mosfet transistor - dc analysis - kfupm · 2011. 4. 6. · mosfet dc analysis...
TRANSCRIPT
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET TransistorDC Analysis
Dr. Alaa El-Din Hussein
March 18, 2008
Dr. Alaa El-Din Hussein — MOSFET Transistor 1/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Outline
1 MOSFET DC Analysis Procedure
2 Examples
3 MOSFET As A Current Source
Dr. Alaa El-Din Hussein — MOSFET Transistor 2/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Outline
1 MOSFET DC Analysis Procedure
2 Examples
3 MOSFET As A Current Source
Dr. Alaa El-Din Hussein — MOSFET Transistor 3/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 5: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/5.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 6: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/6.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 7: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/7.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 8: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/8.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 9: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/9.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 10: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/10.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 11: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/11.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 12: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/12.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 13: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/13.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 14: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/14.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET DC Analysis Procedure
Procedure
1 Apply KVL at the gate source loop to find VGS
2 If VGS < VTN , the transistor is off. Otherwise, assume an operating region(usually saturation)
3 Use VGS from step 1 to calculate ID using the transistor current equation
4 Apply KVL at the drain source loop and use ID from step 3 to find VDS
5 Check the validity of assumed region by comparing VDS to VDSAT
6 Change assumptions and analyze again if required.
An enhancement-mode device with VDS = VGS is always in saturation
If we have a source resistance, we need to solve the equations in steps 1 and 3together to find ID and VGS .
If we include channel length modulation or we are in the triode region, we willsolve the equations in steps 3 and 4 together
If we include channel length modulation or we are in the triode region and wehave a source resistance, we will solve the equations in steps 1, 3, and 4 together
Dr. Alaa El-Din Hussein — MOSFET Transistor 4/24
![Page 15: MOSFET Transistor - DC Analysis - KFUPM · 2011. 4. 6. · MOSFET DC Analysis ProcedureExamplesMOSFET As A Current Source MOSFET DC Analysis Procedure Procedure 1 Apply KVL at the](https://reader036.vdocuments.us/reader036/viewer/2022062608/60b2053cc1a6d070da61eb89/html5/thumbnails/15.jpg)
MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Outline
1 MOSFET DC Analysis Procedure
2 Examples
3 MOSFET As A Current Source
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Biasing in Triode Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 250µA/V 2
Solution
Assumption: Transistor is saturated, and IG = IB = 0
Analysis: From input loop VGS = VDD = 4V
Since the transistor at saturation we can use:ID = Kn
2 (VGS − VTN)2 = 2502µAV 2 (4− 1)2 = 1.13mA
Dr. Alaa El-Din Hussein — MOSFET Transistor 6/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Biasing in Triode Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 250µA/V 2
Solution
Assumption: Transistor is saturated, and IG = IB = 0
Analysis: From input loop VGS = VDD = 4V
Since the transistor at saturation we can use:ID = Kn
2 (VGS − VTN)2 = 2502µAV 2 (4− 1)2 = 1.13mA
Dr. Alaa El-Din Hussein — MOSFET Transistor 6/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Biasing in Triode Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 250µA/V 2
Solution
Assumption: Transistor is saturated, and IG = IB = 0
Analysis: From input loop VGS = VDD = 4V
Since the transistor at saturation we can use:ID = Kn
2 (VGS − VTN)2 = 2502µAV 2 (4− 1)2 = 1.13mA
Dr. Alaa El-Din Hussein — MOSFET Transistor 6/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Biasing in Triode Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 250µA/V 2
Solution
Assumption: Transistor is saturated, and IG = IB = 0
Analysis: From input loop VGS = VDD = 4V
Since the transistor at saturation we can use:ID = Kn
2 (VGS − VTN)2 = 2502µAV 2 (4− 1)2 = 1.13mA
Dr. Alaa El-Din Hussein — MOSFET Transistor 6/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 1Solution
Applying KVL at D-S Loop: VDD = IDRD + VDS
Substitute by the given values:∴ 4 = 1.6 ∗ 1.3 + VDS → VDS = 2.19V
But VDS < VGS − VTN . Hence, saturation region assumption isincorrect and the transistor is in triode region.
Using triode regionequation,4− VDS = 1600 ∗ 250µA
V 2 (4− 1− VDS
2 )VDS
Solving the last equation ∴ VDS = 2.3V , and ID = 1.06mA
VDS < VGS − VTN , transistor is in triode region
Q-pt: (1.06 m.A, 2.3 V) with VGS= 4 V
Dr. Alaa El-Din Hussein — MOSFET Transistor 7/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 2Biasing in Saturation Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN=1V, andKn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, simplify circuit, split VDD into two equal-valuedsources and apply Thevenin’s transformation to find VEQ and REQ
for the gate-bias voltage
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Example 2Biasing in Saturation Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN=1V, andKn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, simplify circuit, split VDD into two equal-valuedsources and apply Thevenin’s transformation to find VEQ and REQ
for the gate-bias voltage
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 2Biasing in Saturation Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN=1V, andKn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, simplify circuit, split VDD into two equal-valuedsources and apply Thevenin’s transformation to find VEQ and REQ
for the gate-bias voltage
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Example 2Biasing in Saturation Region
Example
Find the Q-pt (ID ,VDS)assuming that VTN=1V, andKn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, simplify circuit, split VDD into two equal-valuedsources and apply Thevenin’s transformation to find VEQ and REQ
for the gate-bias voltage
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 2Solution
Apply KVL at G-S Loop:VEQ = VGS + IDRS
Using ID = Kn
2 (VGS − VTN)2
∴ VEQ =
VGS + KnRS
2 (VGS − VTN)2
Substitute by given values 4 = VGS +(25×10−6)(39×103)
2(VGS − 1)2
Solving the quadratic equation results in VGS = −2.71V ,+2.66V
Since VGS < VTN for VGS = -2.71 V, we will ignore it
Substituting with VGS = +2.66 V results in ID = 34.4µA
Applying KVL at D-S loop, VDD = ID(RD + RS) + VDS → VDS = 6.08V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (34.4 µA, 6.08 V) with VGS= 2.66 V
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Example 3Bias Analysis with Body Effect
Example
Find the Q-pt (ID ,VDS)assuming that VTO =1V , 2φF = 0.6V , γ = 0.5
√V ,
and Kn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, using KVL at the G-S loop yields:VGS = VEQ − IDRS = 6− 22, 000ID
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Example 3Bias Analysis with Body Effect
Example
Find the Q-pt (ID ,VDS)assuming that VTO =1V , 2φF = 0.6V , γ = 0.5
√V ,
and Kn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, using KVL at the G-S loop yields:VGS = VEQ − IDRS = 6− 22, 000ID
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Example 3Bias Analysis with Body Effect
Example
Find the Q-pt (ID ,VDS)assuming that VTO =1V , 2φF = 0.6V , γ = 0.5
√V ,
and Kn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, using KVL at the G-S loop yields:VGS = VEQ − IDRS = 6− 22, 000ID
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Example 3Bias Analysis with Body Effect
Example
Find the Q-pt (ID ,VDS)assuming that VTO =1V , 2φF = 0.6V , γ = 0.5
√V ,
and Kn = 25µA/V 2
Solution
Approach: Assume operation region, find Q-point, check to see ifresult is consistent with operation region
Assumption: Transistor is saturated, IG = IB = 0
Analysis: First, using KVL at the G-S loop yields:VGS = VEQ − IDRS = 6− 22, 000ID
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Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
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Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
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Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
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Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
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Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
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Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (88 µA, 6.48 V) with VGS= 4.06 VDr. Alaa El-Din Hussein — MOSFET Transistor 11/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (88 µA, 6.48 V) with VGS= 4.06 VDr. Alaa El-Din Hussein — MOSFET Transistor 11/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (88 µA, 6.48 V) with VGS= 4.06 VDr. Alaa El-Din Hussein — MOSFET Transistor 11/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (88 µA, 6.48 V) with VGS= 4.06 VDr. Alaa El-Din Hussein — MOSFET Transistor 11/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (88 µA, 6.48 V) with VGS= 4.06 VDr. Alaa El-Din Hussein — MOSFET Transistor 11/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 3Solution
Since VSB 6= 0 then use VTN = VTO + γ(√
VSB + 2ϕF −√
2ϕF )
∴ VTN = 1 + 0.5(√
22000ID + 0.6−√
0.6)
Using I ′D =(25×10−6)
2(VGS − VTN)2 we can solve the non-linear equation
to find ID or use the iteration method below
Iteration Method
Estimate value of ID and use it to find VGS and VTN
Find I ′D using VGS and VTN from the last step
If I ′D is not same as original ID estimate, start again.
The iteration sequence leads to ID= 88.0 µA
Applying KVL at D-S loop,VDS = VDD − ID(RD + RS) = 10− 40, 000ID = 6.48V
Since VDS > VGS − VTN . Hence saturation region assumption is correct.
Q-pt: (88 µA, 6.48 V) with VGS= 4.06 VDr. Alaa El-Din Hussein — MOSFET Transistor 11/24
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Bias with Feedback
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 260µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VDS = VGS
Using KVL at the D-S loop yields: VDS = VGS = VDD − IDRD
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Bias with Feedback
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 260µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VDS = VGS
Using KVL at the D-S loop yields: VDS = VGS = VDD − IDRD
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Bias with Feedback
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 260µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VDS = VGS
Using KVL at the D-S loop yields: VDS = VGS = VDD − IDRD
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Bias with Feedback
Example
Find the Q-pt (ID ,VDS)assuming that VTN = 1V , andKn = 260µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VDS = VGS
Using KVL at the D-S loop yields: VDS = VGS = VDD − IDRD
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 4Solution
Since the transistor at saturation we can use: ID = Kn
2 (VGS − VTN)2
Substitute in the last KVL equation yields:VGS = VDD − KnRD
2 (VGS − VTN)2
Substitute by the given values:
∴ VGS = 3.3− (2.6×10−4)(104)2 (VGS − 1)2
Solve the quadratic equation : ∴ VGS = −0.769V ,+2.00V
Since VGS < VTN for VGS= -0.769 V and MOSFET will be cut-off,it will be ignored.
Substitute in the current equation yields: ID = 130µA
Q-pt: (130 µA, 2 V) with VGS= 2 V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 5Enhancement PMOS
Example
Find the Q-pt (ID ,VDS) assumingthat VTP = −2V , andKp = 50µA/V 2
Solution
Assumption: IG = IB = 0
Analysis: Transistor is saturated since VSD = VSG
Applying KVL at D-S loop: −15V + (220kΩ)ID + VSG = 0
∴ 15V − (220kΩ) 502
µAV 2 (VSG − 2)2 − VSG = 0
∴ VSG = 0.369V , 3.45V
Since VSG = 0.369V < |VTP |= 2 V, ∴ VSG = 3.45 V and ID = 52.5 mA.
Q-pt: (52.2 µA, 3.45 V)
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 6
Example
Design the shown circuit so that thetransistor operates at ID = 0.3m.Aand VD = +0.4V . The NMOStransistor hasVt = 1V , µnCox = 60µA/V 2, L =3µm, andW = 120µm.
Answer
RS = 3.3kΩ, and RD = 7kΩ
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 6
Example
Design the shown circuit so that thetransistor operates at ID = 0.3m.Aand VD = +0.4V . The NMOStransistor hasVt = 1V , µnCox = 60µA/V 2, L =3µm, andW = 120µm.
Answer
RS = 3.3kΩ, and RD = 7kΩ
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 7
Example
Design the shown circuit so that thetransistor operates at ID = 80µA.The NMOS transistor hasVt = 0.6V , µnCox = 200µA/V 2, L =0.8µm, andW = 4µm. Also, find thedrain voltage VD .
Answer
R = 25kΩ, and VD = +1V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 7
Example
Design the shown circuit so that thetransistor operates at ID = 80µA.The NMOS transistor hasVt = 0.6V , µnCox = 200µA/V 2, L =0.8µm, andW = 4µm. Also, find thedrain voltage VD .
Answer
R = 25kΩ, and VD = +1V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 8
Example
Design the shown circuit to establishVD of 0.1 V. The NMOS transistorhasVt = 1V , andk ′nW /L = 1mA/V 2.What is the effective resistancebetween drain and source at thisoperating point?.
Answer
RD = 12.4kΩ, and rds = 253Ω
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 8
Example
Design the shown circuit to establishVD of 0.1 V. The NMOS transistorhasVt = 1V , andk ′nW /L = 1mA/V 2.What is the effective resistancebetween drain and source at thisoperating point?.
Answer
RD = 12.4kΩ, and rds = 253Ω
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 9
Example
Analyze the circuit shown todetermine the voltages at all nodesand the currents through allbranches. The NMOS transistor hasVt = 1V , andk ′nW /L = 1mA/V 2.
Answer
IG = 0mA, IRG = 0.5µA, ID = 0.5mA,VG = 5V ,VS =+3V , andVD = +7V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 9
Example
Analyze the circuit shown todetermine the voltages at all nodesand the currents through allbranches. The NMOS transistor hasVt = 1V , andk ′nW /L = 1mA/V 2.
Answer
IG = 0mA, IRG = 0.5µA, ID = 0.5mA,VG = 5V ,VS =+3V , andVD = +7V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 10
Example
Design the shown circuit to obtainthe indicated current and voltagevalues. The NMOS transistor hasVt = 1V , µnCox = 120µA/V 2, λ =0, andL1 = L2 = 1µm.
Answer
R = 12.5kΩ,W1 = 8µm, andW1 = 2µm
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 10
Example
Design the shown circuit to obtainthe indicated current and voltagevalues. The NMOS transistor hasVt = 1V , µnCox = 120µA/V 2, λ =0, andL1 = L2 = 1µm.
Answer
R = 12.5kΩ,W1 = 8µm, andW1 = 2µm
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 11
Example
For the shown circuit calculate theshown current and voltage values forvI = 0V, +2.5V, and -2.5V.Assuming matched transistors withVTN = VTP = 1V , kn = kp =1mA/V 2, andλ = 0.
Answer
IDN = IDP = 1.125mA, andvo = 0V
IDN = 0.244mA, IDP = 0mA, andvo = −2.44V
IDN = 0mA, IDP = 0.244mA, andvo = +2.44V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 11
Example
For the shown circuit calculate theshown current and voltage values forvI = 0V, +2.5V, and -2.5V.Assuming matched transistors withVTN = VTP = 1V , kn = kp =1mA/V 2, andλ = 0.
Answer
IDN = IDP = 1.125mA, andvo = 0V
IDN = 0.244mA, IDP = 0mA, andvo = −2.44V
IDN = 0mA, IDP = 0.244mA, andvo = +2.44V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 11
Example
For the shown circuit calculate theshown current and voltage values forvI = 0V, +2.5V, and -2.5V.Assuming matched transistors withVTN = VTP = 1V , kn = kp =1mA/V 2, andλ = 0.
Answer
IDN = IDP = 1.125mA, andvo = 0V
IDN = 0.244mA, IDP = 0mA, andvo = −2.44V
IDN = 0mA, IDP = 0.244mA, andvo = +2.44V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 11
Example
For the shown circuit calculate theshown current and voltage values forvI = 0V, +2.5V, and -2.5V.Assuming matched transistors withVTN = VTP = 1V , kn = kp =1mA/V 2, andλ = 0.
Answer
IDN = IDP = 1.125mA, andvo = 0V
IDN = 0.244mA, IDP = 0mA, andvo = −2.44V
IDN = 0mA, IDP = 0.244mA, andvo = +2.44V
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Outline
1 MOSFET DC Analysis Procedure
2 Examples
3 MOSFET As A Current Source
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current Source
Ideal current source gives fixedoutput current regardless of thevoltage across it.
MOSFET behaves as as anideal current source if biased inthe pinch-off region (outputcurrent depends on terminalvoltage).
Notes
VDS should be greater than VDSAT for proper operation
If the channel length modulation isn’t neglected, a finite sourceresistance will exist = [λID ]−1
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current Source
Ideal current source gives fixedoutput current regardless of thevoltage across it.
MOSFET behaves as as anideal current source if biased inthe pinch-off region (outputcurrent depends on terminalvoltage).
Notes
VDS should be greater than VDSAT for proper operation
If the channel length modulation isn’t neglected, a finite sourceresistance will exist = [λID ]−1
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current Source
Ideal current source gives fixedoutput current regardless of thevoltage across it.
MOSFET behaves as as anideal current source if biased inthe pinch-off region (outputcurrent depends on terminalvoltage).
Notes
VDS should be greater than VDSAT for proper operation
If the channel length modulation isn’t neglected, a finite sourceresistance will exist = [λID ]−1
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current SourceCurrent Mirror
Assumptions: M1 and M2 haveidentical VTN , K ′n, λ and are insaturation.
Analysis
IREF =K ′n2
(WL
)M1
(VGS1 − VTN)2 (1 + λVDS1)
IO =K ′n2
(WL
)M2
(VGS2 − VTN)2 (1 + λVDS2)
But VGS2 = VGS1, ∴ IO = IREF( W
L )M2
( WL )
M1
(1+λVDS2)(1+λVDS1)
∼= ( WL )
M2
( WL )
M1
IREF
Thus, output current mirrors reference current if VDS1 = VDS2 orλ= 0, and both transistors have the same (W/L).
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current SourceCurrent Mirror
Assumptions: M1 and M2 haveidentical VTN , K ′n, λ and are insaturation.
Analysis
IREF =K ′n2
(WL
)M1
(VGS1 − VTN)2 (1 + λVDS1)
IO =K ′n2
(WL
)M2
(VGS2 − VTN)2 (1 + λVDS2)
But VGS2 = VGS1, ∴ IO = IREF( W
L )M2
( WL )
M1
(1+λVDS2)(1+λVDS1)
∼= ( WL )
M2
( WL )
M1
IREF
Thus, output current mirrors reference current if VDS1 = VDS2 orλ= 0, and both transistors have the same (W/L).
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current SourceCurrent Mirror
Assumptions: M1 and M2 haveidentical VTN , K ′n, λ and are insaturation.
Analysis
IREF =K ′n2
(WL
)M1
(VGS1 − VTN)2 (1 + λVDS1)
IO =K ′n2
(WL
)M2
(VGS2 − VTN)2 (1 + λVDS2)
But VGS2 = VGS1, ∴ IO = IREF( W
L )M2
( WL )
M1
(1+λVDS2)(1+λVDS1)
∼= ( WL )
M2
( WL )
M1
IREF
Thus, output current mirrors reference current if VDS1 = VDS2 orλ= 0, and both transistors have the same (W/L).
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current SourceCurrent Mirror
Assumptions: M1 and M2 haveidentical VTN , K ′n, λ and are insaturation.
Analysis
IREF =K ′n2
(WL
)M1
(VGS1 − VTN)2 (1 + λVDS1)
IO =K ′n2
(WL
)M2
(VGS2 − VTN)2 (1 + λVDS2)
But VGS2 = VGS1, ∴ IO = IREF( W
L )M2
( WL )
M1
(1+λVDS2)(1+λVDS1)
∼= ( WL )
M2
( WL )
M1
IREF
Thus, output current mirrors reference current if VDS1 = VDS2 orλ= 0, and both transistors have the same (W/L).
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
MOSFET As A Current SourceCurrent Mirror
Assumptions: M1 and M2 haveidentical VTN , K ′n, λ and are insaturation.
Analysis
IREF =K ′n2
(WL
)M1
(VGS1 − VTN)2 (1 + λVDS1)
IO =K ′n2
(WL
)M2
(VGS2 − VTN)2 (1 + λVDS2)
But VGS2 = VGS1, ∴ IO = IREF( W
L )M2
( WL )
M1
(1+λVDS2)(1+λVDS1)
∼= ( WL )
M2
( WL )
M1
IREF
Thus, output current mirrors reference current if VDS1 = VDS2 orλ= 0, and both transistors have the same (W/L).
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 12Current Mirror
Example
Find the output current and the minimumoutput voltage vo to maintain the givencurrent mirror in proper operation. Assume,IREF = 50 µ A, VO= 12 V, VTN = 1 V,K ′n= 75 µA/V 2, λ = 0V−1, (W /L)M1 = 2,(W /L)M2=10
Analysis
∴ IO = IREF( W
L )M2
( WL )
M1
= 250µA
VGS = VTN +√
2IREF
K ′n( WL )(1+λVDS1)
= 1V +
√2(50µA)
2∗75 µA
V 2
= 1.82V
Hence, Vomin = VGS–VTN = 0.82V .
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 12Current Mirror
Example
Find the output current and the minimumoutput voltage vo to maintain the givencurrent mirror in proper operation. Assume,IREF = 50 µ A, VO= 12 V, VTN = 1 V,K ′n= 75 µA/V 2, λ = 0V−1, (W /L)M1 = 2,(W /L)M2=10
Analysis
∴ IO = IREF( W
L )M2
( WL )
M1
= 250µA
VGS = VTN +√
2IREF
K ′n( WL )(1+λVDS1)
= 1V +
√2(50µA)
2∗75 µA
V 2
= 1.82V
Hence, Vomin = VGS–VTN = 0.82V .
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 12Current Mirror
Example
Find the output current and the minimumoutput voltage vo to maintain the givencurrent mirror in proper operation. Assume,IREF = 50 µ A, VO= 12 V, VTN = 1 V,K ′n= 75 µA/V 2, λ = 0V−1, (W /L)M1 = 2,(W /L)M2=10
Analysis
∴ IO = IREF( W
L )M2
( WL )
M1
= 250µA
VGS = VTN +√
2IREF
K ′n( WL )(1+λVDS1)
= 1V +
√2(50µA)
2∗75 µA
V 2
= 1.82V
Hence, Vomin = VGS–VTN = 0.82V .
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MOSFET DC Analysis Procedure Examples MOSFET As A Current Source
Example 12Current Mirror
Example
Find the output current and the minimumoutput voltage vo to maintain the givencurrent mirror in proper operation. Assume,IREF = 50 µ A, VO= 12 V, VTN = 1 V,K ′n= 75 µA/V 2, λ = 0V−1, (W /L)M1 = 2,(W /L)M2=10
Analysis
∴ IO = IREF( W
L )M2
( WL )
M1
= 250µA
VGS = VTN +√
2IREF
K ′n( WL )(1+λVDS1)
= 1V +
√2(50µA)
2∗75 µA
V 2
= 1.82V
Hence, Vomin = VGS–VTN = 0.82V .
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