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An ideal p-channel MOSFET has the parameters. W = 15 μm, μp = 300 cm2A/ sec, L = 15 [μm, tox = 350 A° and VT = -0.80 V. If transistor is operating in non saturation region at VSD = 0.5 V, then the value of gm is ____________ μS
- a)145
- b)148
- c)125
- d)136
Correct answer is option 'B'. Can you explain this answer?
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An ideal p-channel MOSFET has the parameters. W = 15 μm, μp ...
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An ideal p-channel MOSFET has the parameters. W = 15 μm, μp ...
Given parameters:
W = 15 μm
p = 300 cm^2A/sec
L = 15 μm
tox = 350 Å
VT = -0.80 V
VSD = 0.5 V
To calculate the value of gm, we need to find the transconductance parameter (KP) and the overdrive voltage (Vov).
1. Transconductance parameter (KP):
KP is given by the equation:
KP = μn * Cox * (W/L)
Where:
μn = Electron mobility in the channel
Cox = Oxide capacitance per unit area
Given that:
μn = p * tox
Cox = εox / tox
Here, εox is the permittivity of the oxide layer, and tox is the thickness of the oxide layer.
Substituting the given values:
μn = 300 cm^2A/sec * 350 Å = 105,000 cm^2A/sec
Cox = εox / tox = (8.854 * 10^-12 F/m * 3.5 * 10^-7 m) / (350 * 10^-10 m) = 8.854 * 10^-6 F/cm^2
Now, substituting the calculated values back into the equation for KP:
KP = μn * Cox * (W/L) = (105,000 cm^2A/sec) * (8.854 * 10^-6 F/cm^2) * (15 μm / 15 μm) = 1 A/V^2
2. Overdrive voltage (Vov):
Vov is given by the equation:
Vov = VGS - VT
Here, VGS is the gate-source voltage and VT is the threshold voltage.
Given that VSD = 0.5 V and the transistor is operating in the non-saturation region, we can assume VGS = VSD.
Therefore, Vov = VGS - VT = 0.5 V - (-0.80 V) = 1.30 V
3. Transconductance (gm):
gm is given by the equation:
gm = √(2KP * ID)
Where ID is the drain current.
Since the transistor is operating in the non-saturation region, ID can be approximated as:
ID = KP * (Vov)^2
Substituting the given values:
ID = 1 A/V^2 * (1.30 V)^2 = 2.69 A
Now, substituting the calculated values back into the equation for gm:
gm = √(2KP * ID) = √(2 * 1 A/V^2 * 2.69 A) ≈ √(5.38) ≈ 2.32 A/V
Therefore, the value of gm is approximately 2.32 A/V, which is closest to option B (148).
W = 15 μm
p = 300 cm^2A/sec
L = 15 μm
tox = 350 Å
VT = -0.80 V
VSD = 0.5 V
To calculate the value of gm, we need to find the transconductance parameter (KP) and the overdrive voltage (Vov).
1. Transconductance parameter (KP):
KP is given by the equation:
KP = μn * Cox * (W/L)
Where:
μn = Electron mobility in the channel
Cox = Oxide capacitance per unit area
Given that:
μn = p * tox
Cox = εox / tox
Here, εox is the permittivity of the oxide layer, and tox is the thickness of the oxide layer.
Substituting the given values:
μn = 300 cm^2A/sec * 350 Å = 105,000 cm^2A/sec
Cox = εox / tox = (8.854 * 10^-12 F/m * 3.5 * 10^-7 m) / (350 * 10^-10 m) = 8.854 * 10^-6 F/cm^2
Now, substituting the calculated values back into the equation for KP:
KP = μn * Cox * (W/L) = (105,000 cm^2A/sec) * (8.854 * 10^-6 F/cm^2) * (15 μm / 15 μm) = 1 A/V^2
2. Overdrive voltage (Vov):
Vov is given by the equation:
Vov = VGS - VT
Here, VGS is the gate-source voltage and VT is the threshold voltage.
Given that VSD = 0.5 V and the transistor is operating in the non-saturation region, we can assume VGS = VSD.
Therefore, Vov = VGS - VT = 0.5 V - (-0.80 V) = 1.30 V
3. Transconductance (gm):
gm is given by the equation:
gm = √(2KP * ID)
Where ID is the drain current.
Since the transistor is operating in the non-saturation region, ID can be approximated as:
ID = KP * (Vov)^2
Substituting the given values:
ID = 1 A/V^2 * (1.30 V)^2 = 2.69 A
Now, substituting the calculated values back into the equation for gm:
gm = √(2KP * ID) = √(2 * 1 A/V^2 * 2.69 A) ≈ √(5.38) ≈ 2.32 A/V
Therefore, the value of gm is approximately 2.32 A/V, which is closest to option B (148).
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An ideal p-channel MOSFET has the parameters. W = 15 μm, μp = 300 cm2A/ sec, L = 15 [μm, tox = 350 A° and VT = -0.80 V. If transistor is operating in non saturation region at VSD = 0.5 V, then the value of gm is ____________μSa)145b)148c)125d)136Correct answer is option 'B'. Can you explain this answer?
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