Recall Lecture 17

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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 gt VDS sat or VSD gt VSD sat
• Confirm your assumption!

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Application of mosfets
3
Digital Logic Gates
NOR gate
NAND gate
NOR gate response NOR gate response NOR gate response   The NAND gate response The NAND gate response The NAND gate response
 
0 0 High   0 0 High
5 0 Low   5 0 High
0 5 Low   0 5 High
5 5 Low   5 5 Low
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CHAPTER 7

• Basic FET Amplifiers

5

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

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AC PARAMETERS
 
 
where
 
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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

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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.

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EXAMPLE
The transistor parameters are VTN 0.8V, Kn
0.2mA/V2 and ? 0.
ID 0.2441 mA
gm 0.442 mA/V
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Steps

• Calculate Rout
• Calculate vo
• __________________________________________________
  ______
• Find vgs in terms of vi
• Calculate the voltage gain, Av

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• The output resistance, Rout RD
• The output voltage
• vo - gmvgs (Rout) - gmvgs (10) -4.42 vgs
• The gate-to-source voltage , Ri RTH
• vgs 198.1 / (198.1 0.5 ) 0.9975 vi ? vi
  1.0025 vgs
• So the small-signal voltage gain

Av vo / vi - 4.42 vgs / 1.0025 vgs ? - 4.41
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Type 2 With Source Resistor, RS
VTN 1V, Kn 1.0mA / V
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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 3ID KVL
at DS loop VDS 10 ID 3ID 3 0 VDS 3 -13
ID Assume biased in saturation mode Hence, ID
1.0 (2 3ID - 1 )2 1.0 (1 3ID )2 ? 9
ID2 7 ID 1 0
VTN 1V, Kn 1.0 mA / V
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ID 0.589 mA
ID 0.19 mA
VGS 2 3ID 0.233 lt VTN
VGS 2 3ID 1.43 V gt VTN
OK
MOSFET is OFF
Not OK
VDS 3 -13 ID 0.53 V
VDS sat VGS - VTN 1.43 1.0 0.43 V
0.53 V gt 0.43 V Transistor in saturation Assumpti
on is correct!
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Steps

• Calculate Rout
• Calculate vo
• __________________________________________________
  ______
• Find v in terms of vgs
• Find v in terms of vi
• Calculate the voltage gain, Av

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V -
RTH
RD 10 k?
66.67 k?
RS 3 k?
gm 0.872 mA/V

• The output resistance, Ro RD
• The output voltage
• Find v
• v vgs gmvgs RS ? v vgs(1 2.616)
  3.616 vgs

vo - gmvgsRD - 0.872 ( vgs) (10) - 8.72
vgs
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V -
RTH
RD 10 k?
66.67 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
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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

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Steps

• Calculate Rout
• Calculate vo
• __________________________________________________
  ______
• Find vgs in terms of vi
• Calculate the voltage gain, Av

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IQ 0.5 mA hence, ID 0.5 mA gm 2 ?Kn ID
1.414 mA/V ro ?
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• The output resistance, Rout RD
• 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
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The MOSFET Amplifier - COMMON DRAIN

• The output is measured at the source terminal
• The gain is positive value

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0.5 k?
150 k?
0.5 k?
113.71 k?
RTH
470 k?
0.75 k?
ID 8 mA , Kn 4 mA /V2 gm 2 ?Kn ID 11.3
mA/V
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Steps

• Calculate Rout
• Calculate vo
• __________________________________________________
  ______
• Find v in terms of vgs
• Find v in terms of vi
• Calculate the voltage gain, Av

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gm 2 ?Kn ID 11.3 mA/V
v -

• The output resistance
• The output voltage
• v in terms of vgs using supermesh
• v in terms of vi
• The voltage gain

Ro ro Rs
vo gmvgs (ro ?? RS) 11.3 vgs (0.70755) 8 vgs
vgs gmvgs (ro ?? RS) v 0
v vgs 8 vgs 9 vgs
v (RTH / RTH RSi) vi 0.9956 vi 9vgs
0.9956 vi ? vi 9.040 vgs
Av vo / vi 8 vgs / 9.040 vgs 0.885
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Output Resistance for Common Drain

• ro Rs 0.708 k?
• vgs in terms of Vx where vgs -Vx

 
- 1.412 Vx 11.3 Vx Ix 0
Ix 12.712 Vx
0.079 k?