA/V DC ANALYSIS Calculate the value of VGS VGS – VTH = 0 Confirm your assumption: VDS > VDSsat, our assumption that the transistor is in saturation region is correct Assume the transistor is biased in the saturation region, the drain current: Use KVL at DS loop IDRD + VDS – VDD = 0 VDS = VDD – IDRD = 2.56 V Calculate the value of VGS VGS – VTH = 0 VGS = 1.905 V Calculate VDSsat = VGS – VTN = 1.905 – 0.8 = 1.105 V gm = 0.442 mA/V

COMMON EMITTER GROUNDED OUTPUT SIDE Get the equivalent resistance at the output side, RO Get the vo equation where vo = - gm vbeRO INPUT SIDE Calculate Ri Get vbe in terms of vi COMMON SOURCE GROUNDED OUTPUT SIDE Equivalent resistance at the output side, RO Get the vo equation where vo = - gm vgs RO ____________________________________________ INPUT SIDE Get vgs in terms of vi vs

The output resistance, Ro = RD The output voltage: RTH 198.1 k 0.5 k RD = 10 k 0.442 vgs + vgs - + vi - vs The output resistance, Ro = RD The output voltage: vo = - gmvgs (Ro) = - gmvgs (10) = -4.42 vgs Get vi in terms of vgs vgs = vi

Equation of vo : vo = - gmvgs (Ro) = - gmvgs (10) = -4.42 vgs vi = vgs Av vi = vo  open circuit voltage Avvi = -4.42 vgs = -4.42 vi Av = -4.42  open circuit voltage gain

RSi = 0.5 kΩ vS vo 10 k Ri = RTH = 198.1 kΩ To find new voltage gain, vo/vs with input signal voltage source, vs vi in terms of vs  use voltage divider: vi = [ Ri / ( Ri + Rs )] * vs = 0.9975 vs vo = Avvi  because there is no load resistor vo = -4.42 (0.9975 vs) vo/vs = -4.41

Type 2: With Source Resistor, RS VTN = 1V, Kn = 1.0 mA /V2 RTH = 66.67 k VTH = ( 200 / 300 ) x 3 = 2 V

Calculate the value of VGS KVL at GS loop: DC ANALYSIS Assume the transistor is biased in the saturation region, the drain current: Calculate the value of VGS KVL at GS loop: 0 + VGS+ 3(ID) - 2 = 0 VGS = 2 - 3ID vs VTN = 1V, Kn = 1.0 mA / V ID = 0.589 mA Replace in VGS equation in step 1 VGS = 2 - 3ID VGS= 0.233 V ID = 0.189 mA VGS = 1.43 V Why choose VGS = 1.43 V ? Because it is bigger than VTN

gm = 0.872 mA/V Use KVL at DS loop IDRD + VDS + IDRS – 3 = 0 Calculate VDSsat = VGS – VTN = 1.43 – 1 = 0.43V Confirm your assumption: VDS > VDS sat , our assumption is correct IDRD + VDS + IDRS – 3 = 0 VDS = 0.543 V gm = 0.872 mA/V  = 0 ro =  VA = 

Equivalent resistance at the output side, RO COMMON EMITTER with RE OUTPUT SIDE Get the equivalent resistance at the output side, Ro Get the vo equation where vo = - gmvbeRo INPUT SIDE Calculate Rib = r + (1+ ) RE Calculate Ri = Rib // RTH vbe in terms of vi COMMON SOURCE with RS OUTPUT SIDE Equivalent resistance at the output side, RO Get the vo equation where vo = - gm vgs RO ________________________________________________________ INPUT SIDE Get vi in terms of vgs : KVL

vi = vgs + gmvgs RS  vi = vgs(1 + 2.616) = 3.616 vgs RTH 66.67 k RD = 10 k RS = 3 k + vi - vs gm = 0.872 mA/V The output resistance, Ro = RD The output voltage: Get vi in terms of vgs : KVL vi = vgs + gmvgs RS  vi = vgs(1 + 2.616) = 3.616 vgs vo = - gmvgsRD = - 0.872 ( vgs) (10) = - 8.72 vgs

Equation of vo : vo = - gmvgs (Ro) = - gmvgs (10) = -8.72 vgs vi = vgs + gmvgs RS  vi = vgs(1 + 2.616) = 3.616 vgs Av vi = vo  open circuit voltage Avvi = -8.72 vgs = -8.72 vi Av = - 2.41  open circuit voltage gain 3.616

10 k Ri = RTH = 66.67 kΩ vS vo To find new voltage gain, vo/vs with input signal voltage source, vs vi in terms of vs vi = vs  in parallel vo = Avvi  because there is no load resistor vo = -2.41 (vs) vo/vs = -2.41

Type 3: With Source Bypass Capacitor, CS Circuit with Source Bypass Capacitor An source bypass capacitor can be used to effectively create a short circuit path during ac analysis hence avoiding the effect RS CS becomes a short circuit path – bypass RS; hence similar to Type 1

COMMON EMITTER with CE OUTPUT SIDE Get the equivalent resistance at the output side, RO Get the vo equation where vo = - gm vbeRO INPUT SIDE Calculate Ri Get vbe in terms of vi COMMON SOURCE with CS OUTPUT SIDE Equivalent resistance at the output side, RO Get the vo equation where vo = - gm vgs RO ____________________________________________ INPUT SIDE Get vgs in terms of vi vs

IQ = 0.5 mA hence, ID = 0.5 mA ro =  gm = 1.414 mA/V + + vgs vi - - vs RG 200 k RD = 7 k 1.414 vgs + vgs - + vi - vs

The output resistance, Ro = RD = 7 k The output voltage: RG 200 k RD = 7 k 1.414 vgs + vgs - vi vs The output resistance, Ro = RD = 7 k The output voltage: vo = - gmvgs (RD) = -1.414 (7) vgs = - 9.898 vgs 3. The gate-to-source voltage: vgs = vi

Equation of vo : vo = - gmvgs (RD) = -1.414 (7) vgs = - 9.898 vgs vi = vgs Av vi = vo  open circuit voltage Avvi = -9.898 vgs = -9.898 vi Av = - 9.898  open circuit voltage gain

7 k Ri = RG = 200 kΩ vS vo To find new voltage gain, vo/vs with input signal voltage source, vs vi in terms of vs vi = vs  in parallel vo = Avvi  because there is no load resistor vo = -9.898 (vs) vo/vs = -9.898