DC analysis of FET - International Islamic University Malaysia
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Transcript DC analysis of FET - International Islamic University Malaysia
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