L20 DC MOSFET Circuits V2 - nanoHUB.org · Outline 2 1) BJT Analysis is Easy 2) DC N-MOSFET Circuit...
Transcript of L20 DC MOSFET Circuits V2 - nanoHUB.org · Outline 2 1) BJT Analysis is Easy 2) DC N-MOSFET Circuit...
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ECE 255:
DC MOSFET Circuits
(Sedra and Smith, 7th Ed., Sec. 5.3)
Mark Lundstrom School of ECE
Purdue University West Lafayette, IN USA
Lundstrom: Fall 2019
ECE 255: Fall 2019 Purdue University
Outline
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1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples
Lundstrom: Fall 2019
BJT circuit analysis
3 Lundstrom: Fall 2019
+V
CE= ? V
−
β = 100
+5 V
−5 V
RE = 4.3 kΩ
RC = 1kΩIC = ? mA Find IC and VCE
VBE on( ) = 0.7 V
Why is it so easy?
4 Lundstrom: Fall 2019
+V
CE= 4.71V
−
β = 100
+5 V
−5 V
RE = 4.3 kΩ
RC = 1kΩIC = 0.99 mA
VBE on( ) = 0.7 V
+V
BE
− Because we can guess this voltage
IC
VBE
IC = ISeVBE VT
VBE≈ 0.7 V
Why is MOSFET circuit analysis harder?
5 Lundstrom: Fall 2019
Because we cannot guess this voltage
ID
VGSV
tn
+VDS = ?−
+5 V
−5 V
RS = 4.3 kΩ
RD = 1kΩID = ?
+VGS−
ID =
kn
2VGS −Vtn( )2
Vtn > 0 V
Outline
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1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples
Lundstrom: Fall 2019
IV Summary (saturation region)
7 Lundstrom: Fall 2019
VDS
IC
act VGS −Vtn
VDSsat
ID =
kn
2VGS −Vtn( )2
IG = 0 kn =
WL
′kn =WLµnCox
Vtn > 0 V( )
VDSsat =VGS −Vtn
VDS >VDSsat
VGS >Vtn
ID = IS
MOSFET circuit analysis
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+VDS = ?−
+5 V
−5 V
RS = 4 kΩ
RD = 3.5 kΩID = ?
ID =
′kn
2WL
VGS −Vtn( )2
Transistor model:
′kn
2WL= 1mA/V2
Vtn = 1.0 V
Lundstrom: Fall 2019
ID = 1 VGS −1.0( )2 (device)
VGS = ? (circuit)
MOSFET circuit analysis (ii)
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+VDS = ?−
+5 V
−5 V
RS = 4 kΩ
RD = 3.5 kΩID = ? ID = 1 VGS −1.0( )2
VGS = ?
VGS =VG −VS
VS = −5+ ID RS
VGS = 5− ID RS
ID = 5− ID RS −1.0( )2
Lundstrom: Fall 2019
VG
VD
VS+IDRS−
MOSFET circuit analysis (iii)
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+VDS = ?−
+5 V
−5 V
RS = 4 kΩ
RD = 3.5 kΩID = ? ID = 5− ID RS −1.0( )2
ID2 − 2.06ID +1= 0
ID = 1.28 / 0.76
Lundstrom: Fall 2019
MOSFET circuit analysis (iv)
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+VDS = ?−
+5 V
−5 V
RS = 4 kΩ
RD = 3.5 kΩID = ? ID = 1.28 mA Does not work.
Lundstrom: Fall 2019
Vtn = 1.0 V
MOSFET circuit analysis (iv)
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+VDS = ?−
+5 V
−5 V
RS = 4 kΩ
RD = 3.5 kΩID = ? ID = 0.76 mA
VD = 5− ID RD = 2.3 V
VGS = −VS = 1.96 V >Vtn
VDS =VD −VS = 4.26 V
VDSsat =VGS −Vtn = 0.96 V
✓
✓
VS = −5+ ID RS = −1.96 V
VDS >VDSsat
Lundstrom: Fall 2019
Vtn = 1.0 V
MOSFET circuit analysis (summary)
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+VDS = ?−
+5 V
−5 V
RS = 4 kΩ
RD = 3.5 kΩID = ?
Vtn = 1.0 V
ID =
′kn
2WL
VGS −Vtn( )2
1) Write the device equation:
VGS =
2) Write the circuit equation:
3) Solve the equations
4) Check solution
Outline
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1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples
Lundstrom: Fall 2019
MOSFET circuit design
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+VDS = 3.0 V−
+5 V
−5 V
RS = ? kΩ
RD = ? kΩID = 0.5 mA
ID =
′kn
2WL
VGS −Vtn( )2
Transistor model:
′kn
2WL= 1mA/V2
Vtn = 1.0 V
ID = 1 VGS −1.0( )2= 0.5
VGS = 1.71
Now design the circuit
MOSFET circuit design (ii)
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+VDS = 3.0 V−
+5 V
−5 V
RS = ? kΩ
RD = ? kΩID = 0.5 mA VGS = 1.71
VGS =VG −VS = 1.71
VS = −1.71
RS = 6.6 kΩ
VD =VS +VDS
RD = 7.4 kΩ
✓
✓
Lundstrom: Fall 2019
VG
VD = 1.29
MOSFET circuit design (summary)
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+VDS = y.yymA−
+5 V
−5 V
RS = ? kΩ
RD = ? kΩID = x.x mA
ID =
′kn
2WL
VGS −Vtn( )2
1) Write the device equation:
2) Solve for VGS:
3) Design the circuit to give that VGS
4) Design the circuit to give the desired VDS
Outline
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1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples
Lundstrom: Fall 2019
IV Summary (NMOS saturation region)
19 Lundstrom: Fall 2019
VDS
IC
act VGS −Vtn
VDSsat
ID =
kn
2VGS −Vtn( )2
kn =
WL
′kn =WLµnCox
Vtn > 0 V( )
VDSsat =VGS −Vtn
VDS >VDSsat
VGS >Vtn
IV Summary (PMOS saturation region)
20 Lundstrom: Fall 2019
VSD
IC
act VSG − Vtp
VSDsat
ID =
kp
2VSG − Vtp( )2
kp =
WL
′kp =WLµ pCox
Vtp < 0 V( )
VSDsat =VSG − Vtp
VSD >VSDsat
VSG > Vtp
P-MOSFET circuit design
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+VSD = 3.0 V−
+5 V
−5 V
RD = ? kΩ
RS = ? kΩ
ID = 0.5 mA
ID =
′kp
2WL
VSG − Vtp( )2
Transistor model:
′kp
2WL= 1mA/V2
Vtp = −1.0 V
Lundstrom: Fall 2019
ID = 1 VSG −1.0( )2= 0.5
VSG = 1.71
P-MOSFET circuit design (ii)
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+VSD = 3.0 V−
+5 V
−5 V
RD = ? kΩ
RS = ? kΩ
ID = 0.5 mA
VSG = 1.71
VS = 1.71 RS = 6.6 kΩ
VD =VS −VSD
VD = −1.29
RD = 7.4 kΩ
✓
✓
Lundstrom: Fall 2019
P-MOSFET circuit analysis
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+VSD = ?−
+5 V
−5 V
RD = 3.5 kΩ
RS = 4 kΩ
ID = ?
ID =
′kp
2WL
VSG − Vtp( )2
Transistor model:
′kp
2WL= 1mA/V2
Vtp = −1.0 V
ID = 1 VSG −1.0( )2
VSG = 5− ID RS
Lundstrom: Fall 2019
P-MOSFET circuit analysis
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ID = 1 VSG −1.0( )2
VSG = 5− ID RS
ID = 5− ID RS −1.0( )2
ID2 − 2.06ID +1= 0
ID = 1.28 or 0.76
ID = 0.76
ID = 5−VSG( ) RS = 5−VSG( ) 4
4VSG2 − 7VSG −1= 0
VSG = 1.88 or − 0.13
VSG = 1.88
ID = 0.77
Option1: Option2: 1
2
Outline
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1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples
Lundstrom: Fall 2019
Example 1: Analysis
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VDD = +5 V
RD = 4.3kΩ
1) Operating region?
+VGS
−
Lundstrom: Fall 2019
ID = 0.1 VGS −1( )2
Example 1: Analysis
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VDD = +5 V
RD = 4.3kΩ
1) Operating region?
VGS =VD
VDS =VD
VDS >VGS −Vtn
+VGS
−Saturation
Lundstrom: Fall 2019
ID = 0.1 VGS −1( )2
Example 1: Analysis
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VDD = +5 V
RD = 4.3kΩ
ID = 0.1 VGS −1( )2
ID = 0.1 VGS −1( )2
+VGS
−
Lundstrom: Fall 2019
VD
Example 1: Analysis
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VDD = +5 V
RD = 4.3kΩ
ID = 0.1 VGS −1( )2
ID = 0.1 VGS −1( )2
VGS = 5− ID RD = 5− 4.3ID
ID2 − 2.86ID +1.35= 0
+VGS
−
ID = 2.26 or 0.60
ID = 0.60
Lundstrom: Fall 2019
VD
Example 2: Design
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VDD = +5 V
RD = ?kΩ
ID =
′kn
2WL
VGS −Vtn( )2
ID = 0.1 VGS −1( )2
Design for: ID = 0.9 mA
Lundstrom: Fall 2019
VD
Example 2: Design
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VDD = +5 V
RD = ?kΩ
ID =
′kn
2WL
VGS −Vtn( )2
ID = 0.1 VGS −1( )2
ID = 0.1 VGS −1( )2= 0.9
Design for: ID = 0.9 mA
VGS = 4.0 V
VGS =VD = 4.0 V
RD = 5− 4
0.9= 1.11kΩ
Lundstrom: Fall 2019
VD
Example 3: Design (i)
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VDD = +5 V
RS = ? kΩ
ID = 0.1 VSG −1( )2
Design for: ID = 0.9 mA
1) Operating region?
Lundstrom: Fall 2019
VDS >VGS −Vtn ?VS
Example 3: Design (i)
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VDD = +5 V
RS = ? kΩ
ID = 0.1 VSG −1( )2
Design for: ID = 0.9 mA
1) Operating region?
VSD >VSG − Vtp ?
saturation
Lundstrom: Fall 2019
VDS >VGS −Vtn ?VS
Example 3: Design (ii)
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VDD = +5 V
RS = ? kΩ
ID =
′kp
2WL
VSG − Vtp( )2
ID = 0.1 VSG −1( )2
Design for: ID = 0.9 mA
Lundstrom: Fall 2019
Example 3: Design (ii)
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VDD = +5 V
RS = ? kΩ
ID =
′kp
2WL
VSG − Vtp( )2
ID = 0.1 VSG −1( )2
ID = 0.1 VSG −1( )2
Design for: ID = 0.9 mA
0.9 = 0.1 VSG −1( )2
VSG = 4.0 V
VSG =VS = 4.0 V
RS =
5− 4.00.9
= 1.11kΩ
Lundstrom: Fall 2019
Example 4
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+5 V
RD = 2 kΩ
ID = 0.1 VSG −1( )2
I = 1.0 mA
−5 V
VS = ?
VD = ?
Lundstrom: Fall 2019
Example 4
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+5 V
RD = 2 kΩ
ID = 0.1 VSG −1( )2
I = 1.0 mA
−5 V
VS = ?
VD = −5+ ID RD
= −5+1× 2= −3 V
VD = ?
1= 0.1 VSG −1( )2
VSG =VS = 4.16 V
VS =VSG
Lundstrom: Fall 2019
Example 5
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+5 V
ID = 0.1 VGSn −1( )2
1) Operating region?
VD = ?
ID = 0.1 VSGp −1( )2
+VGSp
−
+VSGn
− saturation
Lundstrom: Fall 2019
Example 5 (ii)
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+5 V
ID = 0.1 VGSn −1( )2
VDSn =VD
1) Operating region?
VD = ?
VGSn =VD
VDSn > VGSn −Vtn( ) ID = 0.1 VSGp −1( )2
+VGSp
−
+VSGn
−
VSDp = 5−VD
VSGp = 5−VD
VSDp > VSGp − Vtp( )
saturation Lundstrom: Fall 2019
Example 5 (iii)
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+5 V
ID = 0.1 VGSn −1( )2
VD = ?
ID = 0.1 VSGp −1( )2
+VGSn
−
+VSGp
− VGSn =VSGp Why?
Lundstrom: Fall 2019
Example 5 (iii)
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+5 V
ID = 0.1 VGSn −1( )2
VD = ?
ID = 0.1 VSGp −1( )2
+VGSn
−
+VSGp
− VGSn =VSGp
VGSn +VSGp = 5
VD = 2.5
ID = 0.1 2.5−1( )2= 0.225 mA
Why?
Lundstrom: Fall 2019
Example 6: Analysis
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VDD = 10 V
RD = 100 k
RG1
700 k
RG2
300 k
Transistor model:
kn = 25 µA/V2 Vtn = 1.0 V
ID =
kn
2VGS −Vtn( )2
ID = 0.025
2VGS −1( )2
mA VGS = 3
ID = 0.05 mA
VDS = 5 V >VGD -Vtn
Summary
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1) DC MOSFET analysis often involves solving a quadratic equation and throwing away the unphysical solution.
2) DC MOSFET analysis is often more difficult than DC BJT analysis
3) DC MOSFET design is usually easier than design.
Lundstrom: Fall 2019
DC MOSFET Circuits
Lundstrom: Fall 2019 44
1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples