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For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)
Q. What is value of internal pinch-off voltage?
- a)0.5 V
- b)1.0 V
- c)1.5
- d)2.0 V
Correct answer is option 'B'. Can you explain this answer?
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For a P-channel Ge JFET, it is having doping concentration of Na = 1.7...
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For a P-channel Ge JFET, it is having doping concentration of Na = 1.7...
X 10^15 cm^-3. The electron mobility is given as μn = 3800 cm^2/Vs. The channel length is L = 2 μm and the channel width is W = 10 μm. The oxide thickness is Tox = 100 nm and the relative permittivity of the oxide is εr = 3.9. The gate voltage is VGS = -2 V and the drain voltage is VDS = -10 V. The reverse saturation current is given as IDSS = 1 mA. Calculate the drain current, transconductance, and pinch-off voltage for the JFET.
To calculate the drain current (ID), we can use the equation:
ID = IDSS * (1 - (VGS / VP))^2
Where VP is the pinch-off voltage. Rearranging the equation, we can solve for VP:
VP = VGS / (1 - sqrt(ID / IDSS))
Plugging in the given values:
VP = -2 / (1 - sqrt(ID / 1))
Next, we need to calculate the transconductance (gm). The transconductance is given by the equation:
gm = 2 * sqrt(k * ID)
Where k is the process transconductance parameter. Rearranging the equation, we can solve for k:
k = gm^2 / (4 * ID)
Plugging in the given values:
k = gm^2 / (4 * ID)
Finally, we need to calculate the pinch-off voltage (VP). The pinch-off voltage is given by the equation:
VPO = VDS - ID * RD
Where RD is the drain resistance. Rearranging the equation, we can solve for RD:
RD = (VDS - VPO) / ID
Plugging in the given values:
RD = (-10 - VPO) / ID
Now, let's calculate the values:
ID = IDSS * (1 - (VGS / VP))^2
ID = 1 * (1 - (-2 / VP))^2
Using this equation, we need to iterate to find the value of VP that satisfies the equation. Let's start with an initial guess of VP = -3 V:
ID = 1 * (1 - (-2 / -3))^2
ID = 1 * (1 - 0.6667)^2
ID = 1 * (0.3333)^2
ID = 0.1111 mA
Now, let's calculate the transconductance:
gm = 2 * sqrt(k * ID)
gm = 2 * sqrt(k * 0.1111)
Using this equation, we need to iterate to find the value of k that satisfies the equation. Let's start with an initial guess of k = 0.1 mA/V^2:
gm = 2 * sqrt(0.1 * 0.1111)
gm = 2 * 0.3333
gm = 0.6666 mS
Finally, let's calculate the pinch-off voltage:
RD = (-10 - VPO) / ID
RD = (-10 - VPO) / 0.1111
Using this equation, we need to iterate to find the value of VPO that satisfies the equation. Let's start with an initial guess of VPO = -4 V:
RD = (-10 - (-4)) /
To calculate the drain current (ID), we can use the equation:
ID = IDSS * (1 - (VGS / VP))^2
Where VP is the pinch-off voltage. Rearranging the equation, we can solve for VP:
VP = VGS / (1 - sqrt(ID / IDSS))
Plugging in the given values:
VP = -2 / (1 - sqrt(ID / 1))
Next, we need to calculate the transconductance (gm). The transconductance is given by the equation:
gm = 2 * sqrt(k * ID)
Where k is the process transconductance parameter. Rearranging the equation, we can solve for k:
k = gm^2 / (4 * ID)
Plugging in the given values:
k = gm^2 / (4 * ID)
Finally, we need to calculate the pinch-off voltage (VP). The pinch-off voltage is given by the equation:
VPO = VDS - ID * RD
Where RD is the drain resistance. Rearranging the equation, we can solve for RD:
RD = (VDS - VPO) / ID
Plugging in the given values:
RD = (-10 - VPO) / ID
Now, let's calculate the values:
ID = IDSS * (1 - (VGS / VP))^2
ID = 1 * (1 - (-2 / VP))^2
Using this equation, we need to iterate to find the value of VP that satisfies the equation. Let's start with an initial guess of VP = -3 V:
ID = 1 * (1 - (-2 / -3))^2
ID = 1 * (1 - 0.6667)^2
ID = 1 * (0.3333)^2
ID = 0.1111 mA
Now, let's calculate the transconductance:
gm = 2 * sqrt(k * ID)
gm = 2 * sqrt(k * 0.1111)
Using this equation, we need to iterate to find the value of k that satisfies the equation. Let's start with an initial guess of k = 0.1 mA/V^2:
gm = 2 * sqrt(0.1 * 0.1111)
gm = 2 * 0.3333
gm = 0.6666 mS
Finally, let's calculate the pinch-off voltage:
RD = (-10 - VPO) / ID
RD = (-10 - VPO) / 0.1111
Using this equation, we need to iterate to find the value of VPO that satisfies the equation. Let's start with an initial guess of VPO = -4 V:
RD = (-10 - (-4)) /
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For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer?
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For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer?.
For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer?.
Solutions for For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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ample number of questions to practice For a P-channel Ge JFET, it is having doping concentration of Na = 1.77 × 1015 cm−3 and Nd = 3.53 × 1018 per cm3, If channel thickness is a = 1 µ met. (Given Relative permittivity of Ge = 16 at T = 300° K)Q.What is value of internal pinch-off voltage?a)0.5 Vb)1.0 Vc)1.5d)2.0 VCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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