Recall Last Lecture
• The MOSFET has only one current, ID
• Operation of MOSFET
– NMOS and PMOS
– For NMOS,
• VGS > VTN
• VDS sat = VGS – VTN
– For PMOS
• VSG > |VTP|
• VSD sat = VSG + VTP
© Electronics
• ID versus VDS (NMOS) or ID versus VSD (PMOS)
© Electronics
• NMOS
• PMOS
o VTN is POSITIVE
o VGS > VTN to turn on
o Triode/non-saturation
region
o VTP is NEGATIVE
o VSG > |VTP| to turn on
o Triode/non-saturation
region
o Saturation region
o Saturation region
o VDSsat = VGS - VTN
o VSDsat = VSG + VTP
© Electronics
DC analysis of FET
© Electronics
MOSFET DC Circuit Analysis - NMOS

The source terminal is
at ground and common
to both input and output
portions of the circuit.

The CC acts as an open
circuit to dc but it allows
the signal voltage to the
gate of the MOSFET.

In the DC equivalent circuit, the gate current into the transistor is
zero, the voltage at the gate is given by a voltage divider principle:
VG = VTH =
R2
R1 + R2
Use KVL at GS loop:
VGS –VTH + 0 = 0
© Electronics VGS = VTH
VDD
MOSFET DC Circuit Analysis - NMOS
1. Calculate the value of VGS
2.
Assume the transistor is biased in the saturation
region, the drain current:
3. Use KVL at DS loop
IDRD + VDS – VDD = 0
4.
Calculate VDSsat = VGS - VTN
5.
© Electronics
Confirm your assumption:
If VDS > VDS(sat) = VGS – VTN, then the transistor is biased in the
saturation region. If VDS < VDS(sat), then the transistor is biased in the
non-saturation region.
EXAMPLE:
Calculate the drain current and drain to source voltage of a common source
circuit with an n-channel enhancement mode MOSFET. Assume that R1 =
30 k, R2 = 20 k, RD = 20 k, VDD = 5V, VTN = 1V and Kn = 0.1 mA/V2
VTH =
20
5 = 2V hence VGS = VTH = 2V
30 + 20
VDSsat = VGS – VTN = 2 – 1 = 1V, so, VDS > VDSsat, our assumption
that the transistor is in saturation region is correct
© Electronics
EXAMPLE
VDD = 10V
• The transistor has parameters
VTN = 2V and Kn = 0.25mA/V2.
• Find ID and VDS
R1 = 280k
R2 = 160k
© Electronics
RD =
10k
Solution
1. VTH =
160
10 = 3.636 V
160 + 280
KVL at GS loop: VGS – VTH + 0 = 0  VGS = VTH
2. Assume in saturation mode:
ID = Kn(VGS - VTN)2
So, ID = 0.669 mA
3. KVL at DS loop: VDS = VDD – IDRD = 10 – 0.669 (10) = 3.31 V
4. VDS sat = VGS – VTN = 3.636 – 2 = 1.636 V
So, VDS > VDSsat , therefore, assumption is correct!
Answer: ID = 0.669 mA and VDS = 3.31 V
© Electronics
MOSFET DC Circuit Analysis - PMOS
Different notation:
VSG and VSD
Threshold Voltage = VTP
VG = VTH =
R2
R1 + R2
Use KVL at GS loop:
VSG + 0 + VTH – VDD = 0
VSG = VDD - VTH
© Electronics
VDD
MOSFET DC Circuit Analysis - PMOS

Assume the transistor is biased in the saturation
region, the drain current:
ID = Kp (VSG + VTP)2

Calculate VSD:
Use KVL at DS loop:
VSD + IDRD - VDD = 0
VSD = VDD - IDRD

If VSD > VSD(sat) = VSG + VTP, then the transistor is biased in the
saturation region.

If VSD < VSD(sat), then the transistor is biased in the non-saturation
region.
© Electronics
Calculate the drain current and source to drain voltage of a common
source circuit with an p-channel enhancement mode MOSFET.
Also find the power dissipation.
Assume that, VTP = -1.1V and Kp = 0.3 mA/V2
5V
Use KVL at SG loop:
VSG + 0 +2.5 – 5 = 0
VSG = 5 – 2.5 = 2.5 V
VSG > |VTP |
50 k
Assume biased in saturation mode:
50 k
Hence, ID = 0.3 ( 2.5 – 1.1)2 = 0.5888mA
Calculate VSD
Use KVL at SD loop:
VSD + IDRD – 5 = 0
VSD = 5 - IDRD
VSD = 5 – 0.5888 ( 7.5) = 0.584 V
© Electronics
7.5 k
VSD sat = VSG + VTP = 2.5 – 1.1 = 1.4V
Hence, VSD < VSD sat. Therefore assumption is incorrect. The transistor is
in non-saturation mode!
ID = 0.3 2 ( 2.5 – 1.1) (5 – IDRD) – (5 – IDRD)2
ID = 0.3 2.8 (5 – 7.5ID) – (5-7.5ID)2
ID = 0.3 14 – 21ID – (25 – 75ID + 56.25ID2)
ID = 0.3 14 – 21ID -25 +75ID – 56.25ID2
ID = 0.536 mA
56.25 ID2 – 50.67 ID + 11 = 0
ID = 0.365 mA
© Electronics
ID = 0.536 mA
VSD = 5 – IDRD = 0.98 V
ID = 0.365 mA
VSD = 5 – IDRD = 2.26 V
VSD sat = VSG + VTP = 2.5 – 1.1 = 1.4V
0.98V < 1.4V
Smaller than VSD sat : OK!
Answer: ID = 0.536 mA and VSD = 0.98V
Power dissipation = ID x VSD = 0.525 mW
© Electronics
2.26V > 1.4V
Bigger than VSD sat : not OK
LOAD LINE
• 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
© Electronics
ID (mA)
y-intercept
ID
Q-POINTS
VDS
© Electronics
VGS
x-intercept
VDS (V)
• 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.
+5V
RD = 2 k
-1V
RG = 24 k
RS = 1 k
-5V
© Electronics
+5V
ID
1. Get an expression for VGS in terms of ID
use KVL:
-1V
RD = 2 k
RG = 24 k
0 + VGS+ 1(ID) -5 +1 = 0
VGS = 4 - ID
RS = 1 k
ID
2. Assume in saturation
-5V
ID = 0.5 ( 4 - ID – 1)2 = 0.5 ( 3 – ID) 2
2ID = 9 – 6ID + ID2
ID = 1.354 mA
ID2 – 8ID + 9 = 0
ID = 6.646 mA
© Electronics
Replace in
VGS equation
VGS = 4 - ID
Why choose VGS = 2.646 V ?
Because it is bigger than VTN
VGS = 2.646 V
VGS= -2. 646 V
3. Get VDS equation and use the value of ID from step 2
Use KVL:
IDRD + VDS + IDRS – 5 – 5 = 0
1.354 (2) + VDS + 1.354 – 10 = 0
VDS = 10 – 1.354 – 2.708 = 5.938 V
+5V
ID
4. Calculate VDS sat
VDS sat = VGS – VTN = 2.646 – 1 = 1.646 V
5. Confirm your assumption
-1V
RD = 2 k
RG = 24 k
RS = 1 k
ID
VDS > VDS sat  CONFIRMED
-5V
© Electronics