Recall Lecture 17 MOSFET DC Analysis 1.Using GS (SG) Loop to
calculate V GS Remember that there is NO gate current! 2.Assume in
saturation Calculate I D using saturation equation 3.Find V DS (for
NMOS) or V SD (for PMOS) Using DS (SD) loop 4.Calculate V DS sat or
V SD sat 5.Confirm that V DS > V DS sat or V SD > V SD sat
Confirm your assumption! APPLICATION OF MOSFETS Digital Logic Gates
NOR gateNAND gate NOR gate response The NAND gate response 00High
00 50Low 50High 05Low 05High 55Low 55 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, R S with bypass capacitor, C S 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 V
DD = 5V R si R D = 10 k 0.5 k 520 k 320 k The transistor parameters
are: V TN = 0.8V, K n = 0.2mA/V 2 and = 0. I D = mA g m = mA/V
Steps 1.Calculate R out 2.Calculate v o
________________________________________________________ 3.Find v
gs in terms of v i 4.Calculate the voltage gain, Av 1.The output
resistance, R out = R D 2.The output voltage: v o = - g m v gs (R
out ) = - g m v gs (10) = v gs 3.The gate-to-source voltage:, R i =
R TH v gs = [198.1 / ( )] = v i v i = v gs 4.So the small-signal
voltage gain: A v = v o / v i = v gs / v gs R TH k 0.5 k R D = 10 k
v gs Type 2: With Source Resistor, R S V TN = 1V, K n = 1.0mA / V V
G = ( 200 / 300 ) x 3 = 2 V Hence, KVL at GS Loop: V GS + I D R S V
TH = 0 V GS = 2 3I D KVL at DS loop V DS + 10 I D + 3I D 3 = 0 V DS
= I D Assume biased in saturation mode: Hence, I D = 1.0 (2 3I D -
1 ) 2 = 1.0 (1 3I D ) 2 9 I D 2 7 I D + 1 = 0 Perform DC analysis
Assume transistor in saturation V TN = 1V, K n = 1.0 mA / V I D =
mAI D = 0.19 mA V DS sat = V GS - V TN = 1.43 1.0 = 0.43 V 0.53 V
> 0.43 V Transistor in saturation Assumption is correct! V GS =
2 3I D = < V TN V GS = 2 3I D = 1.43 V > V TN MOSFET is OFF
OK V DS = I D = 0.53 V Not OK Steps 1.Calculate R out 2.Calculate v
o ________________________________________________________ 3.Find v
in terms of v gs 4.Find v in terms of v i 5.Calculate the voltage
gain, Av 1.The output resistance, R o = R D 2.The output voltage:
3.Find v v = v gs + g m v gs R S v = v gs ( ) = v gs v o = - g m v
gs R D = ( v gs ) (10) = v gs R TH k R D = 10 k R S = 3 k g m =
mA/V + V - 5. Calculate the voltage gain A V = v o / v i = v gs / v
gs = Find v in terms of v i : using voltage divider R TH k R D = 10
k R S = 3 k + V - v = [R TH / (R si + R TH )] v i But in this
circuit, R si = 0 so, v = v i = v gs 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 R S C S becomes a short circuit path bypass R S ; hence
similar to Type 1 Type 3: With Source Bypass Capacitor, C S Steps
1.Calculate R out 2.Calculate v o
________________________________________________________ 3.Find v
gs in terms of v i 4.Calculate the voltage gain, Av I Q = 0.5 mA
hence, I D = 0.5 mA g m = 2 K n I D = mA/V r o = R G 200 k R D = 7
k v gs 1.The output resistance, R out = R D 2.The output voltage: v
o = - g m v gs (R D ) = (7) v gs = v gs 3. The gate-to-source
voltage: v gs = v i in parallel ( no need voltage divider) 4. So
the small-signal voltage gain: A v = v gs / v gs = The MOSFET
Amplifier - COMMON DRAIN The output is measured at the source
terminal The gain is positive value R TH 150 k 470 k 0.75 k k 0.5 k
I D = 8 mA, K n = 4 mA /V 2 g m = 2 K n I D = 11.3 mA/V Steps
1.Calculate R out 2.Calculate v o
________________________________________________________ 3.Find v
in terms of v gs 4.Find v in terms of v i 5.Calculate the voltage
gain, Av 1.The output resistance: 2.The output voltage 3.v in terms
of v gs using supermesh: 4.v in terms of v i : 5.The voltage gain v
gs + g m v gs (r o R S ) v = 0 v = v gs + 8 v gs = 9 v gs v o = g m
v gs (r o R S ) = 11.3 v gs ( ) = 8 v gs v = (R TH / R TH + R Si )
v i = v i 9v gs = v i v i = v gs + v - g m = 2 K n I D = 11.3 mA/V
A v = v o / v i = 8 v gs / v gs = R o = ro || R s +-+- Vx Ix r o ||
R s = k v gs in terms of V x where v gs = -V x - V x + g m v gs + I
x = V x - g m V x + I x = V x 11.3 V x + I x = 0 I x = V x k Output
Resistance for Common Drain
