Recall Lecture 17 MOSFET DC Analysis 1 Using

Recall Lecture 17 • MOSFET DC Analysis 1. Using GS (SG) Loop to calculate VGS • Remember that there is NO gate current! 2. Assume in saturation • Calculate ID using saturation equation 3. Find VDS (for NMOS) or VSD (for PMOS) • Using DS (SD) loop 4. Calculate VDS sat or VSD sat 5. Confirm that VDS > VDS sat or VSD > VSD sat • Confirm your assumption!

APPLICATION OF MOSFETS

Digital Logic Gates NOR gate response NAND gate The NAND gate response 0 0 High 5 0 Low 5 0 High 0 5 Low 0 5 High 5 5 Low

CHAPTER 7 Basic FET Amplifiers

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

AC PARAMETERS where

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 R 1 and R 2, and the signal frequency is sufficiently large for the coupling capacitor to act essentially as a short circuit.

EXAMPLE The transistor parameters are: VTN = 0. 8 V, Kn = 0. 2 m. A/V 2 and = 0. VDD = 5 V Rsi 520 k 0. 5 k ID = 0. 2441 m. A gm = 0. 442 m. A/V 320 k RD = 10 k

Steps 1. Calculate Rout 2. Calculate vo ____________________________ 3. Find vgs in terms of vi 4. Calculate the voltage gain, Av

0. 5 k RTH 198. 1 k 0. 442 vgs RD = 10 k 1. The output resistance, Rout = RD 2. The output voltage: vo = - gmvgs (Rout) = - gmvgs (10) = -4. 42 vgs 3. The gate-to-source voltage: , Ri = RTH vgs = [198. 1 / (198. 1 + 0. 5 )] = 0. 9975 vi = 1. 0025 vgs 4. So the small-signal voltage gain: Av = vo / vi = - 4. 42 vgs / 1. 0025 vgs - 4. 41

Type 2: With Source Resistor, RS VTN = 1 V, Kn = 1. 0 m. A / V

Perform DC analysis Assume transistor in saturation VG = ( 200 / 300 ) x 3 = 2 V Hence, KVL at GS Loop: VGS + IDRS – VTH = 0 VGS = 2 – 3 ID KVL at DS loop VDS + 10 ID + 3 ID – 3 = 0 VDS = 3 -13 ID Assume biased in saturation mode: Hence, ID = 1. 0 (2 – 3 ID - 1 )2 = 1. 0 (1 – 3 ID )2 9 ID 2 – 7 ID + 1 = 0 VTN = 1 V, Kn = 1. 0 m. A / V

ID = 0. 589 m. A VGS = 2 – 3 ID = 0. 233 < VTN MOSFET is OFF Not OK ID = 0. 19 m. A VGS = 2 – 3 ID = 1. 43 V > VTN OK VDS = 3 -13 ID = 0. 53 V VDS sat = VGS - VTN = 1. 43 – 1. 0 = 0. 43 V 0. 53 V > 0. 43 V Transistor in saturation Assumption is correct!

Steps 1. Calculate Rout 2. Calculate vo ____________________________ 3. Find v’ in terms of vgs 4. Find v’ in terms of vi 5. Calculate the voltage gain, Av

+ V’ RTH 66. 67 k - RD = 10 k RS = 3 k gm = 0. 872 m. A/V 1. The output resistance, Ro = RD 2. The output voltage: vo = - gmvgs. RD = - 0. 872 ( vgs) (10) = - 8. 72 vgs 3. Find v’ v’ = vgs + gmvgs RS v’ = vgs(1 + 2. 616) = 3. 616 vgs

+ V’ - RTH 66. 67 k RD = 10 k RS = 3 k 4. Find v’ in terms of vi : using voltage divider v’ = [RTH / (Rsi + RTH)] vi But in this circuit, Rsi = 0 so, v’ = vi = 3. 616 vgs 5. Calculate the voltage gain AV= vo / vi = - 8. 72 vgs / 3. 616 vgs = - 2. 41

Type 3: With Source Bypass Capacitor, CS Ø 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

Steps 1. Calculate Rout 2. Calculate vo ____________________________ 3. Find vgs in terms of vi 4. Calculate the voltage gain, Av

IQ = 0. 5 m. A hence, ID = 0. 5 m. A gm = 2 Kn ID = 1. 414 m. A/V ro = RG 200 k 1. 414 vgs RD = 7 k

1. The output resistance, Rout = RD 2. The output voltage: vo = - gmvgs (RD) = -1. 414 (7) vgs = - 9. 898 vgs 3. The gate-to-source voltage: vgs = vi in parallel ( no need voltage divider) 4. So the small-signal voltage gain: Av = -9. 898 vgs / vgs = - 9. 898

The MOSFET Amplifier - COMMON DRAIN • The output is measured at the source terminal • The gain is positive value

0. 5 k 150 k 113. 71 RTH k 470 k 0. 75 k ID = 8 m. A , Kn = 4 m. A /V 2 gm = 2 Kn ID = 11. 3 m. A/V

Steps 1. Calculate Rout 2. Calculate vo ____________________________ 3. Find v’ in terms of vgs 4. Find v’ in terms of vi 5. Calculate the voltage gain, Av

gm = 2 Kn ID = 11. 3 m. A/V + v’ - 1. The output resistance: Ro = ro || Rs 2. The output voltage vo = gmvgs (ro RS) = 11. 3 vgs (0. 70755) = 8 vgs 3. v’ in terms of vgs using supermesh: vgs + gmvgs (ro RS) – v’ = 0 v’ = vgs + 8 vgs = 9 vgs 4. v’ in terms of vi: v’ = (RTH / RTH + RSi) vi = 0. 9956 vi 9 vgs = 0. 9956 vi = 9. 040 vgs 5. The voltage gain A = v / v = 8 v / 9. 040 v = 0. 885 v o i gs gs

Output Resistance for Common Drain Ix + - • ro|| Rs = 0. 708 k Vx • vgs in terms of Vx where vgs = -Vx • - Vx + gmvgs + Ix = 0 0. 708 • - Vx - gm. Vx + Ix = 0 0. 708 - 1. 412 Vx – 11. 3 Vx + Ix = 0 Ix = 12. 712 Vx 0. 079 k